1589. Exercises - 1589.1. Brønsted-Lowry Acids and Bases
(c) $\mathrm{NH}_{2}{ }^{-}$
(d) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
(e) $\mathrm{O}^{2-}$
(f) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$
7. What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) $\mathrm{OH}^{-}$
(b) $\mathrm{H}_{2} \mathrm{O}$
(c) $\mathrm{HCO}_{3}{ }^{-}$
(d) $\mathrm{NH}_{3}$
(e) $\mathrm{HSO}_{4}^{-}$
(f) $\mathrm{H}_{2} \mathrm{O}_{2}$
(g) $\mathrm{HS}^{-}$
(h) $\mathrm{H}_{5} \mathrm{~N}_{2}{ }^{+}$
8. What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) $\mathrm{H}_{2} \mathrm{~S}$
(b) $\mathrm{H}_{2} \mathrm{PO}_{4}-$
(c) $\mathrm{PH}_{3}$
(d) $\mathrm{HS}^{-}$
(e) $\mathrm{HSO}_{3}{ }^{-}$
(f) $\mathrm{H}_{3} \mathrm{O}_{2}{ }^{+}$
(g) $\mathrm{H}_{4} \mathrm{~N}_{2}$
(h) $\mathrm{CH}_{3} \mathrm{OH}$
9. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) $\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{NO}_{3}{ }^{-}$
(b) $\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HCN}+\mathrm{OH}^{-}$


1589. Exercises - 1589.1. Brønsted-Lowry Acids and Bases
(b) $\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HCN}+\mathrm{OH}^{-}$
(c) $\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{Cl}^{-} \longrightarrow \mathrm{HCl}+\mathrm{HSO}_{4}^{-}$
(d) $\mathrm{HSO}_{4}{ }^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{SO}_{4}{ }^{2-}+\mathrm{H}_{2} \mathrm{O}$
(e) $\mathrm{O}^{2-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{OH}^{-}$
(f) $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \longrightarrow\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}$
(g) $\mathrm{H}_{2} \mathrm{~S}+\mathrm{NH}_{2}{ }^{-} \longrightarrow \mathrm{HS}^{-}+\mathrm{NH}_{3}$
10. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
(a) $\mathrm{NO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HNO}_{2}+\mathrm{OH}^{-}$
(b) $\mathrm{HBr}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Br}^{-}$


1589. Exercises - 1589.1. Brønsted-Lowry Acids and Bases
(b) $\mathrm{HBr}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Br}^{-}$
(c) $\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{~S}+\mathrm{OH}^{-}$
(d) $\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{HPO}_{4}{ }^{2-}+\mathrm{H}_{2} \mathrm{O}$
(e) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{HCl} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{Cl}^{-}$
(f) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} \longrightarrow\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}$
(g) $\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{O}^{-}+\mathrm{H}_{2}$
11. What are amphiprotic species? Illustrate with suitable equations.
12. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
(a) $\mathrm{H}_{2} \mathrm{O}$
(b) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$
(c) $\mathrm{S}^{2-}$
(d) $\mathrm{CO}_{3}{ }^{2-}$
(e) $\mathrm{HSO}_{4}{ }^{-}$


1589. Exercises - 1589.1. Brønsted-Lowry Acids and Bases
(c) $\mathrm{S}^{2-}$
(d) $\mathrm{CO}_{3}{ }^{2-}$
(e) $\mathrm{HSO}_{4}{ }^{-}$
13. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
(a) $\mathrm{NH}_{3}$
(b) $\mathrm{HPO}_{4}^{-}$
(c) $\mathrm{Br}^{-}$
(d) $\mathrm{NH}_{4}{ }^{+}$
(e) $\mathrm{ASO}_{4}{ }^{3-}$
14. Is the self-ionization of water endothermic or exothermic? The ionization constant for water $\left(K_{\mathrm{w}}\right)$ is $2.9 \times$ $10^{-14}$ at $40^{\circ} \mathrm{C}$ and $9.3 \times 10^{-14}$ at $60^{\circ} \mathrm{C}$.


1590. 2 pH and pOH -
15. Explain why a sample of pure water at $40^{\circ} \mathrm{C}$ is neutral even though $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.7 \times 10^{-7} \mathrm{M}$. $K_{\mathrm{w}}$ is $2.9 \times 10^{-14}$ at $40^{\circ} \mathrm{C}$.
16. The ionization constant for water $\left(K_{\mathrm{w}}\right)$ is $2.9 \times 10^{-14}$ at $40^{\circ} \mathrm{C}$. Calculate $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH}$, and pOH for pure water at $40^{\circ} \mathrm{C}$.
17. The ionization constant for water $\left(K_{\mathrm{w}}\right)$ is $9.311 \times 10^{-14}$ at $60^{\circ} \mathrm{C}$. Calculate $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH}$, and pOH for pure water at $60^{\circ} \mathrm{C}$.
18. Calculate the pH and the pOH of each of the following solutions at $25^{\circ} \mathrm{C}$ for which the substances ionize completely:
(a) 0.200 M HCl
(b) 0.0143 M NaOH
(c) $3.0 \mathrm{M} \mathrm{HNO}_{3}$
(d) $0.0031 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}$
19. Calculate the pH and the pOH of each of the following solutions at $25^{\circ} \mathrm{C}$ for which the substances ionize completely:
(a) $0.000259 \mathrm{M} \mathrm{HClO}_{4}$
(b) 0.21 M NaOH
(c) $0.000071 \mathrm{MBa}(\mathrm{OH})_{2}$


1590. 2 pH and pOH -
(b) 0.21 M NaOH
(c) $0.000071 \mathrm{MBa}(\mathrm{OH})_{2}$
(d) 2.5 M KOH
20. What are the pH and pOH of a solution of 2.0 M HCl , which ionizes completely?
21. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52 ?
22. Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH . See Figure 14.2 for useful information.
23. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH . See Figure 14.2 for useful information.
24. The hydronium ion concentration in a sample of rainwater is found to be $1.7 \times 10^{-6} \mathrm{M}$ at $25^{\circ} \mathrm{C}$. What is the concentration of hydroxide ions in the rainwater?
25. The hydroxide ion concentration in household ammonia is $3.2 \times 10^{-3} \mathrm{M}$ at $25^{\circ} \mathrm{C}$. What is the concentration of hydronium ions in the solution?


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
26. Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
27. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
28. Use this list of important industrial compounds (and Figure 14.8) to answer the following questions regarding: $\mathrm{Ca}(\mathrm{OH})_{2}, \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{HCl}, \mathrm{H}_{2} \mathrm{CO}_{3}$, $\mathrm{HF}, \mathrm{HNO}_{2}, \mathrm{HNO}_{3}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{NH}_{3}, \mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CO}_{3}$. (a) Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
(b) Identify the compounds that can behave as Brønsted-Lowry acids with strengths lying between those of $\mathrm{H}_{3} \mathrm{O}^{+}$and $\mathrm{H}_{2} \mathrm{O}$.
(c) Identify the compounds that can behave as Brønsted-Lowry bases with strengths lying between those of $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{OH}^{-}$.
29. The odor of vinegar is due to the presence of acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
30. Household ammonia is a solution of the weak base $\mathrm{NH}_{3}$ in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
31. Explain why the ionization constant, $K_{\mathrm{a}}$, for $\mathrm{H}_{2} \mathrm{SO}_{4}$ is larger than the ionization constant for $\mathrm{H}_{2} \mathrm{SO}_{3}$.
32. Explain why the ionization constant, $K_{\mathrm{a}}$, for HI is larger than the ionization constant for HF.
33. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
34. Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of $\mathrm{HNO}_{3}$ with CuO .
35. What is the ionization constant at $25^{\circ} \mathrm{C}$ for the weak acid $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}$, the conjugate acid of the weak base $\mathrm{CH}_{3} \mathrm{NH}_{2}, K_{\mathrm{b}}=4.4 \times 10^{-4}$.
36. What is the ionization constant at $25^{\circ} \mathrm{C}$ for the weak acid $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}{ }^{+}$, the conjugate acid of the weak base $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, K_{\mathrm{b}}=5.9 \times 10^{-4}$ ?


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
37. Which base, $\mathrm{CH}_{3} \mathrm{NH}_{2}$ or $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$, is the stronger base? Which conjugate acid, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}{ }^{+}$or $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}$, is the stronger acid?
38. Which is the stronger acid, $\mathrm{NH}_{4}{ }^{+}$or HBrO ?
39. Which is the stronger base, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}^{2}$ or $\mathrm{H}_{2} \mathrm{BO}_{3}-$ ?
40. Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
(a) $\mathrm{H}_{2} \mathrm{O}$ or HF
(b) $\mathrm{B}(\mathrm{OH})_{3}$ or $\mathrm{Al}(\mathrm{OH})_{3}$
(c) $\mathrm{HSO}_{3}{ }^{-}$or $\mathrm{HSO}_{4}^{-}$
(d) $\mathrm{NH}_{3}$ or $\mathrm{H}_{2} \mathrm{~S}$
(e) $\mathrm{H}_{2} \mathrm{O}$ or $\mathrm{H}_{2} \mathrm{Te}$
41. Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
(a) $\mathrm{HSO}_{4}^{-}$or $\mathrm{HSeO}_{4}^{-}$
(b) $\mathrm{NH}_{3}$ or $\mathrm{H}_{2} \mathrm{O}$
(c) $\mathrm{PH}_{3}$ or HI
(d) $\mathrm{NH}_{3}$ or $\mathrm{PH}_{3}$
(e) $\mathrm{H}_{2} \mathrm{~S}$ or HBr
42. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
(a) acidity: $\mathrm{HCl}, \mathrm{HBr}, \mathrm{HI}$


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
(a) acidity: $\mathrm{HCl}, \mathrm{HBr}, \mathrm{HI}$
(b) basicity: $\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}, \mathrm{H}^{-}, \mathrm{Cl}^{-}$
(c) basicity: $\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{Si}(\mathrm{OH})_{4}, \mathrm{ClO}_{3}(\mathrm{OH})$ (Hint: Formula could also be written as $\mathrm{HClO}_{4}$.)
(d) acidity: $\mathrm{HF}, \mathrm{H}_{2} \mathrm{O}, \mathrm{NH}_{3}, \mathrm{CH}_{4}$
43. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
(a) acidity: $\mathrm{NaHSO}_{3}, \mathrm{NaHSeO}_{3}, \mathrm{NaHSO}_{4}$
(b) basicity: $\mathrm{BrO}_{2}{ }^{-}, \mathrm{ClO}_{2}{ }^{-}, \mathrm{IO}_{2}{ }^{-}$
(c) acidity: $\mathrm{HOCl}, \mathrm{HOBr}, \mathrm{HOI}$
(d) acidity: $\mathrm{HOCl}, \mathrm{HOClO}, \mathrm{HOClO}_{2}, \mathrm{HOClO}_{3}$
(e) basicity: $\mathrm{NH}_{2}{ }^{-}, \mathrm{HS}^{-}, \mathrm{HTe}^{-}, \mathrm{PH}_{2}{ }^{-}$
(f) basicity: $\mathrm{BrO}^{-}, \mathrm{BrO}_{2}^{-}, \mathrm{BrO}_{3}^{-}, \mathrm{BrO}_{4}^{-}$
44. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, $\mathrm{F}^{-}$or $\mathrm{CN}^{-}$, is the stronger base?


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
44. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, $\mathrm{F}^{-}$or $\mathrm{CN}^{-}$, is the stronger base?
45. The active ingredient formed by aspirin in the body is salicylic acid, $\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{CO}_{2} \mathrm{H}\right)$. The carboxyl group $\left(-\mathrm{CO}_{2} \mathrm{H}\right)$ acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a $0.001-M$ aqueous solution of $\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{CO}_{2} \mathrm{H}\right)$.
46. Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
47. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?
48. Which of the following will increase the percent of $\mathrm{NH}_{3}$ that is converted to the ammonium ion in water?
(a) addition of NaOH
(b) addition of HCl
(c) addition of $\mathrm{NH}_{4} \mathrm{Cl}$
49. Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?
(a) addition of NaOH
(b) addition of HCl
(c) addition of NaF
50. What is the effect on the concentrations of $\mathrm{NO}_{2}{ }^{-}, \mathrm{HNO}_{2}$, and $\mathrm{OH}^{-}$when the following are added to a solution of $\mathrm{KNO}_{2}$ in water:
(a) HCl
(b) $\mathrm{HNO}_{2}$
(c) NaOH
(d) NaCl
(e) KNO


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
(a) HCl
(b) $\mathrm{HNO}_{2}$
(c) NaOH
(d) NaCl
(e) KNO
51. What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
(a) HCl
(b) KF
(c) NaCl
(d) KOH
(e) HF
52. Why is the hydronium ion concentration in a solution that is 0.10 Min HCl and 0.10 Min HCOOH determined by the concentration of HCl ?
53. From the equilibrium concentrations given, calculate $K_{\mathrm{a}}$ for each of the weak acids and $K_{\mathrm{b}}$ for each of the weak bases.
(a) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]=1.34 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}$;
(b) $\mathrm{ClO}^{-}:\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}$;
[ HClO ] $=2.38 \times 10^{-4} \mathrm{M}$;
$\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}$;
(c) $\mathrm{HCO}_{2} \mathrm{H}:\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 \mathrm{M}$;
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} \mathrm{M}$;


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{HCO}_{2}{ }^{-}\right]=9.8 \times 10^{-3} \mathrm{M}$;
(d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}:\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\right]=0.233 \mathrm{M}$;
$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}$
54. From the equilibrium concentrations given, calculate $K_{\mathrm{a}}$ for each of the weak acids and $K_{\mathrm{b}}$ for each of the weak bases.
(a) $\mathrm{NH}_{3}:\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{NH}_{4}{ }^{+}\right]=3.1 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{NH}_{3}\right]=0.533 \mathrm{M}$;
(b) $\mathrm{HNO}_{2}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.011 \mathrm{M}$;
$\left[\mathrm{NO}_{2}{ }^{-}\right]=0.0438 \mathrm{M}$;
$\left[\mathrm{HNO}_{2}\right]=1.07 \mathrm{M}$;
(c) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}:\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\right]=0.25 \mathrm{M}$;


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
$\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=4.3 \times 10^{-3} \mathrm{M}$;
$\left[\mathrm{OH}^{-}\right]=3.7 \times 10^{-3} \mathrm{M}$;
(d) $\mathrm{NH}_{4}{ }^{+}:\left[\mathrm{NH}_{4}{ }^{+}\right]=0.100 \mathrm{M}$;
$\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} \mathrm{M}$;
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} \mathrm{M}$
55. Determine $K_{\mathrm{b}}$ for the nitrite ion, $\mathrm{NO}_{2}{ }^{-}$. In a $0.10-M$ solution this base is $0.0015 \%$ ionized.
56. Determine $K_{\mathrm{a}}$ for hydrogen sulfate ion, $\mathrm{HSO}_{4}{ }^{-}$. In a $0.10-M$ solution the acid is $29 \%$ ionized.
57. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) $\mathrm{F}^{-}$
(b) $\mathrm{NH}_{4}{ }^{+}$
(c) $\mathrm{AsO}_{4}{ }^{3-}$
(d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}{ }^{+}$
(e) $\mathrm{NO}_{2}{ }^{-}$
(f) $\mathrm{HC}_{2} \mathrm{O}_{4}^{-}$(as a base)
58. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
(a) $\mathrm{HTe}^{-}$(as a base)
(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}$


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
(a) $\mathrm{HTe}^{-}$(as a base)
(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}$
(c) $\mathrm{HAsO}_{4}{ }^{2-}$ (as a base)
(d) $\mathrm{HO}_{2}{ }^{-}$(as a base)
(e) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}+$
(f) $\mathrm{HSO}_{3}{ }^{-}$(as a base)
59. Using the $K_{\mathrm{a}}$ value of $1.4 \times 10^{-5}$, place $\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}$ in the correct location in Figure 14.7.
60. Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.
(a) 0.0092 M HClO , a weak acid
(b) $0.0784 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$, a weak base
(c) 0.0810 M HCN , a weak acid
(d) $0.11 \mathrm{M}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$, a weak base
(e) $0.120 \mathrm{MFe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}$ a weak acid, $K_{\mathrm{a}}=1.6 \times 10^{-7}$


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
61. Propionic acid, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.34 \times 10^{-5}\right)$, is used in the manufacture of calcium propionate, a food preservative. What is the pH of a $0.698-M$ solution of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}$ ?
62. White vinegar is a $5.0 \%$ by mass solution of acetic acid in water. If the density of white vinegar is $1.007 \mathrm{~g} /$ $\mathrm{cm}^{3}$, what is the pH ?
63. The ionization constant of lactic acid, $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CO}_{2} \mathrm{H}$, an acid found in the blood after strenuous exercise, is $1.36 \times 10^{-4}$. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L , what is the concentration of hydronium ion in the solution?
64. Nicotine, $\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}$, is a base that will accept two protons ( $K_{\mathrm{b} 1}=7 \times 10^{-7}, K_{\mathrm{b} 2}=1.4 \times 10^{-11}$ ). What is the concentration of each species present in a $0.050-M$ solution of nicotine?
65. The pH of a $0.23-M$ solution of HF is 1.92 . Determine $K_{\mathrm{a}}$ for HF from these data.
66. The pH of a $0.15-M$ solution of $\mathrm{HSO}_{4}{ }^{-}$is 1.43. Determine $K_{\mathrm{a}}$ for $\mathrm{HSO}_{4}{ }^{-}$from these data.


1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases
67. The pH of a $0.10-M$ solution of caffeine is 11.70 . Determine $K_{\mathrm{b}}$ for caffeine from these data: $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$
68. The pH of a solution of household ammonia, a 0.950 M solution of $\mathrm{NH}_{3}$, is 11.612 . Determine $K_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ from these data.


1590. 2 pH and pOH - 1590.2. Hydrolysis of Salts
69. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$
(b) RbI
(c) $\mathrm{KHCO}_{2}$
(d) $\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Br}$
70. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) $\mathrm{FeCl}_{3}$
(b) $\mathrm{K}_{2} \mathrm{CO}_{3}$
(c) $\mathrm{NH}_{4} \mathrm{Br}$
(d) $\mathrm{KClO}_{4}$
71. Novocaine, $\mathrm{C}_{13} \mathrm{H}_{21} \mathrm{O}_{2} \mathrm{~N}_{2} \mathrm{Cl}$, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is $7 \times 10^{-6}$. Is a solution of novocaine acidic or basic? What are $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right]$, and pH of a $2.0 \%$ solution by mass of novocaine, assuming that the density of the solution is $1.0 \mathrm{~g} / \mathrm{mL}$.


1590. 2 pH and pOH - 1590.3. Polyprotic Acids
72. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a $0.134-M$ solution of $\mathrm{H}_{2} \mathrm{CO}_{3}$, a diprotic acid: $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right],\left[\mathrm{HCO}_{3}{ }^{-}\right]$, $\left[\mathrm{CO}_{3}{ }^{2-}\right]$ ? No calculations are needed to answer this question.
73. Calculate the concentration of each species present in a $0.050-M$ solution of $\mathrm{H}_{2} \mathrm{~S}$.
74. Calculate the concentration of each species present in a $0.010-M$ solution of phthalic acid, $\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}$.


1590. 2 pH and pOH - 1590.3. Polyprotic Acids
$$
\begin{array}{lr}
\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)\left(\mathrm{CO}_{2}\right)^{-}(a q) & K_{\mathrm{a}}=1.1 \times 10^{-3} \\
\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)\left(\mathrm{CO}_{2}\right)(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2}\right)_{2}^{2-}(a q) & K_{\mathrm{a}}=3.9 \times 10^{-6}
\end{array}
$$


1590. 2 pH and pOH - 1590.3. Polyprotic Acids
75. Salicylic acid, $\mathrm{HOC}_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{H}$, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.
(a) Both functional groups of salicylic acid ionize in water, with $K_{\mathrm{a}}=1.0 \times 10^{-3}$ for the $-\mathrm{CO}_{2} \mathrm{H}$ group and $4.2 \times 10^{-13}$ for the -OH group. What is the pH of a saturated solution of the acid (solubility $=1.8 \mathrm{~g} / \mathrm{L}$ ).
(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{H}$. The $-\mathrm{CO}_{2} \mathrm{H}$ functional group is still present, but its acidity is reduced, $K_{\mathrm{a}}=3.0 \times 10^{-4}$. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).
76. The ion $\mathrm{HTe}^{-}$is an amphiprotic species; it can act as either an acid or a base.
(a) What is $K_{\mathrm{a}}$ for the acid reaction of $\mathrm{HTe}^{-}$with $\mathrm{H}_{2} \mathrm{O}$ ?
(b) What is $K_{\mathrm{b}}$ for the reaction in which $\mathrm{HTe}^{-}$functions as a base in water?


1590. 2 pH and pOH - 1590.3. Polyprotic Acids
(b) What is $K_{\mathrm{b}}$ for the reaction in which $\mathrm{HTe}^{-}$functions as a base in water?
(c) Demonstrate whether or not the second ionization of $\mathrm{H}_{2} \mathrm{Te}$ can be neglected in the calculation of [ $\mathrm{HTe}^{-}$] in a 0.10 M solution of $\mathrm{H}_{2} \mathrm{Te}$.


1590. 2 pH and pOH - 1590.4. Buffers
77. Explain why a buffer can be prepared from a mixture of $\mathrm{NH}_{4} \mathrm{Cl}$ and NaOH but not from $\mathrm{NH}_{3}$ and NaOH .
78. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid $\mathrm{H}_{3} \mathrm{PO}_{4}$ and a salt of its conjugate base $\mathrm{NaH}_{2} \mathrm{PO}_{4}$.
79. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base $\mathrm{NH}_{3}$ and a salt of its conjugate acid $\mathrm{NH}_{4} \mathrm{Cl}$.
80. What is $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$in a solution of $0.25 \mathrm{MCH}_{3} \mathrm{CO}_{2} \mathrm{H}$ and $0.030 \mathrm{M} \mathrm{NaCH} \mathrm{NO}_{2}$ ? $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}(a q) \quad K_{\mathrm{a}}=1.8 \times 10^{-5}$
81. What is $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$in a solution of $0.075 \mathrm{M} \mathrm{HNO}_{2}$ and $0.030 M \mathrm{NaNO}_{2}$ ?
$\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \quad K_{\mathrm{a}}=4.5 \times 10^{-5}$


1590. 2 pH and pOH - 1590.4. Buffers
82. What is $\left[\mathrm{OH}^{-}\right]$in a solution of $0.125 \mathrm{MCH}_{3} \mathrm{NH}_{2}$ and $0.130 \mathrm{MCH}_{3} \mathrm{NH}_{3} \mathrm{Cl}$ ?
$\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}+(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{b}}=4.4 \times 10^{-4}$
83. What is $\left[\mathrm{OH}^{-}\right]$in a solution of $1.25 \mathrm{MNH}_{3}$ and $0.78 \mathrm{MNH}_{4} \mathrm{NO}_{3}$ ?
$\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}{ }^{( }(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{b}}=1.8 \times 10^{-5}$
84. What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
(a) HCl
(b) $\mathrm{KCH}_{3} \mathrm{CO}_{2}$
(c) NaCl
(d) KOH
(e) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$
85. What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
(a) KI
(b) $\mathrm{NH}_{3}$
(c) HI
(d) NaOH
(e) $\mathrm{NH}_{4} \mathrm{Cl}$


1590. 2 pH and pOH - 1590.4. Buffers
(a) KI
(b) $\mathrm{NH}_{3}$
(c) HI
(d) NaOH
(e) $\mathrm{NH}_{4} \mathrm{Cl}$
86. What will be the pH of a buffer solution prepared from $0.20 \mathrm{~mol} \mathrm{NH}_{3}, 0.40 \mathrm{~mol} \mathrm{NH}_{4} \mathrm{NO}_{3}$, and just enough water to give 1.00 L of solution?
87. Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of $\mathrm{KH}_{2} \mathrm{PO}_{4}$, and enough water to make 0.500 L of solution.
88. How much solid $\mathrm{NaCH}_{3} \mathrm{CO}_{2} \cdot 3 \mathrm{H}_{2} \mathrm{O}$ must be added to 0.300 L of a $0.50-M$ acetic acid solution to give a buffer with a pH of 5.00 ? (Hint: Assume a negligible change in volume as the solid is added.)
89. What mass of $\mathrm{NH}_{4} \mathrm{Cl}$ must be added to 0.750 L of a $0.100-\mathrm{M}$ solution of $\mathrm{NH}_{3}$ to give a buffer solution with a pH of 9.26 ? (Hint: Assume a negligible change in volume as the solid is added.)
90. A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use $1.80 \times 10^{-5}$ as $K_{\mathrm{a}}$ for acetic acid.
(a) What is the pH of the solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to 0.200 L of the original buffer?
91. A $5.36-\mathrm{g}$ sample of $\mathrm{NH}_{4} \mathrm{Cl}$ was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L .


1590. 2 pH and pOH - 1590.4. Buffers
(a) What is the pH of this buffer solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to the solution?


1590. 2 pH and pOH - 1590.5. Acid-Base Titrations
92. Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
93. Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH .
94. Calculate the pH at the following points in a titration of $40 \mathrm{~mL}(0.040 \mathrm{~L})$ of 0.100 M barbituric acid $\left(K_{\mathrm{a}}=9.8\right.$ $\times 10^{-5}$ ) with 0.100 M KOH .
(a) no KOH added
(b) 20 mL of KOH solution added
(c) 39 mL of KOH solution added
(d) 40 mL of KOH solution added
(e) 41 mL of KOH solution added
95. The indicator dinitrophenol is an acid with a $K_{\mathrm{a}}$ of $1.1 \times 10^{-4}$. In a $1.0 \times 10^{-4}-M$ solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from $10 \%$ ionized (colorless) to $90 \%$ ionized (yellow).

724 14•Exercises


1591. CHAPTER 15 Equilibria of Other Reaction Classes -
Figure 15.1 The mineral fluorite $\left(\mathrm{CaF}_{2}\right)$ is formed when dissolved calcium and fluoride ions precipitate from groundwater within the Earth's crust. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metal ions in the crystal.


1592. CHAPTER OUTLINE - 1592.1. Precipitation and Dissolution
15.2 Lewis Acids and Bases
15.3 Coupled Equilibria

INTRODUCTION The mineral fluorite, $\mathrm{CaF}_{2}$ Figure 15.1, is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble $\mathrm{CaF}_{2}$ in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH , and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions.


1593. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Write chemical equations and equilibrium expressions representing solubility equilibria
- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations

Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal
rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.


1594. The Solubility Product -
Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.

In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (Figure 15.2). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.

FIGURE 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of $\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$ions in equilibrium with undissolved silver chloride.

The equilibrium constant for solubility equilibria such as this one is called the solubility product constant, $\boldsymbol{K}_{\mathbf{s p}}$, in this case

$$
\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}(a q)\right]\left[\mathrm{Cl}^{-}(a q)\right]
$$

Recall that only gases and solutes are represented in equilibrium constant expressions, so the $K_{\text {sp }}$ does not include a term for the undissolved AgCl . A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.


1596. Writing Equations and Solubility Products -
Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:
(a) AgI, silver iodide, a solid with antiseptic properties
(b) $\mathrm{CaCO}_{3}$, calcium carbonate, the active ingredient in many over-the-counter chewable antacids
(c) $\mathrm{Mg}(\mathrm{OH})_{2}$, magnesium hydroxide, the active ingredient in Milk of Magnesia
(d) $\mathrm{Mg}\left(\mathrm{NH}_{4}\right) \mathrm{PO}_{4}$, magnesium ammonium phosphate, an essentially insoluble substance used in tests for
magnesium
(e) $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}$, the mineral apatite, a source of phosphate for fertilizers


1596. Writing Equations and Solubility Products -
Solution
(a) $\mathrm{AgI}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \quad K_{\text {sp }}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]$
(b) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q)$
$K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}{ }^{2-}\right]$
(c) $\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$
$K_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$
(d) $\mathrm{Mg}\left(\mathrm{NH}_{4}\right) \mathrm{PO}_{4}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q)$
$K_{\text {sp }}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{PO}_{4}{ }^{3-}\right]$
(e) $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right) 3 \mathrm{OH}(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}{ }^{3-}(a q)+\mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^{5}\left[\mathrm{PO}_{4}^{3-}\right]^{3}\left[\mathrm{OH}^{-}\right]$


1596. Writing Equations and Solubility Products -
Check Your Learning
Write the dissolution equation and the solubility product for each of the following slightly soluble compounds:
(a) $\mathrm{BaSO}_{4}$
(b) $\mathrm{Ag}_{2} \mathrm{SO}_{4}$
(c) $\mathrm{Al}(\mathrm{OH})_{3}$
(d) $\mathrm{Pb}(\mathrm{OH}) \mathrm{Cl}$


1597. Answer: -
(a) $\mathrm{BaSO}_{4}(s) \rightleftharpoons \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_{4}{ }^{2-}\right]$;
(b) $\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$
$K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{SO}_{4}{ }^{2-}\right] ;$
(c) $\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q)$
$K_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}$;
(d) $\mathrm{Pb}(\mathrm{OH}) \mathrm{Cl}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{Cl}^{-}(a q)$
$K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{OH}^{-}\right]\left[\mathrm{Cl}^{-}\right]$


1598. $K_{\text {sp }}$ and Solubility -
The $K_{\text {sp }}$ of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:

$$
\mathrm{M}_{p} \mathrm{X}_{q}(s) \rightleftharpoons p \mathrm{M}^{\mathrm{m}+}(a q)+q \mathrm{X}^{\mathrm{n}-}(a q)
$$

For cases such as these, one may derive $K_{\text {sp }}$ values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound's molar solubility, measured as moles of dissolved solute per liter of saturated solution.


1600. Calculation of $\boldsymbol{K}_{\text {sp }}$ from Equilibrium Concentrations -
Fluorite, $\mathrm{CaF}_{2}$, is a slightly soluble solid that dissolves according to the equation:

$$
\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)
$$

The concentration of $\mathrm{Ca}^{2+}$ in a saturated solution of $\mathrm{CaF}_{2}$ is $2.15 \times 10^{-4} \mathrm{M}$. What is the solubility product of fluorite?


1601. Solution -
According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a $\mathrm{CaF}_{2}$ solution is equal to twice its calcium ion molarity:

$$
\left[\mathrm{F}^{-}\right]=\left(2 \mathrm{~mol} \mathrm{~F}^{-} / 1 \mathrm{~mol} \mathrm{Ca}^{2+}\right)=(2)\left(2.15 \times 10^{-4} M\right)=4.30 \times 10^{-4} M
$$

Substituting the ion concentrations into the $K_{\mathrm{sp}}$ expression gives

$$
K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}=\left(2.15 \times 10^{-4}\right)\left(4.30 \times 10^{-4}\right)^{2}=3.98 \times 10^{-11}
$$


1602. Check Your Learning -
In a saturated solution of $\mathrm{Mg}(\mathrm{OH})_{2}$, the concentration of $\mathrm{Mg}^{2+}$ is $1.31 \times 10^{-4} \mathrm{M}$. What is the solubility product for $\mathrm{Mg}(\mathrm{OH})_{2}$ ?

$$
\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)
$$


1603. Answer: -
$8.99 \times 10^{-12}$


1605. Determination of Molar Solubility from $\boldsymbol{K}_{\mathbf{s p}}$ -
The $K_{\mathrm{sp}}$ of copper(I) bromide, CuBr , is $6.3 \times 10^{-9}$. Calculate the molar solubility of copper bromide.


1606. Solution -
The dissolution equation and solubility product expression are

$$
\begin{gathered}
\mathrm{CuBr}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q)+\mathrm{Br}^{-}(a q) \\
K_{\mathrm{sp}}=\left[\mathrm{Cu}^{+}\right]\left[\mathrm{Br}^{-}\right]
\end{gathered}
$$

Following the ICE approach to this calculation yields the table

|  | $\mathrm{CuBr}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q) \quad+\mathrm{Br}^{-}(\mathrm{aq})$ |  |  |
| :---: | :---: | :---: | :---: |
| Initial concentration $(M)$ |  | 0 | 0 |
| Change $(M)$ | $+x$ | $+x$ |  |
| Equilibrium concentration $(M)$ |  | $x$ | $x$ |

Substituting the equilibrium concentration terms into the solubility product expression and solving for $x$ yields

$$
\begin{gathered}
K_{\mathrm{sp}}=\left[\mathrm{Cu}^{+}\right]\left[\mathrm{Br}^{-}\right] \\
6.3 \times 10^{-9}=(x)(x)=x^{2} \\
x=\sqrt{\left(6.3 \times 10^{-9}\right)}=7.9 \times 10^{-5} M
\end{gathered}
$$

Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is $7.9 \times 10^{-5} \mathrm{M}$.


1607. Check Your Learning -
The $K_{\mathrm{sp}}$ of AgI is $1.5 \times 10^{-16}$. Calculate the molar solubility of silver iodide.


1608. Answer: -
$1.2 \times 10^{-8} \mathrm{M}$


1610. Determination of Molar Solubility from $\boldsymbol{K}_{\text {sp }}$ -
The $K_{\mathrm{sp}}$ of calcium hydroxide, $\mathrm{Ca}(\mathrm{OH})_{2}$, is $1.3 \times 10^{-6}$. Calculate the molar solubility of calcium hydroxide.


1611. Solution -
The dissolution equation and solubility product expression are

$$
\begin{aligned}
\mathrm{Ca}(\mathrm{OH})_{2}(s) & \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \\
K_{\text {sp }} & =\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}
\end{aligned}
$$

The ICE table for this system is

|  | $\mathbf{C a}(\mathbf{O H})_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq}) \mathbf{+} \mathbf{2 O H}^{-}(\mathrm{aq})$ |  |  |
| :---: | :---: | :---: | :---: |
| Initial concentration $(M)$ |  | 0 | 0 |
| Change $(M)$ |  | $+x$ | $+2 x$ |
| Equilibrium concentration $(M)$ |  | $x$ | $2 x$ |

Substituting terms for the equilibrium concentrations into the solubility product expression and solving for $x$ gives

$$
\begin{gathered}
K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \\
1.3 \times 10^{-6}=(x)(2 x)^{2}=(x)\left(4 x^{2}\right)=4 x^{3} \\
x=\sqrt[3]{\frac{1.3 \times 10^{-6}}{4}}=6.9 \times 10^{-3} M
\end{gathered}
$$

As defined in the ICE table, $x$ is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of $\mathrm{Ca}(\mathrm{OH})_{2}$ is $6.9 \times 10^{-3} \mathrm{M}$.


1612. Check Your Learning -
The $K_{\mathrm{sp}}$ of $\mathrm{PbI}_{2}$ is $1.4 \times 10^{-8}$. Calculate the molar solubility of lead(II) iodide.


1613. Answer: -
$1.5 \times 10^{-3} \mathrm{M}$


1615. Determination of $\boldsymbol{K}_{\text {sp }}$ from Gram Solubility -
Many of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist's pigment chrome yellow, $\mathrm{PbCrO}_{4}$, is $4.6 \times 10^{-6} \mathrm{~g} / \mathrm{L}$. Determine the solubility product for $\mathrm{PbCrO}_{4}$.

FIGURE 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow $\left(\mathrm{PbCrO}_{4}\right)$, examples include Prussian blue $\left(\mathrm{Fe}_{7}(\mathrm{CN})_{18}\right)$, the reddish-orange color vermilion $(\mathrm{HgS})$, and green color veridian $\left(\mathrm{Cr}_{2} \mathrm{O}_{3}\right)$. (credit: Sonny Abesamis)


1616. Solution -
Before calculating the solubility product, the provided solubility must be converted to molarity:

$$
\begin{gathered}
{\left[\mathrm{PbCrO}_{4}\right]=\frac{4.6 \times 10^{-6} \mathrm{~g} \mathrm{PbCrO}_{4}}{1 \mathrm{~L}} \times \frac{1 \mathrm{~mol} \mathrm{PbCrO}_{4}}{323.2 \mathrm{~g} \mathrm{PbCrO}_{4}}} \\
=\frac{1.4 \times 10^{-8} \mathrm{~mol} \mathrm{PbCrO}_{4}}{1 \mathrm{~L}} \\
=1.4 \times 10^{-8} \mathrm{M}
\end{gathered}
$$

The dissolution equation for this compound is

$$
\mathrm{PbCrO}_{4}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)
$$

The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both $\left[\mathrm{Pb}^{2+}\right]$ and $\left[\mathrm{CrO}_{4}{ }^{2-}\right]$ are equal to the molar solubility of $\mathrm{PbCrO}_{4}$ :

$$
\left[\mathrm{Pb}^{2+}\right]=\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.4 \times 10^{-8} \mathrm{M}
$$

$K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{CrO}_{4}{ }^{2-}\right]=\left(1.4 \times 10^{-8}\right)\left(1.4 \times 10^{-8}\right)=2.0 \times 10^{-16}$


1617. Check Your Learning -
The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at $20^{\circ} \mathrm{C}$. What is its solubility product?


1618. Answer: -
$1.69 \times 10^{-4}$


1620. Calculating the Solubility of $\mathrm{Hg}_{2} \mathbf{C l}_{\mathbf{2}}$ -
Calomel, $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$, is a compound composed of the diatomic ion of mercury(I), $\mathrm{Hg}_{2}{ }^{2+}$, and chloride ions, $\mathrm{Cl}^{-}$. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small $K_{\mathrm{sp}}$ :

$$
\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad K_{\mathrm{sp}}=1.1 \times 10^{-18}
$$

Calculate the molar solubility of $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$.


1621. Solution -
The dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$ is equal to the concentration of $\mathrm{Hg}_{2}{ }^{2+}$ ions

Following the ICE approach results in

|  | $\mathbf{H g}_{2} \mathrm{Cl}_{2}(\mathbf{s}) \rightleftharpoons \mathbf{H g}_{2}{ }^{2+}(a q) \mathbf{+} \mathbf{2 C l}(a q)$ |  |  |
| :---: | :---: | :---: | :---: |
| Initial concentration $(M)$ |  | 0 | 0 |
| Change $(M)$ |  | $+x$ | $+2 x$ |
| Equilibrium concentration $(M)$ | $x$ | $2 x$ |  |

Substituting the equilibrium concentration terms into the solubility product expression and solving for $x$ gives

$$
\begin{gathered}
K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2} \\
1.1 \times 10^{-18}=(x)(2 x)^{2} \\
4 x^{3}=1.1 \times 10^{-18} \\
x=\sqrt[3]{\left(\frac{1.1 \times 10^{-18}}{4}\right)}=6.5 \times 10^{-7} M \\
{\left[\mathrm{Hg}_{2}^{2+}\right]=6.5 \times 10^{-7} M=6.5 \times 10^{-7} M} \\
{\left[\mathrm{Cl}^{-}\right]=2 x=2\left(6.5 \times 10^{-7}\right)=1.3 \times 10^{-6} M}
\end{gathered}
$$


1621. Solution -
The dissolution stoichiometry shows the molar solubility of $\mathrm{Hg}_{2} \mathrm{Cl}_{2}$ is equal to $\left[\mathrm{Hg}_{2}{ }^{2+}\right]$, or $6.5 \times 10^{-7} \mathrm{M}$.


1622. Check Your Learning -
Determine the molar solubility of $\mathrm{MgF}_{2}$ from its solubility product: $K_{\mathrm{sp}}=6.4 \times 10^{-9}$.


1623. Answer: -
$1.2 \times 10^{-3} \mathrm{M}$


1625. Using Barium Sulfate for Medical Imaging -
Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the $K_{\text {sp }}$ of barium sulfate is $2.3 \times 10^{-8}$, very little of it dissolves as it coats the lining of the patient's intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).

FIGURE 15.4 A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by "glitzy queen00"/Wikimedia Commons)

Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn's disease, and ulcers in addition to other conditions.

Visit this website (http://openstax.org/l/16barium) for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.


1626. Predicting Precipitation -
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:

$$
\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}^{2-}\right]=8.7 \times 10^{-9}
$$

It is important to realize that this equilibrium is established in any aqueous solution containing $\mathrm{Ca}^{2+}$ and $\mathrm{CO}_{3}{ }^{2-}$ ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, $Q_{s p}$, that exceeds the solubility product, $K_{\text {sp }}$, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established ( $Q_{s p}=K_{\mathrm{sp}}$ ). The comparison of $Q_{s p}$ to $K_{\mathrm{sp}}$ to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:
$Q_{s p}<K_{\mathrm{sp}}$ : the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)
$Q_{s p}>K_{\mathrm{sp}}$ : the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)
This predictive strategy and related calculations are demonstrated in the next few example exercises.


1628. Precipitation of $\mathrm{Mg}(\mathrm{OH})_{\mathbf{2}}$ -
The first step in the preparation of magnesium metal is the precipitation of $\mathrm{Mg}(\mathrm{OH})_{2}$ from sea water by the addition of lime, $\mathrm{Ca}(\mathrm{OH})_{2}$, a readily available inexpensive source of $\mathrm{OH}^{-}$ion:

$$
\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=8.9 \times 10^{-12}
$$

The concentration of $\mathrm{Mg}^{2+}(\mathrm{aq})$ in sea water is 0.0537 M . Will $\mathrm{Mg}(\mathrm{OH})_{2}$ precipitate when enough $\mathrm{Ca}(\mathrm{OH})_{2}$ is added to give a $\left[\mathrm{OH}^{-}\right]$of 0.0010 M ?


1629. Solution -
Calculation of the reaction quotient under these conditions is shown here:

$$
Q=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=(0.0537)(0.0010)^{2}=5.4 \times 10^{-8}
$$

Because $Q$ is greater than $K_{\mathrm{sp}}\left(Q=5.4 \times 10^{-8}\right.$ is larger than $K_{\mathrm{sp}}=8.9 \times 10^{-12}$ ), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that $Q_{s p}=K_{\mathrm{sp}}$.


1630. Check Your Learning -
Predict whether $\mathrm{CaHPO}_{4}$ will precipitate from a solution with $\left[\mathrm{Ca}^{2+}\right]=0.0001 \mathrm{M}$ and $\left[\mathrm{HPO}_{4}{ }^{2-}\right]=0.001 \mathrm{M}$.


1631. Answer: -
No precipitation of $\mathrm{CaHPO}_{4} ; Q=1 \times 10^{-7}$, which is less than $K_{\text {sp }}\left(7 \times 10^{-7}\right)$


1633. Precipitation of AgCl -
Does silver chloride precipitate when equal volumes of a $2.0 \times 10^{-4}-M$ solution of $\mathrm{AgNO}_{3}$ and a $2.0 \times 10^{-4}-M$ solution of NaCl are mixed?


1634. Solution -
The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:

$$
\operatorname{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)
$$

The solubility product is $1.6 \times 10^{-10}$ (see Appendix J).
AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of $\mathrm{AgNO}_{3}$ and NaCl is greater than $K_{\text {sp }}$. Because the volume doubles when equal volumes of $\mathrm{AgNO}_{3}$ and NaCl solutions are mixed, each concentration is reduced to half its initial value

$$
\frac{1}{2}\left(2.0 \times 10^{-4}\right) M=1.0 \times 10^{-4} M
$$

The reaction quotient, $Q$, is greater than $K_{\mathrm{sp}}$ for AgCl , so a supersaturated solution is formed:

$$
Q=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left(1.0 \times 10^{-4}\right)\left(1.0 \times 10^{-4}\right)=1.0 \times 10^{-8}>K_{\mathrm{sp}}
$$

AgCl will precipitate from the mixture until the dissolution equilibrium is established, with $Q$ equal to $K_{\mathrm{sp}}$.


1635. Check Your Learning -
Will $\mathrm{KClO}_{4}$ precipitate when 20 mL of a $0.050-M$ solution of $\mathrm{K}^{+}$is added to 80 mL of a $0.50-M$ solution of $\mathrm{ClO}_{4}{ }^{-}$? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)


1636. Answer: -
No, $Q=4.0 \times 10^{-3}$, which is less than $K_{\text {sp }}=1.05 \times 10^{-2}$


1638. Precipitation of Calcium Oxalate -
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$, for this purpose (Figure 15.5). At sufficiently high concentrations, the calcium and oxalate ions form solid, $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$ (calcium oxalate monohydrate). The concentration of $\mathrm{Ca}^{2+}$ in a sample of blood serum is $2.2 \times 10^{-3} \mathrm{M}$. What concentration of $\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}$ ion must be established before $\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}$ begins to precipitate?

FIGURE 15.5 Anticoagulants can be added to blood that will combine with the $\mathrm{Ca}^{2+}$ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)


1639. Solution -
The equilibrium expression is:

$$
\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)
$$

For this reaction:

$$
K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=1.96 \times 10^{-8}
$$

(see Appendix J)
Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:

$$
\begin{gathered}
Q=K_{\text {sp }}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=1.96 \times 10^{-8} \\
\left(2.2 \times 10^{-3}\right)\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=1.96 \times 10^{-8} \\
{\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=\frac{1.96 \times 10^{-8}}{2.2 \times 10^{-3}}=8.9 \times 10^{-6} \mathrm{M}}
\end{gathered}
$$

A concentration of $\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.9 \times 10^{-6} M$ is necessary to initiate the precipitation of $\mathrm{CaC}_{2} \mathrm{O}_{4}$ under these
conditions.


1640. Check Your Learning -
If a solution contains 0.0020 mol of $\mathrm{CrO}_{4}{ }^{2-}$ per liter, what concentration of $\mathrm{Ag}^{+}$ion must be reached by adding solid $\mathrm{AgNO}_{3}$ before $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.


1641. Answer: -
$6.7 \times 10^{-5} \mathrm{M}$


1643. Concentrations Following Precipitation -
Clothing washed in water that has a manganese $\left[\mathrm{Mn}^{2+}(\mathrm{aq})\right]$ concentration exceeding $0.1 \mathrm{mg} / \mathrm{L}\left(1.8 \times 10^{-6} \mathrm{M}\right)$ may be stained by the manganese upon oxidation, but the amount of $\mathrm{Mn}^{2+}$ in the water can be decreased by adding a base to precipitate $\mathrm{Mn}(\mathrm{OH})_{2}$. What pH is required to keep $\left[\mathrm{Mn}^{2+}\right.$ ] equal to $1.8 \times 10^{-6} \mathrm{M}$ ?


1644. Solution -
The dissolution of $\mathrm{Mn}(\mathrm{OH})_{2}$ is described by the equation:

$$
\mathrm{Mn}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mn}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=2 \times 10^{-13}
$$

At equilibrium:

$$
K_{\mathrm{sp}}=\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}
$$

or

$$
\left(1.8 \times 10^{-6}\right)\left[\mathrm{OH}^{-}\right]^{2}=2 \times 10^{-13}
$$

so

$$
\left[\mathrm{OH}^{-}\right]=3.3 \times 10^{-4} \mathrm{M}
$$

Calculate the pH from the pOH :

$$
\begin{gathered}
\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log (3.3 \times 10-4)=3.48 \\
\mathrm{pH}=14.00-\mathrm{pOH}=14.00-3.48=10.52
\end{gathered}
$$

(final result rounded to one significant digit, limited by the certainty of the $K_{\mathrm{sp}}$ )


1645. Check Your Learning -
The first step in the preparation of magnesium metal is the precipitation of $\mathrm{Mg}(\mathrm{OH})_{2}$ from sea water by the addition of $\mathrm{Ca}(\mathrm{OH})_{2}$. The concentration of $\mathrm{Mg}^{2+}(\mathrm{aq})$ in sea water is $5.37 \times 10^{-2} \mathrm{M}$. Calculate the pH at which $\left[\mathrm{Mg}^{2+}\right]$ is decreased to $1.0 \times 10^{-5} \mathrm{M}$


1646. Answer: -
10.97

In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound's solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.


1648. The Role of Precipitation in Wastewater Treatment -
Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 15.6). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions $\left(\mathrm{PO}_{4}{ }^{3-}\right)$ are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.

FIGURE 15.6 Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: "eutrophication\&hypoxia"/Wikimedia Commons)

One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, $\mathrm{Ca}(\mathrm{OH})_{2}$. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, $\mathrm{Ca}_{5}(\mathrm{PO} 4)_{3} \mathrm{OH}$, which then precipitates out of the solution:

$$
5 \mathrm{Ca}^{2+}+3 \mathrm{PO}_{4}{ }^{3-}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \cdot \mathrm{OH}(s)
$$

Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of $\mathrm{CO}_{2}$ in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.

View this site (http://openstax.org/l/16Wastewater) for more information on how phosphorus is removed from wastewater.


1650. Precipitation of Silver Halides -
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. $\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgBr or solid AgCl ?


1651. Solution -
The two equilibria involved are:

$$
\begin{array}{lll}
\mathrm{AgCl}(s) & \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) & K_{\mathrm{sp}}=1.6 \times 10^{-10} \\
\mathrm{AgBr}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Br}^{-}(a q) & K_{\mathrm{sp}}=5.0 \times 10^{-13}
\end{array}
$$

If the solution contained about equal concentrations of $\mathrm{Cl}^{-}$and $\mathrm{Br}^{-}$, then the silver salt with the smaller $K_{\text {sp }}$ ( AgBr ) would precipitate first. The concentrations are not equal, however, so the $\left[\mathrm{Ag}^{+}\right]$at which AgCl begins to precipitate and the $\left[\mathrm{Ag}^{+}\right]$at which AgBr begins to precipitate must be calculated. The salt that forms at the lower $\left[\mathrm{Ag}^{+}\right]$precipitates first.

AgBr precipitates when $Q$ equals $K_{\mathrm{sp}}$ for AgBr

$$
\begin{gathered}
Q_{\mathrm{sp}}=K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=\left\mathrm{Ag}^{+}\right=5.0 \times 10^{-13} \\
{\left[\mathrm{Ag}^{+}\right]=\frac{5.0 \times 10^{-13}}{0.00010}=5.0 \times 10^{-9} M}
\end{gathered}
$$


1651. Solution -
AgBr begins to precipitate when $\left[\mathrm{Ag}^{+}\right]$is $5.0 \times 10^{-9} \mathrm{M}$.
For AgCl : AgCl precipitates when $Q$ equals $K_{\text {sp }}$ for $\mathrm{AgCl}\left(1.6 \times 10^{-10}\right)$. When $\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M}$ :

$$
\begin{gathered}
Q_{\mathrm{sp}}=K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left\mathrm{Ag}^{+}\right=1.6 \times 10^{-10} \\
{\left[\mathrm{Ag}^{+}\right]=\frac{1.6 \times 10^{-10}}{0.10}=1.6 \times 10^{-9} M}
\end{gathered}
$$

AgCl begins to precipitate when $\left[\mathrm{Ag}^{+}\right]$is $1.6 \times 10^{-9} \mathrm{M}$.
AgCl begins to precipitate at a lower $\left[\mathrm{Ag}^{+}\right]$than AgBr , so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a $K_{\mathrm{sp}}$ greater than that of silver bromide.


1652. Check Your Learning -
If silver nitrate solution is added to a solution which is 0.050 M in both $\mathrm{Cl}^{-}$and $\mathrm{Br}^{-}$ions, at what $\left[\mathrm{Ag}^{+}\right]$would precipitation begin, and what would be the formula of the precipitate?


1653. Answer: -
$\left[\mathrm{Ag}^{+}\right]=1.0 \times 10^{-11} \mathrm{M} ; \mathrm{AgBr}$ precipitates first


1654. Common Ion Effect -
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le ChÂtelier's principle. Consider the dissolution of silver iodide:

$$
\operatorname{AgI}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)
$$

This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved $\mathrm{Ag}^{+}$and $\mathrm{I}^{-}$. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.

This effect may also be explained in terms of mass action as represented in the solubility product expression:

$$
K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]
$$

The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion's concentration must be balanced by a proportional decrease in the other.


1655. LINK TO LEARNING -
View this simulation (http://openstax.org/l/16solublesalts) to explore various aspects of the common ion effect.


1657. Common Ion Effect on Solubility -
What is the effect on the amount of solid $\mathrm{Mg}(\mathrm{OH})_{2}$ and the concentrations of $\mathrm{Mg}^{2+}$ and $\mathrm{OH}^{-}$when each of the following are added to a saturated solution of $\mathrm{Mg}(\mathrm{OH})_{2}$ ?
(a) $\mathrm{MgCl}_{2}$
(b) KOH
(c) $\mathrm{NaNO}_{3}$
(d) $\mathrm{Mg}(\mathrm{OH})_{2}$


1658. Solution -
The solubility equilibrium is

$$
\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)
$$

(a) Adding a common ion, $\mathrm{Mg}^{2+}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.
(b) Adding a common ion, $\mathrm{OH}^{-}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.
(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.

$$
Q=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}
$$

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of $Q$, and no shift is required to restore $Q$ to the value of the equilibrium constant.


1659. Check Your Learning -
What is the effect on the amount of solid $\mathrm{NiCO}_{3}$ and the concentrations of $\mathrm{Ni}^{2+}$ and $\mathrm{CO}_{3}{ }^{2-}$ when each of the following are added to a saturated solution of $\mathrm{NiCO}_{3}$
(a) $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$
(b) $\mathrm{KClO}_{4}$
(c) $\mathrm{NiCO}_{3}$
(d) $\mathrm{K}_{2} \mathrm{CO}_{3}$


1660. Answer: -
(a) mass of $\mathrm{NiCO}_{3}(s)$ increases, $\left[\mathrm{Ni}^{2+}\right]$ increases, $\left[\mathrm{CO}_{3}{ }^{2-}\right]$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid $\mathrm{NiCO}_{3}$; (d) mass of $\mathrm{NiCO}_{3}(s)$ increases, $\left[\mathrm{Ni}^{2+}\right]$ decreases, $\left[\mathrm{CO}_{3}{ }^{2-}\right]$ increases;


1662. Common Ion Effect -
Calculate the molar solubility of cadmium sulfide (CdS) in a $0.010-M$ solution of cadmium bromide $\left(\mathrm{CdBr}_{2}\right)$. The $K_{\mathrm{sp}}$ of CdS is $1.0 \times 10^{-28}$.


1663. Solution -
This calculation can be performed using the ICE approach:

| $\mathrm{CdS}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q)$ |  |  |  |
| :---: | :---: | :---: | :---: |
| $\mathrm{CdS}(\mathrm{s}) \rightleftharpoons \mathrm{Cd}^{2+}(\mathrm{aq})+\mathrm{s}^{2-}(\mathrm{aq})$ |  |  |  |
| Initial concentration $(M)$ |  | 0.010 | 0 |
| Change (M) |  | $+x$ | $+x$ |
| Equilibrium concentration $(M)$ |  | $0.010+x$ | $x$ |

Because $K_{\text {sp }}$ is very small, assume $x \ll 0.010$ and solve the simplified equation for $x$ :

$$
\begin{gathered}
(0.010)(x)=1.0 \times 10^{-28} \\
x=1.0 \times 10^{-26} M
\end{gathered}
$$

The molar solubility of CdS in this solution is $1.0 \times 10^{-26} \mathrm{M}$.


1664. Check Your Learning -
Calculate the molar solubility of aluminum hydroxide, $\mathrm{Al}(\mathrm{OH})_{3}$, in a $0.015-M$ solution of aluminum nitrate, $\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}$. The $K_{\text {sp }}$ of $\mathrm{Al}(\mathrm{OH})_{3}$ is $2 \times 10^{-32}$.

Answer:
$4 \times 10^{-11} \mathrm{M}$


1665. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Explain the Lewis model of acid-base chemistry
- Write equations for the formation of adducts and complex ions
- Perform equilibrium calculations involving formation constants

In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.

A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.

Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is $\mathrm{H}^{+}$. A few examples involving other Lewis acids and bases are described below.

The boron atom in boron trifluoride, $\mathrm{BF}_{3}$, has only six electrons in its valence shell. Being short of the preferred octet, $\mathrm{BF}_{3}$ is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:

In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:

Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:

Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:


1665. LEARNING OBJECTIVES -
Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:

Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands. These ligands can be neutral molecules like $\mathrm{H}_{2} \mathrm{O}$ or $\mathrm{NH}_{3}$, or ions such as $\mathrm{CN}^{-}$or $\mathrm{OH}^{-}$. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry-the topic of another chapter in this text.

The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant ( $\boldsymbol{K}_{\mathbf{f}}$ ) (sometimes called a stability constant). For example, the complex ion $\mathrm{Cu}(\mathrm{CN})_{2}{ }^{-}$

```
[:N三C-Cu-C\equivN:]-
```

is produced by the reaction

$$
\mathrm{Cu}^{+}(a q)+2 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{2}^{-}(a q)
$$

The formation constant for this reaction is

$$
K_{\mathrm{f}}=\frac{\left[\mathrm{Cu}(\mathrm{CN})_{2}^{-}\right]}{\left[\mathrm{Cu}^{+}\right]\left[\mathrm{CN}^{-}\right]^{2}}
$$

Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant ( $\boldsymbol{K}_{\mathbf{d}}$ ). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, $K_{d}=\mathrm{K}_{\mathrm{f}}{ }^{-1}$. A tabulation of formation constants is provided in Appendix K.


1665. LEARNING OBJECTIVES -
As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of $\mathrm{Ag}^{+}\left(\left[\mathrm{Ag}^{+}\right]=1.3 \times 10^{-5} \mathrm{M}\right)$ :

$$
\operatorname{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)
$$

However, if $\mathrm{NH}_{3}$ is present in the water, the complex ion, $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}$, can form according to the equation:

$$
\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)
$$

with

$$
K_{\mathrm{f}}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_{3}\right]^{2}}=1.7 \times 10^{7}
$$

The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with $\mathrm{NH}_{3}$ to form $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}$. As a consequence, the concentration of silver ions, $\left[\mathrm{Ag}^{+}\right.$], is reduced, and the reaction quotient for the dissolution of silver chloride, $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$, falls below the solubility
product of AgCl :

$$
Q=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]<K_{\mathrm{sp}}
$$

More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.


1667. Dissociation of a Complex Ion -
Calculate the concentration of the silver ion in a solution that initially is $0.10 M$ with respect to $\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}$.


1668. Solution -
Applying the standard ICE approach to this reaction yields the following:

|  | $\mathbf{A g}^{+}+\mathbf{2 N H} \mathbf{3}^{c \mid} \rightleftharpoons \mathbf{A g}\left(\mathbf{N H}_{3}\right)_{2}{ }^{+}$ |  |  |
| :---: | :---: | :---: | :---: |
| Initial concentration $(M)$ | 0 | 0 | 0.10 |
| Change $(M)$ | $+x$ | $+2 x$ | $-x$ |
| Equilibrium concentration $(M)$ | $x$ | $2 x$ | $0.10-x$ |

Substituting these equilibrium concentration terms into the $K_{\mathrm{f}}$ expression gives

$$
\begin{aligned}
K_{\mathrm{f}} & =\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_{3}\right]^{2}} \\
1.7 & \times 10^{7}=\frac{0.10-x}{(x)(2 x)^{2}}
\end{aligned}
$$

The very large equilibrium constant means the amount of the complex ion that will dissociate, $x$, will be very small. Assuming $x \ll 0.1$ permits simplifying the above equation:

$$
\begin{gathered}
1.7 \times 10^{7}=\frac{0.10}{(x)(2 x)^{2}} \\
x^{3}=\frac{0.10}{4\left(1.7 \times 10^{7}\right)}=1.5 \times 10^{-9} \\
x=\sqrt[3]{1.5 \times 10^{-9}}=1.1 \times 10^{-3}
\end{gathered}
$$


1668. Solution -
Because only $1.1 \%$ of the $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}$dissociates into $\mathrm{Ag}^{+}$and $\mathrm{NH}_{3}$, the assumption that $x$ is small is justified. Using this value of $x$ and the relations in the above ICE table allows calculation of all species' equilibrium concentrations:

$$
\begin{gathered}
{\left[\mathrm{Ag}^{+}\right]=0+x=1.1 \times 10^{-3} M} \\
{\left[\mathrm{NH}_{3}\right]=0+2 x=2.2 \times 10^{-3} M} \\
{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]=0.10-x=0.10-0.0011=0.099}
\end{gathered}
$$

The concentration of free silver ion in the solution is 0.0011 M .


1669. Check Your Learning -
Calculate the silver ion concentration, $\left[\mathrm{Ag}^{+}\right.$], of a solution prepared by dissolving 1.00 g of $\mathrm{AgNO}_{3}$ and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because $K_{\mathrm{f}}$ is very large, assume the reaction goes to completion then calculate the $\left[\mathrm{Ag}^{+}\right]$produced by dissociation of the complex.)


1670. Answer: -
$2.9 \times 10^{-22} \mathrm{M}$


1671. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe examples of systems involving two (or more) coupled chemical equilibria
- Calculate reactant and product concentrations for coupled equilibrium systems

As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.

An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean's coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is

$$
\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{-2}(a q) \quad K_{\mathrm{sp}}=8.7 \times 10^{-9}
$$

Rising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:


1671. LEARNING OBJECTIVES -
Rising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:

$$
\begin{gathered}
\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(a q) \\
\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \\
\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \quad K_{\mathrm{a} 1}=4.3 \times 10^{-7} \\
\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \quad K_{\mathrm{a} 2}=4.7 \times 10^{-11}
\end{gathered}
$$

Inspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields

$$
\mathrm{CaCO}_{3}(s)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad K=K_{\mathrm{sp}} / K_{\mathrm{a} 2}=180
$$


1671. LEARNING OBJECTIVES -
The equilibrium constant for this net reaction is much greater than the $K_{s p}$ for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy (Figure 15.7).

FIGURE 15.7 Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonate skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by
"prilfish"/Flickr)


1672. LINK TO LEARNING -
Learn more about ocean acidification (http://openstax.org/l/16acidicocean) and how it affects other marine creatures.

This site (http://openstax.org/l/16coralreef) has detailed information about how ocean acidification specifically affects coral reefs.

The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 15.8), a sparingly soluble ionic compound whose dissolution equilibrium is

FIGURE 15.8 Crystal of the mineral hydroxyapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}$, is shown here. The pure compound is white, but like many other minerals, this sample is colored because of the presence of impurities.

This compound dissolved to yield two different basic ions: triprotic phosphate ions

$$
\begin{gathered}
\mathrm{PO}_{4}{ }^{3-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\
\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\
\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
\end{gathered}
$$

and monoprotic hydroxide ions:


1672. LINK TO LEARNING -
and monoprotic hydroxide ions:

$$
\mathrm{OH}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}
$$

Of the two basic productions, the hydroxide is, of course, by far the stronger base (it's the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or $\mathrm{SnF}_{2}$ that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:

$$
\mathrm{NaF}+\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH} \rightleftharpoons \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}+\mathrm{Na}^{+}+\mathrm{OH}^{-}
$$

The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.


1674. Role of Fluoride in Preventing Tooth Decay -
As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, $\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}$. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure 15.9).

FIGURE 15.9 Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).
Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of $4 \mathrm{ppm}(4 \mathrm{mg} / \mathrm{L})$ of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.

The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $\mathrm{Al}(\mathrm{OH})_{4}{ }^{-}$.


1674. Role of Fluoride in Preventing Tooth Decay -
The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of $\mathrm{Al}(\mathrm{OH})_{3}$.

$$
\begin{array}{cl}
\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) & K_{\mathrm{sp}}=2 \times 10^{-32} \\
\mathrm{Al}^{3+}(a q)+4 \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) & K_{\mathrm{f}}=1.1 \times 10^{33} \\
\mathrm{Net}: \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q) & K=K_{\mathrm{sp}} K_{\mathrm{f}}=22
\end{array}
$$


1676. Increased Solubility in Acidic Solutions -
Compute and compare the molar solublities for aluminum hydroxide, $\mathrm{Al}(\mathrm{OH})_{3}$, dissolved in (a) pure water and (b) a buffer containing $0.100 M$ acetic acid and $0.100 M$ sodium acetate.


1677. Solution -
(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:

$$
\begin{aligned}
& \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=2 \times 10^{-32} \\
& \text { molar solubility in water }=\left[\mathrm{Al}^{3+}\right]=\left(2 \times 10^{-32} / 27\right)^{1 / 4}=5 \times 10^{-9} M
\end{aligned}
$$

(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the HendersonHasselbalch equation:

$$
\begin{aligned}
& \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] /\left[\mathrm{CH}_{3} \mathrm{COOH}\right] \\
& \mathrm{pH}=4.74+\log (0.100 / 0.100)=4.74
\end{aligned}
$$

At this pH , the concentration of hydroxide ion is

$$
\begin{aligned}
& \mathrm{pOH}=14.00-4.74=9.26 \\
& {\left[\mathrm{OH}^{-}\right]=10^{-9.26}=5.5 \times 10^{-10}}
\end{aligned}
$$

The solubility of $\mathrm{Al}(\mathrm{OH})_{3}$ in this buffer is then calculated from its solubility product expressions:

$$
K_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}
$$


1677. Solution -
$$
K_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}
$$

molar solubility in buffer $=\left[\mathrm{Al}^{3+}\right]=K_{\text {sp }} /\left[\mathrm{OH}^{-}\right]^{3}=\left(2 \times 10^{-32}\right) /\left(5.5 \times 10^{-10}\right)^{3}=1.2 \times 10^{-4} \mathrm{M}$
Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).


1678. Check Your Learning -
What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?


1679. Answer: -
0.1 M


1681. Multiple Equilibria -
Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate $\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right.$, called hypo) to form the complex ion $\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\left(K_{\mathrm{f}}=4.7 \times 10^{13}\right)$.

What mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}$ ?


1682. Solution -
Two equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the $\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}$ ion:

$$
\begin{aligned}
& \text { dissolution: } \mathrm{AgBr}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Br}^{-}(a q) \quad K_{\mathrm{sp}}=5.0 \times 10^{-13} \\
& \text { complexation: } \mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K_{\mathrm{f}}=4.7 \times 10^{13}
\end{aligned}
$$

Combining these two equilibrium equations yields

$$
\begin{array}{cc}
\mathrm{AgBr}(s)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) & \mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \\
\frac{K=\left[\mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\right]\left[\mathrm{Br}^{-}\right]}{\left[\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\right]^{2}}=K_{\mathrm{sp}} K_{\mathrm{f}}=24
\end{array}
$$

The concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is


1682. Solution -
The concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is

$$
\left[\mathrm{Br}^{-}\right]=\frac{1.00 \mathrm{~g} \mathrm{AgBr} \times \frac{1 \mathrm{~mol} \mathrm{AgBr}}{187.77 \mathrm{~g} / \mathrm{mol}} \times \frac{1 \mathrm{~mol} \mathrm{Br}^{-}}{1 \mathrm{~mol} \mathrm{AgBr}}}{1.00 \mathrm{~L}}=0.00532 \mathrm{M}
$$

The stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M , and the very large value of $K_{\mathrm{f}}$ ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion:

$$
\left[\mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\right]=0.00532 \mathrm{M}
$$

Rearranging the K expression for the combined equilibrium equations and solving for the concentration of thiosulfate ion yields

$$
\left[\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\right]=\frac{\left[\mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\right]\left[\mathrm{Br}^{-}\right]}{K}=\frac{(0.00532 \mathrm{M})(0.00532 \mathrm{M})}{24}=0.0011 \mathrm{M}
$$

Finally, the total mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.


1682. Solution -
Finally, the total mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.

Mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ required to yield $0.00532 \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{2}\right)_{2}{ }^{3-}$

$$
0.00532 \frac{\mathrm{~mol} \mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}}{1.00 \mathrm{~L}} \times \frac{2 \mathrm{~mol} \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}}{1 \mathrm{~mol} \mathrm{Ag}\left(\mathrm{~S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}} \times \frac{1 \mathrm{~mol} \mathrm{NaS}_{2} \mathrm{O}_{3}}{1 \mathrm{~mol} \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}} \times \frac{158.1 \mathrm{~g} \mathrm{NaS}_{2} \mathrm{O}_{3}}{1 \mathrm{~mol} \mathrm{NaS}_{2} \mathrm{O}_{3}}=1.68 \mathrm{~g}
$$


1682. Solution -
Mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ required to yield $0.00110 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}$
$0.0011 \frac{\mathrm{~mol} \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}}{1.00 \mathrm{~L}} \times \frac{1 \mathrm{~mol} \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}}{1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}} \times \frac{158.1 \mathrm{~g} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}}{1 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}}=0.17 \mathrm{~g}$
The mass of $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ required to dissolve 1.00 g of AgBr in 1.00 L of water is thus $1.68 \mathrm{~g}+0.17 \mathrm{~g}=1.85 \mathrm{~g}$


1683. Check Your Learning -
$\mathrm{AgCl}(s)$, silver chloride, has a very low solubility: $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q), K_{\mathrm{sp}}=1.6 \times 10^{-10}$. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:
$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}(a q), K_{\mathrm{f}}=1.7 \times 10^{7}$. What mass of $\mathrm{NH}_{3}$ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}$?


1684. Answer: -
1.00 L of a solution prepared with $4.81 \mathrm{~g} \mathrm{NH}_{3}$ dissolves 2.0 g of AgCl .


1685. Key Terms -
common ion effect effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
complex ion ion consisting of a central atom surrounding molecules or ions called ligands via coordinate covalent bonds
coordinate covalent bond (also, dative bond) covalent bond in which both electrons originated from the same atom
coupled equilibria system characterized the simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product
dissociation constant ( $\boldsymbol{K}_{\mathbf{d}}$ ) equilibrium constant for the decomposition of a complex ion into its components
formation constant ( $\boldsymbol{K}_{\mathbf{f}}$ ) (also, stability constant) equilibrium constant for the formation of a
complex ion from its components
Lewis acid any species that can accept a pair of electrons and form a coordinate covalent bond
Lewis acid-base adduct compound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base
Lewis acid-base chemistry reactions involving the formation of coordinate covalent bonds
Lewis base any species that can donate a pair of electrons and form a coordinate covalent bond
ligand molecule or ion acting as a Lewis base in complex ion formation; bonds to the central atom of the complex
molar solubility solubility of a compound expressed in units of moles per liter ( $\mathrm{mol} / \mathrm{L}$ )
selective precipitation process in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product constant ( $\boldsymbol{K}_{\text {sp }}$ ) equilibrium constant for the dissolution of an ionic compound


1686. Key Equations -
$\mathrm{M}_{p} \mathrm{X}_{q}(s) \rightleftharpoons p \mathrm{M}^{\mathrm{m}+}(a q)+q \mathrm{X}^{\mathrm{n}-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{M}^{\mathrm{m}+}\right]^{p}\left[\mathrm{X}^{\mathrm{n}-}\right]^{q}$


1687. Summary - 1687.1. Precipitation and Dissolution
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, $K_{\mathrm{sp}}$, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid $\mathrm{M}_{p} \mathrm{X}_{q}$ and its ions $\mathrm{M}^{\mathrm{m}+}$ and $\mathrm{X}^{\mathrm{n}-}$ :

$$
\mathrm{M}_{p} \mathrm{X}_{q}(s) \rightleftharpoons p \mathrm{M}^{\mathrm{m}+}(a q)+q \mathrm{X}^{\mathrm{n}-}(a q)
$$

the solubility product expression is:

$$
K_{\mathrm{sp}}=\left[\mathrm{M}^{\mathrm{m}+}\right]^{p}\left[\mathrm{X}^{\mathrm{n}-}\right]^{q}
$$

The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its $K_{\mathrm{sp}}$, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.

A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.


1687. Summary - 1687.2. Lewis Acids and Bases
A Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts and comprise central metal atoms or ions acting as Lewis acids bonded to molecules or ions called ligands that act as Lewis bases. The equilibrium constant for the reaction between a metal ion and ligands produces a complex ion called a formation constant; for the reverse reaction, it is called a dissociation constant.


1687. Summary - 1687.3. Coupled Equilibria
Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.


1688. Exercises - 1688.1. Precipitation and Dissolution
1. Complete the changes in concentrations for each of the following reactions:
(a)
$\mathrm{AgI}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)$
$x$
(b)
$\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q)$
$\qquad$
(c)
$\operatorname{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$
$x$
(d)
$\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \longrightarrow 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{PO}_{4}{ }^{3-}(a q)$

$x$
2. Complete the changes in concentrations for each of the following reactions:
(a)
$\mathrm{BaSO}_{4}(s) \longrightarrow \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$
$x$
(b
$\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$
(c)
$\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q)$
$x$
(d)
$\mathrm{Pb}(\mathrm{OH}) \mathrm{Cl}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{OH}^{-}(a q)+\mathrm{Cl}^{-}(a q)$
(e)
$\mathrm{Ca}_{3}\left(\mathrm{AsO}_{4}\right)_{2}(s) \longrightarrow 3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{AsO}_{4}{ }^{3-}(a q)$