2294. Chapter 11 -
23. (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) $-1.8^{\circ} \mathrm{C}$
24. (a) Determine the molar mass of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$; determine the number of moles of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm
25. (a) Determine the molal concentration from the change in boiling point and $K_{\mathrm{b}}$; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) $2.1 \times 10^{2} \mathrm{~g} \mathrm{~mol}^{-1}$
26. No. Pure benzene freezes at $5.5^{\circ} \mathrm{C}$, and so the observed freezing point of this solution is depressed by $\Delta T_{\mathrm{f}}$ $=5.5-0.4=5.1^{\circ} \mathrm{C}$. The value computed, assuming no ionization of HCl , is $\Delta T_{\mathrm{f}}=(1.0 \mathrm{~m})\left(5.14{ }^{\circ} \mathrm{C} / \mathrm{m}\right)=5.1$ ${ }^{\circ} \mathrm{C}$. Agreement of these values supports the assumption that HCl is not ionized.
27. $144 \mathrm{~g} \mathrm{~mol}^{-1}$
28. $0.870^{\circ} \mathrm{C}$
29. $\mathrm{S}_{8}$


2294. Chapter 11 -
27. $144 \mathrm{~g} \mathrm{~mol}^{-1}$
28. $0.870^{\circ} \mathrm{C}$
29. $\mathrm{S}_{8}$
30. $1.39 \times 10^{4} \mathrm{~g} \mathrm{~mol}^{-1}$
31. 54 g
32. $100.26^{\circ} \mathrm{C}$
33. (a) $X_{\mathrm{CH}_{3} \mathrm{OH}}=0.590 ; X_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}=0.410$; (b) Vapor pressures are: $\mathrm{CH}_{3} \mathrm{OH}$ : 55 torr; $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ : 18 torr; (c) $\mathrm{CH}_{3} \mathrm{OH}: 0.75 ; \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}: 0.25$
34. The ions and compounds present in the water in the beef lower the freezing point of the beef below $-1^{\circ} \mathrm{C}$.
35. $\Delta \mathrm{bp}=K_{\mathrm{b}} m=\left(1.20^{\circ} \mathrm{C} / m\right)\left(\frac{9.41 \mathrm{~g} \times \frac{1 \mathrm{~mol} \mathrm{Hg} \mathrm{Cl}_{2}}{271.496 \mathrm{~g}}}{0.03275 \mathrm{~kg}}\right)=1.27^{\circ} \mathrm{C}$


2294. Chapter 11 -
The observed change equals the theoretical change; therefore, no dissociation occurs.
71.

| Colloidal System | Dispersed Phase | Dispersion Medium |
| :--- | :--- | :--- |
| starch dispersion | starch | water |
| smoke | solid particles | air |
| fog | water | air |
| pearl | water | calcium carbonate $\left(\mathrm{CaCO}_{3}\right)$ |
| whipped cream | air | cream |
| floating soap | air | soap |
| jelly | fruit juice | pectin gel |
| milk | butterfat | water |
| ruby | chromium(III) oxide $\left(\mathrm{Cr}_{2} \mathrm{O}_{3}\right)$ | aluminum oxide $\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)$ |

73. Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.
74. If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate.


2295. Chapter 12 -
1. A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.
2. (a) spontaneous; (b) nonspontaneous; (c) spontaneous; (d) nonspontaneous; (e) spontaneous; (f) spontaneous
3. Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
4. There are four initial microstates and four final microstates.
$\Delta S=k \ln \frac{W_{\mathrm{f}}}{W_{\mathrm{i}}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \times \ln \frac{4}{4}=0$
5. The probability for all the particles to be on one side is $\frac{1}{32}$. This probability is noticeably lower than the $\frac{1}{8}$ result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
6. There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.
$\Delta S=k \ln \left(\frac{W_{\mathrm{f}}}{W_{\mathrm{i}}}\right)=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \times \ln \left(\frac{4}{1}\right)=1.91 \times 10^{-23} \mathrm{~J} / \mathrm{K}$


2295. Chapter 12 -
7. The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, $\mathrm{I}_{2}$ is a solid, $\mathrm{Br}_{2}$ is a liquid, and $\mathrm{Cl}_{2}$ is a gas.
8. (a) $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(I)$ as it is a larger molecule (more complex and more massive), and so more microstates
describing its motions are available at any given temperature. (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)$ as it is in the gaseous state. (c) $2 \mathrm{H}(\mathrm{g})$, since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
9. (a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution.
(b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.
10. $\mathrm{C}_{6} \mathrm{H}_{6}(l)+7.5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g)$


2295. Chapter 12 -
There are 7.5 moles of gas initially, and $3+6=9$ moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and $\Delta S$ is positive.
21. (a) $107 \mathrm{~J} / \mathrm{K}$; (b) $-86.4 \mathrm{~J} / \mathrm{K}$; (c) $133.2 \mathrm{~J} / \mathrm{K}$; (d) $118.8 \mathrm{~J} / \mathrm{K}$; (e) $-326.6 \mathrm{~J} / \mathrm{K}$; (f) $-171.9 \mathrm{~J} / \mathrm{K}$; (g) $-7.2 \mathrm{~J} / \mathrm{K}$
23. $100.6 \mathrm{~J} / \mathrm{K}$
25. (a) $-198.1 \mathrm{~J} / \mathrm{K}$; (b) $-348.9 \mathrm{~J} / \mathrm{K}$
27. As $\Delta S_{\text {univ }}<0$ at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K . It is assumed that these do not change significantly at the higher temperatures used in the problem.
29. (a) $2.86 \mathrm{~J} / \mathrm{K}$; (b) $24.8 \mathrm{~J} / \mathrm{K}$; (c) $-113.2 \mathrm{~J} / \mathrm{K}$; (d) $-24.7 \mathrm{~J} / \mathrm{K}$; (e) $15.5 \mathrm{~J} / \mathrm{K}$; (f) $290.0 \mathrm{~J} / \mathrm{K}$
31. The reaction is nonspontaneous at room temperature.


2295. Chapter 12 -
Above $400 \mathrm{~K}, \Delta G$ will become negative, and the reaction will become spontaneous.
33. (a) 465.1 kJ nonspontaneous; (b) -106.86 kJ spontaneous; (c) -291.9 kJ spontaneous; (d) -83.4 kJ spontaneous; (e) -406.7 kJ spontaneous; (f) -154.3 kJ spontaneous
35. (a) The standard free energy of formation is $-1124.3 \mathrm{~kJ} / \mathrm{mol}$. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.
37. (a) The reaction is nonspontaneous; (b) Above $566^{\circ} \mathrm{C}$ the process is spontaneous.
39. (a) $1.5 \times 10^{2} \mathrm{~kJ}$; (b) -21.9 kJ ; (c) -5.34 kJ ; (d) -0.383 kJ ; (e) 18 kJ ; (f) 71 kJ
41. (a) $K=41$; (b) $K=0.053$; (c) $K=6.9 \times 10^{13}$; (d) $K=1.9$; (e) $K=0.04$
42. (a) 22.1 kJ ;
(b) $98.9 \mathrm{~kJ} / \mathrm{mol}$
44. $90 \mathrm{~kJ} / \mathrm{mol}$


2295. Chapter 12 -
42. (a) 22.1 kJ ;
(b) $98.9 \mathrm{~kJ} / \mathrm{mol}$
44. $90 \mathrm{~kJ} / \mathrm{mol}$
46. (a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) $K_{p}=0.031$; (c) The evaporation of water is spontaneous; (d) $P_{\mathrm{H}_{2} \mathrm{O}}$ must always be less than $K_{p}$ or less than 0.031 atm .0 .031 atm represents air saturated with water vapor at $25^{\circ} \mathrm{C}$, or $100 \%$ humidity.
48. (a) Nonspontaneous as $\Delta G^{\circ}>0$; (b) $\Delta G^{\circ}=\Delta G^{\circ}+R T \ln Q$,
$\Delta G=1.7 \times 10^{3}+\left(8.314 \times 310 \times \ln \frac{28}{120}\right)=-2.1 \mathrm{~kJ}$. The forward reaction to produce F6P is spontaneous under these conditions.
50. $\Delta G$ is negative as the process is spontaneous. $\Delta H$ is positive as with the solution becoming cold, the dissolving must be endothermic. $\Delta S$ must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
52. (a) Increasing $P_{\mathrm{O}_{2}}$ will shift the equilibrium toward the products, which increases the value of $K . \Delta G^{\circ}$ therefore becomes more negative.
(b) Increasing $P_{\mathrm{O}_{2}}$ will shift the equilibrium toward the products, which increases the value of $K . \Delta G^{\circ}$ therefore becomes more negative.
(c) Increasing $P_{\mathrm{O}_{2}}$ will shift the equilibrium the reactants, which decreases the value of $K . \Delta G^{\circ}$ therefore becomes more positive.


2296. Chapter 13 -
1. The reaction can proceed in both the forward and reverse directions.
2. When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates.
3. Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.
4. Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the
bottle because the system is not closed; one of the components of the equilibrium, the $\mathrm{Br}_{2}$ vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.
5. (a) $K_{c}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]<1$. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M ; (b) $K_{c}=\frac{1}{\left[\mathrm{~Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}}>1$ because $\mathrm{PbCl}_{2}$ is insoluble and formation of the solid will reduce the concentration of ions to a low level ( $<1 M$ ).
6. Since $K_{c}=\frac{\left[\mathrm{C}_{6} \mathrm{H}_{6}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{2}\right]^{3}}$, a value of $K_{c} \approx 10$ means that $\mathrm{C}_{6} \mathrm{H}_{6}$ predominates over $\mathrm{C}_{2} \mathrm{H}_{2}$. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.
7. $K_{c}>1$


2296. Chapter 13 -
7. $K_{c}>1$
8. (a) $Q_{c}=\frac{\left\mathrm{CH}_{3} \mathrm{Cl}\right}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{Cl}_{2}\right]}$; (b) $Q_{c}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{O}_{2}\right]}$; (c) $Q_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}$; (d) $Q_{c}=\left[\mathrm{SO}_{2}\right]$; (e) $Q_{c}=\frac{1}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}$; (f) $Q_{c}=\frac{[\mathrm{Br}]^{2}}{\left[\mathrm{Br}_{2}\right]} ;$ (g) $Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{O}_{2}\right]^{2}}$; (h) $Q_{c}=\left[\mathrm{H}_{2} \mathrm{O}\right]^{5}$
9. (a) $Q_{\mathrm{c}} 25$ proceeds left; (b) $Q_{P} 0.22$ proceeds right; (c) $Q_{c}$ undefined proceeds left; (d) $Q_{P} 1.00$ proceeds right; (e) $Q_{P} 0$ proceeds right; (f) $Q_{c} 4$ proceeds left
10. The system will shift toward the reactants to reach equilibrium.
11. (a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous
12. This situation occurs in (a) and (b).


2296. Chapter 13 -
12. This situation occurs in (a) and (b).
13. (a) $K_{P}=1.6 \times 10^{-4}$; (b) $K_{P}=50.2$; (c) $K_{c}=5.34 \times 10^{-39}$; (d) $K_{c}=4.60 \times 10^{-3}$
14. $K_{P}=P_{\mathrm{H}_{2} \mathrm{O}}=0.042$.
15. $Q_{c}=\frac{\left[\mathrm{NH}_{4}{ }^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}$
16. The amount of $\mathrm{CaCO}_{3}$ must be so small that $\mathrm{P}_{\mathrm{CO}_{2}}$ is less than $K_{P}$ when the $\mathrm{CaCO}_{3}$ has completely decomposed. In other words, the starting amount of $\mathrm{CaCO}_{3}$ cannot completely generate the full $P_{\mathrm{CO}_{2}}$ required for equilibrium.
17. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side.
18. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.
19. Add $\mathrm{N}_{2}$; add $\mathrm{H}_{2}$; decrease the container volume; heat the mixture.


2296. Chapter 13 -
19. Add $\mathrm{N}_{2}$; add $\mathrm{H}_{2}$; decrease the container volume; heat the mixture.
20. (a) $T$ increase $=$ shift right, $V$ decrease $=$ shift left; (b) $T$ increase $=$ shift right, $V=$ no effect; (c) $T$ increase $=$ shift left, $V$ decrease $=$ shift left; (d) $T$ increase $=$ shift left, $V$ decrease $=$ shift right.
21. (a) $K_{c}=\frac{\left[\mathrm{CH}_{3} \mathrm{OH}\right]}{\left[\mathrm{H}_{2}\right]^{2}[\mathrm{CO}]}$; (b) $\left[\mathrm{H}_{2}\right]$ increases, $[\mathrm{CO}]$ decreases, $\left[\mathrm{CH}_{3} \mathrm{OH}\right]$ increases; (c), $\left[\mathrm{H}_{2}\right]$ increases, $[\mathrm{CO}]$ decreases, $\left[\mathrm{CH}_{3} \mathrm{OH}\right]$ decreases; (d), $\left[\mathrm{H}_{2}\right]$ increases, $[\mathrm{CO}]$ increases, $\left[\mathrm{CH}_{3} \mathrm{OH}\right]$ increases; (e), $\left[\mathrm{H}_{2}\right]$ increases, [CO] increases, $\left[\mathrm{CH}_{3} \mathrm{OH}\right]$ decreases


2296. Chapter 13 -
22. (a) $K_{c}=\frac{[\mathrm{CO}]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{O}\right]}$; (b) $\left[\mathrm{H}_{2} \mathrm{O}\right]$ no change, $[\mathrm{CO}]$ no change, $\left[\mathrm{H}_{2}\right]$ no change; (c) $\left[\mathrm{H}_{2} \mathrm{O}\right]$ decreases, $[\mathrm{CO}]$ decreases, $\left[\mathrm{H}_{2}\right]$ decreases; (d) $\left[\mathrm{H}_{2} \mathrm{O}\right.$ ] increases, [CO] increases, [ $\mathrm{H}_{2}$ ] decreases; (f) [ $\mathrm{H}_{2} \mathrm{O}$ ] decreases, [CO] increases, $\left[\mathrm{H}_{2}\right]$ increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.
23. Only (b)
24. Add NaCl or some other salt that produces $\mathrm{Cl}^{-}$to the solution. Cooling the solution forces the equilibrium to the right, precipitating more $\mathrm{AgCl}(s)$.
25. Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive $\mathrm{Ag}^{+}$
ions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive $\mathrm{Ag}^{+}$.
26. $K_{c}=\frac{[\mathrm{C}]^{2}}{\mathrm{~A}^{2}}$. $[\mathrm{A}]=0.1 \mathrm{M},[\mathrm{B}]=0.1 \mathrm{M},[\mathrm{C}]=1 \mathrm{M}$; and $[\mathrm{A}]=0.01,[\mathrm{~B}]=0.250,[\mathrm{C}]=0.791$.


2296. Chapter 13 -
27. $K_{c}=6.00 \times 10^{-2}$
28. $K_{c}=0.50$
29. $K_{P}=1.9 \times 10^{3}$
30. $K_{P}=3.06$
31. (a) $-2 x,+2 x$; (b) $+\frac{4}{3} x,-\frac{2}{3} x,-2 x$; (c) $-2 x, 3 x$; (d) $+x,-x,-3 x$; (e) $+x$; (f) $-\frac{1}{4} x$.
32. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.
33. $\left[\mathrm{NH}_{3}\right]=9.1 \times 10^{-2} \mathrm{M}$
34. $P_{\mathrm{BrCl}}=4.9 \times 10^{-2} \mathrm{~atm}$
35. $[\mathrm{CO}]=2.04 \times 10^{-4} \mathrm{M}$
36. $P_{\mathrm{H}_{2} \mathrm{O}}=3.64 \times 10^{-3} \mathrm{~atm}$
37. Calculate $Q$ based on the calculated concentrations and see if it is equal to $K_{c}$. Because $Q$ does equal 4.32, the system must be at equilibrium.
38. (a) $\left[\mathrm{NO}_{2}\right]=1.17 \times 10^{-3} \mathrm{M}$
$\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.128 \mathrm{M}$
(b) The assumption that $x$ is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the $\mathrm{N}_{2} \mathrm{O}_{4}$ to its concentration at equilibrium (they differ by just 1 in the least significant digit's place).
39. (a) $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.810 \mathrm{~atm}$


2296. Chapter 13 -
39. (a) $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.810 \mathrm{~atm}$
$\left[\mathrm{H}_{2}\right]=0.014 \mathrm{~atm}$
$\left[\mathrm{S}_{2}\right]=0.0072 \mathrm{~atm}$
(b) The assumption that $2 x$ is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the $\mathrm{H}_{2} \mathrm{~S}$ to its concentration at equilibrium ( 0.824 atm versus 0.810 atm , a difference of less than $2 \%$ ).
40. 507 g
41. 330 g
42. (a) 0.33 mol .
(b) $\left[\mathrm{CO}_{2}=0.50 \mathrm{M}\right.$. Added $\mathrm{H}_{2}$ forms some water as a result of a shift to the left after $\mathrm{H}_{2}$ is added.
43. (a) $K_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}}$. (b) $\left[\mathrm{NH}_{3}\right]$ must increase for $Q_{c}$ to reach $K_{c}$. (c) The increase in system volume would lower the partial pressures of all reactants (including $\mathrm{NO}_{2}$ ). (d) $P_{\mathrm{O}_{2}}=49$ torr
44. $P_{\mathrm{N}_{2} \mathrm{O}_{3}}=1.90 \mathrm{~atm}$ and $P_{\mathrm{NO}}=P_{\mathrm{NO}_{2}}=1.90 \mathrm{~atm}$


2296. Chapter 13 -
45. In each of the following, the value of $\Delta G$ is not given at the temperature of the reaction. Therefore, we must calculate $\Delta G$ from the values $\Delta H^{\circ}$ and $\Delta S$ and then calculate $\Delta G$ from the relation $\Delta G=\Delta H^{\circ}-T \Delta S^{\circ}$.
(a) $K=1.07 \times 10^{-13}$;
(b) $K=2.42 \times 10^{-3}$;
(c) $K=2.73 \times 10^{4}$;
(d) $K=0.229$;
(e) $K=16.1$
46. The standard free energy change is $\Delta G_{298}^{\circ}=-R T \ln K=4.84 \mathrm{~kJ} / \mathrm{mol}$. When reactants and products are in their standard states ( 1 bar or 1 atm ), $Q=1$. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): $Q<1$, and $\Delta G_{298}$ becomes less positive as it approaches zero. At equilibrium, $Q=K$, and $\Delta G=0$.
47. The reaction will be spontaneous at temperatures greater than 287 K .
48. $K=5.35 \times 10^{15}$


2296. Chapter 13 -
The process is exothermic.
98. $1.0 \times 10^{-8} \mathrm{~atm}$. This is the maximum pressure of the gases under the stated conditions.
100. $x=1.29 \times 10^{-5} \mathrm{~atm}=P_{\mathrm{O}_{2}}$
102. -0.16 kJ


2297. Chapter 14 -
1. One example for $\mathrm{NH}_{3}$ as a conjugate acid: $\mathrm{NH}_{2}{ }^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{3}$; as a conjugate base:
$\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
2. (a) $\mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$; (b) $\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)$; (c)
$\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{NH}_{2}^{-}(a q) ;$ (d) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)$; (e)
$\mathrm{NH}_{4}{ }^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{NH}_{3}(a q) ;$ (f) $\mathrm{HSO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)$
3. (a) $\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)$; (b) $\mathrm{OH}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$; (c)


2297. Chapter 14 -
$\mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{NH}_{4}^{+}(a q) ;$ (d) $\mathrm{CN}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{HCN}(a q)$; (e)
$\mathrm{S}^{2-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{HS}^{-}(a q) ;$ (f) $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)$
4. (a) $\mathrm{H}_{2} \mathrm{O}, \mathrm{O}^{2-}$; (b) $\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}^{-}$; (c) $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{CO}_{3}{ }^{2-}$; (d) $\mathrm{NH}_{4}^{+}, \mathrm{NH}_{2}^{-}$; (e) $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}{ }^{2-}$; (f) $\mathrm{H}_{3} \mathrm{O}_{2}{ }^{+}, \mathrm{HO}_{2}^{-}$; (g) $\mathrm{H}_{2} \mathrm{~S} ; \mathrm{S}^{2-}$; (h) $\mathrm{H}_{6} \mathrm{~N}_{2}{ }^{2+}, \mathrm{H}_{4} \mathrm{~N}_{2}$


2297. Chapter 14 -
5. The labels are $\mathrm{Br} \varnothing$ nsted-Lowry acid $=\mathrm{BA}$; its conjugate base $=\mathrm{CB}$; Brønsted-Lowry base $=\mathrm{BB}$; its conjugate acid = CA. (a) $\mathrm{HNO}_{3}(\mathrm{BA}), \mathrm{H}_{2} \mathrm{O}(\mathrm{BB}), \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{CA}), \mathrm{NO}_{3}{ }^{-}(\mathrm{CB}) ;$ (b) $\mathrm{CN}^{-}(\mathrm{BB}), \mathrm{H}_{2} \mathrm{O}(\mathrm{BA}), \mathrm{HCN}(\mathrm{CA}), \mathrm{OH}^{-}(\mathrm{CB})$; (c) $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{BA}), \mathrm{Cl}^{-}(\mathrm{BB}), \mathrm{HCl}(\mathrm{CA}), \mathrm{HSO}_{4}{ }^{-}(\mathrm{CB})$; (d) $\mathrm{HSO}_{4}^{-}(\mathrm{BA}), \mathrm{OH}^{-}(\mathrm{BB}), \mathrm{SO}_{4}{ }^{2-}(\mathrm{CB}), \mathrm{H}_{2} \mathrm{O}(\mathrm{CA}) ;$ (e) $\mathrm{O}^{2-}(\mathrm{BB})$, $\mathrm{H}_{2} \mathrm{O}(\mathrm{BA}) \mathrm{OH}^{-}(\mathrm{CB}$ and CA$)$; (f) $\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}(\mathrm{OH})\right]^{+}(\mathrm{BB}),\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(\mathrm{BA}),\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}(\mathrm{CA}),\left[\mathrm{Al}\left(\mathrm{H}_{2}


2297. Chapter 14 -
\mathrm{O}\right)_{4}\right]^{2+}(\mathrm{CA}),\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})\right]^{2+}(\mathrm{CB})$; (g) $\mathrm{H}_{2} \mathrm{~S}(\mathrm{BA}), \mathrm{NH}_{2}{ }^{-}(\mathrm{BB}), \mathrm{HS}^{-}(\mathrm{CB}), \mathrm{NH}_{3}(\mathrm{CA})$


2297. Chapter 14 -
6. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is $\mathrm{H}_{2} \mathrm{O}$. As an acid: $\mathrm{H}_{2} \mathrm{O}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{NH}_{4}+(a q)+\mathrm{OH}^{-}(a q)$. As a base: $\mathrm{H}_{2} \mathrm{O}(a q)+\mathrm{HCl}(a q) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q)$
7. amphiprotic: (a) $\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{NH}_{4} \mathrm{OH}+\mathrm{H}_{2} \mathrm{O}, \mathrm{NH}_{3}+\mathrm{OCH}_{3}-\longrightarrow \mathrm{NH}_{2}^{-}+\mathrm{CH}_{3} \mathrm{OH}$; (b) $\mathrm{HPO}_{4}{ }^{2-}+\mathrm{OH}^{-} \longrightarrow \mathrm{PO}_{4}{ }^{3-}+\mathrm{H}_{2} \mathrm{O}, \mathrm{HPO}_{4}{ }^{2-}+\mathrm{HClO}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}+\mathrm{ClO}_{4}{ }^{-}$; not amphiprotic: (c) $\mathrm{Br}^{-}$; (d) $\mathrm{NH}_{4}{ }^{+}$; (e) $\mathrm{AsO}_{4}{ }^{3-}$


2297. Chapter 14 -
8. In a neutral solution $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]$. At $40^{\circ} \mathrm{C},\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\left(2.910 \times 10^{-14}\right)^{1 / 2}=1.7 \times 10^{-7}$.
9. $x=3.051 \times 10^{-7} M=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right] ; \mathrm{pH}=-\log 3.051 \times 10^{-7}=-(-6.5156)=6.5156 ; \mathrm{pOH}=\mathrm{pH}=6.5156$
10. (a) $\mathrm{pH}=3.587 ; \mathrm{pOH}=10.413$; (b) $\mathrm{pOH}=0.68 ; \mathrm{pH}=13.32$; (c) $\mathrm{pOH}=3.85 ; \mathrm{pH}=10.15$; (d) $\mathrm{pOH}=-0.40$; pH $=14.4$
11. $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=3.0 \times 10^{-7} \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=3.3 \times 10^{-8} \mathrm{M}$
12. $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1 \times 10^{-2} \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=1 \times 10^{-12} \mathrm{M}$
13. $\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-12} \mathrm{M}$


2297. Chapter 14 -
13. $\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-12} \mathrm{M}$
14. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms $\mathrm{OH}^{-}$, which causes the solution to be basic.
15. $\left[\mathrm{H}_{2} \mathrm{O}\right]>\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]>\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \approx\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]>\left[\mathrm{OH}^{-}\right]$
16. The oxidation state of the sulfur in $\mathrm{H}_{2} \mathrm{SO}_{4}$ is greater than the oxidation state of the sulfur in $\mathrm{H}_{2} \mathrm{SO}_{3}$.
17. $\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$


2297. Chapter 14 -
BB BA CB CA
35. $K_{\mathrm{a}}=2.3 \times 10^{-11}$
37. The stronger base or stronger acid is the one with the larger $K_{\mathrm{b}}$ or $K_{\mathrm{a}}$, respectively. In these two examples, they are $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$ and $\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}$.
39. triethylamine
41. (a) $\mathrm{HSO}_{4}{ }^{-}$; higher electronegativity of the central ion. (b) $\mathrm{H}_{2} \mathrm{O}$; $\mathrm{NH}_{3}$ is a base and water is neutral, or decide on the basis of $K_{\mathrm{a}}$ values. (c) $\mathrm{HI} ; \mathrm{PH}_{3}$ is weaker than $\mathrm{HCl} ; \mathrm{HCl}$ is weaker than HI . Thus, $\mathrm{PH}_{3}$ is weaker than HI . (d) $\mathrm{PH}_{3}$; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr ; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S , so HBr is the stronger acid.


2297. Chapter 14 -
43. (a) $\mathrm{NaHSeO}_{3}<\mathrm{NaHSO}_{3}<\mathrm{NaHSO}_{4}$; in polyoxy acids, the more electronegative central element-S, in this case-forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) $\mathrm{ClO}_{2}{ }^{-}<\mathrm{BrO}_{2}{ }^{-}<\mathrm{IO}_{2}{ }^{-}$; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br , and I is the least electronegative of the three. (c) $\mathrm{HOI}<\mathrm{HOBr}<\mathrm{HOCl}$; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br , and I is the least


2297. Chapter 14 -
electronegative of the three. (d) $\mathrm{HOCl}<\mathrm{HOClO}<\mathrm{HOClO}_{2}<\mathrm{HOClO}_{3}$; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) $\mathrm{HTe}^{-}<\mathrm{HS}^{-} \ll \mathrm{PH}_{2}^{-}<\mathrm{NH}_{2}^{-} ; \mathrm{PH}_{2}^{-}$and $\mathrm{NH}_{2}-$ are anions of weak bases, so they act as strong bases toward $\mathrm{H}^{+}$. $\mathrm{HTe}^{-}$and $\mathrm{HS}^{-}$are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) $\mathrm{BrO}_{4}{ }^{-}<\mathrm{BrO}_{3}{ }^{-}<\mathrm{BrO}_{2}{ }^{-}<\mathrm{BrO}^{-}$; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
45. $\left[\mathrm{H}_{2} \mathrm{O}\right]>\left[\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{CO}_{2} \mathrm{H}\right)\right]>\left[\mathrm{H}^{+}\right] 0>\left[\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OH}\left(\mathrm{CO}_{2}\right)^{-}\right] \gg\left[\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}\left(\mathrm{CO}_{2} \mathrm{H}\right)^{-}\right]>\left[\mathrm{OH}^{-}\right]$


2297. Chapter 14 -
47. 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$.
48. (b) The addition of HCl
50. (a) Adding HCl will add $\mathrm{H}_{3} \mathrm{O}^{+}$ions, which will then react with the $\mathrm{OH}^{-}$ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of $\mathrm{HO}_{2}$, and decreasing the concentration of $\mathrm{NO}_{2}{ }^{-}$ions. (b) Adding $\mathrm{HNO}_{2}$ increases the concentration of $\mathrm{HNO}_{2}$ and shifts the equilibrium to the left, increasing the concentration of $\mathrm{NO}_{2}{ }^{-}$ions and decreasing the concentration of $\mathrm{OH}^{-}$ions. (c) Adding NaOH adds $\mathrm{OH}^{-}$ions, which shifts the equilibrium to the left, increasing the concentration of $\mathrm{NO}_{2}{ }^{-}$ions and decreasing the concentrations of $\mathrm{HNO}_{2}$. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding $\mathrm{KNO}_{2}$ adds $\mathrm{NO}_{2}{ }^{-}$ions and shifts the equilibrium to the right, increasing the $\mathrm{HNO}_{2}$ and $\mathrm{OH}^{-}$ion concentrations.


2297. Chapter 14 -
52. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the $\mathrm{HCO}_{2} \mathrm{H}$ exists primarily as $\mathrm{HCO}_{2} \mathrm{H}$ molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the $\mathrm{HCO}_{2} \mathrm{H}$ contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl , is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$produced by the stronger acid.
54. (a) $K_{\mathrm{b}}=1.8 \times 10^{-5}$; (b) $K_{\mathrm{a}}=4.5 \times 10^{-4}$; (c) $K_{\mathrm{b}}=6.4 \times 10^{-5}$; (d) $K_{\mathrm{a}}=5.6 \times 10^{-10}$
56. $K_{\mathrm{a}}=1.2 \times 10^{-2}$
58. (a) $K_{\mathrm{b}}=4.3 \times 10^{-12}$ (b) $K_{\mathrm{a}}=1.6 \times 10^{10}$ (c) $K_{\mathrm{b}}=5.9 \times 10^{8}$ (d) $K_{\mathrm{b}}=4.2 \times 10^{-3}$ (e) $K_{\mathrm{b}}=2.3 \times 10^{5}$ (f) $K_{\mathrm{b}}=6.3 \times 10^{-13}$


2297. Chapter 14 -
60. (a) $\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{ClO}^{-}\right]}{[\mathrm{HClO}]}=\frac{(x)(x)}{(0.0092-x)} \approx \frac{(x)(x)}{0.0092}=2.9 \times 10^{-8}$


2297. Chapter 14 -
Solving for $x$ gives $1.63 \times 10^{-5} \mathrm{M}$. This value is less than $5 \%$ of 0.0092 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{ClO}^{-}\right]=1.6 \times 10^{-5} \mathrm{M}$
$\left[\mathrm{HClO}^{-}\right]=0.0092 \mathrm{M}$
$\left[\mathrm{OH}^{-}\right]=6.1 \times 10^{-10} \mathrm{M}$;
(b) $\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}=\frac{(x)(x)}{(0.0784-x)} \approx \frac{(x)(x)}{0.0784}=4.3 \times 10^{-10}$


2297. Chapter 14 -
Solving for $x$ gives $5.81 \times 10^{-6} \mathrm{M}$. This value is less than $5 \%$ of 0.0784 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\right]=\left[\mathrm{OH}^{-}\right]=5.8 \times 10^{-6} \mathrm{M}$
$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=0.0784 \mathrm{M}$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.7 \times 10^{-9} \mathrm{M}$;
(c) $\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}=\frac{(x)(x)}{(0.0810-x)} \approx \frac{(x)(x)}{0.0810}=4.9 \times 10^{-10}$


2297. Chapter 14 -
Solving for $x$ gives $6.30 \times 10^{-6} \mathrm{M}$. This value is less than $5 \%$ of 0.0810 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CN}^{-}\right]=6.3 \times 10^{-6} \mathrm{M}$
$[\mathrm{HCN}]=0.0810 \mathrm{M}$
$\left[\mathrm{OH}^{-}\right]=1.6 \times 10^{-9} \mathrm{M}$;
(d) $\frac{\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\right]}=\frac{(x)(x)}{(0.11-x)} \approx \frac{(x)(x)}{0.11}=6.3 \times 10^{-5}$


2297. Chapter 14 -
Solving for $x$ gives $2.63 \times 10^{-3} \mathrm{M}$. This value is less than $5 \%$ of 0.11 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=\left[\mathrm{OH}^{-}\right]=2.6 \times 10^{-3} \mathrm{M}$
$\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\right]=0.11 \mathrm{M}$
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=3.8 \times 10^{-12} \mathrm{M}$;
(e) $\frac{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{+}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]}=\frac{(x)(x)}{(0.120-x)} \approx \frac{(x)(x)}{0.120}=1.6 \times 10^{-7}$


2297. Chapter 14 -
Solving for $x$ gives $1.39 \times 10^{-4} \mathrm{M}$. This value is less than $5 \%$ of 0.120 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{+}\right]=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.4 \times 10^{-4} \mathrm{M}$
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}\right]=0.120 \mathrm{M}$
$\left[\mathrm{OH}^{-}\right]=7.2 \times 10^{-11} \mathrm{M}$
62. $\mathrm{pH}=2.41$
64. $\left[\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}\right]=0.049 \mathrm{M} ;\left[\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2} \mathrm{H}^{+}\right]=1.9 \times 10^{-4} \mathrm{M} ;\left[\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2} \mathrm{H}_{2}{ }^{2+}\right]=1.4 \times 10^{-11} \mathrm{M} ;\left[\mathrm{OH}^{-}\right]=1.9 \times 10^{-4}$ $M ;\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5.3 \times 10^{-11} \mathrm{M}$
66. $K_{\mathrm{a}}=1.2 \times 10^{-2}$
68. $K_{\mathrm{b}}=1.77 \times 10^{-5}$
70. (a) acidic; (b) basic; (c) acidic; (d) neutral


2297. Chapter 14 -
68. $K_{\mathrm{b}}=1.77 \times 10^{-5}$
70. (a) acidic; (b) basic; (c) acidic; (d) neutral
72. $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$and $\left[\mathrm{HCO}_{3}{ }^{-}\right]$are practically equal
74. $\left[\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}\right] 7.2 \times 10^{-3} \mathrm{M},\left[\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)\left(\mathrm{CO}_{2}\right)^{-}\right]=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] 2.8 \times 10^{-3} \mathrm{M},\left[\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2}\right)_{2}{ }^{2-}\right] 3.9 \times 10^{-6} \mathrm{M}$, $\left[\mathrm{OH}^{-}\right] 3.6 \times 10^{-12} \mathrm{M}$
76. (a) $K_{\mathrm{a} 2}=1.5 \times 10^{-11}$;
(b) $K_{\mathrm{b}}=4.3 \times 10^{-12}$;
(c) $\frac{\left[\mathrm{Te}^{2-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{HTe}^{-}\right]}=\frac{(x)(0.0141+x)}{(0.0141-x)} \approx \frac{(x)(0.0141)}{0.0141}=1.5 \times 10^{-11}$


2297. Chapter 14 -
Solving for $x$ gives $1.5 \times 10^{-11} M$. Therefore, compared with $0.014 M$, this value is negligible $(1.1 \times$ $10^{-7} \%$ ).
78. Excess $\mathrm{H}_{3} \mathrm{O}^{+}$is removed primarily by the reaction: $\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ Excess base is removed by the reaction: $\mathrm{OH}^{-}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
80. $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.5 \times 10^{-4} \mathrm{M}$
82. $\left[\mathrm{OH}^{-}\right]=4.2 \times 10^{-4} \mathrm{M}$


2297. Chapter 14 -
84. (a) The added HCl will increase the concentration of $\mathrm{H}_{3} \mathrm{O}^{+}$slightly, which will react with $\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}$and produce $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ in the process. Thus, $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$decreases and $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]$ increases. (b) The added $\mathrm{KCH}_{3} \mathrm{CO}_{2}$ will increase the concentration of $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$which will react with $\mathrm{H}_{3} \mathrm{O}^{+}$and produce $\mathrm{CH}_{3} \mathrm{CO}_{2}$ H in the process. Thus, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$decreases slightly and $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]$ increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce $\mathrm{OH}^{-}$ions, which will react with the $\mathrm{H}_{3} \mathrm{O}^{+}$, thus reducing $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$. Some additional $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ will dissociate, producing $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$ions in the process. Thus, $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]$ decreases slightly and $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$increases. (e) The added $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ will increase its concentration, causing more of it to dissociate and producing more $\left[\mathrm{CH}_{3}


2297. Chapter 14 -
(e) The added $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ will increase its concentration, causing more of it to dissociate and producing more $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$and $\mathrm{H}_{3} \mathrm{O}^{+}$in the process. Thus, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$increases slightly and $\left[\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right]$increases.


2297. Chapter 14 -
86. $\mathrm{pH}=8.95$
88. $37 \mathrm{~g}(0.27 \mathrm{~mol})$
90. (a) $\mathrm{pH}=5.222$; (b) The solution is acidic. (c) $\mathrm{pH}=5.220$
92. At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
94.
(a) $\mathrm{pH}=2.50$; (b)
(b) $\mathrm{pH}=4.01$;
(c) $\mathrm{pH}=5.60$;
; (d) $\mathrm{pH}=8.35$; (e) $\mathrm{pH}=11.08$


2298. Chapter 15 -
1. $(\mathrm{a})$

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

$x \quad \underline{x}$
(b)
$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q)$
$\underline{x} \quad x$
(c)
$\begin{array}{ll}\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q) & +2 \mathrm{OH}^{-}(a q) \\ x & \underline{2 x}\end{array}$
(d)
$\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{PO}_{4}{ }^{3-}(a q)$

$$
\underline{x} \quad \frac{2}{3} x
$$


2298. Chapter 15 -
(e)
$\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}{ }^{3-}(a q)+\mathrm{OH}^{-}(a q)$
$\underline{5 x} \quad \underline{3 x} \quad x$
3. There is no change. A solid has an activity of 1 whether there is a little or a lot.
5. The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.
7. $\mathrm{CaF}_{2}, \mathrm{MnCO}_{3}$, and ZnS
9. (a) $\mathrm{LaF}_{3}(s) \rightleftharpoons \mathrm{La}^{3+}(a q)+3 \mathrm{~F}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{La}^{3+}\right]\left[\mathrm{F}^{-}\right]^{3}$;
(b) $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \quad K_{\text {sp }}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}{ }^{2-}\right]$;
(c) $\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{SO}_{4}{ }^{2-}\right]$;


2298. Chapter 15 -
(d) $\mathrm{Pb}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$
11. (a) $1.77 \times 10^{-7}$; (b) $1.6 \times 10^{-6}$; (c) $2.2 \times 10^{-9}$; (d) $7.91 \times 10^{-22}$
13. (a) $2 \times 10^{-2} \mathrm{M}$; (b) $1.5 \times 10^{-3} \mathrm{M}$; (c) $2.27 \times 10^{-9} \mathrm{M}$; (d) $2.2 \times 10^{-10} \mathrm{M}$
15. (a) $6.4 \times 10^{-9} M=\left[\mathrm{Ag}^{+}\right],\left[\mathrm{Cl}^{-}\right]=0.025 M$. Check: $\frac{6.4 \times 10^{-9} M}{0.025 M} \times 100 \%=2.6 \times 10^{-5} \%$,an insignificant change;
(b) $2.2 \times 10^{-5} M=\left[\mathrm{Ca}^{2+}\right],\left[\mathrm{F}^{-}\right]=0.0013 M$. Check: $\frac{2.26 \times 10^{-5} M}{0.00133 M} \times 100 \%=1.70 \%$. This value is less than $5 \%$ and can be ignored.


2298. Chapter 15 -
(c) $0.2238 M=\left[\mathrm{SO}_{4}^{2-}\right] ;\left[\mathrm{Ag}^{+}\right]=7.4 \times 10^{-3} M$. Check: $\frac{3.7 \times 10^{-3}}{0.2238} \times 100 \%=1.64 \times 10^{-2}$; the condition is satisfied.
(d) $\left[\mathrm{OH}^{-}\right]=2.8 \times 10^{-3} M$; $5.7 \times 10^{-12} M=\left[\mathrm{Zn}^{2+}\right]$. Check: $\frac{5.7 \times 10^{-12}}{2.8 \times 10^{-3}} \times 100 \%=2.0 \times 10^{-7} \% ; x$ is less than $5 \%$ of $\left[\mathrm{OH}^{-}\right]$and is, therefore, negligible.
17. (a) $\left[\mathrm{Cl}^{-}\right]=7.6 \times 10^{-3} \mathrm{M}$


2298. Chapter 15 -
Check: $\frac{7.6 \times 10^{-3}}{0.025} \times 100 \%=30 \%$
This value is too large to drop $x$. Therefore solve by using the quadratic equation:
$\left[\mathrm{Ti}^{+}\right]=3.1 \times 10^{-2} \mathrm{M}$
$\left[\mathrm{Cl}^{-}\right]=6.1 \times 10^{-3}$
(b) $\left[\mathrm{Ba}^{2+}\right]=7.7 \times 10^{-4} \mathrm{M}$


2298. Chapter 15 -
Check: $\frac{7.7 \times 10^{-4}}{0.0313} \times 100 \%=2.4 \%$
Therefore, the condition is satisfied.
$\left[\mathrm{Ba}^{2+}\right]=7.7 \times 10^{-4} \mathrm{M}$
$\left[\mathrm{F}^{-}\right]=0.0321 \mathrm{M}$;
(c) $\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}=0.02444 \mathrm{M}$
$\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=2.9 \times 10^{-5}$
Check: $\frac{2.9 \times 10^{-5}}{0.02444} \times 100 \%=0.12 \%$
The condition is satisfied; the above value is less than $5 \%$.
$\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=2.9 \times 10^{-5} \mathrm{M}$
$\left[\mathrm{Mg}^{2+}\right]=0.0244 \mathrm{M}$
(d) $\left[\mathrm{OH}^{-}\right]=0.0501 \mathrm{M}$
$\left[\mathrm{Ca}^{2+}\right]=3.15 \times 10^{-3}$
Check: $\frac{3.15 \times 10^{-3}}{0.050} \times 100 \%=6.28 \%$
This value is greater than $5 \%$, so a more exact method, such as successive approximations, must be used.
$\left[\mathrm{Ca}^{2+}\right]=2.8 \times 10^{-3} \mathrm{M}$
$\left[\mathrm{OH}^{-}\right]=0.053 \times 10^{-2} \mathrm{M}$
19. The changes in concentration are greater than $5 \%$ and thus exceed the maximum value for disregarding the change.


2298. Chapter 15 -
19. The changes in concentration are greater than $5 \%$ and thus exceed the maximum value for disregarding the change.
21. $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ is the most soluble Ca salt in $\mathrm{mol} / \mathrm{L}$, and it is also the most soluble Ca salt in $\mathrm{g} / \mathrm{L}$.
23. $4.8 \times 10^{-3} M=\left[\mathrm{SO}_{4}{ }^{2-}\right]=\left[\mathrm{Ca}^{2+}\right]$; Since this concentration is higher than $2.60 \times 10^{-3} M$, "gyp" water does not meet the standards.
25. Mass $\left(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)=0.72 \mathrm{~g} / \mathrm{L}$


2298. Chapter 15 -
25. Mass $\left(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)=0.72 \mathrm{~g} / \mathrm{L}$
27. (a) $\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{I}^{-}\right]=1.3 \times 10^{-5} \mathrm{M}$; (b) $\left[\mathrm{Ag}^{+}\right]=2.88 \times 10^{-2} \mathrm{M},\left[\mathrm{SO}_{4}{ }^{2-}\right]=1.44 \times 10^{-2} \mathrm{M}$; (c) $\left[\mathrm{Mn}^{2+}\right]=3.7 \times 10^{-5} \mathrm{M}$, $\left[\mathrm{OH}^{-}\right]=7.4 \times 10^{-5} \mathrm{M}$; (d) $\left[\mathrm{Sr}^{2+}\right]=4.3 \times 10^{-2} \mathrm{M},\left[\mathrm{OH}^{-}\right]=8.6 \times 10^{-2} \mathrm{M}$; (e) $\left[\mathrm{Mg}^{2+}\right]=1.3 \times 10^{-4} \mathrm{M},\left[\mathrm{OH}^{-}\right]=2.6 \times$ $10^{-4} \mathrm{M}$.
29. (a) $1.45 \times 10^{-4}$; (b) $8.2 \times 10^{-55}$; (c) $1.35 \times 10^{-4}$; (d) $1.18 \times 10^{-5}$; (e) $1.08 \times 10^{-10}$
31. (a) $\mathrm{CaCO}_{3}$ does precipitate. (b) The compound does not precipitate. (c) The compound does not precipitate.
(d) The compound precipitates.
33. $3.03 \times 10^{-7} \mathrm{M}$
35. $9.2 \times 10^{-13} \mathrm{M}$


2298. Chapter 15 -
(d) The compound precipitates.
33. $3.03 \times 10^{-7} \mathrm{M}$
35. $9.2 \times 10^{-13} \mathrm{M}$
37. $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-3} \mathrm{M}$
39. $6.3 \times 10^{-4}$
41. (a) 2.25 L ; (b) $7.2 \times 10^{-7} \mathrm{~g}$
43. $100 \%$ of it is dissolved
45. (a) $\mathrm{Hg}_{2}{ }^{2+}$ and $\mathrm{Cu}^{2+}$ : Add $\mathrm{SO}_{4}{ }^{2-}$. (b) $\mathrm{SO}_{4}{ }^{2-}$ and $\mathrm{Cl}^{-}$: Add $\mathrm{Ba}^{2+}$. (c) $\mathrm{Hg}^{2+}$ and $\mathrm{Co}^{2+}$ : Add $\mathrm{S}^{2-}$. (d) $\mathrm{Zn}^{2+}$ and $\mathrm{Sr}^{2+}$ : Add $\mathrm{OH}^{-}$until $\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}$. (e) $\mathrm{Ba}^{2+}$ and $\mathrm{Mg}^{2+}$ : Add $\mathrm{SO}_{4}{ }^{2-}$. (f) $\mathrm{CO}_{3}{ }^{2-}$ and $\mathrm{OH}^{-}$: Add $\mathrm{Ba}^{2+}$.
47. AgI will precipitate first.
49. $1.5 \times 10^{-12} \mathrm{M}$
51. 3.99 kg
53. (a) $3.1 \times 10^{-11}$; (b) $\left[\mathrm{Cu}^{2+}\right]=2.6 \times 10^{-3} ;\left[\mathrm{IO}_{3}{ }^{-}\right]=5.3 \times 10^{-3}$


2298. Chapter 15 -
55. $1.8 \times 10^{-5} \mathrm{~g} \mathrm{~Pb}(\mathrm{OH})_{2}$
57. $\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{OH}^{-} \quad K_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$ $1.23 \times 10^{-3} \mathrm{~g} \mathrm{Mg}(\mathrm{OH})_{2}$
59. $\mathrm{MnCO}_{3}$ will form first since it has the smallest $K_{\text {sp }}$ value among these homologous compounds and is therefore the least soluble. $\mathrm{MgCO}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}$ will be the last to precipitate since it has the largest K _sp value and is the most soluble. $K_{\mathrm{sp}}$ value.
62. when the amount of solid is so small that a saturated solution is not produced
64. $1.8 \times 10^{-5} \mathrm{M}$
66. $5 \times 10^{23}$
68.


2298. Chapter 15 -
|  | $\left[\mathrm{Cd}(\mathrm{CN})_{4}{ }^{2}\right]$ |  | $[\mathrm{CN}]$ |
| :---: | :---: | :---: | :---: |$]\left[\mathrm{Cd}^{2+}\right]$

$$
\left[\mathrm{Cd}^{2+}\right]=9.5 \times 10^{-5} \mathrm{M} ;\left[\mathrm{CN}^{-}\right]=3.8 \times 10^{-4} \mathrm{M}
$$

70. $\left[\mathrm{Co}^{3+}\right]=3.0 \times 10^{-6} \mathrm{M} ;\left[\mathrm{NH}_{3}\right]=1.8 \times 10^{-5} \mathrm{M}$
71. 1.3 g
72. 0.79 g
73. (a)

(b)

(c)

(d)

Acid Base
(e)

Acid Base
78. (a)

$$
\mathrm{HCl}(g)+\mathrm{PH}_{3}(g) \longrightarrow\left[\mathrm{PH}_{4}\right]^{+}+[\because \ddot{\mathrm{Cl}}:]^{-}
$$

(b) $\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3}{ }^{-} \longrightarrow \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O}$

(c) $\mathrm{CaO}+\mathrm{SO}_{3} \longrightarrow \mathrm{CaSO} 4$

(d) $\mathrm{NH}_{4}{ }^{+}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{NH}_{3}$


2298. Chapter 15 -
80. 0.0281 g
82. $\mathrm{HNO}_{3}(l)+\mathrm{HF}(l) \longrightarrow \mathrm{H}_{2} \mathrm{NO}_{3}{ }^{+}+\mathrm{F}^{-} ; \mathrm{HF}(l)+\mathrm{BF}_{3}(g) \longrightarrow \mathrm{H}^{+}+\mathrm{BF}_{4}$
84. (a) $\mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{4} \mathrm{BO}_{4}^{-}+\mathrm{H}^{+}$; (b) The electronic and molecular shapes are the same-both tetrahedral. (c) The tetrahedral structure is consistent with $s p^{3}$ hybridization.
86. $0.014 M$
88. $7.2 \times 10^{-15} \mathrm{M}$
90. $4.4 \times 10^{-22} \mathrm{M}$
93. $\left[\mathrm{OH}^{-}\right]=4.5 \times 10^{-6} ;\left[\mathrm{Al}^{3+}\right]=2 \times 10^{-16}$ (molar solubility)
95. $\left[\mathrm{SO}_{4}{ }^{2-}\right]=0.049 \mathrm{M} ;\left[\mathrm{Ba}^{2+}\right]=4.7 \times 10^{-7}$ (molar solubility)
97. $\left[\mathrm{OH}^{-}\right]=7.6 \times 10^{-3} \mathrm{M} ;\left[\mathrm{Pb}^{2+}\right]=2.1 \times 10^{-11}$ (molar solubility)
99. 7.66
101. (a) $K_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}=\left(1.21 \times 10^{-3}\right)\left(2 \times 1.21 \times 10^{-3}\right)^{2}=7.09 \times 10^{-9}$


2298. Chapter 15 -
(b) $7.09 \times 10^{-7} \mathrm{M}$
(c) Determine the concentration of $\mathrm{Mg}^{2+}$ and $\mathrm{F}^{-}$that will be present in the final volume. Compare the value of the ion product $\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}$ with $K_{\mathrm{sp}}$. If this value is larger than $K_{\mathrm{sp}}$, precipitation will occur.
$0.1000 \mathrm{~L} \times 3.00 \times 10^{-3} \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}=0.3000 \mathrm{~L} \times M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}$
$M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}=1.00 \times 10^{-3} \mathrm{M}$
$0.2000 \mathrm{~L} \times 2.00 \times 10^{-3} M \mathrm{NaF}=0.3000 \mathrm{~L} \times M \mathrm{NaF}$
$M \mathrm{NaF}=1.33 \times 10^{-3} \mathrm{M}$
ion product $=\left(1.00 \times 10^{-3}\right)\left(1.33 \times 10^{-3}\right)^{2}=1.77 \times 10^{-9}$ This value is smaller than $K_{\mathrm{sp}}$, so no precipitation will occur.
(d) $\mathrm{MgF}_{2}$ is less soluble at $27^{\circ} \mathrm{C}$ than at $18{ }^{\circ} \mathrm{C}$. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier's principle states that the equilibrium will shift to the reactants' side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.


2298. Chapter 15 -
103. $\mathrm{BaF}_{2}, \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$, ZnS ; each is a salt of a weak acid, and the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations
105. Effect on amount of solid $\mathrm{CaHPO}_{4},\left[\mathrm{Ca}^{2+}\right],\left[\mathrm{OH}^{-}\right]$: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect


2299. Chapter 16 -
1. (a) reduction; (b) oxidation; (c) oxidation; (d) reduction
2. (a) $\mathrm{F}_{2}+\mathrm{Ca} \longrightarrow 2 \mathrm{~F}^{-}+\mathrm{Ca}^{2+}$; (b) $\mathrm{Cl}_{2}+2 \mathrm{Li} \longrightarrow 2 \mathrm{Li}^{+}+2 \mathrm{Cl}^{-}$; (c) $3 \mathrm{Br}_{2}+2 \mathrm{Fe} \longrightarrow 2 \mathrm{Fe}^{3+}+6 \mathrm{Br}^{-}$; (d) $\mathrm{MnO}_{4}-+4 \mathrm{H}^{+}+3 \mathrm{Ag} \longrightarrow 3 \mathrm{Ag}^{+}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
3. Oxidized: (a) $\mathrm{Sn}^{2+}$; (b) Hg ; (c) Al; reduced: (a) $\mathrm{H}_{2} \mathrm{O}_{2}$; (b) $\mathrm{PbO}_{2}$; (c) $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$; oxidizing agent: (a) $\mathrm{H}_{2} \mathrm{O}_{2}$; (b) $\mathrm{PbO}_{2}$; (c) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$; reducing agent: (a) $\mathrm{Sn}^{2+}$; (b) Hg ; (c) Al
4. Oxidized = reducing agent: (a) $\mathrm{SO}_{3}{ }^{2-}$; (b) $\mathrm{Mn}(\mathrm{OH})_{2}$; (c) $\mathrm{H}_{2}$; (d) Al; reduced = oxidizing agent: (a) $\mathrm{Cu}(\mathrm{OH})_{2}$; (b) $\mathrm{O}_{2}$; (c) $\mathrm{NO}_{3}^{-}$; (d) $\mathrm{CrO}_{4}{ }^{2-}$


2299. Chapter 16 -
5. In basic solution, $\left[\mathrm{OH}^{-}\right]>1 \times 10^{-7} M>\left[\mathrm{H}^{+}\right]$. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.
6. (a) $\operatorname{Mg}(s)\left|\operatorname{Mg}^{2+}(a q) \| \mathrm{Ni}^{2+}(a q)\right| \mathrm{Ni}(s)$; (b) $\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q) \| \mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s)$; (c)
$\operatorname{Mn}(s)\left|\mathrm{Mn}^{2+}(a q)\left\|\mathrm{Sn}^{2+}(a q)|\operatorname{Sn}(s) ;(\mathrm{d}) \operatorname{Pt}(s)| \mathrm{Cu}^{+}(a q), \mathrm{Cu}^{2+}(a q)\right\| \mathrm{Au}^{3+}(a q)\right| \mathrm{Au}(s)$
7. (a) $\mathrm{Mg}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)$; (b) $2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ag}(s)$


2299. Chapter 16 -
8. Species oxidized = reducing agent: (a) $\mathrm{Al}(\mathrm{s})$; (b) $\mathrm{NO}(g)$; (c) $\mathrm{Mg}(s)$; and (d) $\mathrm{MnO}_{2}(s)$; Species reduced $=$ oxidizing agent: (a) $\mathrm{Zr}^{4+}(a q)$; (b) $\mathrm{Ag}^{+}(a q)$; (c) $\mathrm{SiO}_{3}{ }^{2-}$ (aq); and (d) $\mathrm{ClO}_{3}{ }^{-}(a q)$
9. Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.
10. Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.
11. (a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)
12. $3 \mathrm{Cu}(s)+2 \mathrm{Au}^{3+}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{Au}(s)$; +1.16 V ; spontaneous
13. $3 \mathrm{Cd}(s)+2 \mathrm{Al}^{3+}(a q) \longrightarrow 3 \mathrm{Cd}^{2+}(a q)+2 \mathrm{Al}(s)$; -1.259 V ; nonspontaneous
14. (a) $0 \mathrm{~kJ} / \mathrm{mol}$; (b) $-83.7 \mathrm{~kJ} / \mathrm{mol}$; (c) $+235.3 \mathrm{~kJ} / \mathrm{mol}$


2299. Chapter 16 -
15. (a) standard cell potential: 1.50 V , spontaneous; cell potential under stated conditions: 1.43 V , spontaneous; (b) standard cell potential: 1.405 V , spontaneous; cell potential under stated conditions: 1.423 V, spontaneous; (c) standard cell potential: -2.749 V, nonspontaneous; cell potential under stated conditions: -2.757 V , nonspontaneous
16. (a) $1.7 \times 10^{-10}$; (b) $2.6 \times 10^{-21}$; (c) $4.693 \times 10^{21}$; (d) $1.0 \times 10^{-14}$
17. (a) anode: $\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-}$
cathode: $2 \times\left(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\right)$
$E_{\text {anode }}^{\circ}=0.34 \mathrm{~V}$


2299. Chapter 16 -
$E^{\circ}=0.7996 \mathrm{~V}$$\quad$; (b) $3.5 \times 10^{15}$; (c) $5.6 \times 10^{-9}$

M
34. Batteries are self-contained and have a limited supply of reagents to expend before going dead.


2299. Chapter 16 -
Alternatively, battery reaction byproducts accumulate and interfere with the reaction. Because a fuel cell is constantly resupplied with reactants and products are expelled, it can continue to function as long as reagents are supplied.
36. $E_{\text {cell }}$, as described in the Nernst equation, has a term that is directly proportional to temperature. At low temperatures, this term is decreased, resulting in a lower cell voltage provided by the battery to the device-the same effect as a battery running dead.
38. Mg and Zn
40. Both examples involve cathodic protection. The (sacrificial) anode is the metal that corrodes (oxidizes or reacts). In the case of iron ( -0.447 V ) and zinc ( -0.7618 V ), zinc has a more negative standard reduction potential and so serves as the anode. In the case of iron and copper ( 0.34 V ), iron has the smaller standard reduction potential and so corrodes (serves as the anode).
42. While the reduction potential of lithium would make it capable of protecting the other metals, this high potential is also indicative of how reactive lithium is; it would have a spontaneous reaction with most substances. This means that the lithium would react quickly with other substances, even those that would not oxidize the metal it is attempting to protect. Reactivity like this means the sacrificial anode would be depleted rapidly and need to be replaced frequently. (Optional additional reason: fire hazard in the presence of water.)


2299. Chapter 16 -
46. (a) mass $\mathrm{Ca}=69.1 \mathrm{~g}$; (b) $\begin{aligned} & \text { mass } \mathrm{Li}=23.9 \mathrm{~g} \\ & \text { mass } \mathrm{H}_{2}=3.48 \mathrm{~g}\end{aligned}$; (c) $\begin{aligned} & \text { mass } \mathrm{Al}=31.0 \mathrm{~g} \\ & \text { mass } \mathrm{Cl}_{2}=122 \mathrm{~g}\end{aligned}$; (d) $\begin{aligned} & \text { mass } \mathrm{Cr}=59.8 \mathrm{~g} \\ & \operatorname{mass} \mathrm{Br}_{2}=276 \mathrm{~g}\end{aligned}$
48. 0.79 L


2300. Chapter 17 -
1. The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.
2. rate $=+\frac{1}{2} \frac{\Delta\left[\mathrm{CIF}_{3}\right]}{\Delta t}=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t}=-\frac{1}{3} \frac{\Delta\left[\mathrm{~F}_{2}\right]}{\Delta t}$
3. (a) average rate, $0-10 \mathrm{~s}=0.0375 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$; average rate, $10-20 \mathrm{~s}=0.0265 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$; (b) instantaneous rate, $15 \mathrm{~s}=0.023 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$; (c) average rate for $B$ formation $=0.0188 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$; instantaneous rate for B formation $=0.012 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
4. Higher molarity increases the rate of the reaction. Higher temperature increases the rate of the reaction. Smaller pieces of magnesium metal will react more rapidly than larger pieces because more reactive surface exists.
5. (a) Depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other. (b) Particles of reactant must come into contact with each other before they can react.


2300. Chapter 17 -
6. (a) very slow; (b) As the temperature is increased, the reaction proceeds at a faster rate. The amount of reactants decreases, and the amount of products increases. After a while, there is a roughly equal amount of $B C, A B$, and $C$ in the mixture and a slight excess of $A$.
7. (a) 2 ; (b) 1
8. (a) The process reduces the rate by a factor of 4 . (b) Since CO does not appear in the rate law, the rate is not affected.
9. $4.3 \times 10^{-5} \mathrm{~mol} / \mathrm{L} / \mathrm{s}$
10. $7.9 \times 10^{-13} \mathrm{~mol} / \mathrm{L} /$ year
11. rate $=k$; $k=2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}$ (about $0.9 \mathrm{~g} \mathrm{~L}^{-1} \mathrm{~h}^{-1}$ for the average male); The reaction is zero order.
12. rate $=k[\mathrm{NOCl}]^{2} ; k=8.0 \times 10^{-8} \mathrm{~L} / \mathrm{mol} / \mathrm{h}$; second order
13. rate $=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] ; k=9.1 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~h}^{-1}$; second order in NO; first order in $\mathrm{Cl}_{2}$
14. (a) The rate law is second order in $A$ and is written as rate $=k[A]^{2}$. (b) $k=7.88 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$
15. (a) $2.5 \times 10^{-4} \mathrm{~mol} / \mathrm{L} / \mathrm{min}$


2300. Chapter 17 -
15. (a) $2.5 \times 10^{-4} \mathrm{~mol} / \mathrm{L} / \mathrm{min}$
16. rate $=k\left[\mathrm{I}^{-}\right]\left[\mathrm{OCl}^{-}\right] ; k=6.1 \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$
17. Plotting a graph of $\ln \left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]$ versus $t$ reveals a linear trend; therefore we know this is a first-order reaction:


2300. Chapter 17 -
$k=2.20 \times 10^{-5} \mathrm{~s}^{-1}$
18.

The plot is nicely linear, so the reaction is second order. $k=50.1 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~h}^{-1}$
36. 14.3 d
38. $8.3 \times 10^{7} \mathrm{~s}$
40. 0.826 s
42. The reaction is first order. $k=1.0 \times 10^{7} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~min}^{-1}$
44. $1.16 \times 10^{3} \mathrm{~s} ; 20 \%$ remains
46. 252 days
48.

| $[A]_{0}(M)$ | $k \times 10^{3}\left(\mathrm{~s}^{-1}\right)$ |
| :--- | :--- |
| 4.88 | 2.45 |
| 3.52 | 2.51 |
| 2.29 | 2.53 |
| 1.81 | 2.58 |
| 5.33 | 2.36 |
| 4.05 | 2.47 |
| 2.95 | 2.48 |
| 1.72 | 2.43 |


2300. Chapter 17 -
50. The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.
51. The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.
52. After finding $k$ at several different temperatures, a plot of $\ln k$ versus $\frac{1}{T}$, gives a straight line with the slope $\frac{-E_{\mathrm{a}}}{R}$ from which $E_{\mathrm{a}}$ may be determined.
53. (a) 4-times faster (b) 128-times faster
54. $3.9 \times 10^{15} \mathrm{~s}^{-1}$
55. $43.0 \mathrm{~kJ} / \mathrm{mol}$
56. $177 \mathrm{~kJ} / \mathrm{mol}$
57. $E_{\mathrm{a}}=108 \mathrm{~kJ} ; A=2.0 \times 10^{8} \mathrm{~s}^{-1} ; k=3.2 \times 10^{-10} \mathrm{~s}^{-1}$; (b) $1.81 \times 10^{8} \mathrm{~h}$ or $7.6 \times 10^{6}$ day; (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.
58. The $A$ atom has enough energy to react with $B C$; however, the different angles at which it bounces off of $B C$ without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.
59. No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate law. Yes. If the reaction is an elementary reaction, then doubling the concentration of $A$ doubles the rate.
60. Rate $=kA^{2}$; Rate $=k[A]^{3}$


2300. Chapter 17 -
60. Rate $=kA^{2}$; Rate $=k[A]^{3}$
61. (a) Rate $_{1}=k\left[\mathrm{O}_{3}\right]$; (b) Rate $_{2}=k\left\mathrm{O}_{3}\right$; (c) Rate ${ }_{3}=k\mathrm{ClO}$; (d) Rate ${ }_{2}=k\left\mathrm{O}_{3}\right$; (e) Rate ${ }_{3}=k\left\mathrm{NO}_{2}\right$
62. (a) Doubling $\left[\mathrm{H}_{2}\right]$ doubles the rate. $\left[\mathrm{H}_{2}\right]$ must enter the rate law to the first power. Doubling [NO] increases the rate by a factor of 4 . [ NO ] must enter the rate law to the second power. (b) Rate $=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]$; (c) $k=5.0$ $\times 10^{3} \mathrm{~mol}^{-2} \mathrm{~L}^{-2} \mathrm{~min}^{-1}$; (d) $0.0050 \mathrm{~mol} / \mathrm{L}$; (e) Step II is the rate-determining step. If step I gives $\mathrm{N}_{2} \mathrm{O}_{2}$ in adequate amount, steps 1 and 2 combine to give $2 \mathrm{NO}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2} \mathrm{O}$. This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3 , which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.
63. The general mode of action for a catalyst is to provide a mechanism by which the reactants can unite more
readily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium.


2300. Chapter 17 -
readily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium.
64. (a) Chlorine atoms are a catalyst because they react in the second step but are regenerated in the third step. Thus, they are not used up, which is a characteristic of catalysts. (b) NO is a catalyst for the same reason as in part (a).
65. no changes.
66. The lowering of the transition state energy indicates the effect of a catalyst. (a) B; (b) B
67. The energy needed to go from the initial state to the transition state is (a) 10 kJ ; (b) 10 kJ
68. Both diagrams describe two-step, exothermic reactions, but with different changes in enthalpy, suggesting the diagrams depict two different overall reactions.


2301. Chapter 18 -
1. The alkali metals all have a single $s$ electron in their outermost shell. In contrast, the alkaline earth metals have a completed $s$ subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period.
2.


2301. Chapter 18 -
$\mathrm{Na}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{NaI}$
$2 \mathrm{Na}+\mathrm{Se} \longrightarrow \mathrm{Na}_{2} \mathrm{Se}$
$2 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}$
$\mathrm{Sr}+\mathrm{I}_{2} \longrightarrow \mathrm{SrI}_{2}$
$\mathrm{Sr}+\mathrm{Se} \longrightarrow \mathrm{SrSe}$
$2 \mathrm{Sr}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO}$
$2 \mathrm{Al}+3 \mathrm{I}_{2} \longrightarrow 2 \mathrm{AlI}_{3}$
$2 \mathrm{Al}+3 \mathrm{Se} \longrightarrow \mathrm{Al}_{2} \mathrm{Se}_{3}$
$4 \mathrm{Al}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}$


2301. Chapter 18 -
$4 \mathrm{Al}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}$
5. The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At $20^{\circ} \mathrm{C}, \mathrm{NaCl}$ dissolves to the extent of $\frac{35.7 \mathrm{~g}}{100 \mathrm{~mL}}$ compared with $\frac{53.8 \mathrm{~g}}{100 \mathrm{~mL}}$ for $\mathrm{SrCl}_{2}$. Heating to $100^{\circ} \mathrm{C}$ provides an easy test, since the solubility of NaCl is $\frac{39.12 \mathrm{~g}}{100 \mathrm{~mL}}$, but that of $\mathrm{SrCl}_{2}$ is $\frac{100.8 \mathrm{~g}}{100 \mathrm{~mL}}$. Density determination on a solid is sometimes difficult, but there is enough difference ( $2.165 \mathrm{~g} / \mathrm{mL} \mathrm{NaCl}$ and $3.052 \mathrm{~g} / \mathrm{mL} \mathrm{SrCl}_{2}$ ) that this method would be viable and perhaps the easiest and least expensive test to perform.
7. (a) $2 \mathrm{Sr}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SrO}(s)$; (b) $\mathrm{Sr}(s)+2 \mathrm{HBr}(g) \longrightarrow \mathrm{SrBr}_{2}(s)+\mathrm{H}_{2}(g)$; (c)
$\mathrm{Sr}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{SrH}_{2}(s) ;(\mathrm{d}) 6 \mathrm{Sr}(s)+\mathrm{P}_{4}(s) \longrightarrow 2 \mathrm{Sr}_{3} \mathrm{P}_{2}(s)$; (e)


2301. Chapter 18 -
$\mathrm{Sr}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2}(g)$
9. 11 lb
11. Yes, tin reacts with hydrochloric acid to produce hydrogen gas.
13. In $\mathrm{PbCl}_{2}$, the bonding is ionic, as indicated by its melting point of $501^{\circ} \mathrm{C}$. In $\mathrm{PbCl}_{4}$, the bonding is covalent, as evidenced by it being an unstable liquid at room temperature.


2301. Chapter 18 -
$$
\begin{aligned}
& \text { countercurrent } \\
& \text { fractionating }
\end{aligned}
$$

15. $2 \mathrm{CsCl}(l)+\mathrm{Ca}(g) \xrightarrow{\text { tower }} 2 \mathrm{Cs}(g)+\mathrm{CaCl}_{2}(l)$
16. Cathode (reduction): $2 \mathrm{Li}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Li}(l)$; Anode (oxidation): $2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-}$; Overall reaction: $2 \mathrm{Li}^{+}+2 \mathrm{Cl}^{-} \longrightarrow 2 \mathrm{Li}(l)+\mathrm{Cl}_{2}(g)$
17. $0.5035 \mathrm{~g} \mathrm{H}_{2}$
18. Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning.
19. Extract from ore: $\mathrm{AlO}(\mathrm{OH})(s)+\mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Na}\left\mathrm{Al}(\mathrm{OH})_{4}\right$ Recover: $2 \mathrm{Na}\left\mathrm{Al}(\mathrm{OH})_{4}\right+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$


2301. Chapter 18 -
Sinter: $2 \mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$
Dissolve in $\mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{I})$ and electrolyze: $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s)$
25. $25.83 \%$
27. 39 kg
29. (a) $\mathrm{H}_{3} \mathrm{BPH}_{3}$ :

(b) $\mathrm{BF}_{4}^{-}$:

(c) $\mathrm{BBr}_{3}$ :

(d) $\mathrm{B}\left(\mathrm{CH}_{3}\right)_{3}$ :

(e) $\mathrm{B}(\mathrm{OH})_{3}$ :


2301. Chapter 18 -
31. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} 3 d^{0}$.
33. (a) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiH}$ : $s p^{3}$ bonding about Si ; the structure is tetrahedral; (b) $\mathrm{SiO}_{4}{ }^{4-}: s p^{3}$ bonding about Si ; the structure is tetrahedral; (c) $\mathrm{Si}_{2} \mathrm{H}_{6}: s p^{3}$ bonding about each Si ; the structure is linear along the $\mathrm{Si}-\mathrm{Si}$ bond; (d) $\mathrm{Si}(\mathrm{OH})_{4}: s p^{3}$ bonding about Si ; the structure is tetrahedral; (e) $\mathrm{SiF}_{6}{ }^{2-}: s p^{3} d^{2}$ bonding about Si; the structure is octahedral
35. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
37. (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride
39. Boron has only $s$ and $p$ orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no $d$ orbitals are available in boron.
41. (a) $\Delta H^{\circ}=87 \mathrm{~kJ} ; \Delta G^{\circ}=44 \mathrm{~kJ}$; (b) $\Delta H^{\circ}=-109.9 \mathrm{~kJ} ; \Delta G^{\circ}=-154.7 \mathrm{~kJ}$; (c) $\Delta H^{\circ}=-510 \mathrm{~kJ} ; \Delta G^{\circ}=-601.5 \mathrm{~kJ}$


2301. Chapter 18 -
43. A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond.
45. In the $N_{2}$ molecule, the nitrogen atoms have an $\sigma$ bond and two $\pi$ bonds holding the two atoms together. The presence of three strong bonds makes $\mathrm{N}_{2}$ a very stable molecule. Phosphorus is a third-period element, and as such, does not form $\pi$ bonds efficiently; therefore, it must fulfill its bonding requirement by forming three $\sigma$ bonds.
47. (a) $\mathrm{H}=1+, \mathrm{C}=2+$, and $\mathrm{N}=3-$; (b) $\mathrm{O}=2+$ and $\mathrm{F}=1-$; (c) $\mathrm{As}=3+$ and $\mathrm{Cl}=1-$
49. $\mathrm{S}<\mathrm{Cl}<\mathrm{O}<\mathrm{F}$
51. The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself.
53. Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form.
55. $0.43 \mathrm{~g} \mathrm{H}_{2}$
57. (a) $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$; (b) $\mathrm{CaO}(s)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{CaSO}_{3}(s)$;
(c) $2 \mathrm{NaHCO}_{3}(s)+\mathrm{NaH}_{2} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$
59. (a) $\mathrm{NH}^{2-}$ :


2301. Chapter 18 -
(b) $\mathrm{N}_{2} \mathrm{~F}_{4}$ :

(c) $\mathrm{NH}_{2}{ }^{-}$:

(d) $\mathrm{NF}_{3}$ :

(e) $\mathrm{N}_{3}{ }^{-}$:

61. Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate.
Brønsted base: $\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O}$
Lewis base: $2 \mathrm{NH}_{3}+\mathrm{Ag}^{+} \longrightarrow\left[\mathrm{H}_{3} \mathrm{~N}-\mathrm{Ag}-\mathrm{NH}_{3}\right]^{+}$
63. (a) $\mathrm{NO}_{2}$ :

Nitrogen is $s p^{2}$ hybridized. The molecule has a bent geometry with an ONO bond angle of approximately $120^{\circ}$.
(b) $\mathrm{NO}_{2}{ }^{-}$:

Nitrogen is $s p^{2}$ hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than $120^{\circ}$.
(c) $\mathrm{NO}_{2}{ }^{+}$:

Nitrogen is $s p$ hybridized. The molecule has a linear geometry with an ONO bond angle of $180^{\circ}$.
65. Nitrogen cannot form a $\mathrm{NF}_{5}$ molecule because it does not have $d$ orbitals to bond with the additional two fluorine atoms.
67. (a)

(b)

(c)

(d)

(e)


2301. Chapter 18 -
(b)

(c)

(d)

(e)

69. (a) $\mathrm{P}_{4}(s)+4 \mathrm{Al}(s) \longrightarrow 4 \mathrm{AlP}(s)$; (b) $\mathrm{P}_{4}(s)+12 \mathrm{Na}(s) \longrightarrow 4 \mathrm{Na}_{3} \mathrm{P}(s)$; (c) $\mathrm{P}_{4}(s)+10 \mathrm{~F}_{2}(g) \longrightarrow 4 \mathrm{PF}_{5}(l)$; (d) $\mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(l)$ or $\mathrm{P}_{4}(s)+10 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{5}(l)$; (e) $\mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s)$ or $\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) ;(\mathrm{f}) \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$
71. 291 mL
73. 28 tons
75. (a)

Tetrahedral
(b)

Trigonal bipyramid
(c)

Octahedral
(d)


2301. Chapter 18 -
77. (a) $\mathrm{P}=3+$; (b) $\mathrm{P}=5+$; (c) $\mathrm{P}=3+$; (d) $\mathrm{P}=5+$; (e) $\mathrm{P}=3-$; (f) $\mathrm{P}=5+$
79. $\mathrm{FrO}_{2}$
81. (a) $2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)$; (b) $\mathrm{ZnCO}_{3}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{CO}_{2}(g)$; (c)
$\mathrm{ZnCO}_{3}(s)+2 \mathrm{CH}_{3} \mathrm{COOH}(a q) \longrightarrow \mathrm{Zn}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$; (d)
$\mathrm{Zn}(s)+2 \mathrm{HBr}(a q) \longrightarrow \mathrm{ZnBr}_{2}(a q)+\mathrm{H}_{2}(g)$
83. $\mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}+3 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-} \longrightarrow\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(a q)$


2301. Chapter 18 -
85. (a) $\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)$; (b) $\mathrm{Cs}_{2} \mathrm{CO}_{3}(s)+2 \mathrm{HF}(a q) \longrightarrow 2 \mathrm{CsF}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$;
(c) $\mathrm{Al}_{2} \mathrm{O}_{3}(s)+6 \mathrm{HClO}_{4}(a q) \longrightarrow 2 \mathrm{Al}\left(\mathrm{ClO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$; (d)
$\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{BaCO}_{3}(s) ;(\mathrm{e})$
$\mathrm{TiCl}_{4}(l)+4 \mathrm{Na}(s) \longrightarrow \mathrm{Ti}(s)+4 \mathrm{NaCl}(s)$
87. $\mathrm{HClO}_{4}$ is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid.
89. As $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{H}_{2} \mathrm{SeO}_{4}$ are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, $\mathrm{H}_{2} \mathrm{SO}_{4}$ is the stronger acid.


2301. Chapter 18 -
91. $\mathrm{SO}_{2}, s p^{2} 4+; \mathrm{SO}_{3}, s p^{2}, 6+; \mathrm{H}_{2} \mathrm{SO}_{4}, s p^{3}, 6+$
93. $\mathrm{SF}_{6}: \mathrm{S}=6+; \mathrm{SO}_{2} \mathrm{~F}_{2}: \mathrm{S}=6+$; $\mathrm{KHS}: \mathrm{S}=2-$
95. Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen.
97. There are many possible answers including: $\mathrm{Cu}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{CuSO}_{4}(a q)+\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ and $\mathrm{C}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$
99. $5.1 \times 10^{4} \mathrm{~g}$
101. $\mathrm{SnCl}_{4}$ is not a salt because it is covalently bonded. A salt must have ionic bonds.
103. In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. $\mathrm{HClO}_{3}$ is stronger than $\mathrm{HBrO}_{3} ; \mathrm{Cl}$ is more electronegative than Br .
105. (a)


2301. Chapter 18 -
Square pyramidal
(b)

Linear
(c)

Trigonal bipyramidal
(d)

Seesaw
(e)

T-shaped
107. (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite
109. (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1-; (e) F: 0
111. (a) $s p^{3} d$ hybridized; (b) $s p^{3} d^{2}$ hybridized; (c) $s p^{3}$ hybridized; (d) $s p^{3}$ hybridized; (e) $s p^{3} d^{2}$ hybridized;
113. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar
115. The empirical formula is $\mathrm{XeF}_{6}$, and the balanced reactions are:

$$
\begin{aligned}
& \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \xrightarrow{\Delta} \mathrm{XeF}_{6}(s) \\
& \mathrm{XeF}_{6}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 6 \mathrm{HF}(g)+\mathrm{Xe}(g)
\end{aligned}
$$


2302. Chapter 19 -
1. (a) $\mathrm{Sc}:[\mathrm{Ar}] 4 s^{2} 3 d^{1}$; (b) $\mathrm{Ti}:[\mathrm{Ar}] 4 s^{2} 3 d^{2}$; (c) $\mathrm{Cr}:[\mathrm{Ar}] 4 s^{1} 3 d^{5}$; (d) $\mathrm{Fe}:[\mathrm{Ar}] 4 s^{2} 3 d^{6}$; (e) $\mathrm{Ru}:[\mathrm{Kr}] 5 s^{2} 4 d^{6}$
2. (a) La: $[\mathrm{Xe}] 6 s^{2} 5 d^{1}, \mathrm{La}^{3+}$ : $[\mathrm{Xe}]$; (b) $\mathrm{Sm}:[\mathrm{Xe}] 6 s^{2} 4 f^{6}, \mathrm{Sm}^{3+}$ : $[\mathrm{Xe}] 4 f^{5}$; (c) $\mathrm{Lu}:[\mathrm{Xe}] 6 s^{2} 4 f^{14} 5 d^{1}, \mathrm{Lu}^{3+}$ : $[\mathrm{Xe}] 4 f^{14}$
3. Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La (III) into La.
4. Mo
5. The $\mathrm{CaSiO}_{3}$ slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag
layer creates a barrier that prevents the molten iron from exposure to $\mathrm{O}_{2}$, which would oxidize the Fe back to $\mathrm{Fe}_{2} \mathrm{O}_{3}$.
6. $2.57 \%$
7. 0.167 V
8. $E^{\circ}=-0.6 \mathrm{~V}, E^{\circ}$ is negative so this reduction is not spontaneous. $E^{\circ}=+1.1 \mathrm{~V}$


2302. Chapter 19 -
8. $E^{\circ}=-0.6 \mathrm{~V}, E^{\circ}$ is negative so this reduction is not spontaneous. $E^{\circ}=+1.1 \mathrm{~V}$
9. (a) $\mathrm{Fe}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{H}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$; (b)
$\mathrm{FeCl}_{3}(a q)+3 \mathrm{Na}^{+}(a q)+3 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+3 \mathrm{Na}^{+}(a q)+3 \mathrm{Cl}^{+}(a q) ;$ (c)
$\mathrm{Mn}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 \mathrm{Br}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{Br}^{-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) ;$ (d)
$4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s) ;(\mathrm{e}) \mathrm{Mn}_{2} \mathrm{O}_{3}(s)+6 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+6 \mathrm{Cl}^{-}(a q) \longrightarrow 2 \mathrm{MnCl}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(l)$;
(f) $\mathrm{Ti}(s)+x s \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TiF}_{4}(g)$
10. (a)


2302. Chapter 19 -
(f) $\mathrm{Ti}(s)+x s \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TiF}_{4}(g)$
10. (a)
$\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2 \mathrm{Zn}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow 2 \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cr}^{2+}(a q)+3 \mathrm{SO}_{4}{ }^{2-}(a q) ;$
(b) $4 \mathrm{TiCl}_{3}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q)+8 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Ti}^{4+}(a q)+\mathrm{Cr}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)+12 \mathrm{Cl}^{-}(a q)$; (c) In acid solution between pH 2 and $\mathrm{pH} 6, \mathrm{CrO}_{4}{ }^{2-}$ forms $\mathrm{HCrO}_{4}{ }^{-}$, which is in equilibrium with dichromate ion. The reaction is $2 \mathrm{HCrO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$. At other acidic pHs, the reaction is $3 \mathrm{Cr}^{2+}(a q)+\mathrm{CrO}_{4}{ }^{2-}(a q)+8 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \longrightarrow 4 \mathrm{Cr}^{3+}(a q)+12 \mathrm{H}_{2} \mathrm{O}(l)$; (d)


2302. Chapter 19 -
$8 \mathrm{CrO}_{3}(s)+9 \mathrm{Mn}(s) \xrightarrow{\Delta} 4 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{Mn}_{3} \mathrm{O}_{4}(s) ;(\mathrm{e})$
$\mathrm{CrO}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cr}^{2+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) ;$ (f)
$\mathrm{CrCl}_{3}(s)+3 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)+3 \mathrm{Na}^{+}(a q)+3 \mathrm{Cl}^{-}(a q)$
11. (a) $3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$; (b)
$3 \mathrm{NaOH}(a q)+\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q) \xrightarrow{\mathrm{H}_{2} \mathrm{O}} \mathrm{Fe}(\mathrm{OH})_{3}(s)+3 \mathrm{Na}^{+}(a q)+3 \mathrm{NO}_{3}-(a q) ;$ (c)
$\mathrm{MnO}^{4-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}_{3}+4 \mathrm{H}_{2} \mathrm{O}$; (d)


2302. Chapter 19 -
$\mathrm{Fe}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{H}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$; (e)
$4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{HNO}_{3}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{NO}_{3}{ }^{-}(a q) ;(\mathrm{f})$
$\mathrm{FeCO}_{3}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) ;(\mathrm{g}) 3 \mathrm{Fe}(s)+2 \mathrm{O}_{2}(g) \xrightarrow{\Delta} \mathrm{Fe}_{3} \mathrm{O}_{4}(s)$
12. $\mathrm{As} \mathrm{CN}^{-}$is added,
$\mathrm{Ag}^{+}(a q)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{AgCN}(s)$
As more $\mathrm{CN}^{-}$is added,
$\mathrm{Ag}^{+}(a q)+2 \mathrm{CN}^{-}(a q) \longrightarrow\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(a q)$
$\mathrm{AgCN}(s)+\mathrm{CN}^{-}(a q) \longrightarrow\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(a q)$


2302. Chapter 19 -
$\mathrm{AgCN}(s)+\mathrm{CN}^{-}(a q) \longrightarrow\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(a q)$
13. (a) $\mathrm{Sc}^{3+}$; (b) $\mathrm{Ti}^{4+}$; (c) $\mathrm{V}^{5++}$; (d) $\mathrm{Cr}^{6+}$; (e) $\mathrm{Mn}^{4+}$; (f) $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$; (g) $\mathrm{Co}^{2+}$ and $\mathrm{Co}^{3+}$; (h) $\mathrm{Ni}^{2+}$; (i) $\mathrm{Cu}^{+}$
14. (a) $4,\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$; (b) $6,\left[\mathrm{Pd}(\mathrm{CN})_{6}\right]^{2-}$; (c) $2,\left[\mathrm{AuCl}_{2}\right]^{-}$; (d) $4,\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$; (e) $6, \mathrm{~K}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{4}\right]$; (f) 6 , $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}(\mathrm{CN})_{6}\right] ;(\mathrm{g}) 6,\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Br}_{2}\right] \mathrm{NO}_{3}$
15. (a) $\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Br}_{2}\right]$ :


2302. Chapter 19 -
cis

trans
(b) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{py})(\mathrm{Cl})(\mathrm{Br})\right]$ :

(c) $\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$:

(d) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right]^{+}$:

(e) $\left[\mathrm{Ni}^{\left.\left(\mathrm{H}_{2} \mathrm{O}\right){ }_{4} \mathrm{Cl}_{2}\right] \text { : }}\right.$

(f) $\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2} \mathrm{Cl}_{2}\right]^{3-}$ :

16. (a) tricarbonatocobaltate(III) ion; (b) tetraaminecopper(II) ion; (c) tetraaminedibromocobalt(III) sulfate; (d) tetraamineplatinum(II) tetrachloroplatinate(II); (e) tris-(ethylenediamine)chromium(III) nitrate; (f) diaminedibromopalladium(II); (g) potassium pentachlorocuprate(II); (h) diaminedichlorozinc(II)
17. (a) none; (b) none; (c) The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.
18.

37.


2302. Chapter 19 -
37.

39. $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ with three unpaired electrons.
41. (a) 4 ; (b) 2 ; (c) 1 ; (d) 5 ; (e) 0
43. (a) $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$; (b) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$; (c) $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$
45. The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.
47. No. $\mathrm{Au}^{+}$has a complete $5 d$ sublevel.


2303. Chapter 20 -
1. (a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194
2. (a) ${ }_{14}^{34} \mathrm{Si}$; (b) ${ }_{15}^{36} \mathrm{P}$; (c) ${ }_{25}^{57} \mathrm{Mn}$; (d) ${ }_{56}^{121} \mathrm{Ba}$
3. (a) ${ }_{25}^{45} \mathrm{Mn}^{+1}$; (b) ${ }_{45}^{69} \mathrm{Rh}^{+2}$; (c) ${ }_{53}^{142} \mathrm{I}^{-1}$; (d) ${ }_{97}^{243} \mathrm{Bk}$
4. Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Nuclear reactions involve much larger energies than chemical reactions and have measureable mass changes.
5. (a), (b), (c), (d), and (e)
6. (a) A nucleon is any particle contained in the nucleus of the atom, so it can refer to protons and neutrons.
(b) An $\alpha$ particle is one product of natural radioactivity and is the nucleus of a helium atom. (c) A $\beta$ particle is a product of natural radioactivity and is a high-speed electron. (d) A positron is a particle with the same mass as an electron but with a positive charge. (e) Gamma rays compose electromagnetic radiation of high energy and short wavelength. (f) Nuclide is a term used when referring to a single type of nucleus. (g) The mass number is the sum of the number of protons and the number of neutrons in an element. (h) The atomic number is the number of protons in the nucleus of an element.


2303. Chapter 20 -
7. (a) ${ }_{13}^{27} \mathrm{Al}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{15}^{30} \mathrm{P}+{ }_{0}^{1} \mathrm{n}$; (b) ${ }_{94}^{239} \mathrm{Pu}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{96}^{242} \mathrm{Cm}+{ }_{0}^{1} \mathrm{n}$; (c) ${ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{8}^{17} \mathrm{O}+{ }_{1}^{1} \mathrm{H}$; (d) ${ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{37}^{96} \mathrm{Rb}+{ }_{55}^{135} \mathrm{Cs}+4{ }_{0}^{1} \mathrm{n}$
8. (a) ${ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{8}^{17} \mathrm{O}+{ }_{1}^{1} \mathrm{H}$; (b) ${ }_{7}^{14} \mathrm{C}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}$; (c) ${ }_{90}^{232} \mathrm{Th}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{90}^{233} \mathrm{Th}$; (d) ${ }_{92}^{238} \mathrm{U}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{92}^{239} \mathrm{U}+{ }_{1}^{1} \mathrm{H}$


2303. Chapter 20 -
9. (a) 148.8 MeV per atom; (b) $7.808 \mathrm{MeV} /$ nucleon
10. $\alpha$ (helium nuclei), $\beta$ (electrons), $\beta^{+}$(positrons), and $\eta$ (neutrons) may be emitted from a radioactive element, all of which are particles; $\gamma$ rays also may be emitted.
11. (a) conversion of a neutron to a proton: ${ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{1}^{1} \mathrm{p}+{ }_{+1}^{0} \mathrm{e}$; (b) conversion of a proton to a neutron; the positron has the same mass as an electron and the same magnitude of positive charge as the electron has negative charge; when the n :p ratio of a nucleus is too low, a proton is converted into a neutron with the emission of a positron: ${ }_{1}^{1} \mathrm{p} \longrightarrow{ }_{0}^{1} \mathrm{n}+{ }_{+1}^{0} \mathrm{e}$; (c) In a proton-rich nucleus, an inner atomic electron can be absorbed. In simplest form, this changes a proton into a neutron: ${ }_{1}^{1} \mathrm{p}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{0}^{1} \mathrm{p}$
12. The electron pulled into the nucleus was most likely found in the $1 s$ orbital. As an electron falls from a higher energy level to replace it, the difference in the energy of the replacement electron in its two energy levels is given off as an X-ray.


2303. Chapter 20 -
13. Manganese-51 is most likely to decay by positron emission. The n:p ratio for $\mathrm{Cr}-53$ is $\frac{29}{24}=1.21$; for $\mathrm{Mn}-51$, it is $\frac{26}{25}=1.04$; for $\mathrm{Fe}-59$, it is $\frac{33}{26}=1.27$. Positron decay occurs when the n :p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, ${ }_{24}^{53} \mathrm{Cr}$ is a stable
isotope, and ${ }_{26}^{59} \mathrm{Fe}$ decays by beta emission.
14. (a) $\beta$ decay; (b) $\alpha$ decay; (c) positron emission; (d) $\beta$ decay; (e) $\alpha$ decay
15. ${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathrm{He} ;{ }_{90}^{234} \mathrm{Th} \longrightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} \mathrm{e} ;{ }_{91}^{234} \mathrm{~Pa} \longrightarrow{ }_{92}^{234} \mathrm{U}+{ }_{-1}^{0} \mathrm{e} ;{ }_{92}^{234} \mathrm{U} \longrightarrow{ }_{90}^{230} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}$


2303. Chapter 20 -
${ }_{90}^{230} \mathrm{Th} \longrightarrow{ }_{88}^{226} \mathrm{Ra}+{ }_{2}^{4} \mathrm{He}{ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{22} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} ;{ }_{86}^{22} \mathrm{Rn} \longrightarrow{ }_{84}^{218} \mathrm{Po}+{ }_{2}^{4} \mathrm{He}$
16. Half-life is the time required for half the atoms in a sample to decay. Example (answers may vary): For C-14, the half-life is 5770 years. A 10-g sample of C-14 would contain 5 g of C-14 after 5770 years; a $0.20-\mathrm{g}$ sample of $\mathrm{C}-14$ would contain 0.10 g after 5770 years.
17. $\left(\frac{1}{2}\right)^{0.04}=0.973$ or $97.3 \%$
18. $2 \times 10^{3} \mathrm{y}$
19. $0.12 \mathrm{~h}^{-1}$
20. (a) 3.8 billion years; (b) The rock would be younger than the age calculated in part (a). If Sr was originally in the rock, the amount produced by radioactive decay would equal the present amount minus the initial amount. As this amount would be smaller than the amount used to calculate the age of the rock and the age is proportional to the amount of Sr , the rock would be younger.
21. $c=0$; This shows that no Pu-239 could remain since the formation of the earth. Consequently, the plutonium now present could not have been formed with the uranium.
22. 17.5 MeV


2303. Chapter 20 -
21. $c=0$; This shows that no Pu-239 could remain since the formation of the earth. Consequently, the plutonium now present could not have been formed with the uranium.
22. 17.5 MeV
23. (a) ${ }_{83}^{212} \mathrm{Bi} \longrightarrow{ }_{84}^{212} \mathrm{Po}+{ }_{-1}^{0} \mathrm{e}$; (b) ${ }_{5}^{8} \mathrm{~B} \longrightarrow{ }_{4}^{8} \mathrm{Be}+{ }_{-1}^{0} \mathrm{e}$; (c) ${ }_{92}^{238} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{93}^{239} \mathrm{~Np}+{ }_{-1}^{0} \mathrm{~Np}$, ${ }_{93}^{239} \mathrm{~Np} \longrightarrow{ }_{94}^{239} \mathrm{Pu}+{ }_{-1}^{0} \mathrm{e}$; (d) ${ }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \mathrm{e}$


2303. Chapter 20 -
24. (a) ${ }_{95}^{241} \mathrm{Am}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{97}^{244} \mathrm{Bk}+{ }_{0}^{1} \mathrm{n}$; (b) ${ }_{94}^{239} \mathrm{Pu}+15{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{100}^{254} \mathrm{Fm}+6{ }_{-1}^{0} \mathrm{e}$; (c) ${ }_{98}^{250} \mathrm{Cf}+{ }_{5}^{11} \mathrm{~B} \longrightarrow{ }_{103}^{257} \mathrm{Lr}+4{ }_{0}^{1} \mathrm{n}$; (d) ${ }_{98}^{249} \mathrm{Cf}+{ }_{7}^{15} \mathrm{~N} \longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}$
25. Two nuclei must collide for fusion to occur. High temperatures are required to give the nuclei enough kinetic energy to overcome the very strong repulsion resulting from their positive charges.
26. A nuclear reactor consists of the following:
27. A nuclear fuel. A fissionable isotope must be present in large enough quantities to sustain a controlled chain reaction. The radioactive isotope is contained in tubes called fuel rods.
28. A moderator. A moderator slows neutrons produced by nuclear reactions so that they can be absorbed by the fuel and cause additional nuclear reactions.
29. A coolant. The coolant carries heat from the fission reaction to an external boiler and turbine where it is transformed into electricity.
30. A control system. The control system consists of control rods placed between fuel rods to absorb neutrons and is used to adjust the number of neutrons and keep the rate of the chain reaction at a safe level.
31. A shield and containment system. The function of this component is to protect workers from radiation produced by the nuclear reactions and to withstand the high pressures resulting from hightemperature reactions.


2303. Chapter 20 -
31. A shield and containment system. The function of this component is to protect workers from radiation produced by the nuclear reactions and to withstand the high pressures resulting from hightemperature reactions.
32. The fission of uranium generates heat, which is carried to an external steam generator (boiler). The resulting steam turns a turbine that powers an electrical generator.
33. Introduction of either radioactive $\mathrm{Ag}^{+}$or radioactive $\mathrm{Cl}^{-}$into the solution containing the stated reaction, with subsequent time given for equilibration, will produce a radioactive precipitate that was originally devoid of radiation.
34. (a) ${ }_{53}^{133} \mathrm{I} \longrightarrow{ }_{54}^{133} \mathrm{Xe}+{ }_{-1}^{0} \mathrm{e}$; (b) 37.6 days
35. Alpha particles can be stopped by very thin shielding but have much stronger ionizing potential than beta particles, X -rays, and $\gamma$-rays. When inhaled, there is no protective skin covering the cells of the lungs, making it possible to damage the DNA in those cells and cause cancer.
36. (a) $7.64 \times 10^{9} \mathrm{~Bq}$; (b) $2.06 \times 10^{-2} \mathrm{Ci}$