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DnySy87He-A-000|Many compounds have several acidic protons, so-called polyprotic acids.
DnySy87He-A-002|The titration curve of polyprotic acids look stepwise.
DnySy87He-A-003|The protons are titrated individually.
DnySy87He-A-004|Usually in a polyprotic system, the pKa of the first proton is significantly different than the pKa of the second proton.
DnySy87He-A-005|Now I'm using pKas to describe protons, which is common parlance.
DnySy87He-A-006|We know, of course, that the pKas refer to K's, which are equilibrium constants.
DnySy87He-A-007|But I talk about the pKa of the first proton as a property of the first proton.
DnySy87He-A-008|It's the pH at which that proton is removed from the molecule in a titration.
DnySy87He-A-009|Let's look at that.
DnySy87He-A-010|So here's the titration curve of a polyprotic acid.
DnySy87He-A-013|So I always get a point in the titration curve for free where the pH is equal to the pKa at halfway to the equivalence point.
DnySy87He-A-014|In this case, it's halfway to the first equivalence point, the first proton coming off.
DnySy87He-A-015|So I can look at this and I say, well, below the first pKa, the acid form predominates.
DnySy87He-A-016|Above the first pKa, the base form predominates.
DnySy87He-A-017|So at the equivalence point, half equivalence point, I have an equal mixture of the acid and base form.
DnySy87He-A-018|So let's go along the titration curve, mostly acid form to start.
DnySy87He-A-019|As I get to half equivalence, equal concentration of the acid and base.
DnySy87He-A-020|As I go beyond that, the base form starts to predominate.
DnySy87He-A-021|And when I get to the first equivalence point, I'm all the base form.
DnySy87He-A-022|But now I can start another titration of this proton.
DnySy87He-A-024|So let's look at the species that are available.
DnySy87He-A-025|So right here at the equivalence point, I had all HCO 3-.
DnySy87He-A-026|And then HCO 3- starts to act like an acid and its proton is titrated.
DnySy87He-A-027|So as I go along the titration curve, the predominant form is the acid form.
DnySy87He-A-028|I'm below the second pKa.
DnySy87He-A-029|As I get to the second pKa, equal amounts of the acid and base form.
DnySy87He-A-030|And as I pass through the second pKa, the base form predominates, CO3 -2.
DnySy87He-A-033|Let's look at a polyprotic acid, carbonic acid, buffering your blood.
DnySy87He-A-034|Now your blood is buffered with carbonic acid, but only the first pKa-- that is, the buffer around the first pKa-- is used.
DnySy87He-A-035|And that first pKa is around 6.
DnySy87He-A-036|So how is it done?
DnySy87He-A-037|Well, carbon dioxide is produced in yourselves by metabolism and then exhaled through your lungs.
DnySy87He-A-042|Now the base form, HCO 3- is maintained at about 24 millimolar by your kidneys.
DnySy87He-A-043|Your kidneys filter your blood, and they have the remarkable property of being able to remove selectively specific ions.
DnySy87He-A-044|So they can maintain the H3O- concentration at 24 millimolar.
DnySy87He-A-045|So what does that mean for the pH of your blood?
DnySy87He-A-046|Well, you can use and apply your Henderson-Hasselbalch expression.
DnySy87He-A-051|So if you solve, you get a pH of around 7.4.
DnySy87He-A-052|So pH is maintained in your blood by a buffer system, and it's maintained around pH 7.4.
DnySy87He-A-053|And that turns out to be vitally important.
DnySy87He-A-054|If your pH changes by just half a unit in either direction, it's essentially fatal.
DnySy87He-A-055|And it's fatal because you changed the protonation states.
DnySy87He-A-056|Remember, the pH determines which forms are present in solution, the acid or the base form.
DnySy87He-A-057|If you arbitrarily change that, that changes the potential difference and the selectivity of the membranes in your system, in your body.
DnySy87He-A-058|It changes the structure of the proteins in your body.
DnySy87He-A-059|So very dramatic changes can occur by just a small change in pH.
DnySy87He-A-060|So it's vital to buffer your blood and resist changes in pH.
DnySy87He-A-062|If you reduce the partial pressure of carbon dioxide in your lungs, your blood will become slightly alkaline.
DnySy87He-A-063|You're reducing the acid form.
DnySy87He-A-064|So if your blood becomes slightly alkaline by hyperventilating, that can be dangerous.
DnySy87He-A-065|But you know how to cure that.
DnySy87He-A-066|If you find someone that's hyperventilating, what do you do?
DnySy87He-A-067|Well, you have them breathe into a bag.
zsk1S2r2r_o-005|Because the intercept is determined by delta-H.
zsk1S2r2r_o-006|So an endothermic reaction will have a positive intercept.
zsk1S2r2r_o-007|The slope is determined by delta-S.
zsk1S2r2r_o-008|So if delta-S is positive, the slope will be negative.
zsk1S2r2r_o-012|So for an endothermic reaction-- that's a positive delta-H-- you get a negative slope.
zsk1S2r2r_o-015|Let's summarize the relationship between the free energy in temperature and the equilibrium constant in temperature.
zsk1S2r2r_o-016|We can plot them both in a linear fashion if we plot the free energy versus temperature directly and natural log of the equilibrium constant versus 1 over temperature.
zsk1S2r2r_o-017|There's a couple of situations to consider depending on the relative sign of the enthalpy and the entropy.
zsk1S2r2r_o-021|And delta-S negative gives me a negative intercept.
zsk1S2r2r_o-022|We can continue.
zsk1S2r2r_o-026|Negative intercept for delta-G because delta-H determines the intercept.
zsk1S2r2r_o-027|Negative intercept for lnK versus 1 over T because delta-S determines the intercept.
zsk1S2r2r_o-035|So that's a summary of the various lnK versus 1 over T and delta-G versus T situations.
MySri1HHJ_0-000|When a chemical reaction occurs, there's an enthalpy change, or an energy change.
MySri1HHJ_0-002|Either way, the enthalpy of a reaction is a state function.
MySri1HHJ_0-003|It only depends on where I start and where I finish, and not the pathway to go between the products and the reactants.
MySri1HHJ_0-008|It's downhill.
MySri1HHJ_0-009|Energy is released.
MySri1HHJ_0-013|So if I'm going from reactant to their atoms, that's adding energy for all the bonds in the reactants.
MySri1HHJ_0-014|Now, I can just deform all the bonds in the products.
MySri1HHJ_0-015|And that will always release energy.
MySri1HHJ_0-016|So there'll always be an exothermic step, and endothermic step to break, and then exothermic step to make the bonds.
MySri1HHJ_0-017|Now, it just depends.
MySri1HHJ_0-018|Do I get more energy back when I make the product bonds than when I broke the reactant bonds?
MySri1HHJ_0-019|That will determine whether the overall reaction is exothermic-- in this case, I put in some energy to break the react in bonds.
MySri1HHJ_0-020|But the product bonds were more stable overall.
MySri1HHJ_0-021|So I got an overall release in energy.
MySri1HHJ_0-022|But you could imagine a case where you break the reactant bonds and form the product bonds, but you don't get as much energy back.
MySri1HHJ_0-023|And that would be an overall endothermic reaction.
MySri1HHJ_0-028|Hydrogen, as in hydrogen gas, in its standard state-- diatomic molecules.
MySri1HHJ_0-032|Simply accounting, to keep track of the entropy of a chemical reaction.
ancOLDlrc88-001|So how does K vary with temperature for water gas going to water liquid?
ancOLDlrc88-009|We're talking about the equilibrium between water liquid and water gas at various temperatures.
ancOLDlrc88-010|K, the equilibrium constant for chemical reactions and physical processes, varies with temperature.
ancOLDlrc88-011|So how does it vary for this reaction?
ancOLDlrc88-014|And this is varying with temperature.
ancOLDlrc88-015|So this line in the phase diagram corresponds to the equilibrium values.
ancOLDlrc88-016|These are the values of the equilibrium constant with temperature.
ancOLDlrc88-017|So the equilibrium constant looks just like that line as it varies with temperature.
ancOLDlrc88-018|Now, another way you could have done this is said, well, I know as temperature increases, this favors the products.
ancOLDlrc88-019|So K should get larger.
ancOLDlrc88-020|It's more likely to have gaseous water at equilibrium at high temperatures than the liquid water.
ancOLDlrc88-021|So you could also say well, K increases with T because I know something about this physical process.
ancOLDlrc88-022|Either way, you arrive at the same answer, C.
eoFw8xEy-m4-000|We're talking about the orbitals about a hydrogen atom, and we're using the three quantum numbers-- n, l, and m sub l-- to describe wave functions.
eoFw8xEy-m4-001|Each wave function describes an orbital.
eoFw8xEy-m4-002|We're going to use those terms interchangeably-- wave function, orbital.