I agree with what Ruken said. Here are some of the others:

4.0 #16: I get x=0.0169 and [OH]-=0.0179 M, so pH=12.25, but I have to use the quadratic equation (and a calculator).

5.0 #4: Use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

You start with 0.1 moles of weak acid HA. It reacts completely with the 0.05 moles of the strong base.

This produces 0.05 moles of A- and leaves 0.05 moles of HA. It’s a weak acid, so assume it doesn’t ionize further. Now just plug in to the H-H equation - you get pKa=4.2 and Ka=6.3 x 10^-5

(you listed this question twice)

Ch. 5 #10:

Think of what species would be present at various pH values. At very low pH (<1), it’s all H2SO3.

At pH=2, half the H2SO3 has ionized to make H+ (or H3O+) and [HSO3]- because K1 is 10^-2

As the pH goes up (towards 7), more and more of the H2SO3 is converted to [HSO3]- and some of the [HSO3]- starts to ionize to make [SO3]2-.

At pH=7, the H2SO3 is essentially all gone, and half the [HSO3]- has ionized to [SO3]2- (because K2 is 10^-7). You’re told that the pH is 7, so that’s a given. Yes, if you just add this acid to water, you won’t get a pH of 7. Maybe they’re doing a titration and have added some amount of a strong base. Don’t worry about how the pH became 7, that’s irrelevant.

<sorry, I didn’t realize Ruken already answered this one, much more concisely>

Ch. 5 #11:

Use the Henderson-Hasselbalch equation; this is a buffer (strong acid + weak base), that’s why the titration curve flattens out near the middle.

The half-equivalence point is when [A-]=[HA], so pH=pKa

Ch 6, #2:

I agree with what you wrote (and don’t agree with their answer).

For the condensation, delta H is negative (it’s -44.2 kJ/mol at 25C)

delta U = delta H - delta(pV) = delta H - delta(nRT)

Usually the second term is much smaller than the first (and delta n would be negative here), so delta U should be negative. The temperature isn’t given, but this should be true at any reasonable T.