Serious Chemistry tutor wanted

I’m a new-ish AP teacher, and some of the sample problems I have are beating me, because it’s been too long since I thought about some of these topics. I just paid to have a phone conversation with a recent college grad (I guess) who had the nerve to call and “tutor” when he didn’t actually know how to explain the problems.

Here are the ones I currently need help with:

I will happily pay to have a phone conversation that explains these so I can actually understand the logic. I’m writing out the “gory details” for my students ahead of time, so they have something to read at home that’s actually helpful.

If you REALLY understand these, use the email in my contact info to tell me that you can totally do it, and are ready to answer questions from an inquisitive “student” who wants to truly understand the guts of the problem.

I can use PayPal or Venmo.

I’m in the middle of a business trip, I haven’t thought about high-school level acid/base stuffs in years, and my rates are deliberately unattractive. But this is good procrastination material.

This isn’t all of them, because sleep:

#5 Kw varies extensively with temperature. Water is complicated. You have not only dissociation to hydronium and hydroxide, but also rearrangement of the associated hydration sphere. For high school it might be sufficient to say the ionization is endothermic, so lower T pushed the equilibrium to the left. There are of course entropic considerations, but you can get an entire PhD studying small groups of water molecules.

Scientific notation can’t really be “violated”. I wouldn’t generally write a single number that way, but if it’s part of a series or in a table column, it’s easier to compare if all the decimal points line up. So yes I would actually do that.

I’m not sure what the problem is here. I wouldn’t have written the first step twice like they did. Not sure what you mean about not liking to have all even coefficients. The AP folks are probably hoping students will avoid any math here and just figure it out by inspection.

Roughly, an acid will be mostly protonated at pH below its pKa and be mostly deprotonated at pH above it. Since we have pkas of 2 and 7, the first proton is long gone and the second is 50/50 because pH = pKa. How can the pH be 7? With buffer:

Ch6 #2
I guess the initial system must not be at equilibrium? My initial assumption was that they were describing isobaric compression.

Ch6 #5
First balance propane + O2 --> H2O + CO2
Then sum the heats of formation of the products and subtract propane(s)

Ch6 #15
DeltaG = -RTlnK[sub]eq[/sub]

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.

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