Jokerine's Acid-Base Problem

Jokerine recently posted a first-year problem here. I read it last night...and it ended up being my first waking thought this morning. Haunting, that was. Anyway.

Let me simplify her problem a bit:
How much 10 M HCl should one add to a 100 L solution with pH = 8.0 to decrease the pH to 7.5?
(I changed the volume. Yep, lazy me...)

Assumptions: there are no weak acids or bases in the solution, i.e. all pH's are directly due to H⁺ or OH^- (Yes, I know it's a swimming pool, but I am too lazy to deal with the chemistry of ClO^-). Actually, let's just make the initial solution one of NaOH. Also, temperature = 25 ^oC.

Oh, and to make my life even easier, I am crassly ignoring significant figures.

In the problem, Jokerine mentioned two approaches:
1. Think of the system as having too much base, and we are neutralizing it. Then we need [10^{-(14.0 - 8.0)} - 10^-7.5]*100/10 L = 9.7*10^-6 L of 10 M HCl.
2. Think of the problem as having not enough acid, and we are adding to that. Then we need [10^-7.5 - 10^-8.0]*100/10 L = 2.2*10^-7 L of 10 M HCl.

Awkwardly obvious: the two answers are different.

(Aside: Which reminds me...where'd my LaTeX button go? Anyway...)

Let me solve this problem in a rather convoluted way, mainly because I think this will give me an actual, correct answer. Approach:

1. Neutralize excess NaOH in solution, such that the result is of pH 7.
2. Add NaOH to the intermediate, neutral solution, such that the final pH is 7.5
3. Back-calculate the quantity of HCl required to go from pH = 8.0 to pH = 7.5 in one step.

Now, sub-question: What concentration of NaOH will give me a solution with pH = 8.0, if we include the autoionization of water and all that?
Consider charge equality:
[H⁺] + [Na⁺] = [OH^-]
[H⁺] + [Na⁺] = K_w/[H⁺]
[Na⁺] = K_w/[H⁺] - [H⁺]

Since pH = 8.0,
[H⁺] = 10^-8
[Na⁺] = 10^-14/10^-8 - 10^-8 = 9.9*10^-7 = [NaOH]

Note: if we ignore the effects of water, the pH of this solution is not exactly 8.0, more like 7.9956... (etc).

To neutralize this NaOH solution, we need (9.9*10^-7)*100/10 = 9.9*10^-6 L (or 9.9 &mu L) of 10 M HCl (just stoichiometry here).

Now we need to re-basify this to pH = 7.5. Once again, we consider charge equality:
[H⁺] + [Na⁺] = [OH^-] + [Cl^-]
Before adding any hydroxide, [Na⁺] = [Cl^-]. So, if we consider only the [Na⁺] contribution from NaOH added in rebasification, the charge equality equation becomes
[H⁺] + [Na⁺] = [OH^-]
Once again, this becomes
[Na⁺] = K_w/[H⁺] - [H⁺]

Now we want pH = 7.5.
[H⁺] = 10^-7.5
[Na⁺] = 10^-14/10^-7.5 - 10^-7.5 = 2.8*10^-7 = [NaOH]

This requires (2.8*10^-7)*(100 + 9.9*10^-6 L)/10 = 2.8*10^-6 L or 2.8 &mu L of 10 M HCl to neutralize.

All in all, we need 9.9 - 2.8 = 7.1 &mu L of 10 M HCl to get from pH = 8.0 to pH = 7.5.

Am I right? Am I wrong? You decide. However, I must admit that this solution is not elegant by any means.

Personally, I like to think of this as a titration: we must consider the stoichiometry before considering any equilibria. In dilute solutions like this, though, the autoionization of water is important, so we end up having to build that particular equilibrium into the calculations, anyway.