Question...

An email question...

Qu: Is the pH of the first equivalence point of a polyprotic acid necessarily basic? And vice versa, is the pH of the first equivalence point of a polyprotic base necessarily basic? For example, in the winter 2006-exam 3 practice exam (#5), we are drawing a titration curve for the 1M potassium sulfite (kb=1.6*10^-7) with 1M HClO4. The answer key seems to show the first equivalence point being slightly acidic, but when I actually go through the calculations I find the pH at the first equivalence point to be 7.204 (I got that using an ICE table, with the assumption that at the 1st equiv point [k2SO3]=[KHSO3], if that is not correct could you go through the calculations?).

There may be a labelling problem here, so let me work through this. If the concentration of sulfite (conjugate base) and hydrogen sulfite (conjugate acid) are equal, this would be a buffer, it would not be at an equivalence point. {The potassium ion is a neutral spectator, so I'll leave it out...} When this titration reaches the equivalence point, all of the sulfite ion has been converted to hydrogen sulfite ion, so at that first equivalence point, this can be considered to be a solution of hydrogen sulfite ions. If you want to calculate a pH for this, you could use a table (not exactly an equilibrium table, but similar...), but let's try to talk through it instead.

Let's say that to start our titration we have V mL of 1M sulfite solution. To reach the first equivalence point, we have to add V mL of 1M perchloric acid, so the total volume of the mixture is 2V mL. (This is probably not strictly true, but it's a close enough assumption for our purposes here.) So if the initial concentration of sulfite ions was 1M, the concentration of hydrogen sulfite ions at the first equivalence point will be 0.5M. This means that at the first equivalence point, the pH of this mixture would be the same as an authentically-prepared 0.5M solution of HSO₃^-(aq). From the K_b given in the problem, we can calculate that K_a = 6.25x10^-8 for HSO₃^-(aq). Now we should set up an equilibrium table for the reaction:
HSO₃^-(aq) + H₂O(l) <=> H₃O⁺(aq) + SO₃^-2(aq)
Go ahead, set it up on paper right now before you read the rest of this.

OK, what assumptions can we make to simplify this problem? Based upon its K_a, HSO₃^-(aq) is a weak acid, so it's probably safe to assume that "x" will be small compared to 0.5M. That means your math simplfies to:
6.25x10^-8 = x² / 0.5
Solving, x = 1.77x10^-4 = [H₃O⁺], so pH = 3.75. This may seem kind of low, but sulfurous acid is one of the strongest weak acids, so its titration curve should tend toward the acidic side. If you really wanted to analyze this titration and draw a super-accurate titration curve, you could also calculate the pH of the "half equivalence points". These are the points where [conj acid] = [conj base], so when you plug them into the Henderson-Hasselbalch equation they are the points where pH = pK_a of the conjugate acid. For sulfite/hydrogen sulfite/sulfurous acid, these fall at pH = 7.2 and 1.9.

After all of that discussion, let's take a step back. For the exam problem mentioned, I was not expecting that anyone would calculate exact pH's of the equilivalence and half-equivalence points, I was more interested in qualitatively reasonable equivalence points. All of those calculations would have taken a lot of time and wouldn't have really resulted in "better" picture of the titration curve. In this case, sulfite ion is a weak base being titrated with a strong acid, so the solution should initially start out basic. The first equivalence point in this titration should be on the acidic side so as long as the first equivalence point was below 7, I was satisfied. Obviously, the second equivalence point should be at an even lower pH than the first, and it should take twice as much titrant to reach the second equivalence point. Some of the more common (and frustrating) errors in this question are that people don't always label their axes, or don't label them correctly, or titrate in the wrong direction.

Good luck with your preparation, and again, the exam will take place on the second class after we return. If you have questions before then, email me and I'll answer them to the blog.

Be safe.