What is the pH of a Weak Acid/Base Mixture?

[cncbmb]

Homework Statement

We have two solutions: One solution that has $$0.300$$ L of $$0.100$$ M acetic acid, and the other with $$0.100$$ L of $$0.200$$ M $$CN^-$$.

We mix the two solutions. At equilibrium, what is the pH of the resulting solution?

Homework Equations

For acetic acid, $$K_a=1.8*10^{-5}$$
For $$CN^-$$, $$K_b=2.5*10^{-5}$$

The Attempt at a Solution

$$CH_3COOH + CN^- <=> CH_3COO- + HCN$$...
The K values aren't useful here since there's no water in the reaction.
We can't assume the reaction goes to completion, in either direction since the acids and bases are weak. Not sure if solution is basic even though the $$K_b$$ for cyanide anion is greater than the $$K_a$$ of acetic acid since we have more acetic acid than base.

...
My textbook seems to dodge this issue in its explanation of "the reaction of a weak acid with a weak base." The textbook considers an equilmolar mixture of acetic acid and ammonia and says that the solution in that case would be neutral because the Ka of acetic acid the Kb of NH3 are the same.

Whoa why is the tex so thin?..

 

[Borek]

The K values aren't useful here since there's no water in the reaction.

Plenty of water around.

We can't assume the reaction goes to completion, in either direction since the acids and bases are weak.

There is an excess of one reagent... Try to assume that reaction goes to completion (find limiting reagent).

 

[cncbmb]

Why can we assume that the reaction goes to completion?

 

[Borek]

In general - we can't. But you have huge excess of one of the reagents, that shifts equilibrium to the products side.

Note that when you mix acetic acid with stoichiometric amount of ammonia, neutralization goes ALMOST to completion with both acid and base neutralized in over 99%. We are just so close to the neutral solution, that tiny equilibrium shift can give a large effect in terms of pH change. However, here we are very far from this kind of situation.

Some reading:
general case of pH calculation
pH of salts