Consider a solution of a weak acid at a pH equal to its pKa. By how much would the pH change, and in which direction, if we added to this solution enough base to neutralize 10% of the total acid?

The pH of a weak acid solution is equal to its #pK_a# when you have equal concentrations of weak acid and of its conjugate base.

A solution that contains a weak acid and its conjugate base in comparable amounts, not necessarily in equal amounts, is called a buffer solution.

The pH of a buffer is described by the Henderson - Hasselbalch equation, which looks like this

#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#

As you can see here, equal amounts of weak acid and of conjugate base will make the log term equal to zero, and thus the pH equal to the #pK_a#.

Right from the start, wou can predict that adding more base will increase the pH of the solution.

Now, a weak acid cannot be neutralized by a weak base and a weak base cannot be neutralized by a weak acid.

Judging by the information provided in the problem, I will assume that you're adding enough strong base to neutralize #10%# of the weak acid.

A generic weak acid - strong base reaction looks like this

#"HA"_text((aq]) + "BOH"_text((aq]) -> "BA"_text((aq]) + "H"_2"O"_text((l])#

Here #"BA"# is the salt of the conjugate base of the weak acid, which will be #"A"^(-)#.

Now, assuming that you have a #1:1# mole ratio between the weak acid and the conjugate base, it's important to notice that the reaction consumes one mole of weak acid and produces one mole of conjugate base.

Let's say that before adding the strong base, you have #x# moles of weak acid and #x# moles of conjugate base in a volume #V#.

Now you're adding a volume #v# of strong base. If #10%# of the acid is neutralized, the number of moles you'll be left with is

#x_"acid" = x - 10/100x = 9/10x#

The number of moles of conjugate base will increase by the same amount, so you have

#x_"base" = x + 10/100x = 11/10x#

The new concentrations of weak acid and conjugate base will be

#["HA"] = 9/10x * 1/(V + v)#

#["A"^(-)] = 11/10x * 1/(V + v)#

Plug this into the H-H equation to get

#"pH" = pK_a + log( (11/10x * color(red)(cancel(color(black)(1/(V + v)))))/(9/10x * color(red)(cancel(color(black)(1/(V + v))))))#

#"pH" = pK_a + log(11/9) = pK_a + 0.087#

The pH of the solution will thus increase by about #0.09#.