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?

Question

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?

Olivia Anton · Accepted Answer

Here's what I got.

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 buffer solution is one that has comparable, though not necessarily equal, concentrations of a weak acid and its conjugate base. The Henderson-Hasselbalch equation, which looks like this, describes the pH of a buffer. #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#. It is obvious from the outset that adding more base will raise the solution's pH. It is now impossible for a weak base to neutralize a weak acid, and vice versa for 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. This is an example of a common weak acid-strong base reaction. #"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 conjugate base's moles will rise by the same amount, so you have #x_"base" = x + 10/100x = 11/10x# The modified weak acid and conjugate base concentrations will be #["HA"] = 9/10x * 1/(V + v)# #["A"^(-)] = 11/10x * 1/(V + v)# Enter this into the H-H formula to obtain #"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#.

Naomi Aronson · Answer

undefined The pH would increase by 0.48 units. The direction of the change is towards basic.