Question

Question #6f00a

Answer 1

The pH of the buffer will be `2.28`.

Explanation:

When dealing with buffer solutions, you always have to be aware of the fact that you can use the Henderson-Hasselbalch equation to solve for pH if you know the concentrations of the weak acid and its conjugate base.

`"pH"_"sol" = "p"K_a + log((["conjugate base"])/(["weak acid"]))`

In your case, the weak acid will be sodium bisulfate, `"NaHSO"_4`, and its conjugate base will be the sulfate anion, `"SO"_4^(2-)`, delivered to the solution by sodium sulfate, its salt.

In other words, you're going to be dealing with hydrogen sulfate, `"HSO"_4^(-)`, and the sulfate ion, `"SO"_4^(2-)`.

The acid dissociation constant will give you `"p"K_a`

`"p"K_a = -log(K_a) = -log(1.2 * 10^(-2)) = 1.92`

Now just plug and play

`"pH"_"sol" = "p"K_a + log((["SO"_4^(2-)])/(["HSO"_4^(-)]))`

`"pH"_"sol" = 1.92 + log((0.230cancel("M"))/(0.1cancel("M"))) = color(green)(2.28)`