# Question 6f00a

Apr 19, 2015

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, ${\text{NaHSO}}_{4}$, and its conjugate base will be the sulfate anion, ${\text{SO}}_{4}^{2 -}$, delivered to the solution by sodium sulfate, its salt.

In other words, you're going to be dealing with hydrogen sulfate, ${\text{HSO}}_{4}^{-}$, and the sulfate ion, ${\text{SO}}_{4}^{2 -}$.

The acid dissociation constant will give you $\text{p} {K}_{a}$

$\text{p} {K}_{a} = - \log \left({K}_{a}\right) = - \log \left(1.2 \cdot {10}^{- 2}\right) = 1.92$

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)#