# Question #6f00a

##### 1 Answer

#### 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,

In other words, you're going to be dealing with *hydrogen sulfate*, *sulfate ion*,

The acid dissociation constant will give you

#"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)#