Question #6f00a

1 Answer
Apr 19, 2015

Answer:

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)#