Question #6f00a

1 Answer
Stefan V.
Apr 19, 2015

The pH of the buffer will be 2.282.28.

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"]))pHsol=pKa+log([conjugate base][weak acid])

In your case, the weak acid will be sodium bisulfate, "NaHSO"_4NaHSO4, and its conjugate base will be the sulfate anion, "SO"_4^(2-)SO2−4, delivered to the solution by sodium sulfate, its salt.

In other words, you're going to be dealing with hydrogen sulfate, "HSO"_4^(-)HSO−4, and the sulfate ion, "SO"_4^(2-)SO2−4.

The acid dissociation constant will give you "p"K_apKa

"p"K_a = -log(K_a) = -log(1.2 * 10^(-2)) = 1.92pKa=−log(Ka)=−log(1.2⋅10−2)=1.92

Now just plug and play

"pH"_"sol" = "p"K_a + log((["SO"_4^(2-)])/(["HSO"_4^(-)]))pHsol=pKa+log⎛⎜⎝[SO2−4][HSO−4]⎞⎟⎠

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

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