The of the will be 2.28.

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

##pH_”sol” = pK_a + log(([“conjugate base”])/([“weak acid”]))##

In your case, the weak will be sodium bisulfate, ##NaHSO_4##, and its conjugate base will be sodium sulfate, ##NaSO_4##. More accurately, 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 ##pK_a##

##pK_a = -log(K_a) = -log(1.2 * 10^(-2)) = 1.92##

Now just plug and play

##pH_”sol” = pK_a + log(([SO_4^(2-)])/([HSO_4^(-)]))##

##pH_”sol” = 1.96 + log((0.230cancel(“M”))/(0.1cancel(“M”))) = color(green)(2.28)##

The of the will be 2.28.

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

##pH_”sol” = pK_a + log(([“conjugate base”])/([“weak acid”]))##

In your case, the weak will be sodium bisulfate, ##NaHSO_4##, and its conjugate base will be sodium sulfate, ##NaSO_4##. More accurately, 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 ##pK_a##

##pK_a = -log(K_a) = -log(1.2 * 10^(-2)) = 1.92##

Now just plug and play

##pH_”sol” = pK_a + log(([SO_4^(2-)])/([HSO_4^(-)]))##

##pH_”sol” = 1.96 + log((0.230cancel(“M”))/(0.1cancel(“M”))) = color(green)(2.28)##