Question

What is the pH of a solution obtained by adding 15 g deK2SO4em 150 ml NaHSO4 with Molarity of 0, 2 m. HSO4 Ka = 1, 3.10^-2

Answer 1

[H+] fm Potassium Sulfate is an hydrolysis calculation using (15g/150ml) = 0.575M sulfate ion in 150 ml soln. Calculate equivalence point [H+](K2SO4) = 1.8 x 10^-8M

[H+] fm 0.20M Bisulfate ion is a weak acid in solution calculation in which [H+] is determined using the quadratic equation b/c Ka of the Bisulfate ion is quite large. [HSO-]/Ka < 100 and the 'x' in the I.C.E. table can not be dropped.
[H+](HSO4-) = 5.0 X 10-2M

[H+](K2SO4) = 1.5 X 10^-8
[H+](HSO4-) = 5.0 X 10^-2
___________________________
Sum [H+] = 0.05M (The presence of K2SO4 makes very little contribution to the [H+] and the dominant source for [H+] is from the Bisulfate ionization giving 5.0 x 10^2M in H+) >>> 1.5 x 10^-8.

Therefore, using [H+](sum) = 0.05M => pH = -log[H+] = -log(0.05) = -(-1.3) = 1.3

Answer 2

I think the correct solution is to recognize this as a buffer problem use the Hendersib-Hasselbalch equation.
pH = pK2 + log (base)/(acid)
pH = 1.88 + log (0.574/0.2) = ?
pH approx 2.3
OR you can work it as a common ion; i.e.,
HSO4^- ==> H^+ + SO4^- k2 = ?
K2SO4 ==> 2K^+ + SO4^2-

K2 = (H^+)(SO4^2)/(HSO4^-)
You know SO4^- is o.574
You know HSO4^- is 0.2M
Solve for H^+ = ?
This gives the same answer as the HH solution; it also allows you to use the quadratic equation. I did both and came out with 2.3 for the HH equation, 2.3 for the other and 2.4 if the second method uses the quadratic.