Question

H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons. A) 1.00 B) 2.05 C) 0.955 D) 0.963 I thought the answer was C, but it was incorrect when I submitted it.

Answer 1

after 1st proton ionisation,
[H+] = 0.100 M
[HSO4-] = 0.100 M

for 2nd proton ionisation:
HSO4- ----> H+ + SO4 2-
0.100 0.100 0 (initial)
0.100-x 0.100+x x (at equilibrium)

Ka = [H+] [SO42-]/[HSO4-]
1.1*10^-2 = (0.100+x)(x) / (0.100-x)
1.1*10^-3 - 1.1*10^-2*x = 0.1*x + x^2
x^2 + 0.111*x - 1.1*10^-3 = 0
solving above quadratic equation for positive x, we get x = 0.0092 M

So,[H+] = 0.100 M + 0.0092 M
= 0.1092 M

pH = -log [H+]
= -log (0.1092)
= 0.962

I will go with D