Question

Calculate the pH of 0.15 M H2SO4(aq) at 25◦C, given that Ka1 is very large and Ka2 = 1.2 × 10−2

Answer 1

H2SO4 is a strong acid for H+;

H2SO4(aq) --> H+(aq) + HSO4-(aq)

this [H+] = 0.15 M for the first ionizaiton since it is very large

now, for HSO4- which is a weak acid

so

HSO4-(aq) --> H+ + SO4-2 which will have a ionization constant Ka2 = 1.2*10^-2

so...

Ka2 = [H+][SO4-2]/[HSO4-]

initially, we know:

[H+] = 0.15

[SO4-2] = 0

[HSO4-] = 0.15 (from previous equilibrium)

then, in equilibrium

[H+] = 0.15 + x

[SO4-2] = 0 + x

[HSO4-] = 0.15 - x (Account for that in solution)

so...

substitute in KA2:

Ka2 = [H+][SO4-2]/[HSO4-]

1.2*10^-2 = (0.15+x)(x) / (0.15-x)

now, solve for x... which is H+ from the second ionization

(1.2*10^-2)(0.15-x) = (0.15x + x^2)

0.0018 - 0.012x = 0.15x + x^2

x^2 + (0.15+0.012)x - 0.0018 = 0

x^2 + 0.162x - 0.0018 = 0

solve for x

x = 0.010438

this is the secon dionization of H+

[H+] = 0.15 + x

[H+] = 0.15 + 0.010438= 0.160438

so

pH = -log(H+) = -log(0.160438) = 0.7946