Question

What is the pH of a 0.260 M solution of H2SO4? Ka2 = 1.20×10^–2?

Please write out steps if possible so that I can understand, thank you!

Answer 1

i) The dissociation steps of H2SO4 is as follows

H2SO4 --------> H⁺ + HSO4^-

HSO4^- <------> H⁺ + SO4^2-

The first step is complete dissociation. So, 0.260M of H2SO4 gives 0.260M of H⁺ and 0.260M H⁺

ii) The second step is not complete and equillibrium expression for the second step is as follows

Ka2 = [H⁺] [SO4^2-]/[HSO4^-]

Value of Ka2 =1.20×10^-2

Make ICE for this equillibrium

Initial concentration

[HSO4^-] = 0.260

[H⁺] = 0

[SO4^2-] = 0

change in concentration

[HSO4^-] = -x

[H⁺] = +x

[SO4^{2-] = +x}

Equillibrium concentration

[HSO4^-]= 0.260 - x

[SO4^2-] = x

[H⁺] = x

Therefore,

x²/(0.260 - x) = 1.20×10^-2

solving for x

x = 0.05018

Therefore,

[H⁺] = 0.05018M

iii) Total H⁺ concentration = H⁺ concentration from dissociatio1 + H⁺ concentration from dissociation 2

Total H⁺ = 0.260M + 0.05018M = 0.3102M

iv) pH = -log[H⁺]

pH = -log(0.3102)

pH = 0.51