H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second...

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. 0.955

B. 2.05

C. 0.963

D. 1.00

Solutions

Expert Solution

The pH will be due to both the dissociation of strong acid

1) Due to first dissociation (almost 100%) concentration of hydrogen ion = [H+ ]= [H2SO4 = 0.100 M

H2SO4 ----> HSO₄^- + H⁺

2) second dissociation

HSO₄^- ----->SO₄^-2 + H⁺

Initial 0.1 0 0.1

Change -x +x +x

Equilibrium 0.1-x x 0.1+ x

Ka = [H+][SO4^-2] / [HSO4^-]

0.011 = x (0.1+x) / 0.1-x

0.0011 - 0.011x = 0.1x + x²

x² + 0.111x - 0.0011 = 0

On solving for x

x = 0.00916

Total [H+] = 0.1 + 0.00916 = 0.10916

pH = -log[H+] = 0.962

queen_honey_blossom answered 1 year ago

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