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

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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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HSO₄^- ----->SO₄^-2 + H⁺

InitialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â 0.1

ChangeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â +x

EquilibriumÂ Â Â Â Â Â Â Â Â Â 0.1-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â 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

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