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,

x^{2}/(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

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