Question

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

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!

i) The dissociation steps of H2SO4 is as follows

H2SO4 --------> H+ + HSO4-

HSO4- <------> H+ + SO42-

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+] [SO42-]/[HSO4-]

Value of Ka2 =1.20×10-2

Make ICE for this equillibrium

Initial concentration

[HSO4-] = 0.260

[H+] = 0

[SO42-] = 0

change in concentration

[HSO4-] = -x

[H+] = +x

[SO42-] = +x

Equillibrium concentration

[HSO4-]= 0.260 - x

[SO42-] = 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

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