[simpleton]

Question:
To calculate the pH of 1M H2SO4. Given, Ka2 = 1.2 x 10-2

Solution:

First dissociation: H2SO4 + H2O --> HSO4- + H3O+

Second dissociation (partial): HSO4- + H2O <---> SO4 2- + H3O+
Ka = [H3O+][SO4 2-]/[HSO4]

Assuming the molarity of H2SO4 = HSO4 and let x be the amount of the conjugate acid and base.

Ka = [x ][x ]/1M
x2 = (1.20 x 10-2) x 1M
x = sq root (0.012)
x = 0.11M

pH = -log (H+)
    = -log (1M + 0.11M)
    = -0.05

Have I done it correctly? Thank you guys for checking my answer.

[Borek]

Ka = [x ][x ]/1M

Nope, you forgot there was already 1M of H⁺ before HSO₄^- started to dissociate.

Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

[simpleton]

Hi Borek, I didn't forget about it. I included that 1M in the final pH calculation where I sum up all the H+ (including those in H2SO4 and HSO4-).

hmm am I right?

[Borek]

You can't calculate [H⁺] and then say that [H⁺] is in fact 1+[H⁺]...

[H⁺] <> 1+[H⁺]