[nozo]

Can someone explain why the answer is not equal to Ka2? According to my book,

For diprotic acids, if Ka1 > Ka2, then we can assume that the concentration of H⁺ ions is the product of only the first stage of ionization. Further, the concentration of the conjugate base for the second-stage ionization is numerically equal to Ka2

H₂C₆H₆O₆ <----> HC₆H₆O₆^- + H⁺   Ka1 = 7.9 x 10^-5

HC₆H₆O₆^- <----> C₆H₆O₆^-2 + H⁺   Ka2 = 1.65 x 10^-12


After using ICE, the answer apparently is x = H⁺ = .0028  ...but shouldn't it be 1.65 x 10^-12? Someone please shed some light

[Borek]

For diprotic acids, if Ka1 > Ka2, then we can assume that the concentration of H⁺ ions is the product of only the first stage of ionization.

Only if the difference is large enough.

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

Further, the concentration of the conjugate base for the second-stage ionization is numerically equal to Ka2

Using K_a2 definition:

K_a2 = [A^2-][H⁺]/[HA^-]

and solving it for [A^2-] we get:

[A^2-] = K_a2 [HA^-]/[H⁺]

If all H⁺ is from the first dissociation step we can assume [HA^-] = [H⁺] - so these cancel out, leaving only K_a2 on the right side of the equation.

After using ICE, the answer apparently is x = H⁺ = .0028  ...but shouldn't it be 1.65 x 10^-12?

H⁺ and A^2- are completely different things...