[licamine]

The K_sp for BaCO₃ is 5.5٠10^-10. Ignoring the hydrolysis of CO₃^2-, this would mean it has a molar solubility of 2.35٠10^-5 mol٠L^-1. However, taking into account the hydrolysis of CO₃^2-, how does one calculate the actual molarsolublity of BaCO₃?
K_a1 and K_a2 for carbonic acid are, respectively, 4.4٠10^-7 and 5.6٠10^-11.

So I've tried adding up the equations and calculating the K_sp for the resulting equation, like in problems where there's formation of complex ions, but it doesn't seem to work.
BaCO₃(s) --> Ba²⁺(aq) + CO₃^2-(aq) K_sp = 5.5٠10^-10
CO₃^2- (aq) + H₂O(l) --> HCO₃^- (aq) + OH^-(aq) K_h2 = K_w/K_a2 = 1.79٠10^-4
HCO₃^-(aq) + H₂O(l) --> H₂CO₃(aq) + OH^-(aq) K_h1 = K_w/K_a1 = 2.27٠10^-8
---------------------------------------------------------------
BaCO₃(s) + 2H₂O(l) --> Ba²⁺(aq) + H₂CO₃(aq) + 2OH^-(aq) K= K_sp٠K_h2٠K_h1 = 2.24٠10^-21
I don't know what to do with this number, since the concentration of H₂CO₃ is far smaller than that of Ba²⁺, because the hydrolyses have such small constants, unlike the K_f of a complex ion.

Answer: the solubility increases to 5.32٠10^-5

Thank you very much!

[Borek]

Have you tried the standard approach - that is, writing down all involved equilibria, all involved mass balances and charge balances, and trying to solve the set of equations?

My bet is that you can safely ignore the H₂CO₃/HCO₃^- equilibrium, only the first protonation of the CO₃^2- counts.