Question

Calculate the molar solubility of BaSO4 in a solution in which [H3O+] is 0.50 M. Ksp(BaSO4) = 1.1×10-10,
Ka(HSO4-)=1.02×10-2

Answer 1

.......BaSO4 ==> Ba^2+ + SO4^2-
I......solid......0.......0
C......solid......x.......x
E......solid......x.......x

But with [HO^+] added, that combines with the SO4^2- to give HSO4^- as this:
......SO4^2- + H^+ ==> HSO4^- and increases the solubility of BaSO4.
In other words part of the SO4^2- is missing.
alpha1 is the fraction of SO4^2- actually there. That is
alpha 1 = k2/(k2+H^+) =
0.0102/(0.5102) = very close to 0.02.
So Ksp = (Ba^2+)(SO4^2-)
If we call (Ba^2+) = solubility = S, the Ksp = (S)(S*0.02).
Plug in values for Ksp and solve for S. I get approx 7E-5M but that's just a close estimate. To show the added solubility due to the H^+, note that in pure water the solubility of BaSO4 is sqrt(1.1E-10) = 1.05E-5M so the solubility has been increased by about 7x by adding acid to make 0.5M H^+.