Calculate the molar solubility of manganese(II) carbonate in a solution that is
buffered to a pH of 3.00.
I set up the equations:
MnCO3(s)<-->Mn + Co3 ksp = 2.2*10^‐11
CO3 + H3O <--> HCO3 + H2O = 1/Ka2=1/5.6*10^‐11
--------------------------------------------------------------
MnCO3 + H3O <--> Mn + HCO3 + H20 final k = .375
1*10^-3 0 0
-1*10^-3 +1*10^-3 +1*10^-3 (completion)
+X -X -X (back to eq.)
((1*10^-3)-x)^2
-------------------=.375 >>> X=2.7*10^-6
X
So this means (1*10-3)-(2.7*10^-6) = [Mn] and [HCO3]= 9.9*10^-4
Now how do I find the molar solubility of MnCO3 from this. Please help.