Question

Calculate the molar solubility of a solution of BaCO3 (Kip = 5x10^-9) in a solution buffered at pH= 6.5.

For H2CO3, K1=4.45 x 10^-7 and K3 = 4.69 x 10^-11.

Answer 1

Calculate the molar solubility of a solution of BaCO3 (Ksp = 5x10^-9) in a solution buffered at pH= 6.5.

For H2CO3, K1=4.45 x 10^-7 and K2 = 4.69 x 10^-11.

pH= 6.5
[H+] = 10^(-6.5) = 3.16E-7

HCO3- = CO3(-2)(aq) + H+(aq); K2 = [CO3(-2)]*[H+]/[HCO3-] = 4.69E-11
[HCO3-] = [CO3(-2)]*[H+]/K2 = [CO3(-2)]*(3.16E-7)/(4.69E-11) = 6376*[CO3(-2)]
[HCO3-] = 6376*[CO3(-2)]

H2CO3 = HCO3-(aq) + H+(aq); K1 = [HCO3-]*[H+]/[H2CO3] = 4.45E-7
[H2CO3] = [HCO3-]*[H+]/K1 = 6376*[CO3(-2)]*[H+]/K1 = 6375.56*[CO3(-2)]*(3.16E-7)/(4.45E-7)
[H2CO3] = 4527*[CO3(-2)]

Mass balance: [Total Barium] = [Total carbonate];
Solubility, [Ba(+2)] = S
S = [H2CO3]+[HCO3-]+[CO3(-2)] =   4527*[CO3(-2)] + 6376*[CO3(-2)] + [CO3(-2)] = 10904*[CO3(-2)]
[CO3(-2)] = S/10904

Substitute [Ba(+2)] = S and [CO3(-2)] = S/10904 into solubility equation
BaCO3(s) = Ba(+2)(aq) + CO3(-2)(aq); Ksp = [Ba(+2)]*[CO3(-2)] = 5.0E-9
[Ba(+2)]*[CO3(-2)] = 5.0E-9
S*S/10904 = 5.0E-9
solubility, S = (10904*5.0E-9)^0.5 = 0.0074 M