Question

The Ksp of BaCO3 is 5.0 x 10-9. What is the soliubility of barium carbonate in a solution that contains 1.18 M of Na2CO3?

Answer 1

Consider dissolution of BaCO₃ in water as BaCO_3(s) Ba ²⁺_(aq) + CO₃^2-_(aq)

K sp = [Ba ²⁺ ] [CO₃^2-] = 5.0 x 10 ^-09  

Consider S be the solubility of BaCO₃ in mol / L , then [Ba ²⁺ ] =[CO₃^2-]= S mol / L.

But solution contain 1.18 M Na₂CO₃ hence [CO₃^2-] = 1.18 + S.

K sp = [Ba ²⁺ ] [CO₃^2-] = 5.0 x 10 ^-09  

S x ( 1.18+S) = 5.0 x 10 ^-09  

Here S is very small as compared to 1.18 hence we can write 1.18 + S = 1.18

S x ( 1.18) = 5.0 x 10 ^-09  

^{S =} 5.0 x 10 ^-09 / 1.18

S = 4.237 x 10 ^-09 mol / L

ANSWER: Solubility of BaCO₃ in water = 4.237 x 10 ^-09 mol / L