Excess solid Na₂CO₃ is added to a solution containing 0.500 M (each) Mg²⁺ and Zn²⁺ ions.

Question

Ksp for MgCO₃ is 3.50 × 10⁻⁸ and Ksp for ZnCO₃ is 1.00 × 10⁻¹⁰. ZnCO₃, with the smaller Ksp, will be the least soluble and will begin precipitating first. What will be the [Zn²⁺] concentration when MgCO₃ just begins to precipitate? (Assume no volume change upon addition of the solid Na₂CO₃).

Answer 1

MgCO3 will begin precipitating when the CO3^2- ion concentration reaches the value predicted by

K_sp = [Mg²⁺] [CO3^2-] is satisfied. The [CO3^2-] = 3.50 x 10^-8 / 0.5 M = 7.00 x 10^-8 M.

At this point the ZnCO3 has already started to precipitate, but the K_sp relation for ZnCO3 still holds, so [Zn²⁺] = 1.00 x 10^-10 / 7.00 x 10^-8 = 1.43 x 10^-3 M.