Question

The solubility-product constants, Ksp, at 25 ∘C for two compounds [iron(II) carbonate, FeCO3, and cadmium(II) carbonate, CdCO3] are given by the table

Substance	Ksp
FeCO3	2.10×10−11
CdCO3	1.80× 10−14

Part C

What will the concentration of Cd2+ at the moment before Fe2+ begins to precipitate?

Express your answer with the appropriate units.

Answer 1

FeCO₃ dissolves and precipitaes according to the reaction equation
FeCO₃(s) ⇌ Fe²⁺(aq) + CO₃²⁻(aq)
So the ionic molarities in a solution saturated with iron(II) carbonate satisfy the solubility product equation:
K_sp₁ = [Fe²⁺] x [CO₃²⁻]
The precipitation of FeCO₃ starts, when the product of iron(II) ion molarity and carbonate ion molarity on right hand side reaches the value K_sp on left hand side. So for given a value of [Fe²⁺] precipitation starts when carbonate ion reaches the level:
[CO₃²⁻] = K_sp₁ / [Fe²⁺]
Reaction equation and solubility product equation for CdCO₃ are:
CdCO₃(s) ⇌ Cd²⁺(aq) + CO₃²⁻(aq)
K_sp₂ = [Cd²⁺] x [CO₃²⁻]
Due to the smaller value of its K_sp CdCO₃ precipitates at lower level of [CO₃²⁻]. When you increase the [CO₃²⁻] concentration CdCO₃ precipiates such that [Cd²⁺] decreases and solubility product equation is still satisfied. The [Cd²⁺] concentration for certain level of carbonate is given by:
[Cd²⁺] = K_sp₂ / [CO₃²⁻]
So when FeCO₃ starts to precipitate the [Cd²⁺] concentration is:
[Cd²⁺] = K_sp₂ / ( K_sp₁ / [Fe²⁺] )
[Cd²⁺] = ( K_sp₂ / K_sp₁ ) x [Fe²⁺]
[Cd²⁺] = ( 1.80×10^-14 / 2.10×10^-11 ) × [Fe²⁺]
[Cd²⁺] = 0.86 ×10^-3 M × [Fe²⁺]

After putting the value of concentration of [Fe²⁺] having you you will get your answer.