Question

Calculate the solubility in mol/L of Lead(II) carbonate (PbCO3) at 298 K.

Ksp = 7.4 x 10-14
02.7 107 mol/L
O 7.2 x 10-7 mol/L
o 3.5 x 10-4 mol/L
02.7 x 10 5 mol/L
o 7.4 x 10-7 mol/L

Answer 1

2.7x10⁻⁷mol/L

Step-by-step explanation:

Based on the equilbrium:

PbCO₃(s) ⇄ Pb²⁺(aq) + CO₃²⁻(aq)

The solubility product, Ksp, is defined as:

Ksp = 7.4x10⁻¹⁴ = [Pb²⁺] [CO₃²⁻]

Now, solubility is defined as the amount of PbCO₃ that is dissolved, that is the concentration of the ions. Thus:

Solubility = S = [Pb²⁺] = [CO₃²⁻]

Replacing:

7.4x10⁻¹⁴ = [S] [S]

7.4x10⁻¹⁴ = [S]²

S = 2.7x10⁻⁷mol/L