Solubility from Ks

Calculate the solubility, in g L-1, of CaCO3, given

Ks(CaCO3) = 5 × 10-9. M(CaCO3) = 100 g mol–1.

1 Write the equilibrium equation.

2 Write the expression for Ks.

Ks = [Ca2+][CO32–]

3 Write the concentrations of each ion present at equilibrium, using the symbol s to represent the solubility of this substance in mol L-1

[Ca2+] = s [CO32–] = s

4 Rewrite the expression for Ks using these concentrations.

Ks = [Ca2+][CO32–]

= s 2

5 Use the value for Ks to calculate the value of s.

6 Convert the solubility from mol L–1 into g L–1.

Calculate the solubility, in g per 100 mL, of PbI2, given

Ks(PbI2) = 1 × 10-9. M(PbI2) = 461 g mol–1.1 Write the equilibrium equation.

2 Write the expression for Ks.

Ks = [Pb2+][I–]2

3 Write the concentrations of each ion present at equilibrium, using the symbol s to represent the solubility of this substance in mol L-1

[Pb2+] = s [I–] = 2s

4 Rewrite the expression for Ks using these concentrations.

Ks = [Pb2+][I–]2

= (s)(2s)2

= 4s3

5 Use the value for Ks to calculate the value of s.

6 Convert the solubility from mol L–1 to g L–1 and then g per 100 mL.

s = 6.3 × 10–4 mol L–1

s = 6.3 × 10–4 mol L–1 × 461 g mol–1

= 0.29 g L–1

= 0.029 g per 100 mL