The Ksp of strontium carbonate, SrCO3, is 5.60 x 10-10. Calculate the molar solubility, S, of this compound. S=? M

Question

The K_sp of strontium carbonate, SrCO_3, is 5.60 x 10^-10. Calculate the molar solubility, S, of this compound.

S=? M

Answer 1

SrCO₃ when placed in water will dissolve to a small extent as follows:

SrCO₃(s) <==> Sr²⁺(aq) + CO₃^2-(aq)

The Ksp value, 5.60x10^-10 gives us some idea of just how much Sr²⁺ and CO₃^2- are formed. Like other equilibrium constants, Ksp is written as products raised to a power of their coefficient, divided by the reactants raise to a power of their coefficient. In this case, the reactant gets a value of 1 since it is a solid, and hence it is not included in the equilibrium expression. Thus, ...

Ksp = [Sr²⁺][CO₃^2-]

Since we don't know these concentrations, we assign them a value of s

Ksp = (s)(s)

5.60x10^-10 = s²

s = √5.60x10^-10

s = 2.37x10^-5 M

Since each mol SrCO₃ produces 1 mol of Sr²⁺ and 1 mol CO₃^2-, we can say the molar solubility of SrCO₃ is

2.37x10^-5 M

Answer 2

Assume molar solubility of SrCO3 is x mol/L

SrCO3 <-----> Sr2+ + CO32-

x x x

KSP = [Sr2+][CO32-]

= [x][x]

= [x]2

5.60 x 10-10 = [x]2

[x] = 2.4 x 10-5 mol/L