1. Solubility from K s

2. Calculate the solubility, in g L -1, of CaCO 3, given K s (CaCO 3 ) = 5 × 10 -9. M(CaCO 3 ) = 100 g mol –1. 1 Write the equilibrium equation. 2 Write the expression for K s. K s = [Ca 2+ ][CO 3 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 [Ca 2+ ] = s [CO 3 2– ] = s

3. 4 Rewrite the expression for Ks using these concentrations. K s = [Ca 2+ ][CO 3 2– ] = s 2 5 Use the value for K s to calculate the value of s.

4. 6 Convert the solubility from mol L –1 into g L –1.

5. Calculate the solubility, in g per 100 mL, of PbI 2, given K s (PbI 2 ) = 1 × 10 -9. M(PbI 2 ) = 461 g mol –1. 1 Write the equilibrium equation. 2 Write the expression for K s. K s = [Pb 2+ ][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 [Pb 2+ ] = s [I – ] = 2s

6. 4 Rewrite the expression for K s using these concentrations. K s = [Pb 2+ ][I – ] 2 = (s)(2s) 2 = 4s 3 5 Use the value for K s to calculate the value of s.

7. 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