Question

At 25 ^oC the solubility of calcium sulfate is 4.90 x 10^-3 mol/L. Calculate the value of K_sp at this temperature

At 25 ^oC the solubility of zinc sulfide is 1.26 x 10^-12 mol/L. Calculate the value of K_sp at this temperature.

Answer 1

The K_sp expression is:

K_sp = [Ca²⁺] [SO₄^2-]

There is a 1:1 molar ratio between the CaSO4 that dissolves and Ca²⁺ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved CaSO4 and SO₄^2- in solution. This means that, when 4.90 x 10-3mole per liter of CaSO4 dissolves, it produces 4.90 x 10-3 mole per liter of Ca²⁺ and 4.90 x 10-3 mole per liter of SO₄^2- in solution.

Putting the values into the K_sp expression, we obtain:

K_sp = (4.90 x 10^-3) (4.90 x 10^-3) = 2.401 x 10¯⁵

Question 2

The K_sp expression is:

K_sp = [Zn²⁺] [S^2-]

There is a 1:1 molar ratio between the ZnS that dissolves and Zn²⁺ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved ZnS and S^2- in solution. This means that, when 1.26 x 10-12 mole per liter of ZnS dissolves, it produces 1.26 x 10-12 mole per liter of Zn²⁺ and 1.26 x 10-12 mole per liter of S^2- in solution.

Putting the values into the K_sp expression, we obtain:

K_sp = (1.26 x 10^-12) (1.26 x 10^-12) = 1.5876 x 10¯²⁴