Question

3) A saturated solution of PbCl2 in water was prepared and filtered. From the filtrate (solution collected after filtration), 500 mL was measured out into a beaker and evaporated to dryness. The solid PbCl2 residue recovered in the beaker amounted to 2.0 grams. a) Calculate the molar solubility of PbCl2. b) Calculate the solubility product constant, Ksp, for PbCl2

Answer 1

Answer: a) 0.0144mol/L

b) $1.19\times 10^{-5}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the $PbCl_2$ is given as:

We are given:

Solubility of $PbCl_2$ = $\frac{2.0g}{0.5L}=4g/L$

Molar Solubility of $PbCl_2$ = $\frac{4g/L}{278.1g/mol}=0.0144mol/L$

1 mole of $PbCl_2$ gives 1 mole of $Pb^{2+}$ and 2 moles of $Cl^-$ ions

Solubility product of $PbCl_2$ = $[Pb^{2+}][Cl^-]^2$

$K_{sp}=[0.0144][2\times 0.0144]^2$

$K_{sp}=1.19\times 10^{-5}$

Thus the solubility product constant is $1.19\times 10^{-5}$