Question

Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. \[\ce{(K_{sp} of PbCl2}\] = 3.2 × 10^–8, atomic mass of \[\ce{Pb = 207 u}\]).

Answer 1

\[\ce{K_{sp} of PbCl2}\] = 3.2 × 10^–8 

Let S be the solubility of \[\ce{PbCl2}\].

	\[\ce{PbCl2 (s)}\]	\[\ce{⇌ Pb^{2+} (aq)}\]	\[\ce{+ 2Cl- (aq)}\]
Concentration of species at t = 0	1	0	0
Concentration of various species at equilibrium	1 – S	S	2S

\[\ce{K_{sp} = [Pb^{2+}] [Cl-]^2 = (S) (2S)^2 = 4S^3}\]

\[\ce{K_{sp} = 4S^3}\]

S³ = `K_(sp)/4 = (3.2 xx 10^(-8))/4` mol L^–1 = 8 × 10^–9 mol L–1

S = `root(3)(8 xx 10^-9)` = 2 × 10^–3 mol L^–1

∴ S = 2 × 10^–3 mol L^–1

Molar mass of \[\ce{PbCl2}\] = 278

∴ Solubility of \[\ce{PbCl2}\] in g L^–1 = 2 × 10^–3 × 278 g L^–1

= 556 × 10^–3 g L^–1

= 0.556 g L^–1

To get saturated solution,0.556 g of \[\ce{PbCl2}\] is dissolved in 1 L water.

0.1 g \[\ce{PbCl2}\] is dissolved in `0.1/0.556` L = 0.1798 L water.

To make a saturated solution, dissolution of 0.1 g \[\ce{PbCl2}\] in 0.1798 L ≈ 0.2 L of water will be required.