Question

Calculate the volume of water(in mL) required to dissolve 0.1 g lead (II) chloride to get a saturated solution (K_sp of PbCl₂ = 3.21 × 10^-8, atomic mass of Pb = 207 u).Report your answer by rounding it up to nearest whole number.

Answer 1

Suppose solubility of PbCl₂ in water is s mol L^-1.

K_sp = [Pb²⁺] · [Cl^-]²
K_sp = [s] [2s]² = 4s³
3.2 × 10^-8 = 4s³
$\text{s}^{3} = \frac{\text{3.2} \times \text{10}^{- 8}}{4} = \text{0.8} \times \text{10}^{- 8}$
s³ = 8.0 × 10^-9
Solubility of PbCl₂, = s = 2 × 10^-3 mol L^-1
Solubility of PbCl₂ in gL^-1 = 278 × 2 × 10^-3 = 0.556 g L^-1
(∵ Molar mass of PbCl₂ = 207 + (2 × 35.5) = 278)
0.556 g of PbCl₂ dissolve in 1 L of water.
$\therefore \text{0.1 g of PbCl}_{2} \text{will dissolve in} = \frac{1 \times \text{0.1}}{\text{0.556}} \text{L of water}$
​= 0.1798 L = 179.8 mL