Ncert Exercises & Solutions

Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution.

( $K_{sp} of {PbCl}_{2} = 3.2 \times 10^{- 8}, atomic mass of Pb = 207 u)$

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

\underset{(1 - s)}{{PbCl}_{2} (s)} ⇌ \underset{s}{{Pb}^{2 +} (aq)} + \underset{2 s}{2 {Cl}^{-} (aq)}

K_{sp} = [{Pb}^{2 +}] . {[{Cl}^{-}]}^{2}

K_{sp} = [s] {[2 s]}^{2} = 4 s^{3}

32 \times 10^{- 8} = 4 s^{3}

s^{3} = \frac{3.2 \times 10^{- 8}}{4} = 0.8 \times 10^{- 8}

s^{3} = 8.0 \times 10^{- 9}

Solubility of

{PbCl}_{2}, s = 2 \times 10^{- 3} mol L^{- 1}

Solubility of

{PbCl}_{2} in {gL}^{- 1} = 278 \times 2 \times 10^{- 3} = 0.556 g L^{- 1}

(

∵

Molar mass of

{PbCl}_{2} = 207 + (2 \times 35.5) = 278

)

0.556 g of PbCl₂ dissolve in 1 L of water.

∴ 0.1 g of {PbCl}_{2}

will dissolve in

= \frac{1 \times 0.1}{0.556} L of water = 0.1798 L

To make a saturated solution, dissolution of 0.1 g PbCl₂ in

0.1798 L \approx 0.2 L

of water will be required.