Question

The minimum 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 $Pb=207\, u$) is :

• A. 0.36 L
• B. 17.98 L
• C. 0.18 L
• D. 1.798 L

Answer 1

Answer: (C)

The reaction at saturation point is $PbCl _{2}( s ) \rightleftharpoons Pb ^{2+}( aq )+2 Cl ^{-}( aq )$

If the solubility of the salt is $S mol L ^{-1}$, then $\left[ Pb ^{2+}\right]=S M$ and $\left[ Cl ^{-}\right]=S M$

So $K_{\text {sp }} =\left[ Pb ^{2+}\right]\left[ Cl ^{-}\right]^{2}$

$3.2 \times 10^{-8} =S \times(2 S)^{2}$

$4 S^{3} =3.2 \times 10^{-8}$

$S^{3} =\frac{3.2 \times 10^{-8}}{4}$

$\Rightarrow S=2 \times 10^{-3}$

Molecular weight of $PbCl _{2}=207+35.5 \times 2=278\, u$

Let $x$ L be the volume required to dissolve $0.1\, g$ of lead (II) chloride

Therefore $2 \times 10^{-3} =\frac{0.1 g }{278\, g\, mol ^{-1}} \times \frac{1}{x L }$

$x =\frac{0.1}{278 \times 2 \times 10^{-3}}=0.18\, L$