Question

The minimum volume of water required to dissolve $0.1g$ lead(II) chloride to get a saturated solution ($K_{sp}$ of $PbC{l}_2=3.2\times 10^{-8}$ ; atomic mass of $Pb=207u$) 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 $PbC{l}_2(s)\rightleftharpoons P{b}^{2+}(aq)+2C{l}^{-}(aq)$

If the solubility of the salt is $Smol{L}^{-1}$, then $\left[P{b}^{2+}\right]=SM$ and $\left[C{l}^{-}\right]=SM$

So $K_{\text{sp }}=\left[P{b}^{2+}\right]{\left[C{l}^{-}\right]}^2$

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

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

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

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

Molecular weight of $PbC{l}_2=207+35.5\times 2=278u$

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

Therefore $2\times 10^{-3}=\frac{0.1g}{278gmo{l}^{-1}}\times \frac{1}{xL}$

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