Question

The solubility product of $PbBr _{2}$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:

• A. $ 4.18\times 10^{-2} $
• B. $ 6.76\times 10^{-3} $
• C. $ 3.4\times 10^{-4} $
• D. $ 5.44\times 10^{-2} $

Answer 1

Answer: (A)

First find relationship between solubility and solubility product.
$P b B r_{2} P b^{2}+2 B r^{-}$
$K_{s p}=\left[P b^{2+}\right]\left[B r^{-}\right]^{2}$
given $K_{s p} P b B r_{2}=10.8 \times 10^{-5} \alpha=70$
Solubility $=x \,mol / L $
$\therefore \left[ Pb ^{2+}\right]=0.7 x,\left[ Br ^{-}\right]=2 \times 0.7 \times x=1.4 x $
$\therefore K_{s p}=\left[P b^{2+}\right]\left[B r^{-}\right]^{2}=(0.7 x)(1.4 x)^{2}$
or $10.8 \times 10^{-5}=1.372 x^{3}$
or $x=\sqrt[3]{\frac{10.8 \times 10^{-5}}{1.372}}$
$=4.18 \times 10^{-2}$