Question

Solubility product $(K_{sp}) $ of saturated $PbCl_2$ in water is $1.8 \times 10^{-4} \, mol^3 \, dm^{-9}$. What is the concentration of $Pb^{2+}$ in the solution?

• A. $(0.45 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}$
• B. $(1.8 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}$
• C. $(0.9 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}$
• D. $(2.0 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}$
• E. $(2.45 \times 10^{-4} )^{1/3} \, mol \, dm^{-3}$

Answer 1

Answer: (A)

For the reaction of the $A B_{2}$

i.e. $\left( PbCl _{2}\right)$

$K_{SP} =4 S^{3}$

or, $S =\left[\frac{K_{S P}}{4}\right]^{1 / 3} $

Given, $K_{SP}=1.8 \times 10^{-4}\, mol ^{3} \,dm ^{-9}$

$\therefore $ Solubility of $Pb ^{+2}$ ions will be

$\therefore S=\left[\frac{1.8 \times 10^{-4}}{4}\right]^{\frac{1}{3}} $

$=\left[0.45 \times 10^{-4}\right]^{\frac{1}{3}} mol.\, dm ^{-3}$