Question

Solubility product $(K_{sp})$ of saturated $PbC{l}_2$ in water is $1.8\times 10^{-4}mo{l}^{3}d{m}^{-9}$. What is the concentration of $P{b}^{2+}$ in the solution?

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

Answer 1

Answer: (A)

For the reaction of the $A{B}_2$

i.e. $\left(PbC{l}_2\right)$

$K_{SP}=4S^3$

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

Given, $K_{SP}=1.8\times 10^{-4}mo{l}^{3}d{m}^{-9}$

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

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

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