Question

Solubility product $\left({\left(\text{K}\right)}_{\text{s}\text{p}}\right)$ of saturated $\text{P}\text{b}\text{C}{\text{l}}_2$ in water is $1.8\times 10^{-4}\text{m}\text{o}{\text{l}}^3\text{d}{\text{m}}^{-9}.$ What is the concentration of $\text{P}{\text{b}}^{2+}$ in the solution?

• A. ${\left(0.45\times {\left(10\right)}^{-4}\right)}^{\frac{1}{3}}\text{m}\text{o}\text{l}\text{d}{\left(\text{m}\right)}^{-3}$
• B. ${\left(1.8\times {\left(10\right)}^{-4}\right)}^{\frac{1}{3}}\text{m}\text{o}\text{l}\text{d}{\left(\text{m}\right)}^{-3}$
• C. ${\left(0.9\times {\left(10\right)}^{-4}\right)}^{\frac{1}{3}}\text{m}\text{o}\text{l}\text{d}{\left(\text{m}\right)}^{-3}$
• D. ${\left(2.0\times {\left(10\right)}^{-4}\right)}^{\frac{1}{3}}\text{m}\text{o}\text{l}\text{d}{\left(\text{m}\right)}^{-3}$

Answer 1

Answer: (A)

For the reaction of the $\text{A}{\text{B}}_2$ i.e. $\left(\text{P}\text{b}\text{C}{\left(\text{l}\right)}_2\right)$
${\text{K}}_{\text{s}\text{p}}=4{\text{s}}^3$
Or, $\text{S}={\left[\frac{{\text{K}}_{\text{s}\text{p}}}{4}\right]}^{\frac{1}{3}}$
Given, ${\text{K}}_{\text{s}\text{p}}=1.8\times 10^{-4}\text{m}\text{o}{\text{l}}^3\text{d}{\text{m}}^{-9}$
$\therefore$ Solubility of $\text{P}{\text{b}}^{+2}$ ions will be
$\therefore \text{S}={\left[\frac{1.8\times 10^{-4}}{4}\right]}^{\frac{1}{3}}$
$={\left[0.45\times 10^{-4}\right]}^{\frac{1}{3}}\text{m}\text{o}\text{l}.\text{d}{\text{m}}^{-3}$