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
$\therefore \, \, \, \text{S}=\left[\frac{1.8 \times 10^{- 4}}{4}\right]^{\frac{1}{3}}$
$=\left[0.45 \times 10^{- 4}\right]^{\frac{1}{3}} \, \text{m}\text{o}\text{l}. \, \text{d}\text{m}^{- 3}$