Question

If the maximum concentration of $PbCl_{2}$ in water is 0.01 M at $25^\circ C$ , its maximum concentration in 0.1 M NaCl will be

• A. $2\times 10^{- 3}M$
• B. $1.6\times 10^{- 2}M$
• C. $1\times 10^{- 4}M$
• D. $4\times 10^{- 4}M$

Answer 1

Answer: (D)

$PbCl_{2}\rightleftharpoonsPb^{2 +}+2Cl$
$K_{s p}=\left[\right.Pb^{2 +}\left]\right.\left[\right.Cl^{-}\left]\right.^{2}$
$=\left(\right.0.01\left.\right)\left(\right.2\times 0.01\left(\left.\right)^{2}$
$K_{s p}=4\times 10^{- 6}$
$PbCl_{2}\rightleftharpoons\underset{\underset{S}{0}}{P b^{2 +} \left(\right. a q \left.\right)}+\underset{\underset{\left(\right. 2 S + 0.1 \left.\right)}{0.1}}{2 C l^{-} \left(\right. a q \left.\right)}$
$\text{K}_{\text{sp}} \, \text{=} \, \left[\right. \text{Pb}^{\text{2+}} \left]\right. \text{[Cl}^{-} \text{]}^{\text{2}}$
$\left[0.1 + 2 \text{S} \approx 0.1\right]$
$4\times 10^{- 6}=S\times \left(\right.0.1\left(\left.\right)^{2}$
$S=4\times 10^{- 4}M$
Maximum concentration of $PbCl_{2}$ in 0.1 M NaCl is $4\times 10^{- 4}M$ .