Question

If the maximum concentration of $PbC{l}_2$ in water is $0.01M$ at $298K$, Its maximum concentration in $0.1MNaCl$ will be:

• A. $4\times {10}^{-3}M$
• B. $0.4\times {10}^{-4}M$
• C. $4\times {10}^{-2}M$
• D. $4\times {10}^{-4}M$

Answer 1

Answer: (A)

$4\times {10}^{-3}M$

Concentrations of the ions:

The dissociation reaction for $PbC{l}_2$ in water is write as,

$PbC{l}_2\to P{b}^{2+}\left(aq\right)+2C{l}^{-}\left(aq\right)$

The maximum concentration of $PbCl2$ in water is 0.01 M

We can find concentration of the ions using the stoichiometry from the above reaction.

The concentration of $Pb²⁺$ ion is 0.01 M and concentration of Cl⁻ ion is $2\times 0.01=0.02M$.

Solubility product:

The solubility product of $PbC{l}_2$ can be written as,

$K_{SP}=\left[P{b}^{2+}\right]\left[Cl\right]{}^{-2}$

Let us insert in the concentrations of Pb and Cl.

$K_{SP}=\left(0.01\right)\left(0.02\right)$

$Ksp$ =$4.0\times {10}^{-6}$

Concentration of $P{b}^{2+}$ using common ion effect:
When$PbC{l}_2$ is dissolved in $1MNaCl$, there is a common ion $C{l}^{-}$. This decreases the solubility of $PbC{l}_2$ due to the common ion effect.
The concentration of $C{l}^{-}$ in $0.1MNaCl$ is 0.1. Let us plug in this value to find the new concentration of $P{b}^{2+}$ ion.
$Ksp=\left[P{b}^{2+}\right][C{l}^{-}{]}^2$
$4\times {10}^{-6}=\left[P{b}^{2+}\right](0.1{)}^2$
$\left[P{b}^{2+}\right]$=$\frac{4\times {10}^{-6}}{0.01}$
$\left[P{b}^{2+}\right]=0.0004M=4\times {10}^{-4}$ M
$P{b}^{2+}$ ion comes from dissolved $PbC{l}_2$. Therefore the concentration of $PbCl₂$ is the same as that of $P{b}^{2+}$
The maximum concentration of $PbCl₂$ in $0.1MNaCl$ is $0.0004M$
Hence, the correct option is $D$