Question

Two solutions of following compounds having given volume are mixed.
$\left[Pb\right(N{O}_3{)}_2{]}_1=3.0\times {10}^{-2}M$ $[NaCl{]}_2=8.0\times {10}^{-2}M$
$V_1=0.15L$ $V_2=300mL$
$K_{sp}\left(PbC{l}_2\right)=2.4\times {10}^{-4}$
What will happen to $PbC{l}_2$:

• A. $PbC{l}_2$ will precipitate out from the resultant solution.
• B. $PbC{l}_2$ will not precipitate out from the resultant solution.
• C. Nothing can be predicted about the precipitation of $PbC{l}_2$.
• D. None of the above.

Answer 1

The correct option is B $PbC{l}_2$ will not precipitate out from the resultant solution.

We know,
$\left[c\right]=\frac{n}{V}$
$n=\left[c\right]\times V$

$c_1=3.0\times {10}^{-2}$ $M$
$V_1=0.15L$
Putting the values,
$n_1=3.0\times {10}^{-2}\times 0.15=4.5\times {10}^{-3}mol$
$MolesofPb(N{O}_3{)}_2=4.5\times {10}^{-3}mol$

$c_2=8.0\times {10}^{-2}$ $M$
$V_2=0.30L$
Putting the values,
$n_2=8.0\times {10}^{-2}\times 0.30=2.4\times {10}^{-2}mol$
Moles of $NaCl=2.4\times {10}^{-2}mol$

Total volume$=V_1+V_2=0.45L$

$\left[P{b}^{2+}\right]=\frac{n_1}{V}$
$\left[P{b}^{2+}\right]=\frac{4.5\times {10}^{-3}}{0.45}=0.010M$

$\left[C{l}^{-}\right]=\frac{n_2}{V}$
$\left[C{l}^{-}\right]=\frac{2.4\times {10}^{-2}}{0.45}=0.053M$

For $PbC{l}_2$,
$\begin{array}{cccccc}& PbC{l}_2\left(s\right)& \rightleftharpoons & P{b}^{2+}\left(aq\right)& +& 2C{l}^{-}\left(aq\right)\\ [{]}_i& & & 0& & 0\\ [{]}_e& & & +0.010& & +0.053\\ [{]}_e& & & 0.010& & 0.053\end{array}$

$Q_{sp}=\left[P{b}^{2+}\right]\times [C{l}^{-}{]}^2$
$Q_{sp}=0.010\times {0.053}^2=2.8\times {10}^{-5}$
But, given:
$K_{sp}=2.4\times {10}^{-4}$

No precipitation of $PbC{l}_2$ will occur because $Q_{sp}<K_{sp}.$