Question

Upon mixing $45.0mL$ of $0.25M$ lead nitrate solution with $25mL$ of $0.1M$ chromic sulphate solution, precipitation of lead sulphate takes place. How many $moles$ of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Answer 1

The reaction is:
$\underset{3mole}{3Pb(N{O}_3{)}_2}+\underset{1mole}{C{r}_2(S{O}_4{)}_3}\to \underset{3mole}{3PbS{O}_4}+\underset{2mole}{2Cr(N{O}_3{)}_3}$
No. of moles of $Pb(N{O}_3{)}_2=45\times {10}^{-3}\times 0.25$
$=11.25\times {10}^{-3}mole$
No. of moles of $C{r}_2(S{O}_4{)}_3=25\times {10}^{-3}\times 0.1=2.5\times {10}^{-3}mole$
Thus, $C{r}_2(S{O}_4{)}_3$ has limiting concentration. It shall be consumed fully and the number of moles of lead sulphate produced will be
$=3\times 2.5\times {10}^{-3}=7.5\times {10}^{-3}mole$
No. of mole of lead nitrate left $=11.25\times {10}^{-3}-7.5\times {10}^{-3}$
$=3.75\times {10}^{-3}mole$
Total volume $=(450+25.0)=70mL$ or $70\times {10}^{-3}litre$
$Molarity=\frac{3.75\times {10}^{-3}}{70\times {10}^{-3}}=0.0536M$
No. of moles of $Cr(N{O}_3{)}_3$ formed $=2\times 2.5\times {10}^{-3}=5\times {10}^{-3}mole$
$Molarity=\frac{5\times {10}^{-3}}{70\times {10}^{-3}}=0.0714M$
$Pb(N{O}_3{)}_2$ and $Cr(N{O}_3{)}_3$ will be present in solution in ionic form.
Thus, $\left[P{b}^{2+}\right]=0.0536M$
$\left[C{r}^{3+}\right]=0.0714M$
$\left[N{O}_3^{-}\right]=(2\times 0.0536)+(3\times 0.0714)$
$=0.3214M$.