Question

Upon mixing $45.0mL0.25M$ lead nitrate solution with $25.0mL$ of a $0.10M$ chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar concentrations of species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Answer 1

The reaction involved is
$3Pb(N{O}_3{)}_2+C{r}_2(S{O}_4{)}_3\Rightarrow 3PbS{O}_4(s)\downarrow +2Cr(N{O}_3{)}_3$
millimol of $Pb(N{O}_3{)}_2$ taken $=45\times 0.25=11.25$
millimol of $C{r}_2(S{O}_4{)}_3$ taken $=2.5$
Here, chromic sulphate is the limiting reagent, it will determine the amount of product.
$\because 1$ mole $C{r}_2(S{O}_4{)}_3$ produces $3$ moles $PbS{O}_4$.
$\therefore 2.5$ millimol $C{r}_2(S{O}_4{)}_3$ will produce $7.5$ millimol $PbS{O}_4$
Hence, mole of $PbS{O}_4$ precipitate formed = $7.5\times 10^{-3}$
Also, millimol of $Pb(N{O}_3{)}_2$ remaining unreacted
$11.25-7.50=3.75$
$\Rightarrow$ Molarity of $Pb(N{O}_3{)}_2$ in final solution
$=\frac{\text{millimol of }Pb(N{O}_3{)}_2}{\text{Total volume }}=\frac{3.75}{70}=0.054M$
Also, millimol of $Cr(N{O}_3{)}_2$ formed
$=2\times$ millimol of $C{r}_2(S{O}_4{)}_3$ reacted
$\Rightarrow$ Molarity of $Cr(N{O}_3{)}_2=\frac{5}{70}=0.071M$