Question

Upon mixing $50$ mL of $0.1$ M lead nitrate solution with $50$ mL of $0.05$ M chromic sulphate solution, precipitation of lead sulphate solution takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?

• A. 0.005, 0.0084
• B. 0.0084, 0.005
• C. 0.005, 0.00084
• D. 0.05, 0.00084

Answer 1

Answer: (A)

0.005, 0.0084

$\underset{\underset{\underset{=5mmol}{50\times 0.1}}{3Pb\left(N{O}_3{)}_2\right)}}{3mmol}+\underset{C{r}_2)S{O}_4{)}_3}{1mmol}\to \underset{\underset{\underset{=2.5mmol}{50\times 0.05}}{3PbS{O}_4)}}{3mmol}\downarrow +\underset{2mmol}{2Cr(N{O}_3{)}_2}$

First find the limiting reagent.
$3$ mmol of $Pb(N{O}_3{)}_2\Rightarrow$ $1$ mmol of $C{r}_2(S{O}_4{)}_3$
$5$ mmol of $Pb(N{O}_3{)}_2\Rightarrow \frac{1}{3}\times 5$
$\Rightarrow \frac{5}{3}=1.66mmol$

So, $Pb(N{O}_3{)}_2$ is the limiting reagent.
(i) $3$ mmol of $Pb(N{O}_3{)}_2\Rightarrow 3$ mmol of $PbS{O}_4$
$5$ mmol of $Pb(N{O}_3{)}_2\Rightarrow 5mmol$
$\Rightarrow \frac{5}{1000}mol\Rightarrow 0.005mol$
(ii) Species left in the solution are $C{r}_2(S{O}_4{)}_3$ and $Cr(N{O}_3{)}_3$.

To calculate the concentration of $C{r}_2(S{O}_4{)}_3$:
Initial mmol $=2.5$
Reacted mmol $=1.65$
Left mmoles $=2.5-1.66=0.84mmol$
Total Volume $=50+50=100mL$

$Concentration=\frac{0.84}{100}=0.0084M$