Question

Upon mixing 45 ml of 0.25M lead nitrate solution with 25 ml of 0.10M chromic sulphate, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar conc. of the species left behind in final solution. Assume that lead sulphate is completely insoluble.

Answer 1

First write the balanced chemical equation involved and find out the moles of lead nitrate and chromic sulphate, then the limiting reagent. Accordingly calculate the moles of lead sulphate formed. In accordance with the concentration of limiting reagent find out the concentration of the reactant left in the solution and of chromic nitrate also. Then calculate the concentration of the ionic components left behind in the final solution.

Formulas used:
-Molarity: $M = \dfrac{n}{V}$
n = M × V (1)
where, n = moles and V = volume in litres

Complete step by step answer:
-First of all we will see the reaction involved in this question:
$3Pb{(N{O_3})_2} + C{r_2}{(S{O_4})_3} \to 3PbS{O_4} + 2Cr{(N{O_3})_3}$
-According to the question the concentration of lead nitrate [$Pb{(N{O_3})_2}$] is 0.25M and its volume is 45ml of 0.045L. Using these values we can find out its number of moles with the help of equation (1):
n = M × V
= 0.25 × 0.045 = 0.01125 = $11.25 \times {10^{ - 3}}$
So, the number of moles of $Pb{(N{O_3})_2}$ are $11.25 \times {10^{ - 3}}$.
-Similarly, the concentration of chromic sulphate [$C{r_2}{(S{O_4})_3}$] is 0.10M with a volume of 25ml or 0.025L and we will find out its number of moles using equation (1):
n = M × V
= 0.10 × 0.025 = 0.0025 = $2.5 \times {10^{ - 3}}$
So, the number of moles of $C{r_2}{(S{O_4})_3}$ are $2.5 \times {10^{ - 3}}$.
-Here we can see that since the concentration of $C{r_2}{(S{O_4})_3}$ is minimum, it acts as a limiting reagent and so it will be fully consumed during the reaction and the products are formed according to its concentration.
-According to the stoichiometry of the reaction for 1 mole of $C{r_2}{(S{O_4})_3}$ will form 3 moles of lead sulphate ($PbS{O_4}$).
So, $2.5 \times {10^{ - 3}}$ moles of $C{r_2}{(S{O_4})_3}$ will form:
= $3 \times (2.5 \times {10^{ - 3}})$
= $7.5 \times {10^{ - 3}}$ moles of $PbS{O_4}$
Hence the moles of lead sulphate formed are $7.5 \times {10^{ - 3}}$.
-We will now work on calculating the molar concentration of species left behind in the final solution.
-If for 1 mole of $C{r_2}{(S{O_4})_3}$, 3 moles of $Pb{(N{O_3})_2}$ were used up. Then for $2.5 \times {10^{ - 3}}$ moles of $C{r_2}{(S{O_4})_3}$, the number of moles of $Pb{(N{O_3})_2}$ used up would be = $3 \times (2.5 \times {10^{ - 3}})$
= $7.5 \times {10^{ - 3}}$ moles
Hence the moles of $Pb{(N{O_3})_2}$ left in the solution were:
= Total moles – used up moles
= $11.25 \times {10^{ - 3}}$ - $7.5 \times {10^{ - 3}}$
= $3.75 \times {10^{ - 3}}$
Total volume of solution = 25 + 45 = 70ml = 0.07L
Molarity of $Pb{(N{O_3})_2}$ will be: $M = \dfrac{n}{V}$ = $\dfrac{{3.75 \times {{10}^{ - 3}}}}{{0.07}}$ = 0.053M
- If for 1 mole of $C{r_2}{(S{O_4})_3}$, 2 moles of $Cr{(N{O_3})_3}$ will be formed. Then for $2.5 \times {10^{ - 3}}$ moles of $C{r_2}{(S{O_4})_3}$, the number of moles of $Cr{(N{O_3})_3}$ formed will be = $2 \times (2.5 \times {10^{ - 3}})$
= $5 \times {10^{ - 3}}$ moles
Molarity of $Cr{(N{O_3})_3}$ will be: $M = \dfrac{n}{V}$ = $\dfrac{{5 \times {{10}^{ - 3}}}}{{0.07}}$ = 0.0714M
-Since the question says that lead sulphate is completely insoluble, then only $Pb{(N{O_3})_2}$ and $Cr{(N{O_3})_3}$ will be present in the solution in ionic form. Concentration of $Pb{(N{O_3})_2}$ is 0.053M and that of $Cr{(N{O_3})_3}$ is 0.0714M.
Thus, $\left[ {P{b^{ + 2}}} \right]$ = 0.053M
$\left[ {C{r^{ + 3}}} \right]$ = 0.07
$\left[ {N{O_3}^ - } \right]$ = ( 2 × 0.053 ) + ( 3 × 0.0714 )
= 0.3202M

Note: The final solution will not have ionic components of lead sulphate because it is completely insoluble. Also limiting reagent is the reactant that will be totally consumed after the completion of the reaction and the amount of products formed will be limited by the reagent as without it the reaction cannot proceed.