Question

A solution contains $N{a}_{2}C{O}_3$ and $NaHC{O}_3,10mL$ of this solution required $2.5mL$ of $0.1M{H}_{2}S{O}_4$ for neutralisation using phenolphthalein indicator. Methyl orange is added after first end point, further titration required $2.5mL$ of $0.2M{H}_{2}S{O}_4$. The amount of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ in $1$ litre of the solution is:

• A. $5.3g$ and $4.2g$
• B. $3.3g$ and $6.2g$
• C. $4.2g$ and $5.3g$
• D. $6.2g$ and $3.3g$

Answer 1

Answer: (A)

$5.3g$ and $4.2g$

a) ${2Na}_2{CO}_3+H_2{SO}_4\rightleftharpoons 2{NaHCO}_3+{Na}_2{SO}_4$

b) $2{NaHCO}_3+H_2{SO}_4\rightleftharpoons {Na}_2{SO}_4+2H_2{CO}_3$

$2.5$ ml of $0.1M$ $H_2{SO}_4=2.5\times 0.1\times 2\times {10}^{-3}$ moles of $H^{+}$.

$=0.5\times {10}^{-3}$ moles of $H^{+}$

$\therefore$ $0.5\times {10}^{-3}$ moles of ${Na}_2{CO}_3$ is present in the solution.

$2.5$ ml of $0.2M$ $H_2{SO}_4\equiv 2.5\times 0.2\times 2\times {10}^{-3}$ moles of $H^{+}$

$=1.0\times {10}^{-3}$ moles

So total amount of ${NaHCO}_3$ after first end $=1\times {10}^{-3}$ moles

$\therefore$ The mixture contains $=(1\times {10}^{-3}-0.5\times {10}^{-3})$ moles of ${NaHCO}_3$.

The amount of ${Na}_2{CO}_3$ in $1$ litre solution $=\frac{0.5\times {10}^{-3}}{10}\times {10}^3\times 106=5.3gm$

The amount of ${NaHCO}_3=\frac{0.5\times {10}^{-3}}{10}\times {10}^3\times 84=4.2gm$