Question

A solution contains $N{a}_{2}C{O}_3$ and $NaHC{O}_3$. $10\text{mL}$ of this solution required $2.5\text{mL}$ of $0.1\text{M}{H}_{2}S{O}_4$ for neutralization using phenolphthalein as an indicator. Methyl orange is then added when a further $2.5\text{mL}$ of $0.2\text{M}{H}_{2}S{O}_4$ was required. The amount of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ respectively, in 1 litre of solution is:

• A. $5.3\text{g and 4.2 g}$
• B. $\text{3.3 g and 6.2 g}$
• C. $\text{4.2 g and 5.3 g}$
• D. $\text{6.2 g and 3.3 g}$

Answer 1

The correct option is A $5.3\text{g and 4.2 g}$

Lets say moles of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ in 10 ml mixture is x and y respectively
Phenolphthalein will reach end point after all the $N{a}_{2}C{O}_3$ is converted to $NaHC{O}_3$ so
$x\times 1=2.5\times {10}^{-3}\times 0.1\times 2$
$x=5\times {10}^{-4}$
Methyl orange will reach to the end point after all the $NaHC{O}_3$ is reacted that is $x+y$ mole,
Hence $x+y=2.5\times {10}^{-3}\times 0.2\times 2$
$x+y={10}^{-3}soy=5\times {10}^{-4}$
Mass of $N{a}_{2}C{O}_3$ in 1000 ml solution $=\frac{2.5\times {10}^{-3}\times 0.1\times 2\times 106\times 1000}{10}=5.3g$
mass of $NaHC{O}_3$ in 1000 ml $=\frac{5\times {10}^{-4}\times 84\times 1000}{10}=4.2g$