Solve the following

Question:

$0.4 \mathrm{~g}$ mixture of $\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CO}_{3}$ and some inert impurities was first titrated with $\frac{\mathrm{N}}{10} \mathrm{HCl}$ using phenolphthalein as an indicator, $17.5 \mathrm{~mL}$ of $\mathrm{HCl}$ was required at the end point. After this methyl orange was added and titrated. $1.5 \mathrm{~mL}$ of same $\mathrm{HCl}$ was required for the next end point. The weight percentage of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ in the mixture is ___________________(Rounded-off to the nearest integer)

Solution:

(3)

$1^{\text {st }}$ end point reaction

$\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

$n f=1$

$\mathrm{NaCO}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NaHCO}_{3}$

$n f=1$

Eq of $\mathrm{HCl}$ used $=\mathrm{n}_{\mathrm{NaOH}} \times 1+\mathrm{n}_{\mathrm{Na}_{2} \mathrm{CO}_{3}} \times 1$

$17.5 \times \frac{1}{10} \times 10^{-3}=\mathrm{n}_{\mathrm{NaOH}}+\mathbf{n}_{\mathrm{Na}_{2} \mathrm{CO}_{3}}$

$2^{\text {nd }}$ end point

$\mathrm{NaHCO}_{3}+\mathrm{HCl} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}$

$1.5 \times \frac{1}{10} \times 10^{-3}=\mathrm{n}_{\mathrm{NaHCO}_{3}} \times 1=\mathrm{n}_{\mathrm{NaHCO}_{3}}$

$0.15 \mathrm{mmol}=\mathrm{n}_{\mathrm{Na}_{2} \mathrm{CO}_{3}}$

$0.15=\mathrm{n}_{\mathrm{Na}_{2} \mathrm{CO}_{3}}$

$\mathrm{W}_{\mathrm{Na}_{2} \mathrm{CO}_{3}}=\frac{0.15 \times 106 \times 10^{-3}}{0.5} \times 100 \times 10$

$=3 \times 106 \times 10^{-2}$

$=3 \times 1.06=3.18 \%$