# 0.4g mixture of

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:

Upto first end point

gm equi. of $\left(\mathrm{NaOH}+\mathrm{Na}_{2} \mathrm{CO}_{3}\right)=\mathrm{HCl}$

$x+y \times 1=\frac{1}{10} \times 17.5$

$x+y=1.75$..(1)

Upto second end point

$\mathrm{NaOH}+\mathrm{Na}_{2} \mathrm{CO}_{3} \equiv \mathrm{HCl}$

$x+y \times 2=\frac{1}{10} \times 19$

$x+2 y=1.9$..(2)

$y=0.15$

$\% \mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{0.15 \times 10^{-3} \times 106}{0.4} \times 100$

$=3.975 \%$

$=4 \%$