Question

$0.2gm$ of solution of mixture of $NaOH$ and $N{a}_{2}C{O}_3$ and inert impurities was first titrated with phenolphthalein and $N/10$ $HCl$ $17.5ml$ of $HCl$ was required at the endpoint. After this methyl orange was added and $2.5ml$ of same $HCl$ was again required for next endpoint. Find out percentage of $NaOH$ and $N{a}_{2}C{O}_3$ in the mixture.

• A. % $NaOH=13.5$ , % $N{a}_{2}C{O}_3=35$
• B. % $NaOH=30$ , % $N{a}_{2}C{O}_3=25$
• C. % $NaOH=18$ , % $N{a}_{2}C{O}_3=40$
• D. % $NaOH=35$ , % $N{a}_{2}C{O}_3=13.5$

Answer 1

The correct option is D % $NaOH=35$ , % $N{a}_{2}C{O}_3=13.5$

$1.$ In the first titration when we use phenopthalein, $NaOH$ will be neutralized by $HCl$

Equivalent of $HCl=$ Equivalent of $NaOH$

$\to$ Equivalent of $NaOH=\frac{0.1\times 17.5}{1000}=1.75\times {10}^{-3}$

Weight of $NaOH=$ Equivalent $\times$ Equivalent Weight $=1.75\times {10}^{-3}\times 40=0.07$ $gm$

$\%NaOH=\frac{0.07}{0.2}\times 100=35\%$

Equivalent of $HCl=$ Equivalent of $N{a}_{2}C{O}_3$

$\to$ Equivalent of $N{a}_{2}C{O}_3=\frac{2.5\times 0.1}{1000}=0.25\times {10}^{-3}$

Weight of $N{a}_{2}C{O}_3$ in the sample $=$ Equivalent $\times$ Equivalent Weight $=\frac{0.25}{1000}\times 106=0.0265$ $gm$

$\%N{a}_{2}C{O}_3=\frac{0.0265}{2}\times 100=13.5\%$