Question

A commercial sample (2.013 g) of NaOH containing $N{a}_{2}C{O}_3$ as an impurity was dissolved to give 250 mL of solution. A 10 mL portion of the solution required 20 mL of 0.1 N $H_{2}S{O}_4$ solution for complete neutralisation of $NaOH$. Calculate the percentage by weight of $N{a}_{2}C{O}_3$ in the sample.[Write up to 2 decimal places]

Answer 1

Given, Mass of Sample $=2.013g$

Volume of $250ml$ solution of sample used by neutralization $=10ml$

Voluem of $H_{2}S{O}_4$ required for it $=20ml$

Normality of $H_{2}S{O}_4=0.1N$

At equivalence,

No. of gm equivalents of $NaOH=$ No. of gm equivalents of $H_{2}S{O}_4$

$[\because H_{2}S{O}_4$ reacts only with $NaOH$ in sample solution not with salt $N{a}_{2}C{O}_3]$

$\Rightarrow \underset{NaOH}{N_1{V}_1}=\underset{H_{2}S{O}_4}{N_2{V}_2}$

$N_1=\frac{N_2{V}_2}{V_1}=\frac{0.1\times 20}{10}=0.2N$

$\therefore$ Normality of $NaOH$ in solution$=0.2N$

$\Rightarrow$ Molarity of $NaOH$ in solution=$\frac{0.2}{nfactor}$

$=\frac{0.2}{1}=0.2M$

$[\because$ Here $nfactor$ for $NaOH=1$ as acidity of $NaOH=1]$

$\Rightarrow$ Molarity of $NaOH=0.2M$

That means,

$0.2$ moles of $NaOH$ dissolved in $1000ml$ solution

$\therefore$ No. of moles of $NaOH$ in $250ml$ solution is $\frac{0.2\times 250}{1000}=0.05$ moles

Now, for mass calculation,

$1$ mole $NaOH=40g$

$\Rightarrow 0.05$ mole $NaOH=40\times 0.05$

$=2g$

$\therefore$ Mass of $NaOH=2g$

$\Rightarrow$ Mass of impurity $N{a}_{2}C{O}_3=$ Total mass - mass of $NaOH$

$=2.013g-2g$

$=0.013g$

Now, % calculation

Mass % of $N{a}_{2}C{O}_3=\frac{MassofN{a}_{2}C{O}_3}{Totalmass}\times 100$

$=\frac{0.013}{2.013}\times 100$

$=0.6458$%