Question

$1.325g$ of anhydrous sodium carbonate are dissolved in water and the solution made up to $250mL$. On titration $25mL$ of this solution neutralise $20mL$ of a solution of sulphuric acid. How much water should be added to $450mL$ of this acid solution to make it exactly $N/12$?

Answer 1

Eq. mass of $N{a}_{2}C{O}_3=\frac{Mol.mass}{2}=\frac{106}{2}=53$
$250mL$ of the sodium carbonate solution contains $=1.325g$
$1000mL$ of the sodium carbonate solution contains
$=\frac{1.325g}{250}\times 1000=5.300$
Normality of $N{a}_{2}C{O}_3$ solution $=\frac{Strength\left(g/L\right)}{Eq.mass}$
$=\frac{5.30}{53}=\frac{1}{10}N$
Applying $\underset{\left(N{a}_{2}C{O}_3\right)}{N_1{V}_1}\equiv \underset{\left(H_{2}S{O}_4\right)}{N_2{V}_2}$
$\frac{1}{10}\times 25=N_2\times 20$
$N_2=\frac{25}{10\times 20}=\frac{1}{8}$
Applying $\underset{\left(Beforedilution\right)}{N_B{V}_B}\equiv \underset{\left(Afterdilution\right)}{N_A{V}_A}$
$\frac{1}{8}\times 450=\frac{1}{12}\times V_A$
$V_A=\frac{450\times 12}{8}=674mL$
Water to be added for dilution $=(675-450)=225mL$.