Question

$26mL$ of a $1NN{a}_{2}C{O}_3$ solution is neutralised by the acids $A$ and $B$ in different experiments. The volumes of the acids $A$ and $B$ required were $10mL$ and $40mL$ respectively. How many volumes of $A$ and $B$ are to be mixed in order to prepare $1$ litre of normal acid solution?

Answer 1

${Na}_2{CO}_3+A\to$
$V=26mL$ $10mL$
$N=1N$

From the principle of chemical equivalence
meq. of ${Na}_2{CO}_3=$ meq of $A$
$26\times 1=10\times N_A$
$N_A=\frac{26}{10}=2.6N$

${Na}_2{CO}_3+B\to$
$V=26ml$ $40mm$
$N=1N$
$26\times 1=40\times N_B$
$N_B=\frac{26}{40}N=\frac{2.6}{4}N$

Let $V_1$ml of $A$ and $V_2$ ml of $B$ is required to make $1$ lit of $1N$ solution then
$V_1+V_2=1lit=1000ml.....\left(i\right)$

Also $V_1{N}_A+V_2{V}_B=V_f{N}_f$
$V_1\times 2.6+(1000-V_1)\frac{2.6}{4}=1000\times 1$
$V_1\times 2.6\times \frac{3}{4}=1000-\frac{2.6}{4}\times 1000$
$V_1=\frac{1000\times 1.4}{2.6\times 3}=179.5mL$

volume of $A$ required $=179.4mL$
volume of $B$ required $=1000-179.4=820.6mL$