Question

$1$ litre of solution contains $18.9$g of $HN{O}_3$ and $1$ litre of another solution contains $3.2$g of NaOH. In what volume ratio must these solutions be mixed to obtain a solution having a neutral reaction? $(1:3.75)$.

Answer 1

$HN{O}_3+NaOH⟶NaN{O}_3+H_{2}O$

$HN{O}_3$

$1L$, $18.9g$

$m=1+14+48$

$=63g/mol$

$n_1=\frac{18.9}{63}$

$\frac{Molarity}{M_1}=0.3$

$NaOH$

$1L$,$3.2g$

$m=23+16+1$

$=40g/mol$

$n_2=\frac{3.2}{40}$

$M_2=0.08$

$NaN{O}_3+H_{2}O$

$\therefore V_1:V_2$

$=1:3.75$

For neutralisation,

$M_1{V}_1=M_2{V}_2$

$\frac{V_1}{V_2}=\frac{0.08}{0.3\times 10}=\frac{8}{30}=\frac{1}{3.75}$