Question

The $pH$ of pure water at ${25}^{o}C$ and ${60}^{o}C$ are $7$ and $6.5$ respectively. $HCl$ gas is passed through water at ${25}^{o}C$ till the resulting $1$ litre solution aquires a $pH$ of $3$. Now $4\times {10}^{-3}$ mole of $NaCN$ are added into this solution. Also a fresh $0.1M$ $HCN$ solution has $pH$ $5.1936$. Now in the one part of solution obtained after addition of $NaCN$, one millimole of $NaOH$ are added and in the second part of this solution $0.5$ miili mole of $HCl$ are added.

The volume of $HCl$ passed through the solution at ${25}^{o}C$ and $1$ atm is :

• A. $24.46mL$
• B. $2.446mL$
• C. $244.6mL$
• D. $0.2446mL$

Answer 1

The correct option is A $24.46mL$

$pH$ of solution after passage of $HCl=3$
$\therefore$ $\left[H^{+}\right]={10}^{-3}M$ or $\left[HCl\right]={10}^{-3}M$
From $PV=nRT$
$1\times V={10}^{-3}\times 0.0821\times 298$
$V=24.46\times {10}^{-3}L=24.46mL$