Question

A $100$ ml solution of $0.1NHCl$ was titrated with $0.2$ N $NaOH$ solution. The titration was discontinued after adding $30ml$ of $NaOH$ solution. The remaining solution was titrated with $0.25NKOH$ solution. The volume of $KOH$ required to complete the titration is:

• A. $70$ ml
• B. $32$ ml
• C. $35$ ml
• D. $16$ ml

Answer 1

The correct option is D $16$ ml

In this titration, one acid and two bases are used. The appropriate formula is $N_1{V}_1=N_2{V}_2+N_3{V}_3$.

Here, $1$ represents $HCl$, $2$ represents $NaOH$ and $3$ represents $KOH$.

Substituting values in the above expression, we get

$100\times 0.1=0.2\times 30+0.25\times V_3$

$\therefore 10=6+0.25\times V_3$

Hence, $V_3=\frac{4}{0.25}=16$ ml.