Question

What volume of a $0.345\text{M}$ phosphoric acid solution is required to neutralize $125\text{mL}$ of a $0.123\text{M}KOH$ solution?

• A. $6.65\text{mL}$
• B. $10.75\text{mL}$
• C. $14.85\text{mL}$
• D. $12.82\text{mL}$

Answer 1

The correct option is C $14.85\text{mL}$

Let the volume of phosphoric acid $H_{3}P{O}_4$ be $V\text{mL}$
n-factor of phosphoric acid = 3;
n-factor of $KOH$ = 1
$\text{n-factor}\times (MV{)}_{H_{3}P{O}_4}=\text{n-factor}\times (MV{)}_{KOH}$
$(3\times 0.345\times V)=(1\times 0.123\times 125)$
On solving, volume $=14.86\text{mL}$