Question

If $50mL$ of $0.1MCa(OH{)}_2$ solution is required to neutralize $25mL$ of $H_{3}P{O}_4$ solution then what will be the strength of $H_{3}P{O}_4$ in $g/L$ ?

• A. $1.64g/L$
• B. $0.2g/L$
• C. $8.2g/L$
• D. $12.74g/L$

Answer 1

The correct option is D $12.74g/L$

$H_{3}P{O}_4\text{is tribasic acid.}$
$\therefore$ n-factor for $H_{3}P{O}_4=3$
$Ca(OH{)}_2\text{is diacidic base.}$
$\therefore$ n-factor for $Ca(OH{)}_2=2$

The condition of neutralization is :
Equivalents of $H_{3}P{O}_4=$ Equivalents of $Ca(OH{)}_2$
$N_{H_{3}P{O}_4}\times V_{H_{3}P{O}_4}=N_{Ca(OH{)}_2}\times V_{Ca(OH{)}_2}$

$\text{Normality}=\text{Molarity}\times n_f$

$\therefore M_{H_{3}P{O}_4}\times {n_f}_{H_{3}P{O}_4}\times V_{H_{3}P{O}_4}=M_{Ca(OH{)}_2}\times {n_f}_{Ca(OH{)}_2}\times V_{Ca(OH{)}_2}$

Putting the values,

$M_{H_{3}P{O}_4}\times 3\times 25=0.1\times 2\times 50$
$\therefore M_{H_{3}P{O}_4}=0.13M=0.13mol/L$
Molar mass of $H_{3}P{O}_4=98g/mol$
Strength of $H_{3}P{O}_4=0.13\times 98=12.74g/L$