If $50 m L$ of $0.1 M C a (O H)_{2}$ solution is required to neutralize $25 m L$ of $H_{3} P O_{4}$ solution then what will be the strength of $H_{3} P O_{4}$ in $g / L$ ?

1.64 g / L

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0.2 g / L

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8.2 g / L

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12.74 g / L

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Solution

The correct option is D $12.74 g / L$
$H_{3} P O_{4} is tribasic acid.$
$∴$ n-factor for $H_{3} P O_{4} = 3$
$C a (O H)_{2} is diacidic base.$
$∴$ n-factor for $C a (O H)_{2} = 2$

The condition of neutralization is :
Equivalents of $H_{3} P O_{4} =$ Equivalents of $C a (O H)_{2}$
$N_{H_{3} P O_{4}} \times V_{H_{3} P O_{4}} = N_{C a (O H)_{2}} \times V_{C a (O H)_{2}}$

$Normality = Molarity \times n_{f}$

$∴ M_{H_{3} P O_{4}} \times {n_{f}}_{H_{3} P O_{4}} \times V_{H_{3} P O_{4}} = M_{C a (O H)_{2}} \times {n_{f}}_{C a (O H)_{2}} \times V_{C a (O H)_{2}}$

Putting the values,

$M_{H_{3} P O_{4}} \times 3 \times 25 = 0.1 \times 2 \times 50$
$∴ M_{H_{3} P O_{4}} = 0.13 M = 0.13 m o l / L$
Molar mass of $H_{3} P O_{4} = 98 g / m o l$
Strength of $H_{3} P O_{4} = 0.13 \times 98 = 12.74 g / L$

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