The volume of $1.5 M H_{3} P O_{4}$ solution required to neutralize exactly $90$ mL of a $0.5 M B a (O H)_{2}$ solution is :

10

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30

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20

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60

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Solution

The correct option is B $20$ mL
We know,
$N = (Basicity or acidity) \times M$
Acidity of $H_{3} P O_{4} = 3$
Basicity of $B a (O H)_{2} = 2$
$N_{1} V_{1} = N_{2} V_{2}$
$∴ 1.5 \times 3 \times V_{1} = 0.5 \times 0.09 \times 2$
$V_{1} = 0.02$ L or $20$ mL

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