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$V_{1} m l$ of NaOH of molarity X and $V_{2} m l$ of $B a (O H)_{2}$ of molarity $y / 2$ are mixed together. Mixture is completely neutralized by $100 m l 0.1 / 2 M H_{2} S O_{4}$ . If $V_{1} / V_{2} = 1 / 4$ and $x / y = 4,$ what fraction of acid is neutralized by $B a (O H)_{2}$ ?

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0.5

0.25

0.33

0.67

answer is A.

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Detailed Solution

$V_{1} \times x + V_{2} \times \frac{y}{2} \times 2 = 100 \times \frac{0.1}{2} \times 2$

Also, $\frac{V_{2} \times \frac{y}{2} \times 2}{V_{1} \times x + V_{2} \times \frac{y}{2} \times 2} =$ fraction.

Put the values we get,

$= \frac{5}{10} = 0.5$

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