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$25$ ml of the given $H C l$ solution requires $30$ mL of $0.1$ M sodium carbonate solution. What is the volume of this $H C l$ solution required to titrate $30$ mL of $0.2$ M aqueous $N a O H$ solution?

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$25 m L$

$50 m L$

$12.5 m L$

$75 m L$

answer is A.

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Detailed Solution

${(VN)}_{HCl} = {(VN)}_{{Na}_{2} {CO}_{3}}$

$25 \times N = 2 \times 0.1 \times 30$

$N_{HCl}$ = $0.24 N$

${(VM)}_{HCl} = {(VM)}_{NaOH} V \times 0.24 = 30 \times 0.2$

$V = 30 \times 0.2 / 0.24$

$V = 25 m l$

Hence the correct option is A.

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