25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution requried to titrate 30 mL of 0.2 M aqueous NaOH solution?

Engineering

Chemistry

Law of Chemical Equivalence

Titration

Question

25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution requried to titrate 30 mL of 0.2 M aqueous NaOH solution?

A

50 mL

B

75 mL

C

25 mL

D

12.5 mL

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Apply law of equivalence:

25 × N = 30 × 0.1 × 2

$N_{HCl} = \frac{30 \times 0 .2}{25} = \frac{6}{5} \times 0 .2 = \frac{1 .2}{5}$

For the 2nd titration

$\frac{1 .2}{5} \times V_{HCl} = 30 \times 0 .2$

$V_{HCl} = \frac{6 \times 5}{1 .2} = \frac{30}{1 .2} = 25 ml$