1.325 g of anhydrous sodium carbonate is dissolved in water and the solution made upto 250 mL. On titration 25 mL of this solution neutralize 20 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N/12 ?

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answer is 225.

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Detailed Solution

Eq. mass of $N a_{2} C O_{3} = \frac{M o l . m a s s}{2} = \frac{106}{2} = 53$
250 mL of the sodium carbonate solution contains = 1.325 g
1000 mL of the sodium carbonate solution contains $\frac{1.325}{250} \times 1000 = 5.3 g$
Normality of $N a_{2} C O_{3}$ solution
$= \frac{S t r e n g t h (g / L)}{E q . m a s s} = \frac{5.30}{53} = \frac{1}{10} N$
Applying, $\underset{(N a_{2} C O_{3})}{N_{1} V_{1}} = \underset{(H_{2} S O_{4})}{N_{2} V_{2}}$
$\frac{1}{10} \times 25 = N_{2} \times 20 \Rightarrow N_{2} = \frac{25}{10 \times 20} = \frac{1}{8}$
Applying, $\underset{B e f o r e d i l u t i o n}{N_{B} V_{B}} = \underset{A f t e r d i l u t i o n}{N_{A} V_{A}}$
$\frac{1}{8} \times 450 = \frac{1}{2} \times V_{A} \Rightarrow V_{A} \frac{450 \times 12}{8} = 675 m L$
Water to be added for dilution
= (675 – 450) = 225 mL