Question

1.325 g of anhydrous sodium carbonate is dissolved in water and the solution is made up to 250 mL. On titration, 25 mL of this solution neutralised 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 $\dfrac{N}{{12}}$?

Answer 1

Hint: When a solution is diluted, the product of its normality and its volume remains constant before and after the dilution.

Complete step by step answer:
Normality of a solution is defined as the number of gram or mole equivalents of solute present in one litre of a solution. In other words, it is the number of moles of reactive units in a compound. It is denoted by ‘N’. It is a concentration term. Mathematically, it is represented as:
Normality (N) = Molarity (M) $\times$ n – factor
The n – factor in case of an acid is its basicity while in case of a base is its acidity.
Now, in the given question
Molar mass of Sodium carbonate = 106 g
Moles of $N{a_2}C{O_3}$ = $\dfrac{{1.325}}{{106}}$
= 0.0125

And, molarity of $N{a_2}C{O_3}$ = $\dfrac{{0.0125}}{{0.25}}$ = 0.05 M

Thus, normality of $N{a_2}C{O_3}$ = 0.05 $\times$ 2 = 0.1 N

Now, on titration,
${N_1}{V_{1(N{a_2}C{O_3})}} = {N_2}{V_{2({H_2}S{O_4})}}$
$\Rightarrow 0.1 \times 25 = {N_2} \times 20$
$\Rightarrow {N_2} = \dfrac{1}{8}$
Now, when the solution is diluted,
${N_i}{V_{i\left( {before\,dilution} \right)}} = {N_j}{V_{j\left( {after\,dilution} \right)}}$
$\Rightarrow \dfrac{1}{8} \times 450 = \dfrac{1}{{12}} \times {V_j}$
$\Rightarrow {V_j} = 675mL$

Hence, the amount of water to be added = (675 – 450) mL
= 225 mL

Note: Remember that the product of normality and volume remains constant only when the moles of the solute added remain constant, as in case of titration and dilution.