Question

20 ml of $ 0.1{\text{M}} $ NaOH solution is completely neutralised by $ 0.05{\text{ M}} $ HCl solution. The volume of the acidic solution is:
(A) 40ml
(B) 10ml
(C) 15ml
(D) 25ml

Answer 1

Hint: Normality is defined as the number of equivalents of solute dissolved per litre of solution. A 1N solution is one in which one equivalent of solute is dissolved in 1 litre of solution. We shall use the property that the product of normality and volume is constant and use that to find the volume of HCl.

Complete Step by step solution:
NaOH is neutralised by HCl. This is an acid-base reaction where NaOH is base and
HCl is acid. Also, it is a redox reaction. Whenever redox reaction takes place, the number of equivalents of each reactant and each product are the same.
 $ {\text{Number of equivalents = Normality }} \times {\text{ Volume}} $
Normality can also be written as molarity by multiplying by n factor.
Molarity is moles divided by volume. Moles calculated are of solute and volume is of solution in litres.
n factor for acid is the number of $ {H^ + } $ ions. It releases during a reaction and for a base, it is the number of $ O{H^ - } $ ions it releases during a reaction.
Here n factor for NaOH is 1 as it has 1 OH group reacting and for HCl is also for 1.
 $ {N_1}{V_1} = {N_2}{V_2} $
 $ {N_1} $ is normality of NaOH. $ {V_1} $ is volume of NaOH in litres, $ {N_2} $ is normality of HCl and $ {V_2} $ is volume of HCl.
 $ {n_1}{M_1}{V_1} = {n_2}{M_2}{V_2} $
where $ {n_1} $ and $ {n_2} $ are n factors of NaOH and HCl respectively and $ {M_1} $ and $ {M_2} $ are their respective molarities.
  $ 1 \times \dfrac{1}{{10}} \times \dfrac{{20}}{{1000}} = 0.1 \times \dfrac{5}{{100}} \times {V_2} $
 $ {V_2} = 0.04 $ litres = 40 ml
Hence, the volume of acidic solution is 40ml. Therefore, the correct option is A.

Note:
n factor calculation is different for different species. For salts it is based on charge of either cation or anion. For acids and bases, it is the number of hydrogen and hydroxide ions released per molecule. The number of electrons exchanged in a reaction can be calculated with the help of n factor.