Question

100 mL of 0.1N HCl is mixed with 100mL of 0.2N NaOH, then the resulting concentration of the solution is:

Answer 1

In acid base neutralisation reactions we equate the milliequivalents of acid and base. Find the mass of pure$N{{a}_{2}}C{{O}_{3}}$ required for neutralisation and then find the actual mass required considering the purity of $N{{a}_{2}}C{{O}_{3}}$ given. The milliequivalents of acid or base can be found by the formula given below:
$\text{mEq = N x V}$
Where,
mEq is the milliequivalents of acid/base,
N is the normality of the acid/base,
V is the volume of acid/base (in mL).

Complete step-by-step answer:
A neutralization reaction is when an acid and a base react to form salt and water as the products. It involves the combination of ${{\text{H}}^{+}}$ions and $\text{O}{{\text{H}}^{-}}$ ions to generate water.
The neutralization reaction of a strong acid and a strong base result in complete neutralization and the pH is 7.
An equivalent is the amount of substance that is equivalent to or reacts with an arbitrary amount of another substance in a given chemical reaction.
Alternatively equivalent is defined as the number of moles of an ion in a solution, multiplied by the valence of that ion. Milli equivalence is the equivalence of a substance when the volume is taken in milliliters.
Equivalence is also equal to the product of molarity of the solution and n-factor of the molecule.
Given:
Normality of NaOH = 0.2
Normality of HCl = 0.1
Volume of NaOH = 100 mL
Volume of HCl = 100 mL
Milli equivalence of HCl = 100 x 0.1 = 10
Milli equivalence of NaOH = 100 x 0.2 = 20
The n-factor of HCl and NaOH is 1.
So, after neutralization, there will be 10 milli equivalents of NaOH left in the solution. But now the volume changes from 100 mL to 200 mL due to the neutralization reaction.
Normality of resulting solution = $\dfrac{\text{No}\text{. of moles left in solution}}{\text{Total volume of solution}}$
$\Rightarrow \text{ }\dfrac{\text{10}}{100\text{ +}\text{100}}\text{ = 0}\text{.05N}$

The resulting concentration is 0.05N. Therefore, the correct answer is option (A).

Note: n-factor stands for the number of electrons gained or lost by one mole of reactant. For acids, n -factor is the number of${{H}^{+}}$ions that can be replaced in a neutralisation reaction. At time n - factor is equal to the valence of the atom, however it does not hold true when an atom shows variable valency.