Question

The mass of $N{{a}_{2}}C{{O}_{3}}$of 95% purity that would be required to neutralise 45.5 mL of 0.235N acid is:
(write in multiples of 10)

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 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.
Let us now calculate the mass of pure $N{{a}_{2}}C{{O}_{3}}$required to neutralise the acid given in the question.
Volume of acid = 45.50 mL
Normality of acid = 0.235 N
The milliequivalents of acid = 45.5 x 0.235 = 10.6925
Mass of pure $N{{a}_{2}}C{{O}_{3}}$ needed = X g
Molar mass of $N{{a}_{2}}C{{O}_{3}}$ = 106 g
n-factor of $N{{a}_{2}}C{{O}_{3}}$ = 2
The milliequivalents of base = $\dfrac{\text{X}}{106}\text{ x 2 x 1000}$
Equating the milliequivalents of base and acid, we get
$\dfrac{\text{X}}{106}\text{ x 2 x 1000}$ = 10.6925
Hence the value of X is 0.5667 g
Since the purity of $N{{a}_{2}}C{{O}_{3}}$ is 95%, the actual mass needed is,
$\dfrac{100}{95}\text{ x 0}\text{.5667 = 0}\text{.5965 }\approx \text{ 5}\text{.6 x 1}{{\text{0}}^{-1}}\text{ g}$
Therefore, the mass of $N{{a}_{2}}C{{O}_{3}}$ required is $\text{5}\text{.6 x 1}{{\text{0}}^{-1}}\text{ g}$for neutralisation of the acid.

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.