Question

How many mL of 0.1M HCl are required to react completely with 1g mixture of \[N{{a}_{2}}C{{O}_{3}}\] and $NaHC{{O}_{3}}$ containing equimolar amounts of both?

Answer 1

Hint: In acid base neutralisation reactions we equate the milliequivalents of acid and base. Find the respective composition of $N{{a}_{2}}C{{O}_{3}}$ and $NaHC{{O}_{3}}$ in 1g. 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.
Let the mixture contain x g of $N{{a}_{2}}C{{O}_{3}}$ and 1-x g of $NaHC{{O}_{3}}$.
The molar masses of $N{{a}_{2}}C{{O}_{3}}$ and $NaHC{{O}_{3}}$ are 106 g and 84 g respectively.
Since it is an equimolar mixture,
$\dfrac{\text{x}}{\text{106}}\text{ = }\dfrac{\text{1-x}}{\text{84}}$
190x = 106
x = 0.5579 ; 1-x = 0.4421
Number of moles of sodium carbonate = $\dfrac{0.5579}{106}\text{ = 0}\text{.005263}$
Number of moles of sodium bicarbonate = $\dfrac{0.4421}{84}\text{ = 0}\text{.005263}$
One mole of sodium carbonate reacts with 2 moles of HCl. Similarly, one mole of sodium bicarbonate reacts with 1 mole of HCl. This is because the n - factor of sodium carbonate, sodium bicarbonate and hydrochloric acid is 2,1,1 respectively.
Total number of moles of HCl that will react with the mixture of bases = 2 x 0.005263 + 1 x 0.005263 = 0.01578 moles.
Volume of 0.1M HCl required = $\dfrac{0.01578}{0.1}$ = 0.158L = 158mL.

Therefore, the amount of HCl required is 158mL.

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.