Question

$100{\text{ }}c{m^3}$ of a solution of an acid (molar mass= 98) containing 29.4 g of the acid per litre were completely neutralized by $90{\text{ }}c{m^3}$ of aq. $NaOH$ containing 20 g of $NaOH$ per $500{\text{ }}c{m^3}$ . The basicity of the acid is:

Answer 1

According to the law of equivalence of an acid and a base, the product of the normality and volume of the acid is equal to the product of the normality and volume of the base. This law is used in the titration of a base with an acid or the reverse is also possible and along with that, the neutralization point is achieved.

Complete step by step answer:
The basicity of an acid (n) is defined as the number of free hydrogen ions produced by a single molecule of the acid in the solution.
In the given problem, we need to equalize the molar equivalents of the acid with the molar equivalents of the base.
So, ${M_{eq(acid)}} = {M_{eq(base)}}$
Now, as we know that, ${M_{eq}} = N \times V$
Now applying the law of equivalence, we have:
${N_A}{V_A} = {N_B}{V_B}$
Here, {N_A} = \dfrac{w}{{{E_w}}} = \dfrac{{29.4}}{{\dfrac{{98}}{n}}} \\\ {N_B} = \dfrac{w}{{{E_w}}} = \dfrac{{20}}{{40}} \times \dfrac{{1000}}{{500}} \\\ \
On solving, we have:
$\dfrac{{29.4}}{{\dfrac{{98}}{n}}} \times 100 = \dfrac{{20}}{{40}} \times \dfrac{{1000}}{{500}} \times 90 \\\ n = 3 \\\$
Thus, the value of basicity of the acid is equal to 3.

Note:
The law of equivalence is employed in a neutralization reaction of an acid with a base or vice versa during the titration process. The numbers of equivalents are equalized on both the sides and the desired parameters are determined.