Question

$5{\text{ mL}}$ of ${\text{N HCl}}$, $20{\text{ mL}}$ of ${\text{N/2 }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and $30{\text{ mL}}$of ${\text{N/3 HN}}{{\text{O}}_3}$ are mixed together and volume made to $1{\text{L}}$. The normality of the resulting solution is:
A.${\text{N/5}}$
B.${\text{N/10}}$
C.${\text{N/20}}$
D.${\text{N/40}}$

Answer 1

Hint: To answer this question you must recall the concept of equivalents. When solutions are mixed, it is much easier to deal with if it is considered to be the mixing of equivalents of different acids as it is the only quantity that remains unchanged unlike the volume and concentration of the solution.
Formula used: $NV = {N_1}{V_1} + {N_2}{V_2} + {N_3}{V_3}$
Where, $N$ is the normality of the resultant mixture
$V$ is the volume of the resultant mixture
${N_1},{N_2},{N_3}$ are the normalities of the solutions mixed together
And ${V_1},{V_2},{V_3}$ are the volumes of the solutions mixed together.

Complete step by step answer:
Let the normalities of hydrochloric acid, sulphuric acid and nitric acid be ${N_1},{N_2},{N_3}$ respectively and the volumes of three solutions be ${V_1},{V_2},{V_3}$ respectively.
We are given in the question that the resultant mixture obtained is made up to one liter volume which gives us the volume of the resultant mixture as $V = 1{\text{ L}}$
The milli- equivalents of the acid solutions are given as $meq = {N_i}{V_i}$
The sum of the milli- equivalents of all the acid solutions will be the same as the milli- equivalents of acid present in the resultant mixture.
So we can write, $NV = {N_1}{V_1} + {N_2}{V_2} + {N_3}{V_3}$
Substituting the values, we get,
$N \times 1000{\text{ mL}} = 1 \times 5 + \dfrac{1}{2} \times 20 + \dfrac{1}{3} \times 30$
$ \Rightarrow N \times 1000 = 5 + 10 + 10$
$N = \dfrac{{25}}{{1000}} = \dfrac{1}{{40}}$
Thus, the normality of the solution formed by mixing $5{\text{ mL}}$ of ${\text{N HCl}}$, $20{\text{ mL}}$ of ${\text{N/2 }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and $30{\text{ mL}}$of ${\text{N/3 HN}}{{\text{O}}_3}$ solutions making the volume of the mixture to one liter is given as $\dfrac{N}{{40}}$

Hence, option D is correct.

Note:
It must be noted that the volume of the final mixture is given in liters and that of the individual mixtures is in milli- liters. So, the necessary conversions must be made in order to obtain the correct answer.