Question

100 mL of 0.3 N HCl is mixed with 200 mL of 0.6 N ${H_2}S{O_4}$ Solution. The final normality of acid is-
A) 0.4 N
B) 0.5 N
C) 0.6 N
D) 0.9 N

Answer 1

We can define normality of a solution as the equivalent concentration of the solution. It is basically a measurement of reactive species in solution.
We can describe normality as the number of gram-equivalent or mole-equivalent of solute present in $1000{\text{ }}liter$ of solution.
We need to use this formula to solve the question-
${N_1}{V_1} = {N_2}{V_2}$
Where,
${N_1}$ is the normality of Acid.
${V_1}$ is the volume of the Acid.
${N_2}$ is the normality of the solution.
${V_2}$ is the volume of solution

Complete answer:
Normality of $HCl = 0.3N$ in $100{\text{ }}mL$ solution.
Normality of ${H_2}S{O_4} = 0.6N$ in $200{\text{ }}mL$ solution.
$100{\text{ }}mL{\text{ }}0.3{\text{ }}N{\text{ }}HCl$ means,
Normality of 100{\text{ }}mL{\text{ }}HCl$$$ = 0.3N$ ${\text{Normality of 200 mL HCl = 0}}{\text{.3 $\times$ }}\dfrac{{{\text{200}}}}{{{\text{100}}}} \\\ {\text{ = 0}}{\text{.6 N}} \\\$ 200{\text{ }}mL${H_2}S{O_4} = 0.6N$ means, Normality of200{\text{ }}mL ${H_2}S{O_4} = 0.6{\text{N}}$ $\therefore $Normality of Acid $({N_1}) = 0.6\;{\text{N}}$ Volume of Acid (${V_1}$) = {\text{ }}200{\text{ }}mL. Now, the total volume of solution ${V_3}$ $= 100 + 200 \\\ = 300{\text{ mL}} \\\ $ Let us consider, normality of solution $ = {N_3}$ Now, we know, ${N_1}{V_1} = {N_3}{V_3}$ $0.6 \times 200 = {N_3} \times 300$ ${N_3} = \dfrac{{0.6 \times 200}}{{300}}$ $\therefore $Normality of solution is 0.4{\text{ }}N$$

So, The correct answer is Option A).

Note: Normality of solution means the concentration of solution per volume.
To solve this type of problem we have to find the normality of the Acids or Base given and volume of them. Then by using the above formula ${N_1}{V_1} = {N_2}{V_2}$ we can solve the problem.
We often get confused between molarity or molality with normality. We need to understand the conceptual difference between these three terms.