Question

If $100ml$ $0.1N$ $HCl$ is mixed with $100ml$ of $0.2N$ $NaOH$, then the resulting concentration of the solution is:
(a) $0.05N$
(b) $5N$
(c) $2N$
(d) $0.1N$

Answer 1

Hint:For these questions, we have to apply the mili-equivalents concept of the normality and determine the equivalents which don't react in the solution and the number of the moles left in the solution.
Formula used:
The number of equivalents of any solution can be defined as:
$n = N \times V$
Where, $n$ is the number of equivalents, $N$ is the normality of the solution and $V$ is the volume of the solution.
The concentration of the solution when the equivalents are given:
$N = \dfrac{n}{V}$
Where, $n$ is the number of equivalents, $N$ is the concentration of the solution and $V$ is the volume of the solution.

Complete step by step answer:
Here, the data of the reaction is given. We have to find out the final resulting concentration i.e, the normality of the solution after the reaction.
The reaction given from the question:
$HCl + NaOH \to NaCl + {H_2}O$
So, these reactions as strong acid and strong base respectively are present. So, they react with each other and we have to find the number of moles left.
Given,
Normality of $HCl = 0.1N$
Volume of $HCl = 100ml$
Normality of $NaOH = 0.2N$
Volume of $NaOH = 100ml$
Now, finding the number of equivalents
The number of mili-equivalents of $HCl$ present in the solution are $0.1 \times 100 = 10$
And the number of mili-equivalents of $NaOH$ present in the solution are $0.2 \times 100 = 20$
So, the $10$ mili-equivalents of the $HCl$ reacts with $10$ mili-equivalents of $NaOH$ and hence, the mili-equivalents left in the solution will be only $NaOH$ .
No. of mili-equivalents left in the solution $ = 10$
So, the concentration of the solution,
$N = \dfrac{n}{V}$
Putting milli-equivalents as the mili-equivalents left in the solution i.e, $10$ and the volume as the total volume of the solution, i.e, $100 + 100 = 200$
Concentration of the solution, $N = \dfrac{{10}}{{200}} = 0.05N$
Hence, the concentration of the resulting solution will be $0.05N$ .

Hence, the correct option is (a) $0.05N$ .

Note:
While doing these type of question, in the formula $n = N \times V$where, $n$ is the number of equivalents, $N$ is the normality of the solution and $V$ is the volume of the solution if the volume is present in the litres then they are called equivalents and if the volume is present in the millilitres then they are called mili-equivalents.