Question

20 ml of $\dfrac{N}{10}HCl$ is mixed with 40 ml $\dfrac{N}{10}NaOH $ solution. the normality of the resultant acidic/basic solution will be:
(a) $\dfrac{N}{20}$
(b) $\dfrac{N}{30}$
(c) $\dfrac{N}{40}$
(d) $\dfrac{N}{15}$

Answer 1

Hint: By the term normality we means the no of moles of gram equivalent to the total volume of solution and we know the normalities and volume of the HCl and NaOH solutions and now we can easily calculate the normality of the solution by applying the formula as; $normality=\dfrac{{{N}_{2}}{{V}_{2}}-{{N}_{1}}{{V}_{1}}}{\begin{align}
& total\text{ }volume \\
& of\text{ }the\ solution \\
\end{align}}$. Now solve it.

Complete step by step answer:
First of all, let’s discuss what is normality. By the term normality we mean the number of moles of gram equivalent of the solute to the total volume of the solution in litres and it is denoted by the symbol as N. Now, considering the numerical ;
Normality of HCl solution =$\dfrac{N}{10}=0.1N$ (given)
And it is taken as ${{N}_{1}}$.
Similarly, Normality of NaOH solution =$\dfrac{N}{10}=0.1N$ (given)
And it is taken as ${{N}_{2}}$.
Now the volume of HCl solution=20 ml (given)
And it is taken as ${{V}_{1}}$.
Similarly, Normality of NaOH solution =40 ml (given)
And it is taken as ${{V}_{2}}$.
Now, the total volume of the solution i.e. of HCl and NaOH=${{V}_{1}}+{{V}_{2}}=20+40=60\text{ }ml$
Now, we can easily calculate the normality of the solution by applying the formula as;
$normality=\dfrac{{{N}_{2}}{{V}_{2}}-{{N}_{1}}{{V}_{1}}}{\begin{align}
& total\text{ }volume \\
& of\text{ }the\ solution \\
\end{align}}$ (1)
We know the values of ${{N}_{1}}$,${{N}_{2}}$,${{V}_{1}}$and ${{V}_{2}}$and the total volume of the solution and by putting these values in equation (1), we can calculate the normality of the solution as;
$\begin{align}
& normality=\dfrac{[0.1\times 40-0.1\times 20]N }{60} \\
& \text{ =}\dfrac{[4-2]N}{60} \\
& \text{ =}\dfrac{2N}{60} \\
& \text{ =}\dfrac{N}{30} \\
\end{align}$
Hence, when 20 ml of $\dfrac{N}{10}HCl$ is mixed with 40 ml $\dfrac{N}{10}NaOH$ solution , then the normality of the resultant acidic/basic solution is $\dfrac{N}{30}$.

So, option (b) is correct.

Note: Don’t get confused in the word’s molarity and normality. By the term molarity we mean the number moles of the solute to the total volume of the solution in litres whereas normality means the number moles of gram equivalent of the solute to the total volume of the solution in litres. So, $molarity=\dfrac{no\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{total\text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }liters}$ And, $normality=\dfrac{no\text{ }of\text{ }moles\text{ }of\ gram\text{ }equivalent\text{ }of\text{ }the\text{ }solute}{total\text{ }volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }liters}$