Question

The volumes of 4N ${\text{HCl}}$ and 10N ${\text{HCl}}$ required to make 1 litre of 6N ${\text{HCl}}$ are:
A.$0.75$ litre of 10N ${\text{HCl}}$ and $0.25$ litre of 4N ${\text{HCl}}$
B.$0.25$ litre of 4N ${\text{HCl}}$ and $0.75$ litre of 10N ${\text{HCl}}$
C.$0.67$ litre of 4N ${\text{HCl}}$ and $0.33$ litre of 10N ${\text{HCl}}$
D.$0.80$ litre of 4N ${\text{HCl}}$ and $0.20$ litre of 10N ${\text{HCl}}$

Answer 1

Hint: The term ‘normality of a solution’ is used to refer to the number of gram equivalents of the solute dissolved per litre or ${\text{d}}{{\text{m}}^{\text{3}}}$ of the solution. It is represented by the letter N.
If the volume ${{\text{V}}_{\text{1}}}$ of a solution of normality ${{\text{N}}_{\text{1}}}$is mixed with volume ${{\text{V}}_2}$ of another non reacting solution of normality ${{\text{N}}_2}$ , then the normality N of the final resulting solution can be calculated as follows:
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{N}}\left( {{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{V}}_{\text{2}}}} \right)$

Complete step by step answer:
Given the normalities of two hydrochloric acid solutions are 4N ${\text{HCl}}$ and 10N ${\text{HCl}}$ .
Also given, the volume and the normality of the resulting solution to be 1 litre and 6N respectively.
We need to find out the volumes of the two parent solutions.
Let us consider the volume and normality of the 4N ${\text{HCl}}$ solution to be ${{\text{V}}_{\text{1}}}$ and ${{\text{N}}_{\text{1}}}$ respectively and the volume and normality of the 10N ${\text{HCl}}$ solution to be ${{\text{V}}_2}$ and ${{\text{N}}_2}$ respectively.
Also, let normality of the final solution be N.
Thus, we have,
${{\text{N}}_{\text{1}}} = 4{\text{N,}}{{\text{N}}_2} = 10{\text{N,N = 6N,}}{{\text{V}}_1} + {{\text{V}}_2} = 1$
Therefore, substituting all these in the expression mentioned in hint, we get:
${\text{4}} \times {{\text{V}}_{\text{1}}}{\text{ + 10}} \times {{\text{V}}_{\text{2}}} = 6 \times 1$
Again,
$
  {{\text{V}}_1} + {{\text{V}}_2} = 1 \\
   \Rightarrow {{\text{V}}_2} = 1 - {{\text{V}}_1} \\
 $
So, we can write:
$
  {\text{4}} \times {{\text{V}}_{\text{1}}}{\text{ + 10}} \times \left( {1 - {{\text{V}}_{\text{1}}}} \right) = 6 \times 1 \\
   \Rightarrow {\text{4}} \times {{\text{V}}_{\text{1}}}{\text{ + 10}} - {\text{10}} \times {{\text{V}}_{\text{1}}} = 6 \\
   \Rightarrow 6{{\text{V}}_{\text{1}}} = 4 \\
   \Rightarrow {{\text{V}}_{\text{1}}} = \dfrac{2}{3} \\
   \Rightarrow {{\text{V}}_{\text{1}}} = 0.67 \\
 $
Thus,
 $
   \Rightarrow {{\text{V}}_2} = 1 - 0.67 \\
   \Rightarrow {{\text{V}}_2} = 0.33 \\
 $
Therefore, $0.67$ litre of 4N ${\text{HCl}}$ and $0.33$ litre of 10N ${\text{HCl}}$ are required to make 1 litre of 6N ${\text{HCl}}$.

So, C is the correct option.

Note:
The volumes of the solutions can also be determined when molarity instead of normality is given.
Since normality of acid is equal to the product of molarity and the basicity of the acid and since basicity of ${\text{HCl}}$ is 1, so, the normality equation ${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$ becomes ${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$ where ${{\text{M}}_{\text{1}}}$ and ${{\text{M}}_2}$ represents the molarity of 4N ${\text{HCl}}$ and 10N ${\text{HCl}}$ respectively.
Then the equation ${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{N}}\left( {{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{V}}_{\text{2}}}} \right)$ becomes ${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}} = {\text{M}}\left( {{{\text{V}}_{\text{1}}}{\text{ + }}{{\text{V}}_{\text{2}}}} \right)$and then the volumes of the parent solutions can be found out following the same way.