Question

What volume of ${\text{N}}/{2}\;$ and ${\text{N}}/{10\text{ HCl}}\;$ should be taken in order to make a 2 litre solution of ${\text{N}}/{5}\;$ strength?
A. 0.5 litre ${\text{N}}/{2}\;\text{ HCl}$ and 1.5 litre ${\text{N}}/{10}\;\text{ HCl}$
B. 1 litre ${\text{N}}/{2}\;\text{ HCl}$ and 1 litre ${\text{N}}/{10}\;\text{ HCl}$
C. 1.5 litre ${\text{N}}/{2}\;\text{ HCl}$ and 0.5 litre ${\text{N}}/{10}\;\text{ HCl}$
D. 0.7 litre ${\text{N}}/{2}\;\text{ HCl}$ and 1.3 litre ${\text{N}}/{10}\;\text{ HCl}$

Answer 1

Hint: The molarities of both the solutions is given, both solutions are of hydrochloric acid only. We need to add the equivalents of both the solutions of hydrochloric acid and then, equalize them with the equivalents of resultant solution of $\text{HCl}$. The direct formula can prove to be effective here, which is ${{\text{N}}_{1}}{{\text{V}}_{1}}+\text{ }{{\text{N}}_{2}}{{\text{V}}_{2}}=\text{N}\left( {{\text{V}}_{1}}+{{\text{V}}_{2}} \right)$.

Complete step by step answer:
Let us solve this question step by step using the direct formula:
Step (1)- Let the volume of the first solution be $\text{V}$.
Then, the value of the second solution be $\left( 2-\text{V} \right)$ litres because the net volume will make 2 litres of the solution.

Step (2)- The value of ${{\text{N}}_{1}}$ = $\dfrac{1}{2}$ N, ${{\text{V}}_{1}}=\text{V}$ litres, ${{\text{N}}_{2}}$ = $\dfrac{1}{10}$ N and ${{\text{V}}_{2}}$= $\left( 2-\text{V} \right)$ litre and N= $\dfrac{1}{5}$ N.
Using the formula, ${{\text{N}}_{1}}{{\text{V}}_{1}}+\text{ }{{\text{N}}_{2}}{{\text{V}}_{2}}=\text{N}\left( {{\text{V}}_{1}}+{{\text{V}}_{2}} \right)$.
The value of $\left( {{\text{V}}_{1}}+{{\text{V}}_{2}} \right)$ is 2 litres.

Step (3)- Putting the values in the formula, the volume will be $\dfrac{1}{2}\times \text{V}+\dfrac{1}{10}\times \left( \text{2}-\text{V} \right)=\dfrac{1}{5}\times 2$.

Step (4)-Solving one equation one variable by taking the LCM, we will get, $\dfrac{5\times \text{V}+\text{2}-\text{V}}{10}=\dfrac{2}{5}$.
On solving it further, we get, $\dfrac{4\text{V}+\text{2}}{10}=\dfrac{2}{5}$.
Cross multiplying, the equation will be $20\text{V}+\text{10=20}$.
The volume is $\text{V=}\dfrac{10}{20}=\dfrac{1}{2}$ or 0.5 litres.

Step (5)- The volume of ${\text{N}}/{2}\;$ solution is 0.5 litres,
The volume of ${\text{N}}/{10}\;$ solution of $\text{HCl}$ is $\left( 2-\text{V} \right)$;
V is 0.5 litres, so $\left( 2-\text{V} \right)$ will be (2-0.5) is 1.5 litres.
0.5 litres ${\text{N}}/{2}\;\text{ HCl}$ and 1.5 litre ${\text{N}}/{10}\;\text{ HCl}$ is the volume of ${\text{N}}/{2}\;$ and ${\text{N}}/{10\text{ HCl}}\;$

That is taken in order to make a 2 litre solution of ${\text{N}}/{5}\;$ strength.
So, the correct answer is “Option A”.

Note: The formula of this question is different and need not to be confused with the formula ${{\text{N}}_{1}}{{\text{V}}_{1}}\text{=}{{\text{N}}_{2}}{{\text{V}}_{2}}$; as in the question, the solution 1 adds to solution 2 to ‘make’ up a new solution of $\text{HCl}$. Here we are ‘not’ equating the equivalents of any solution with other to use ${{\text{N}}_{1}}{{\text{V}}_{1}}\text{=}{{\text{N}}_{2}}{{\text{V}}_{2}}$.