Question

You are supplied with 500 ml of each 2 N HCl and 5 N HCl. What is the maximum volume of 3 M HCl that you can prepare using only these two solutions?
A. 250 ml
B. 500 ml
C. 750 ml
D. 1166.7 ml

Answer 1

Hint: There is a formula to calculate the volume of the resulting solution and it is as follows.
\[{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}=N({{V}_{1}}+{{V}_{2}})\]
Here, ${{V}_{1}}$ = volume of the first solution
${{V}_{2}}$ = volume of the second solution
${{N}_{1}}$ = normality of the initial solution
${{N}_{2}}$ = normality of the final solution
N = normality of the mixed solution

Complete Solution :
- In the question it is given that there are two HCl solutions which concentrations are 2 N and 3 N.
- The volume of 2 N HCl solution is 500 ml and the volume of the 5 N HCl solution is also 500 ml.
- We have to prepare a HCl solution whose concentration should be 3 N.
- We have to find the volume of the 3 N HCl solution which is formed by mixing 2 N and 5 N solutions.
- We can find the volume of the 3 N HCl solution by using the following formula.
\[{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}=N({{V}_{1}}+{{V}_{2}})\]

Here, ${{V}_{1}}$ = volume of the initial solution = 500 ml
${{V}_{2}}$ = volume of the final solution = x
${{N}_{1}}$ = normality of the first HCl solution = 2 N
${{N}_{2}}$ = normality of the second HCl solution = 5 N
N = normality of the mixed solution = 3N

- Substitute all the known values in the above formula to get the volume of the 3 N solution and it is as follows.
\[\begin{align}
 & {{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}=N({{V}_{1}}+{{V}_{2}}) \\
 & 2\times 500+5x=3(500+x) \\
 & 1000+5x=1500+3x \\
 & x=750ml \\
\end{align}\]
- The maximum volume of the 3 N HCl solution which can be prepared by using 500 ml of each 2N and 5 N solution is 750 ml.
So, the correct answer is “Option C”.

Note: If we are going to mix two different concentrations of the solutions then the concentration of the resulting solution obtained is different from the concentration of the solutions what we have mixed.