Question

How to prepare 0.2% of HCl (36%) in a water tank whose volume is 5000 liter?

Answer 1

Attempt this question by the help of dilution concept as we have to prepare a diluted solution according to the question. Dilution is the process of adding solvent to a solution to decrease its concentration. In this process the amount of solute remains constant whereas the amount of the whole solution increases; hence decreasing the concentration of solution.

Formula used: We will require the following formula:-
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$

Complete step by step answer:
As we are asked to prepare a solution of 0.2%concentration out of 36% HCl, which means we have to dilute the solution to obtain the desired solution concentration.

-Dilution can be described as the process of adding additional solvent to a solution to decrease its concentration. During dilution, the amount of solute remains constant but increases the total amount of solution, thereby decreasing its final concentration.

-To prepare a solution of 0.2% of HCl:-
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
where,
${{M}_{1}}$= concentration of original solution
${{V}_{1}}$= volume of original solution
${{M}_{2}}$ = concentration of diluted solution
${{V}_{2}}$ = volume of diluted (final) solution
-Values are given as follows:-
${{M}_{1}}$= concentration of original solution = 36%
${{V}_{1}}$= volume of original solution =?
${{M}_{2}}$ = concentration of diluted solution = 0.2%
${{V}_{2}}$ = volume of diluted (final) solution = 5000L
On substituting these values, we get:-
$\begin{aligned} & \Rightarrow {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}} \\\ & \Rightarrow 0.36\times {{V}_{1}}=0.002\times 5000 \\\ & \Rightarrow {{V}_{1}}=\dfrac{10}{0.36} \\\ & \Rightarrow {{V}_{1}}=27.78L \\\ \end{aligned}$
Hence we will require 27.78L of 36% HCl to prepare a 0.2%concentrated HCl in a 5000L tank.

Note: -Dilution can also be obtained by mixing a solution of higher concentration with an identical solution of lesser concentration.
-While calculating dilution factors, it is important to make sure that units of volume and concentration are the same for both sides of the equation.