Question

How many milliliters of \[12.0{\text{ }}M{\text{ }}HCl\](aq) are needed to prepare \[215.0{\text{ }}mL\] of \[1.00{\text{ }}M{\text{ }}HCl\](aq)?

Answer 1

Hint: We need to know the concept of concentration of a solution and also the concept of dilution. Dilution is a process of adding solvent to a high concentration solution to obtain a low concentration solution. The volume of the solution increases while the concentration decreases. Since the volume is increasing and molarity is decreasing, we can call it dilution. Hence the concept of dilution and molarity can help us solve this problem.

Complete step by step answer:
The formula used to calculate the concentration and volume is given by \[{M_1}{V_1} = {\text{ }}{M_2}{V_2}\]
Where \[{M_1}\]- the concentration in molarity of the concentrated solution,
\[{V_1}\] is the volume of the concentrated solution,
\[{M_2}\] is the concentration in molarity of the diluted solution and
\[{V_2}\] is the volume of the dilute solution.
Using \[{M_1}{V_1} = {\text{ }}{M_2}{V_2}\] , we can calculate milliliters of \[12.0MHCl\] (aq) which are needed to prepare \[215.0mL\] of \[1.00MHCl\] (aq).
Given,
\[{M_1} = 12.0M\]
\[{V_2} = 215.0mL\]
\[{M_2} = 1.00M\]
We are to calculate the volume \[{V_1}\]
We know, \[{M_1}{V_1} = {\text{ }}{M_2}{V_2}\]
Now we can substitute the known values we get,
\[12.0 \times {V_1} = 1.00 \times 215.0\]
On simplification we get,
Or \[{V_1} = 17.91mL\]
Therefore, \[17.91mL\] of \[12.0MHCl\] is added to \[197.09mL\] of water to prepare \[215.0mL\] of \[1.00MHCl\].

Note:
We must remember that the dilution is the process of decreasing the concentration of a stock solution by adding more solvent to the solution. The solvent added is usually the universal solvent, known as water. The more solvent you add, the more diluted the solution will get. Basically, dilution calculations involve figuring out the final concentration or volume after a volume or concentration has been changed. In dilution equations, you are given three things and you need to find the fourth component.