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Question: How many milliliters of \[12.0{\text{ }}M{\text{ }}HCl\](aq) are needed to prepare \[215.0{\text{ }}...

How many milliliters of 12.0 M HCl12.0{\text{ }}M{\text{ }}HCl(aq) are needed to prepare 215.0 mL215.0{\text{ }}mL of 1.00 M HCl1.00{\text{ }}M{\text{ }}HCl(aq)?

Explanation

Solution

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 M1V1= M2V2{M_1}{V_1} = {\text{ }}{M_2}{V_2}
Where M1{M_1}- the concentration in molarity of the concentrated solution,
V1{V_1} is the volume of the concentrated solution,
M2{M_2} is the concentration in molarity of the diluted solution and
V2{V_2} is the volume of the dilute solution.
Using M1V1= M2V2{M_1}{V_1} = {\text{ }}{M_2}{V_2} , we can calculate milliliters of 12.0MHCl12.0MHCl (aq) which are needed to prepare 215.0mL215.0mL of 1.00MHCl1.00MHCl (aq).
Given,
M1=12.0M{M_1} = 12.0M
V2=215.0mL{V_2} = 215.0mL
M2=1.00M{M_2} = 1.00M
We are to calculate the volume V1{V_1}
We know, M1V1= M2V2{M_1}{V_1} = {\text{ }}{M_2}{V_2}
Now we can substitute the known values we get,
12.0×V1=1.00×215.012.0 \times {V_1} = 1.00 \times 215.0
On simplification we get,
Or V1=17.91mL{V_1} = 17.91mL
Therefore, 17.91mL17.91mL of 12.0MHCl12.0MHCl is added to 197.09mL197.09mL of water to prepare 215.0mL215.0mL of 1.00MHCl1.00MHCl.

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.