Question

How would you prepare \[1.0L\] of a $0.10M$ solution of sulfuric acid from a $3.0M$ solution of sulfuric acid?

Answer 1

Hint: To solve this problem we can use the dilution calculation method. First, we need to be thorough with what we mean by dilution and concentration. Dilution is meant by the addition of solvent, which decreases the concentration of the solute in the solution. Concentration is defined as the removal of solvent, which can increase the concentration of the solute in the solution. In both the cases of dilution as well as concentration, the amount of solute will always be the same.

Complete step-by-step answer:
Dilution is the process of adding additional solvent to a solution in order to decrease its concentration. This process keeps the amount of solute constant, but it increases the total amount of solution, hence decreasing its final concentration. Dilution can also be achieved by mixing a solution of higher concentration with a similar solution of lesser concentration.

The volume of solvent needed to prepare the desired concentration of a new, diluted solution can be calculated mathematically. The relationship is as follows:
${V_1}{C_1} = {V_2}{C_2}$
Where,
${V_1} = $ volume of the starting solution which is required for making the new solution
${C_1} = $ concentration of the starting solution
${V_2} = $ final volume of the new solution
${C_2} = $ final concentration of the new solution
Since we have to find out ${V_1}$ by substituting the values which are given in the question, we get
${V_1} \times 3 = 1 \times 0.1$
${V_1} = \dfrac{{0.1}}{3}$
${V_1} = 0.033L$
${V_1} = 33mL$

Therefore by adding $33mL$ of the concentrated solution of sulfuric acid in a $1L$ volumetric flask and then making up to the mark by adding the desired amount of distilled water we can prepare \[1.0L\] of a $0.10M$solution of sulfuric acid from a $3.0M$ solution of sulfuric acid.

Note: Another method of solving the problem is by calculating the moles where first we have to find the value of ${V_2}{C_2}$ .
${V_2}{C_2} = 1 \times 0.1$
${V_2}{C_2} = 0.1mol$
Since we know that ${V_1}{C_1} = {V_2}{C_2}$
${V_1}{C_1} = 0.1mol$
${V_1} = \dfrac{{0.1}}{3}$
${V_1} = 0.033L$
${V_1} = 33mL$

Hence, as we discussed above, by adding $33mL$ of the concentrated solution of sulfuric acid in a $1L$ volumetric flask and then making up to the mark by adding the desired amount of distilled water we can prepare \[1.0L\] of a $0.10M$ solution of sulfuric acid from a $3.0M$ solution of sulfuric acid.