How many mL of water must be added to 300 mL of 0.75 M HCl to dilute the solution to 0.25 M?

To determine how much water must be added to dilute a solution, you can use the concept of dilution, which is based on the principle that the number of moles of solute remains constant before and after dilution. The formula that describes this relationship is:

C1V1 = C2V2,

where:

• C1 is the initial concentration (0.75 M),

• V1 is the initial volume (300 mL),

• C2 is the final concentration (0.25 M), and

• V2 is the final volume after dilution.

First, you calculate the total number of moles of HCl in the original solution:

Moles of HCl = C1 × V1 = 0.75 M × 0.300 L = 0.225 moles.

Next, set this equal to the moles in the diluted solution:

0.225 moles = C2 × V2.

By substituting the final concentration into the equation:

0.225 moles = 0.25 M × V2,

V2 = 0.225 moles / 0.25 M = 0.9 L = 900 mL.

This means the final volume of the solution after dilution