If 200. mL of water is added to 300. mL of an aqueous solution that is 0.725 M in potassium sulfate, what is the concentration of potassium ions in the final solution?

0.435 M

Explanation:

In case of dilution, the following formula can be used -

M₁V₁ = M₂V₂

where,

M₁ = initial concentration,

V₁ = initial volume,

M₂ = final concentration, i. e., concentration after dilution,

V₂ = final volume.

from, the question,

M₁ = 0.725 M

V₁ = 300 mL

M₂ = ?

V₂ = 300 mL + 200 mL = 500 mL

Since, the final volume of solution would be the summation of the initial and final volume.

Using the above formula, the molarity of the final solution after dilution, can be calculated as,

M₁V₁ = M₂V₂

0.725 M * 300 mL = M₂ * 500mL

M₂ = 0.435 M