Question

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?

Answer 1

Answer:

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