A *simple* concentration/volume question.

What volume of 0.200 mol dm^-3 potassium sulfate solution is required to make, by dilution with water, 1.00 dm^3 of a solution with potassium ion concentration of 0.100 mol dm^-3

A 100 cm^3
B 250 cm^3
C 400 cm^3
D 500 cm^3

Some explanation, would be appreciated as well.

For this question, you have to work backwards and use the formula n=CV

If you have a K⁺ con^c of 0.100 mol dm^-3 and a volume of 1.00 dm³, that means you have 0.100 mol K⁺ ions (since n=CV).

K₂SO_{4 ==>} 2K⁺ + SO₄^2-

One mole of potassium sulphate will dissociate in water to produce two moles of K⁺ ions and so you have to work backwards. If you have 0.100 mol of K⁺ ions, you must have had 0.0500 mol of K₂SO₄ to begin with.

Now, to find the volume you have to rearrange n=CV, so V=n/C.

Therefore the volume of K₂SO₄ = 0.0500/0.200 = 0.250 dm³ = 250cm³. Therefore the answer is B.

Hope this helps you out