How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) Sulfate solution?

Question

I seem to get confused on what to convert to what and how to set up this equation? Step by step how to solve.

Answer 1

This is a dilution problem where you are diluting the 5.0 M solution with water to obtain a 0.30 M solution. Volume x Molarity = Volume x Molarity

V1M1 = V2M2 or ViMi = VfMf (where i = initial and f = final)

(x ml)(5.0 M) = (160 + x ml)(0.30 M) ... you use 160+x as the final volume since you are adding x ml to 160 ml

5x = 48 + 0.3x

4.7x = 48

x = 10.2 mls of 5.0 M CuSO₄ needs to be added to 160 ml H₂O to obtain 170.2 mls of 0.30 M CuSO₄

As a check, you can see if V1M1 does in fact equal V2M2

V1M1 = (10.2)(5.0) = 51

V2M2 = (170.2)(0.3) = 51