Question

A 115 mL stock solution contains 1.2M CuF2. When 11mL of the stock solution is diluted to 21mL, what will the concentration of F

Answer 1

V = 115 ml of soltion

M = 1.2M of CuF2

V2 = 11 ml diluted to 21 ml

V3 = 21 ml

First, find the amount of F- ions in the 11 ml solution

moles of F- = [F-] * Volume

Note that [F-] = 2x1.2 = 2.4 M (as 1 mol of CuF2 will form 2 mol of F-)

moles of F- = (2.4)(11ml) = 26.4 mmol of F-

By definition

[F-] = moles of F- / volume = 26.4*10^-3 moles / 0.021 = 1.25 M

OR we could have used the next formula:

Now, we have a new volume, it was diluted, but moles stay the same, there is no reaction

M1V1 = M2V2

M2 = M1*V1/V2 = 2.4*11 ml / 21 ml = 1.25 M

[F-] = 1.25 M

Both give you the same value