First, we have to get moles of MgCl2 = molarity * volume
= 1.9 x 10^-3 M* 0.2L
= 3.8 x 10^-4 moles
and moles of naf = molarity * volume
= 1.4 x 10^-2M * 0.3 L
= 4.2 x 10^-3 Moles
when the total volume = 0.2 L* 0.3L = 0.5 L
[Mg2+] = moles Mg2+ / total volume
= 3.8 x 10^-4 mol / 0.5 L
= 0.00076 M
[NaF] = moles NaF / total volume
= 4.2 x 10^-3 mol / 0.5 L
= 0.0084 M
According to the reaction equation:
MgF2 ↔ Mg2+ + 2F-
when Qsp expression = [Mg2+] [F-]^2
∴Qsp = (0.00076) * (0.0084)^2
= 5.4 x 10^-8
by comparing the value of Qsp and the Ksp value, we will find
Qsp > Ksp
∴ the precipitate of magnesium fluoride will form.
