Question

Will a precipitate of magnesium fluoride form when 200. ml of 1.9 × 10–3 m mgcl2 are added to 300. ml of 1.4 × 10–2 m naf? [ksp (mgf2) = 6.9 × 10–9]

Answer 1

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.