A solution is prepared by mixing 150.0 ml of 1.00*10^-2 m mg (no3) 2 and 250.0 ml of 1.00*10^-1m naf calculate the concentrations of mg2 + and f at equilibrium with solid mgf2 ksp = 6.4*10^-9

First, we have to get moles of Mg2 + = molarity * volume

= 1 x 10^-2 * 0.15 L

= 0.0015 mol

initial [Mg2+] = 0.0015 / 0.4 L (total volume)

= 0.00375 M

then, moles of F - = molarity * volume

= 1 x 10^-1 * 0.25

= 0.025 Mol

initial [F-] = 0.025 / 0.4 = 0.0625 M

by using ICE table:

MgF2 (s) → Mg2 + 2F-

initial 0.00375 0.0625

change + X + 2 X

Equ 0.00375+X 0.0625+2X

when Ksp = [Mg2+][F-]^2

by substitution:

6.4 x 10^-9 = (0.00375+X) (0.0625 + 2X) ^2 by solving for X

∴X = 0.00024 M

∴[Mg2+] = 0.00375 + X

= 0.00375 + 0.00024

= 0.004 M

and [F-] = 0.0625 + 2X

= 0.0625 + 2 * 0.00024

= 0.06 M