Question

If 1.2 moles of CaF2 (s) are added to 2.7 L of water what will the concentration of F- be at equilibrium? The Kop - 3,9 x 10-11 CaF, (s) = Ca (aq) + 2F (aq) O(A) 5.38 x 10 M (B) 3. 12 x 10-6 M (C) 2.14 x 10- M (D) 7.91 x 10-M (E) 4.27 x 10-4 M

Answer 1

1.2 moles of ammonia are added to 2.7 L of water

thus concentration = moles/volume = 1.2/2.7 =0.44M

CaF2 $\rightleftharpoons$ Ca2+ +2F-

	CaF2(M)	Ca2+(M)	F-(M)
initially	0.44	0	0
change	-x	x	2x
finally	0.44-x	x	2x

as Ksp=3.9*10^(-11)

Ksp = [Ca2+] [F-]^2 = 3.9*10^(-11)

(x)(2x)^2 = 39*10^(-12)

4*x^(3) = 39 *10^(-12)

x^3 = 9.75*10^(-12)

from here x = (9.75*10^(-12))^(1/3)

x = 2. 136 * 10^(-4)

but we habe to find concentration of Cl- = 2x

so 2x = 2*2.136 *10^(-4) = 4.27 *10^(-4)M

thus option E is correct.

based on above calculation other options are incorrect.