�-� --------,-I x 10 g
l mol Ag+
+
AgBr(s) would maintain [Ag ] in the correct range.
L
To determine precipitation conditions, we must know K sp for CaF2 (s) and calculate Q
under the specified conditions. K s p = 3.9 x 10- 11 = [Ca2 +] [F-] 2
[Ca 2 +] and [F-J. The term 1 ppb means 1 part per billion or 1 g solute per billion g
solution. Assume that the density of this very dilute solution is the density of water.
1 g solute
--1 ppb = -------'} x 109 g solution
1
10-6 g solute
lL solution
x
8 ppb Ca2+
x
x
X
1 g solution
1 mL solution
1
x
-6
103 mL 1 x 10----"=-g solute
--1L
1 Lsolution
X ---- -
1 µg
= 1 µg/l L
1 x 10-6 g
l mol Ca2+
1 µg 8 µgCa 2+ 8 x 10-6 g Ca2+
= 2 x 10-1 M Ca 2+
- = -"-- = ---------- x
lL
lL
lL
40 g
1 mol F1 µg 1 µg p- _ 1 x 10-6 g px
=
= 5 x 10_8 M plL
lL
lL
19.0 g
•
+
2
2
7
2
Q = [Ca J[F-] = (2 X 10- )(5 X 10- 8) = 5 X 10- 22
1 ppb p- x
5 x 10- 22 < 3.9 x 10- 11, Q < Ksp, no CaF2 will precipitate
549