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The concept of solubility product constants (Ksp), their role in the solubility of ionic substances, and how common ions can affect solubility. It includes examples of calculating molar solubility and the common ion effect for various salts.
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An equilibrium can exist between a partially soluble substance and its solution:
The Solubility Expression
AaBb(s) aAb+^ (aq) + bBa-^ (aq)
Ksp = [Ab+]a^ [Ba-]b
Example: PbI 2 (s) Pb2+^ + 2 I-
Ksp = [Pb2+] [I-]^2
The greater the ksp the more soluble the solid is in H 2 O.
solubility: quantity of a substance that dissolves to form a saturated solution
molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution
Ksp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution
PbI 2 (s) Pb^2 +(aq) + 2I–(aq)
Common ion : “The ion in a mixture of ionic substances that is common to the formulas of at least two.”
Common ion effect : “The solubility of one salt is reduced by the presence of another having a common ion”
PbI 2 (s) Pb2+(aq) I – (aq) 1 2 0 0. x 2x x 0.10 + 2x
Ksp = [x] [0.10 + 2x]^2 = 7.9 x 10 –^9
Ksp = [Pb2+(aq)] [I – (aq)]^2
What is the Molar solubility of PbI 2 if the concentration of NaI is 0.10? Ksp = 7.9 x 10 –^9 So [I-] = 0.10 M
x is small, thus we can ignore 2x in 0.10 + 2x Ksp = [x] [0.10]^2 = 7.9 x 10 –^9 , x = 7.9 x 10 –^7 M
AgI(s) Ag+(aq) I – (aq) 1 1 0 0. x x x 0.20 + x
Ksp = [x] [0.20 + x] = 8.3 x 10 –^17
Ksp = [Ag+(aq)] [I – (aq)]
Molar solubility of AgI? Ksp = 8.3 x 10–^17 Concentration of NaI is 0.20, thus [I–] = 0.
x is small, thus we can ignore it in 0.20 + x Ksp = [x] [0.20] = 8.3 x 10 –^17 , x = 4.2 x 10 –^16
Common Ion Effect
Example: AgCl ksp = [Ag+][Cl-] = 1. x10- [Ag+][Cl-] = 1.6 x 10- X^2 = 1.6 x 10- X = [Ag+] = [Cl-] = 1.3 x 10-5^ M
Will a Precipitation Occur?
If 1.00 mg of Na 2 CrO 4 is added to 225 ml of 0.00015 M AgNO 3 , will a precipitate form?
Ag 2 CrO 4 (s) 2Ag+^ + CrO 4 2-
Determine the initial concentration of ions. Ag+^ = 1.5 x 10-
CrO 4 2-^ = 1.00 x 10-3^ g / MM = 6.17 x 10-6^ mol CrO 4 2-/ .225 L = 2.74 x 10-5^ M
Will a Precipitation Occur?
Initial concentration of ions: (Ag+)^2 (CrO 4 2-) (1.5 x 10-4)^2 (2.74 x 10-5^ M)= 6.2 x 10-
Ag 2 CrO 4 Ksp = 1.1 x 10-
[Cu2+][NH 3 ]^4
The solution of a slightly soluble salt increase when one of its ions can be changed to a soluble complex ion.
AgBr (s) Ag+^ + Br -^ Ksp = 5.0 x 10-
Add NH 3
Ag+^ + 2NH 3 Ag(NH 3 ) 2 +^ kform = 1.6 x 10^7
Kc = kform x ksp = (1.6 x 10^7 )(5.0x10-13) = 8.0 x 10-