Understanding Solubility Product Constants (Ksp) and their Impact on Solubility, Lecture notes of Chemistry

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Solubility Product Constant

Ksp

Ksp, the solubility-product

constant.

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 and Ksp

Three important definitions:

1. solubility: quantity of a substance that dissolves to form a saturated solution

2. molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution

3. Ksp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

Common ion Effect

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

R

I

C

E

Ksp = [x] [0.10 + 2x]^2 = 7.9 x 10 –^9

Ksp = [Pb2+(aq)] [I – (aq)]^2

Example

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

R

I

C

E

Ksp = [x] [0.20 + x] = 8.3 x 10 –^17

Ksp = [Ag+(aq)] [I – (aq)]

Example

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

• When two salt solutions that share a common ion are mixed the salt with the lower ksp will precipitate first.

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?

• Compare the initial concentration to the solubility product constant

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-

• No precipitation will occur because the initial concentration is less than the Ksp.

Formation Constants

for Complex Ions

• Kform = [Cu(NH 3 ) 4 2+]

[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

Formation Constants

for Complex Ions

• The very soluble silver complex ion removes Ag+ from the solution and shifts the equilibrium to the right increasing the solubility of AgCl. AgBr + 2NH 3 Ag(NH 3 )2+^ + Br - Kc = 8.0 x 10-6^ = [Ag(NH 3 )2+][Br - ] [NH 3 ]

Kc = kform x ksp = (1.6 x 10^7 )(5.0x10-13) = 8.0 x 10-