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

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.

Typology: Lecture notes

2021/2022

Uploaded on 09/12/2022

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Solubility Product Constant
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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-