Solubility product ## Solubility product-

### Solution )

S of Mg(OH)2 = 9.57 x 10-3 gm /litre

mol.wt. of Mg(OH)2 = 24 + (16 + 1)2 = 24 + 34  = 58

S in mole /litre = ( S in gm /litre) / mol.wt.= 9.57 x 10 -3 / 58

S = 1.65 x 10-4 mole/litre

Mg(OH)2    ⇌  Mg++    +   2 OH

S                 2S

Mg(OH)2 is 1: 2 type salt therefore,

sp = 4 S 3

sp = 4 x (1.65 x 10-4)3

#### K sp = 1.79 x 10-11Ans.

In 0.02 M Mg(NO3)2 solution ,

Mg(OH)2    ⇌  Mg++    +   2 OH

S                 2S

Mg(NO3)2  —> Mg++    + 2NO3

0.02

Total [Mg++] = ( S + 0.02)

[OH] = 2S

K sp = [Mg++] [OH]2

= ( S + 0.02) (2S)2

= 4 S3 + 0.08 S2= 1.79 x 10-11

S is less than 0.02 , so 4S3 is neglected

0.08 S2 = 1.79 x 10-11

S2= 1.79 x 10-11/0.08

S = 2.2375 x 10-10

#### S = 1.49 x 10-5 mole / litre Ans.

S in gm / litre = 1.49 x 10 -5 x 58 = 86.42 x 10-5

S in gm / litre = 8.64 x 10-4

### Solution )

#### a) Solubility of AgCl in pure water-

Solubility product ‘K sp‘ = 1.5 x 10-10

sp =  1.5 x 10 -10

AgCl is 1: 1 type salt , therefore

sp = S2

S = √ K sp√ 1.5x 10 -10

#### b) Solubility of AgCl in 0.1 M AgNO3

AgCl   ⇌  Ag+     +    Cl

S                 S

AgNO3  —>  Ag+     +    NO3

0.1                 0.1

[Ag+ ] = 0.1 + S

[Cl] = S

K sp = [Ag+] [Cl]

= (0.1 + S) S= 0.1 S + S2

because S <<< 0.1 and solubility of AgCl decreases in presence of AgNO3 due to common ion effect(common ion is Ag+).Therefore,

0.1 S = 1.5 x 10-10

S = 1.5 x 10-10 /0.1

#### c) Solubility of AgCl in 0.01 M NaCl

AgCl   ⇌  Ag+     +    Cl

S                 S

NaCl  —>  Na+     +    Cl

0.01        0.01

[Ag+ ] =  S

[Cl] = 0.01+ S

K sp = [Ag+] [Cl]

= S (0.01 + S) = 0.01 S + S2

because S <<< 0.01 and solubility of AgCl decreases in presence of  NaCl due to common ion effect(common ion is Cl).Therefore,

0.01 S = 1.5 x 10-10

S = 1.5 x 10-10 /0.01

### Solution )

K sp of BaSO4 =  1.08 x 10-10

[SO4– –] = ?

K sp = [Ba++][SO4– –]

Precipitation will occur when ,

ionic product of [Ba++] [SO4– –] > K sp of BaSO4

[Ba++] = 0.01 M

[SO4] = K sp / [Ba++]

= 1.08 x 10-10 / 0.01

[SO4] = 1.08 x 10-8 mole/litre

For the precipitation of BaSO4 from the solution of BaCl2, the conc. of  [SO4] must be greater than  1.08 x 10-8 mole/litre.