Solubility product

source : toppr

## Solubility product Numerical-

### Solution )

S of Mg(OH)2 =  0.95 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.=  0.95 / 58

S =  0.01637 = 0.0164  mole/litre

Mg(OH)2    ⇌  Mg++    +   2 OH

Ksp = [Mg++][OH]2

1.8 x 10-11 = 0.0164 [OH]2

[OH]2 = 1.8 x 10-11 /  0.0164 = 109.75 x 10-11 =10.975 x 10-10

[OH] = √ 10.975 x 10-10 = 3.31 x 10-5 mole/litre

pOH = – log [OH]= -log [3.31 x 10-5]

= -log 3.31 + 5 log 10= -0.5198 + 5

pOH = 4.48

pH = 14 – pOH= 14 – 4.48

### Solution )

CaCl2 + Na2SO4 ——–> CaSO4 + 2NaCl

Milli moles     0.02V    0.0004V                    0                0

Assume, V ml of both are mixed.

Total volume = V + V = 2V

[Ca++] = 0.02 V/2V = 0.01

[SO4] = 0.0004 V/2V = 0.0002 = 2 x 10-4

ionic product of [Ca++] [SO4] =0.01 x 2 x 10-4= 2 x 10-6

K sp of CaSO4 = 2.4 x 10-5

If  ionic product of [Ca++] [SO4] is greater than K sp of CaSO4 , then precipitation takes place But,

ionic product of [Ca++] [SO4]     <  K sp of CaSO4

2 x 10-6    <    2.4 x 10-5

### Solution )

K sp of CaSO4 = 2.4 x 10-5

CaSO4  ⇌ Ca++  +  SO4

[SO4] = y mole/litre is sufficient to precipitate CaSO4 from a solution having ,

[Ca++] = 0.005 mole /litre

Ksp = [Ca++] [SO4]

2.4 x 10-5 = 0.005 y

y = 2.4 x 10-5  / 0.005= 0.48 x 10-2

### Solution )

Suppose, solubility of salt = S mole / litre

PbBr2    ⇌  Pb++    +      2 Br

0.80 S            2 x 0.80 S

Ksp = [Pb++] [Br]2

Solubility product of PbBr2 = 8 x 10-5

8 x 10-5 = 0.80 S ( 2 x 0.80 S)2

S3 = 3.906 x 10-5 = 39.06 x 10-6

S =  339.06 x 10-6

#### S = 3.39 x 10-2 mole /litre

mol.wt of PbBr2 = 207 + 2 x 80 = 367

S in gm /litre = (S in mole /litre) x mol.wt.= 3.39 x 10-2 x 367
< h4>S = 12.44 gm / litre Ans.