Solubility product

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## Solubility product- Numerical

### Solution )

AgCl   ⇌  Ag+     +    Cl

S                 S

sp =  1.5625 x 10 -10

AgCl is 1: 1 type salt , therefore

sp = S2

S = √ K sp√ 1.5625 x 10 -10

S = 1.25 x 10 -5 mole / litre

Solubility in gm / litre = (Solubility in mole / litre) x molecular weight

molecular weight of AgCl = 108 + 35.5 = 143.5

S in gm / litre = 1.25 x 10-5 x 143.5

### Solution )

Ag2CrO4    ⇌  2Ag+     +    CrO4– –

2S                 S

sp =  ?

Ag2CrO4 is 2 : 1 type salt .

[Ag+] = 1.5 x 10-4  mole/litre

[CrO4– – ] = 1.5 x 10-4 / 2 = 0.75 x 10 -4 mole/litre

sp = [Ag+] 2 [CrO4– – ]

sp = [1.5 x 10-4] 2 [0.75 x 10-4]

### Solution )

solubility of CaF2 =  1.7 x 10-3 gm / 100 ml = 1.7 x 10-3  x 1000/100 gm / litre

S = 1.7 x 10-2 gm / litre

Molecular weight of CaF2 = 40 + 19 x 2= 78

(solubility in gm /litre) = (solubility in mole / litre) x mol. wt.

1.7 x 10 -2 =( S in mole / litre) x 78

S in mole / litre = 1.7 x 10-2 / 78=0.0218 x 10-2 = 2.18 x 10-4

CaF2   ⇌  Ca++    +   2 F

S                 2S

CaF2 is 1: 2 type salt therefore,

sp = 4 S3

sp = 4 x (2.18 x 10-4)3

### Solution )

PbI2  ⇌ Pb++  + 2 I

S              2S

[Pb++] = S mole/litre

[I] = 2S

KI                            –>       K+         +     I

strong electrolyte

conc. of I = 0.1 M

[I] = [2S + 0.1]

sp = [Pb++] [I]2

= [S] [2S + 0.1]2

sp = [S] [4S2 + 0.01 +0.4S]

sp =  [4S3 + 0.01S +0.4S2]

Neglecting S3 and S2

1.4 x 10-8 = 0.01 S

### Solution)

AgBr  ⇌ Ag+   + Br

sp = 4 x 10-13

[Ag+] = 1x 10-6

sp = [Ag+][Br]

4 x 10-13 = 1x 10-6 [Br]

[Br] = 4 x 10-13 / 10-6= 1.4 x 10-7