Solubility product- Numerical part 1

Solubility product

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Solubility product- Numerical

Question 1) The solubility product of AgCl is 1.5625 x 10 ^-10 at 25⁰ C. Find its solubility in mole/litre and

gm /litre ?

Solution )

AgCl   ⇌  Ag⁺     +    Cl^–

                   S                 S

K _sp =  1.5625 x 10 ^-10

AgCl is 1: 1 type salt , therefore

K _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

S = 1.79 x 10^-3 gm / litre Ans.

Question 2) The concentration of Ag⁺ in saturated solution of Ag₂CrO₄ at 20⁰C is 1.5 x 10^-4 mole/litre.Find K _sp of Ag₂CrO₄ at 20⁰C ?

Solution )

Ag₂CrO₄    ⇌  2Ag⁺     +    CrO₄^{– –}

                          2S                 S

K _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

K _sp = [Ag⁺] ² [CrO₄^{– –} ]

K _sp = [1.5 x 10^-4] ² [0.75 x 10^-4]

K _sp = 1.6875 x 10 ^-12 Ans.

Question 3) The solubility of CaF₂ in water at 25⁰C is 1.7 x 10^-3 gm / 100 ml. Calculate solubility product of CaF₂ at 25⁰C ?

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,

K _sp = 4 S³

K _sp = 4 x (2.18 x 10^-4)³

K _sp = 4.14 x 10^-11Ans.

Question 3) The solubility  product of PbI₂ is 1.4 x 10^-8. Calculate its molar solubility in 0.1 M KI solution ?

( K _sp = 1.4 x 10^-8 )

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]

K _sp = [Pb⁺⁺] [I^–]²

= [S] [2S + 0.1]²

K _sp = [S] [4S² + 0.01 +0.4S]

K _sp =  [4S³ + 0.01S +0.4S²]

Neglecting S³ and S²

1.4 x 10^-8 = 0.01 S

S = 1.4 x 10^-6 mole/litre

Question 4)  Solubility product of AgBr is 4 x 10^-13 and conc. of   [Ag⁺] in a solution is 1 x 10^-6 mole / litre. Calculate the [Br^–] in that solution ?

Solution)

AgBr  ⇌ Ag⁺   + Br^–

K _sp = 4 x 10^-13

[Ag⁺] = 1x 10^-6

K _sp = [Ag⁺][Br^–]

4 x 10^-13 = 1x 10^-6 [Br^–]

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

[Br^–] = 1.4 x 10^-7 mole/litre Ans.