Solubility product
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Solubility product-
Question 1) Solubility of Mg(OH)2 in pure water is 9.57 x 10-3 gm /litre.Calculate its solubility product in water and solubility in 0.02 M Mg(NO3)2 solution?
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,
K sp = 4 S 3
K 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
Solubility of Mg(OH)2 in presence of 0.02 M Mg(NO3) 2 is 8.64 x 10-4 gm / litre Ans
Question 2)Solubility product of AgCl is 1.5 x 10-10 at 250C . Calculate solubility of AgCl in
a) pure water
b) 0.1 M AgNO3
c) 0.01 M NaCl
Solution )
a) Solubility of AgCl in pure water-
Solubility product ‘K sp‘ = 1.5 x 10-10
K sp = 1.5 x 10 -10
AgCl is 1: 1 type salt , therefore
K sp = S2
S = √ K sp = √ 1.5x 10 -10
S = 1.22 x 10 -5 mole / litre Ans.
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
S = 1.5 x 10-9 mole/litre Ans.
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
S = 1.5 x 10-8 mole/litre Ans.
Question 3) The solubility product of BaSO4 at 250C is 1.08 x 10-10. What is the minimum conc. of SO4– – ions required to precipitate BaSO4 from a 0.01 M solution of BaCl2 ?
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.
