To determine the solubility, we apply the formula

K_sp = [A⁺][B^-]

Where K_sp is the solubility product which is equal to the product of ionic concentrations in a saturated solution.

a) Silver chromate

ionization of silver chromate:

Ag₂CrO₄ ↔ 2Ag⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ag⁺]

⇒[B^-] = [CrO₄^2-]

∴ K_sp = [Ag⁺]² + [CrO₄^2-]

As K_sp of Ag₂CrO₄ =1.1 × 10^-12 (given)

Let ‘s’ be the solubility of Ag₂CrO₄

[Ag⁺] = 2s

[CrO₄^2-] =s

∴ 1.1 × 10^–12 = (2s)² s

⇒1.1 × 10^-12 = 4(s)³

^⇒0.275 × 10^–12 = s³

⇒s =

⇒s = 0.65 × 10^-4

Thus, Molarity of Ag⁺ = 2s= 2× 0.65× 10^-4 = 1.30× 10^-4 M

Molarity of CrO₄²- = s = 0.65 × 10^-4 M

b) Barium chromate

ionization of barium chromate:

BaCrO₄ ↔ Ba²⁺ + CrO₄^2-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Ba²⁺]

[B^-] = [CrO₄^2-]

∴ K_sp = [Ba²⁺] + [CrO₄^2-]

As K_sp of BaCrO₄ =1.2 × 10^-10 (given)

Let ‘s’ be the solubility of BaCrO₄

[Ba²⁺] =s

[CrO₄²-] =s

∴ 1.2 × 10^–10 = (s× s)

^⇒1.2 × 10^-10 = (s)²

⇒s =

⇒s = 1.09 × 10^-5

Thus, Molarity of Ba²⁺ and CrO₄^2- = s= 1.09× 10^-5M

c) Ferric hydroxide

ionization of ferric hydroxide

Fe(OH)₃ ↔ Fe³⁺ + 3OH^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Fe³⁺]

⇒[B^-] = [OH^-]

∴ K_sp = [Fe³⁺] + [OH^-]

As K_sp of Fe(OH)₃ =1.0 × 10^-38 (given)

Let ‘s’ be the solubility of Fe(OH)₃

[Fe³⁺] =s

[3OH^-] = 3s

∴ 1.0× 10^–38 = (s)(3s)³

^⇒1.0 × 10^-38 = (27s)⁴

^⇒0.37 × 10^–38 = s³

⇒s =

⇒s = 1.39 × 10^-10

Thus, Molarity of Fe³⁺ = s= 1.39× 10^-10 M

Molarity of OH^- = 3s = 4.17 × 10^-10M

d) Lead chloride

ionization of lead chloride:

PbCl₂ ↔ Pb²⁺ + 2Cl^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Pb²⁺]

⇒[B^-] = [Cl^-]

∴ K_sp = [Pb²⁺] + [Cl^-]²

As K_sp of PbCl₂ =1.6 × 10^-5 (given)

Let ‘s’ be the solubility of PbCl₂

[Pb²⁺] =s

[Cl^-] =2 s

∴ 1.6 × 10^–5 = (s)(2s)²

^⇒1.6 × 10^-5 = (4s)³

^⇒0.4 × 10^–5 = s³

⇒s =

⇒s = 1.58 × 10^-2

Thus, Molarity of Pb²⁺ = s= 1.58× 10^-2 M

Molarity of Cl^- = 2s = 3.16 × 10^-2 M

e) Mercurous Iodide

ionization of mercurous iodide:

Hg₂I₂ ↔ Hg₂²⁺ + 2I^-

As we know that,

Ks_p = [A⁺] [B^-]

Where A and B are the ions dissolved

In the above reaction,

⇒[A⁺] = [Hg₂²⁺]

⇒[B^-] = [I^-]²

∴ K_sp = [Hg₂²⁺] + [I^-]²

As K_sp of Hg₂I₂ = 4.5 × 10^–29

Let ‘s’ be the solubility of Hg₂I₂

[Hg₂²⁺] =s

[I^-] 2s

∴ 4.5 × 10^–29 = (s)(2s)²

⇒4.5 × 10^–29 = (4s)³

⇒4.5 × 10^–29 = s³

⇒s =

⇒s = 2.24× 10^-10

Thus, Molarity of Hg₂²⁺ = s= 2.24× 10^-10 M

Molarity of I- = 2s = 4.48 × 10^-10 M

Note: K_sp increases with increase in temperature.

⇒In a saturated solution, K_sp = [A⁺][B^-]

⇒In an unsaturated solution of AB, K_sp> [A⁺][B^-] means more solute can be dissolved.

⇒In a super saturated solution of AB, K_sp< [A⁺][B^-] means precipitation will start to occur