Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.

Question

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.

Philoid · Accepted Answer

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^-] = [CrO₄²^-]

∴ 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