Question

Write the balanced chemical reactions for the dissolution reactions AND the corresponding solubility product expressions for each of the following solids. Lastly, please calculate the solubility of each compound in pure water ignoring any acid-base properties.

a. Fe(OH)3

b. CuS

c. CaF2

Answer 1

a.

Balanced dissolution reaction is

Fe(OH)₃ (aq) Fe³⁺ (aq) + 3 OH^-(aq)

Solubility product (Ksp) = [Fe³⁺] [OH^-]³

if solubility of Fe(OH)₃ is S (mol/L).

Then, [Fe³⁺] = S

and [OH^-] = 3S

So, Ksp = S × (3S)³ = 27S⁴ .

Or, S= (Ksp)^1/4

Now, at 25 ^oc Ksp of Fe(OH)₃ = 6×10^-38.

Or, Solubility = × (6 × 10^-38)^1/4

= 1.83 × 10^-11 (mol/L).

b.

Balanced dissolution reaction is

CuS (aq) Cu²⁺ (aq) + S^-2 (aq)

Ksp = [Cu²⁺][S^-2]

Let solubility of CuS at 25⁰c = S (mol/L)

Then, [Cu²⁺] = [S^-2] = S (mol/L).

Now, Ksp = S × S = S²

Or, S = ✓(Ksp)

Ksp of CuS at 25⁰ c = 1× 10^-36 .

Or, S = ✓(1×10^-36) = 1× 10^-18 (mol/L).

C.

Dissolution reaction is

CaF₂ (aq) Ca²⁺ (aq) + 2F^-(aq)

So,

Ksp = [Ca²⁺][F^-]²

Let solubility of CaF₂ at 25⁰c = S (mol/L).

Then [ Ca²⁺] = S

[F^-] = 2S.

Hence, Ksp = S × (2S)² = 4S³ .

Or, S = (Ksp)^1/3.

Now, at 25⁰ c Ksp of CaF₂ = 4× 10^-11.

So, S = × ( 4×10^-11)^1/3

= 8.55 × 10^-5 (mol/L).