Question

The solubility product, K_sp, of Al(OH)₃(s) is 1.0 x 10^-33. What is its solubility (in g/L) in an aqueous solution of NaOH with a pH of 12.54 ? The temperature is 25^oC.

Answer 1

Solution:

Given:

pH of NaOH solution = 12.54

pH = 12.54

Using the relation,

pH + pOH = 14

or

pOH = 14 - pH

pOH = 14 - (12.54)

pOH = 1.46

-log₁₀[OH^-] = 1.46

or

log₁₀[OH^-] = -1.46

Taking Antilog on both the sides,

[OH^-] = 10^-(1.46)

[OH^-] = 0.0346737 M

This is the concentration of OH^- in mol/litre from the dissolution of NaOH.

Let the molar solubility of Al(OH)₃ in mol/litre be S. Therefore, molar solubility of Al³⁺ and OH^- will be S and 3S mol/litre respectively.

Al(OH)₃ dissociates in the following manner:

Al(OH)₃ (s)   Al³⁺ (aq.) + OH^- (aq.)

Expression for K_sp :

K_sp = [Al³⁺] * [OH^-]³

where the OH^- concentration will include OH^- ions from both Al(OH)₃ and NaOH.

Total OH^- concentration = OH^- from aqueous NaOH + OH^- from dissolution of Al(OH)₃

Total OH^- concentration = (0.0346737) M + (3S) M

K_sp = (S) * (0.0346737 + 3S)³

(1.0 * 10^-33 ) = (S) * (0.0346737 + 3S)³ .....(1)

Since, K_sp is of the order of 10^-33 , therefore, the concentration of OH^- will be very low from dissolution of Al(OH)₃ .

Therefore, we can approximate, 0.0346737 + 3S   0.0346737

The equation (1) can be written as:

(1.0 * 10^-33 ) = (S) * (0.0346737)³

or

This is the solubility of Al(OH)₃ in mol/litre.

1 mol of Al(OH)₃ weighs 78 g.

Therefore,

Solubility of Al(OH)₃ in g/L = (2.399 * 78) * 10^-29

= (1.871 * 10^-27) g/ L  

Hence, the solubility of Al(OH)₃ in g/L will be equal to ( 1.871 * 10^-27 ) g/ L (Ans.)