The solubility product of Al(OH) 3 is 2.7 × 10 –11 . Calculate its solubility in g L –1 and also find out pH of this solution. (Atomic mass of Al is 27 u).

Question

Philoid · Accepted Answer

Let S be the solubility.

Initially, assume 1 mole of Al(OH)₃ was present. From which S moles of it was dissolved to give S and 3S moles of Al³⁺ and 3OH^- respectively. This can be understood from the following equation.

Given, = 2.7 × 10^-11

We know,

= [Al³⁺][OH^-]³

= (S)(3S)³ = 27S⁴

S⁴ = = 1 ×

S = 1 × 10^-3 mol L^-1

Molar mass of Al(OH)₃ = 27 + 163 + 13 = 78 g/mol

Solubility of Al(OH)₃ in g L^-1 = 1 × × 78 g/L

= 78 × = 7.8 × 10^-2 g L^-1

Hence, solubility of Al(OH)₃ is 7.8 × 10^-2 g L^-1.

Now to find pH of the solution:

From above equation,

[OH^-] = 3S = 3 × 1 × = 3 ×

We know,

We also know, pH + pOH = 14

Therefore, pH of the solution is 11.47

The solubility product of Al(OH)3 is 2.7 × 10–11. Calculate its solubility in g L–1 and also find out pH of this solution. (Atomic mass of Al is 27 u).

The solubility product of Al(OH)₃ is 2.7 × 10^–11. Calculate its solubility in g L^–1 and also find out pH of this solution. (Atomic mass of Al is 27 u).