Question

What pH is required to remove all but 0.3 mg/L of the Aluminum ion?

Prairie View, TX groundwater contains 2.2 mg/L of Aluminum (Al3). What pH is required to removed all but 0.3 mg/L of this ion at 25°C. (use pKs 7. 32.9)

Answer 1

Given the pK_sp of the reaction as 32.9

pK_sp = -log₁₀ (K_sp)

Thus K_sp = 10^-32.9

Al(OH)_{3 (s)} $\rightleftharpoons$ Al⁺³ + 3OH^- K_sp = 10^-32.9

[We assigned K_sp to forward reaction since Al(OH)_{3 (s)} formation is more favourable than dissociation in to Al⁺³, OH^- ]

K_sp = ([Al⁺³]* [OH^-]³)/Al(OH)_{3 (s)}

Since [Al(OH)_{3 (s)}] is a precipitate and its concentration is taken as 1

Thus  K_sp = [Al⁺³]* [OH^-]³

10^-32.9 = [Al⁺³]* [OH^-]³ -----------(1)

Given the ground water contains 2.2 mg/l of Al⁺³

and we want to remove 1.9 mg/l (2.2-0.3) of Al⁺³

Al⁺³ = 1.9 mg/l = (1.9 * 10^-3 / 26.981) mol/l = 7.042*10^-5 mol/l [since molecular weight of Al⁺³ = 26.981 g/mol]

Substitute Al⁺³ =  7.042*10^-5 mol/l in equation (1) (since this amount of aluminium ion has to react with OH^- to produce Al(OH)₃ precipitate)

10^-32.9 =  7.042*10^-5 mol/l * [OH^-]³

[OH^-]³ = 10^-32.9/ 7.042*10^-5

[OH^-] = (10^-32.9/ 7.042*10^-5)^(1/3)

[OH^-] = 2.6147 * 10^-10 mol/l ------------(2)

As we know for water Reaction K_w = 10^-14 = [H⁺] * [OH^-] ---------------(3)

Substituting (2) in (3)

we get [H⁺] = 3.8244 * 10^-5 mol/l

pH = -log₁₀[H⁺] = -log₁₀[3.8244 * 10^-5] = 4.417

Thus the required pH of ground water to remove 1.9 mg/l of Al⁺³ is 4.417