Question

28. Al(OH)3 has a solubility product constant of 3.95 x 10-'1. A water sample containing aluminum has a pH of 8.9. Which species will dominate in this water sample? a) Al+(aq) b) Al(OH)3 (S) c) A12+(aq) d) Al(s)

Answer 1

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Given that, pH of the solution is 8.9.

Therefore pOH of the solution is given by,

pOH = 14 – 8.9 = 5.1

We have,

pOH = -log(OH

Therefore, Concentration of the OH^- ions in the solution is given by,

JOH-] = 10-pOH = 10-5.1

..OH-] = 7.943 x 10-6 M

The dissociation of Al(OH)₃ in the solution is given by,

Al(OH)ts) - Al3+ + 3OH-

Therefore, solubility product is given by,

Ksp = [1134].[OH-]} = 3.95 x 10-11

We have,

OH-] = 7.943 x 10-6 M

Therefore, Concentration of Al³⁺ ion in the solution is given by:

Ksp 3.95 x 10-11 LATE TOH-13 – 7.943 x 10-6 [48+] =

: A13+1 = 4.973 x 10-6 M

+/V] <l-H0]

Since , concentration of OH^- is greater than concentration of Al³⁺, Al³⁺ will dominate in this sample.

So the correct option is (a) Al³⁺(aq)

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