The solubility product of aluminium hydroxide at 298 K is 4.2 × 10^ -14 . Calculate its solubility in mol/L at same temperature.

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 1.987 × 10^ -4

The dissociation of aluminium hydroxide is given by:

Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-

Let the solubility of

Al(OH)_3

be

s

mol/L.

[Al^{3+}] = s

[OH^-] = 3s

The solubility product

K_{sp}

is given by:

K_{sp} = [Al^{3+}][OH^-]^3

K_{sp} = (s)(3s)^3 = 27s^4

Given

K_{sp} = 4.2 \times 10^{-14}

:

27s^4 = 4.2 \times 10^{-14}

s^4 = \dfrac{4.2 \times 10^{-14}}{27} = 15.55 \times 10^{-16}

s = (15.55)^{1/4} \times 10^{-4} \approx 1.987 \times 10^{-4}

mol/L

Answer:

1.987 \times 10^{-4}

The solubility product of aluminium hydroxide at 298 K is 4.2 × 10^...