Question

3. The Kp of calcium sulfate, Caso. is 7.1 x 10* while that of calcium hydroxide, Catoi is 4.68 x 10 (a) Calculate the solubility of each salt. (b) Which salt is more soluble in water?

Answer 1

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(a)

1. The solubility equilibrium for CaSO₄ is

$CaSO_{4} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)}$

The solubility product is,

If S is the molar solubility then we have,

....(1)

for CaSO₄ x = y = 1

Therefor from eq(1) we have

...(2)

Given:- for CaSO₄  $K_{sp} = 7.1 \times 10^{-5}$

from eq(2) we have,

$7.1 \times 10^{-5} = S^{2}$

taking square root of both sides we get

$S = \sqrt{7.1 \times 10^{-5}$

solubility in g/dm³

$S = molar solubility mol/dm^{3} \times molar mass g/mol$

$S = 0.00842 mol/dm^{3} \times 136.14 g/mol = 1.1462 g/dm^{3}$

2.

The solubility equilibrium for Ca(OH)₂ is

​​​​$Ca(OH)_{3} \rightleftharpoons Ca^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

The solubility product is

if S is the molar solubility then we have,

...(1)

for Ca(OH)₂ we have, x = 1 y = 2

Therefor from eq(1) we have,

...(2)

Given :- for Ca(OH)₂ $K_{sp} = 4.68 \times 10^{-6}$

From eq(2) we have,

$4.68 \times 10^{-6} = 4S^{3}$

$1.17 \times 10^{-6} = S^{3}$

taking cube root of both sides we get,

$S = molar solubility mol/dm^{3} \times molar mass g/mol$

$S = 0.01053 mol/dm^{3} \times 74.1 g/mol = 0.7802 g/dm^{3}$.

(b)

The solubility of CaSO₄ is 1.1462 g/dm³ while the solubility of Ca(OH)₂ is 0.7802 g/dm³.

Hence CaSO₄ is more soluble salt in water than Ca(OH)₂.