Question

The solubility product of aluminum sulphate is given by the expression:
(A) $4{s^3}$
(B) $6912{s^7}$
(C) ${s^2}$
(D) $108{s^5}$

Answer 1

Hint: The solubility product is the product of the concentration of the ions produced upon dissolution of the salt. The molecular formula of aluminum sulphate is $A{l_2}{(S{O_4})_3}$. We can find the solubility product from the equilibrium constant of the dissolution reaction of the salt.

Complete answer:
We know that solubility of a salt can be broadly divided into three categories; soluble, sparingly soluble and insoluble.
- Here, we will talk about sparingly soluble salts as aluminum sulphate is a sparingly soluble salt. Now, we know that the soluble salt completely dissociates into ions when it gets dissolved. We need to take the equilibrium into consideration in the dissolution of sparingly soluble salts as they are not completely soluble.
- So, in case of aluminum sulphate, we can write the reaction of its dissolution as
     \[A{l_2}{(S{O_4})_3} \to 2A{l^{3 + }}_{(aq)} + 3S{O_4}{^{2 - }_{(aq)}}\]
Now, we can give the equilibrium constant of the above reaction as
     \[K = \dfrac{{{{[A{l^{3 + }}]}^2}{{[S{O_4}^{2 - }]}^3}}}{{[A{l_2}{{(S{O_4})}_3}]}}\]
Now, $A{l_2}{(S{O_4})_3}$ is a pure solid and we will take its concentration as constant. So, we can write the solubility product as
     \[{K_{sp}} = K \times A{l_2}{(S{O_4})_3} = {[A{l^{3 + }}]^2}{[S{O_4}^{2 - }]^3}\]
- Here, ${K_{sp}}$ is the solubility product constant.
- Now, we can say that as two aluminum ions are formed upon dissolution, its solubility can be taken as (2S) and the solubility of sulphate ions can be taken as (3S). Now, we can write the solubility product equation as
     \[{K_{sp}} = {(2S)^2}{(3S)^3}\]
Thus, we can write that
     \[{K_{sp}} = 4{S^2} \times 27{S^3} = 108{S^5}\]
Thus, we can say that its solubility product can be given as $108{S^5}$.

Therefore, the correct answer is (D).

Note:
Remember that as there are two aluminum ions are there, we need to take their solubility as 2S. We cannot take their solubility as S. Same way we took solubility of sulphate ions as 3S as three ions are produced upon dissolution.