Question

The solubility product, \[{K_{sp}}\] , of \[A{g_2}Cr{0_4}\] is \[3.2 \times {10^{ - 11}}\] . Its solubility in \[mol{L^{ - 1}}\] is:
A. \[4 \times {10^{{\text{ }}4}}\]
B. \[2 \times {10^{ - 4}}\]
C. \[2 \times {10^{ - 3}}\]
D. \[4 \times {10^{ - 6}}\]

Answer 1

Hint: To calculate the solubility(concentration) of an ion when the solubility product of its compound is given, we can determine the solubility of the respective ion as the solubility is equal to the concentration of the ion. \[{K_{sp}}\] or the solubility product is the equilibrium between a solid and its corresponding ions in its solution. The value \[{K_{sp}}\] varies according to the degree to which the compounds can dissolute or dissociate in a given aqueous solution.

Complete step by step answer:
The solubility product constant \[({K_{sp}})\] is the equilibrium constant for the dissolution of a solid substance into an aqueous solution of it. This further tells us that the solid compound when in equilibrium with its saturated solution, the product of concentrations of ions of both the products is equal to the solubility product constant.
Solubility products then can be expressed in terms of the solubility of the products formed. It is defined as a property of a substance in which solute is dissolved in a solvent to form a solution. The chemical equation involved in the reaction is-
 \[A{g_2}Cr{O_4} \rightleftharpoons 2A{g^ + } + CrO_4^{2 - }\]
Let us consider the solubility of \[A{g_2}Cr{O_4}\] and \[CrO_4^{2 - }\] is S for both and for \[A{g^ + }\] is 2S after equilibrium.
 \[{K_{sp}} = {[A{g^ + }]^2}[CrO_4^{2 - }]\]
 \[{K_{sp}} = {(2S)^2}S = 4{S^3}\]
 \[S = {(\dfrac{{{K_{sp}}}}{4})^{\dfrac{1}{3}}}\]
Putting the value of \[{K_{sp}}\] in the above formula, we get
 \[ = {(\dfrac{{3.2 \times {{10}^{ - 11}}}}{4})^{\dfrac{1}{3}}}\]
 \[ = 2 \times {10^{ - 4}}mol{l^{ - 1}}\]

Therefore the correct option is B.

Note: One must associate the stoichiometric coefficient of every ion or compound with the solubility product while solving for concentration and solubility. The higher the value \[{K_{sp}}\] , the higher will be the solubility. The solubility product constant can vary due to the molecular size of the constituent atoms, the net applied pressure, temperature. It is also used for understanding the conditions under which a particular precipitate will be formed, and also helps to understand the common ion effect.