The solubility product of $A{l_2}{\left( {S{O_4}} \right)_3}$ is given by the expression:
A.$\left[ {A{l^{3 + }}} \right].\left[ {SO_4^{ - 2}} \right]$
B.${\left[ {A{l^{3 + }}} \right]^2}.\left[ {SO_4^{ - 2}} \right]$
C.${\left[ {A{l^{3 + }}} \right]^3}.{\left[ {SO_4^{ - 2}} \right]^2}$
D.${\left[ {A{l^{3 + }}} \right]^2}.{\left[ {SO_4^{ - 2}} \right]^3}$
Answer
597.6k+ views
Hint: The solubility product for a sparingly soluble salt at a certain fixed temperature is termed as the product of the molar concentration of its ions in a saturated solution and each concentration term raised to the power to the number of ions present in the chemical reaction (stoichiometric coefficients) representing the dissociation of one mole of the salt.
Complete step by step solution:
We are given the salt aluminium sulphate in the question. The formula is $A{l_2}{\left( {S{O_4}} \right)_3}$.
The chemical reaction depicting the dissociation of one mole of aluminium sulphate into its constituent ion is given as:
$A{l_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons 2A{l^{3 + }} + 3SO_4^{2 - }$
From the reaction, it is clear that one mole of aluminium sulphate dissociates in solution to produce two moles of aluminium ions and three moles of sulphate ions
Using the definition of solubility product, we can write the solubility product for the reaction as
${K_{sp}} = {\left[ {A{l^{3 + }}} \right]^2}.{\left[ {SO_4^{ - 2}} \right]^3}$
Hence, the correct answer is D.
Note:
Solubility product is a type of equilibrium constant for the dissolution of a salt in an aqueous solution. Its value generally increases with temperature as the solubility increases with increase in temperature. The solubility of different ionic compounds varies a lot.
Both ionic products and solubility products are given by the same type of formula. The difference is that ionic products can be applied to a solution of any concentration whether it is saturated or unsaturated whereas the solubility product can be applied only for a saturated solution. Therefore, we can say that solubility products are the ionic product of a saturated solution.
Complete step by step solution:
We are given the salt aluminium sulphate in the question. The formula is $A{l_2}{\left( {S{O_4}} \right)_3}$.
The chemical reaction depicting the dissociation of one mole of aluminium sulphate into its constituent ion is given as:
$A{l_2}{\left( {S{O_4}} \right)_3} \rightleftharpoons 2A{l^{3 + }} + 3SO_4^{2 - }$
From the reaction, it is clear that one mole of aluminium sulphate dissociates in solution to produce two moles of aluminium ions and three moles of sulphate ions
Using the definition of solubility product, we can write the solubility product for the reaction as
${K_{sp}} = {\left[ {A{l^{3 + }}} \right]^2}.{\left[ {SO_4^{ - 2}} \right]^3}$
Hence, the correct answer is D.
Note:
Solubility product is a type of equilibrium constant for the dissolution of a salt in an aqueous solution. Its value generally increases with temperature as the solubility increases with increase in temperature. The solubility of different ionic compounds varies a lot.
Both ionic products and solubility products are given by the same type of formula. The difference is that ionic products can be applied to a solution of any concentration whether it is saturated or unsaturated whereas the solubility product can be applied only for a saturated solution. Therefore, we can say that solubility products are the ionic product of a saturated solution.
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