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Question: The solubility of sparingly soluble salt \({{A}_{X}}{{B}_{Y}}\)in water at \(25{}^\circ C\) \(1.4\ti...

The solubility of sparingly soluble salt AXBY{{A}_{X}}{{B}_{Y}}in water at 25∘C25{}^\circ C 1.4×104M1.4\times {{10}^{4}}M.The solubility product is 1.1×10111.1\times {{10}^{11}}. The possibilities are:
A. x=1,y=2x=1,y=2
B. x=2,y=1x=2,y=1
C. x=1,y=3x=1,y=3
D. x=3,y=1x=3,y=1

Explanation

Solution

Solubility is the amount of the substance soluble generally known as solute to make a saturated solution at a particular temperature and in a fixed volume of solvent. Solubility varies with the variation in nature of the substance, solvent, temperature and due to the presence of common ions.

Complete step by step solutions:
The solubility product constant is the equilibrium constant defined for the dissolution of a solid substance into an aqueous solution. The solubility product is a kind of equilibrium constant which is denoted by the term Ksp{{K}_{sp}} and its value depends on temperature. It usually increases with an increase in temperature due to increased solubility. Solubility depends on a number of parameters amongst which lattice enthalpy of salt and solvation enthalpy of ions in the solution are of most importance.
Here according to question the reaction can be considered as:
AXBY⇄XAV++XBV−{{A}_{X}}{{B}_{Y}}\rightleftarrows X{{A}^{V+}}+X{{B}^{V-}}
NowKsp{{K}_{sp}} is given by: (XS)X+(YS)Y=SX+YXXYY{{(XS)}^{X}}+{{(YS)}^{Y}}={{S}^{X+Y}}{{X}^{X}}{{Y}^{Y}}
The value of Ksp{{K}_{sp}}is given 1.1×10111.1\times {{10}^{11}}and solubility s = 1.4×104M1.4\times {{10}^{4}}M
Now if we consider x=1,y=2x=1,y=2
Then Ksp=(1.4×10−4)1+21.22≅1.1×10−11{{K}_{sp}}={{(1.4\times {{10}^{-4}})}^{1+2}}{{1.2}^{2}}\cong 1.1\times {{10}^{-11}}
Now if we consider x=2,y=1x=2,y=1
Then Ksp=(1.4×10−4)1+21.22≠1.1×10−11{{K}_{sp}}={{(1.4\times {{10}^{-4}})}^{1+2}}{{1.2}^{2}}\ne 1.1\times {{10}^{-11}}
If x=1,y=3x=1,y=3
Then Ksp=(1.4×10−4)1+21.22≠1.1×10−11{{K}_{sp}}={{(1.4\times {{10}^{-4}})}^{1+2}}{{1.2}^{2}}\ne 1.1\times {{10}^{-11}}
If x=3,y=1x=3,y=1
Then Ksp=(1.4×10−4)1+21.22≠1.1×10−11{{K}_{sp}}={{(1.4\times {{10}^{-4}})}^{1+2}}{{1.2}^{2}}\ne 1.1\times {{10}^{-11}}

Hence we can say that option A is the correct answer.

Note: Solubility product constant are dependent on the common-ion effect in which the presence of a common ion lowers the value of Ksp{{K}_{sp}}, diverse-ion effect which tells that if the ions of the solutes are uncommon then the value of Ksp{{K}_{sp}} will be high and also the presence of ion pairs.