Question

Determine the solubility of Ag_{​​​​​2} ​​​SO_{​​​​​4} ​​​​​​. Calculate molarities of individual ions and also the solubilities of salts in gL^{​​​​​-1}.
(K_{​​​​​​Sp} = 1.4 x 10^-5)
BOOK SOLUTION
K_{​​​​​​sp} = [Ag⁺]² [SO_{​​​​​4}^​2-]
= (2S)² (S) = 2² x 1 x S^{​​​​​​3} M^{​​​​​​3} = 4S³M³
(I) S = 1.52 x 10^-2 mol L^{​​​​​​-1}
MY PROBLEM
In my book in theory portion it is given a headline as ' RELATION BETWEEN K_{​​​​​​SP} AND MOLAR SOLUBILITY OF A SPARINGLY SOLUBLE SALT'
Type of Salt : M_{​​​​​​x}​​​​​X_​y
Relation : K_{​​​​​​sp} = x^{​​​​​​x} y^{​​​​​​y}.(S)^x+y
If we apply this rule to above problem then we will get 2² x 4⁴(S)²⁺⁴
K_{​​​​​​sp} = 1024(S)^{6
​​​​​​}See, this 'book rule' is working on every problem except this case.
NOTE: I am understanding book Solution but I want to know that when I am using the rule why it is not making correct sense.

Answer 1

Here you don't take y as 4. Because in Ag​​​​​2 ​​​SO​​​​​4 ​​​​​, cation is Ag+ and in Ag2SO4 two Ag are there. Therefore we take x=2. And anion is S04^2-. Here only one SO4 is present. So y=1
RELATION BETWEEN K​​​​​​SP AND MOLAR SOLUBILITY OF A SPARINGLY SOLUBLE SALT'
Type of Salt : M​​​​​​x​​​​​X​y
Relation : K​​​​​​sp = x​​​​​​x y​​​​​​y.(S)x+y
If we apply this rule to above problem
then we will get 22 x1^4(S )^2+1
K​​​​​​sp = 4(S)^3
.Hope you understand this..