Asked by Sara
The solubility product of silver sulfate is 1.6* 10^-5. What is the molar solubility of this compound?
Ksp= [Ag]^2[So4^2-]
1.6*10^-5 = x^3
x= .0252
but the answer in my book is
(16/4)^1/3 *10^-2 which equals to 0.158
Ksp= [Ag]^2[So4^2-]
1.6*10^-5 = x^3
x= .0252
but the answer in my book is
(16/4)^1/3 *10^-2 which equals to 0.158
Answers
Answered by
DrBob222
Ag2SO4 ==> 2Ag^+ + SO4^=
(Ag^+)^2(SO4^=) = 1.6 x 10^-5
If you let molar solubility Ag2SO4 = x then
Ag^+ = 2x and SO4 = x so the expression should be
(2x)^2(x) = Ksp and 4x^3 = Ksp.
That will get the right answer.
(Ag^+)^2(SO4^=) = 1.6 x 10^-5
If you let molar solubility Ag2SO4 = x then
Ag^+ = 2x and SO4 = x so the expression should be
(2x)^2(x) = Ksp and 4x^3 = Ksp.
That will get the right answer.
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