The K_sp of silver acetate Ag(CH₃COO) is 4.40x10^-3. What is the solubility (S) of this compound

Question

Quizplus · Accepted Answer

The equilibrium equation for the dissociation of silver acetate is: 
Ag(CH3COO) ↔ Ag+ + CH3COO-
The Ksp expression is: 
Ksp = [Ag+][CH3COO-]
Since the molar ratio of Ag+ to Ag(CH3COO) is 1:1, the [Ag+] is equal to the solubility (S) of silver acetate. 
Using the given Ksp value, we can solve for S: 
Ksp = [Ag+][CH3COO-]
4.4x10^-3 = (S)(S)  (since [Ag+] = [CH3COO-] = S)
S = sqrt(Ksp)
S = sqrt(4.4x10^-3) = 0.0663
Therefore, the solubility of silver acetate is 0.0663 M. The correct answer is B.

The Ksp of Silver Acetate Ag(CH3COO) Is 4

The K_sp of Silver Acetate Ag(CH₃COO) Is 4