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
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.
To find the molar solubility of a compound using the solubility product constant (Ksp), you need to follow these steps:
1. Write the balanced chemical equation for the dissolution of the compound. In this case, the equation is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO4^2-(aq)
2. Write the expression for the solubility product constant (Ksp) based on the balanced equation:
Ksp = [Ag+]^2[SO4^2-]
3. Substitute the value of Ksp given in the problem: Ksp = 1.6 × 10^-5.
4. Assume that the compound dissolves completely, so the initial amount of Ag+ and SO4^2- ions is x M.
5. Since Ag2SO4 dissociates into 2Ag+ ions and 1 SO4^2- ion, the concentration of both Ag+ and SO4^2- ions will be 2x M.
6. Substitute these values into the Ksp expression:
1.6 × 10^-5 = (2x)^2 * (x) = 4x^3
7. Solve for x by rearranging the equation and taking the cube root:
x^3 = (1.6 × 10^-5) / 4
x^3 = 4 × 10^-6
x = (4 × 10^-6)^(1/3)
x = (4/4)^(1/3) × 10^-2
x = 1/∛4 × 10^-2 ≈ 0.158 M, which matches the answer in your book.
Therefore, the molar solubility of silver sulfate is approximately 0.158 M.