What is the solubility of Ag2SO4 if the Ksp of Ag2SO4 is 1.4 × 10-5?
7.4 × 10^-3 M
1.5 × 10^-2 M
2.4 × 10^-2 M
3.7 × 10^-3 M
It’s 1.5 * 10^-2M
Bobby cmon man. Do better
Cmon Bob,do better
To determine the solubility of Ag2SO4, we can use the solubility product constant (Ksp) value. The Ksp is the equilibrium constant for the dissociation of a compound in a solution and gives us information about its solubility.
The balanced equation for the dissociation of Ag2SO4 is:
Ag2SO4 ⇌ 2 Ag+ + SO4 2-
From this equation, we can see that the molar solubility of Ag2SO4 is "x" (in M), and the concentrations of Ag+ and SO4 2- ions are both "2x" (in M) because the coefficient in front of Ag+ is 2.
The expression for the Ksp of Ag2SO4 is given by the product of the concentrations of the dissociated species raised to their respective powers:
Ksp = [Ag+]^2 * [SO4 2-]
Since we know the Ksp value of Ag2SO4 is 1.4 × 10^-5, we can substitute the concentrations into the equation and solve for "x":
1.4 × 10^-5 = (2x)^2 * x
Simplifying the equation:
1.4 × 10^-5 = 4x^3
Rearranging the equation to solve for "x":
x^3 = (1.4 × 10^-5) / 4
x^3 = 3.5 × 10^-6
Taking the cube root of both sides:
x ≈ 0.0037 M
Therefore, the solubility of Ag2SO4 is approximately 3.7 × 10^-3 M.
.........Ag2SO4 ==> 2Ag^+ + [SO4]^2-
I..........solid..............0.............0
C,,,,,,,,solid.............2x.............x
E........solid.............2x.............x
Write the Ksp expression for Ag2SO4, plug in the E line and solve for x = (Ag2SO4)