The Ksp of Ag2CrO4(s) is 1.12 x 10^-12. Calculate the molar solubility of AgCrO4(s)
a) in pure water
b)in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s)
Worked the same way as the previous problems. In a, there is no common ion.
In b, the common ion is CrO4^=.
Post your work if you get stuck.
To calculate the molar solubility of AgCrO4(s), we need to use the concept of Ksp (solubility product constant). The Ksp expression for Ag2CrO4(s) is as follows:
Ag2CrO4(s) ⇌ 2 Ag⁺ + CrO4²⁻
a) Molar solubility in pure water:
In pure water, we assume that x moles per liter of Ag2CrO4(s) dissociate into 2x moles per liter of Ag⁺ and CrO4²⁻ ions. Therefore, the Ksp expression becomes:
Ksp = [Ag⁺]² [CrO4²⁻]
Since 2x moles per liter of Ag⁺ ions are formed for every x moles per liter of Ag2CrO4(s) dissolved, we can substitute these values into the Ksp expression:
Ksp = (2x)² (x) = 4x³
Now we can solve for x:
4x³ = 1.12 x 10^-12
x³ = (1.12 x 10^-12) / 4
x = [(1.12 x 10^-12) / 4]^(1/3)
So, the molar solubility of AgCrO4(s) in pure water is [(1.12 x 10^-12) / 4]^(1/3) mol/L.
b) Molar solubility in a solution of 0.10 mol/L sodium chromate (Na2CrO4):
When Na2CrO4 is added to the solution, it provides a common ion (CrO4²⁻) that can affect the solubility of AgCrO4(s). The presence of this common ion will result in a decrease in the solubility of AgCrO4(s).
To calculate the molar solubility in the presence of Na2CrO4, we can follow the same steps as before, but we now consider that the initial concentration of CrO4²⁻ is 0.10 mol/L. Since Ag2CrO4(s) dissociates into 2 Ag⁺ and 1 CrO4²⁻, we will calculate the new concentration of Ag⁺ and CrO4²⁻ ions and substitute them into the Ksp expression:
Ksp = [Ag⁺]² [CrO4²⁻]
Let's assume that the solubility of AgCrO4(s) is x mol/L. In the presence of 0.10 mol/L sodium chromate, the concentration of CrO4²⁻ is still 0.10 mol/L, but now we have 2x mol/L of Ag⁺ ions. Therefore, the Ksp expression becomes:
Ksp = (2x)² (0.10) = 4x² (0.10)
Now solve for x:
4x² (0.10) = 1.12 x 10^-12
x² = (1.12 x 10^-12) / (4 (0.10))
x = [(1.12 x 10^-12) / (4 (0.10))]^(1/2)
So, the molar solubility of AgCrO4(s) in a solution of 0.10 mol/L sodium chromate is [(1.12 x 10^-12) / (4 (0.10))]^(1/2) mol/L.