what must be the concentration of Cl- to just start precipitation of CuCl from a solution which is 4.896x10^-6M in CuNO3? Ksp = 1.9x10^-7
To determine the concentration of Cl- required for the precipitation of CuCl, we need to consider the solubility product constant (Ksp) of CuCl.
The equation for the dissolution of CuCl in water is:
CuCl (s) ↔ Cu+ (aq) + Cl- (aq)
The Ksp expression for this equilibrium is:
Ksp = [Cu+][Cl-]
Given that the Ksp value is 1.9x10^-7 and the concentration of CuNO3 is 4.896x10^-6 M, we can assume that the concentration of Cu+ is also 4.896x10^-6 M since CuNO3 dissociates completely to Cu+ and NO3-.
Let's assume that the initial concentration of Cl- is x M. After precipitation occurs, the equilibrium concentration of Cu+ ions will be x M, and the equilibrium concentration of Cl- ions will also be x M.
Plugging in the values into the Ksp expression, we get:
1.9x10^-7 = (4.896x10^-6)(x)
Now we solve for x.
x = (1.9x10^-7) / (4.896x10^-6)
x ≈ 3.89x10^-2 M
Therefore, the concentration of Cl- ions needed to just start the precipitation of CuCl from the solution is approximately 3.89x10^-2 M.