Suppose that 2.25 grams of CuSO4 are dissolved in 930 mLs of 0.30 M ammonia. What concentration (Molarity) of Cu2+ ion is expected to be free in the aqueous solution and not bound to ammonia in the complex ion Cu(NH3)42+? The Kf for Cu(NH3)42+ is 5.0 x 1013. (Hint: assume before you set up the equilibrium that all the copper ion from the copper (II) sulfate forms a complex with the available ammonia in solution.)
To determine the concentration of Cu2+ ion that is free in the aqueous solution, we can use the concept of equilibrium in complex ion formation.
First, let's write the balanced equation for the complex formation between copper (II) ion (Cu2+) and ammonia (NH3):
Cu2+ + 4NH3 ⇌ Cu(NH3)42+
According to the problem, we have 2.25 grams of CuSO4 dissolved in 930 mLs of 0.30 M ammonia. We need to determine the concentration (in Molarity) of Cu2+ ion.
To do that, we need to calculate the concentration of Cu2+ ion in the final solution using the equilibrium expression and the given Kf (formation constant).
The equilibrium expression for the formation of Cu(NH3)42+ is:
Kf = [Cu(NH3)42+] / ([Cu2+][NH3]^4)
Since we are assuming that all the copper ions form a complex with the available ammonia in solution, the concentration of Cu(NH3)42+ is equal to the concentration of Cu2+ ion.
Let's denote the concentration of Cu2+ ion as [Cu2+] and the initial concentration of NH3 as [NH3]0 (0.30 M).
Using the information given, we can set up the equilibrium expression as:
5.0 x 10^13 = [Cu2+] / ([Cu2+](0.30)^4)
Now, let's solve for [Cu2+]:
5.0 x 10^13 = [Cu2+] / (0.30^4[Cu2+])
5.0 x 10^13 = 1 / (0.30^4)
5.0 x 10^13 = 1 / 0.0081
[Cu2+] = 5.0 x 10^13 x 0.0081
[Cu2+] ≈ 4.05 x 10^11 M
Therefore, the expected concentration of Cu2+ ion that is free in the aqueous solution and not bound to ammonia in the complex ion Cu(NH3)42+ is approximately 4.05 x 10^11 M.