During the lab, we assumed that the complexation reaction between Cu^2+ and NH_3
ID: 1018105 • Letter: D
Question
During the lab, we assumed that the complexation reaction between Cu^2+ and NH_3 went to completion (K_f = 4.8 times 10^12). Let's check that assumption. Assuming that you used 5.00 mL of 0.010 M Cu(NO_3)_2 and 1.00 mL of 3.0 M NH_3 and that the reaction goes to completion, complete the following reaction table. Now consider the reverse reaction, Cu(NH_3)_4^2+ Cu^2+ + 4NH_3. What is the equilibrium constant for this reaction? Beginning with the "final" concentrations from the table above and taking equilibrium into account this time, please complete the following reaction table. What percentage of the original Cu^2+ is still present as free Cu^2+ after addition of the NH_3?Explanation / Answer
From the above reaction the expression for K is given as
K = [NH3]^4 [Cu2+] / [Cu(NH3)4^2+]
2.1*10^(-13) = x^2/(0.0083-x)
x = 4.18*10^(-8)
The free Cu2+ is 4.18*10^(-8) M
% of Cu2+ = (4.18*10^(-8)/0.0083)*100%
= 5.0*10^(-4)%
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