Calculate the concentration of the free copper (II) that is in equilibrium with

Question

Calculate the concentration of the free copper (II) that is in equilibrium with the complexed copper (II) ion, Cu(NH3)42+ in the solution. Does the calculated value make sense and what does it imply?

B. EFFECT OF CONCENTRATION ON CELL PO TENTIALS 1. Complex lon Formation: Cu(NH3)42+ E0cell, before adding anything Ecell, after adding 6 M NHs to the copper cell 0.414 V 0.812 V Calculate the concentration of the free copper (II) that is in equilibrium with the complexed copper (Il) ion, Cu(NH3)42+ in the solution 0592 Vlog cell -log 0.0592 V Cu2+ E , , = 0.0592V, ([Cu21] 10.1 M] 0.812 V 0.414 V-_-log 2(0.812V - 0.414 V) [Cu2+] -0.0592 V [Cu2+] [0.1 MI2 10 13.45 [Cu2+] = 3.58 x 10-16M

Explanation / Answer

It seems you have already calculated the concentration of free Cu (II) from the cell potential values. Are you looking for the concentration of the Cu-ammonia complex? If so, then this may be the solution.

The reaction that we need to consider is

Cu (s) + 4 NH3 (aq) -----------> [Cu(NH3)4]2+ (aq) + 2e-

K = [Cu(NH3)42+]/[Cu2+][NH3]4

At equilibrium, the emf of the cell, E = 0 and from the Nernst equation, we have,

0 = E0 – 2.303RT/nF log10K

or, E0 = 2.303RT/nF log10K

Now, E0 = 0.414 V. Therefore,

0.414 V = (8.314 J/mol.K)(298 K)/(2 mol)(96500 C) log10K

or, 0.414 = 0.059/2 log10K

or, 0.414 = 0.0295 log10K

or, log10 K = 14.034

or, K = 1014.034 = 1.08*1014

The equilibrium constant is 1.08*1014. It is seen that the equilibrium constant has a large positive value, indicating that virtually all the Cu is complexed with the ammonia solution. Also, we have calculated free [Cu2+] as 3.58*10-16 M. The concentration of free Cu is extremely small, so that’s also an indication that almost the Cu is complexed to ammonia. This is the uncomplexed Cu2+ that is in equilibrium with the copper tetra-amine complex.

Let the concentration of [Cu(NH3)42+] be x M. Also, it is important to note that the concentration of ammonia is virtually equal to that of the copper tetra-amine complex, hence, [NH3] = x M.

Therefore,

K = (x)/(3.58*10-16)(x)4

or, 1.08*1014 = 1/(3.58*10-16).x3

or, 0.0387 = 1/x3

or, x3 = 25.8397

or, x = 325.8397 = 2.956

The concentration of free Cu(II) is 3.58*10-16 M and that of the complexed species is 2.96 M.

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