The overall formation constant for Cu(NH3)42+ is 1.1x1013 at 25oC. What is [Cu2+

Question

The overall formation constant for Cu(NH3)42+ is 1.1x1013 at 25oC. What is [Cu2+], in moles per litre, at equilibrium when 8.3 g Cu(NO3)2 is dissolved in 0.500 L of 1.00 mol/L NH3(aq)? Assume that there is no volume change when the Cu(NO3)2 is added to the solution. Give your answer accurate to two signficant figures. Enter 1.23x10-10 as 1.23e-10. Give your answer in mol/L. Molar masses (in g/mol) N, 14.01 O, 16.00 Cu, 63.55 Hint: Assume that Cu2++4NH3 --> Cu(NH3)42+ is the only important reaction. Notice that the equilibrium constant for this reaction is very large. One way to approach the problem is to imagine the reaction proceeds to equilibrium by first going 100% to completion and then backing up a little..

Explanation / Answer

You have to realize with problems like these that the [Cu2+] at equilibrium is minuscule; it has to be to result in a Kf of 1.1*10^13. This allows us to make an assumptions that is completely justified and avoids nasty quadratic equations (cf pH of a weak acid). [Cu2+] + 4NH3 Cu(NH3)42+ Kf = [Cu(NH3)42+]/ [Cu2+][NH3]^4 = 1.1*10^13 mols of Cu(NO3)2 = 5.3/(Mol Wt of Cu(NO3)2) = 5.3/187.57 = 2.826*10^-2 mol this is in 0.500L so [Cu(NO3)2] = 2.826*10^-2/0.500 = 5.65 10^-2 mol L-1 At equilibrium let [Cu2+] = xM [Cu(NH3)42+] = (5.65 10^-2) - x = (5.65 10^-2)M since x small [NH3] at equilibrium = 0.700 - 4*(5.65 10^-2) = 0.474 M since it takes 4 mols of NH3 to make 1 mol of cmplx Kf = [Cu(NH3)42+]/ [Cu2+][NH3]^4 = 1.1*10^13 [(5.65 10^-2)]/[x][0.474]^4 = 1.1*10^13 (Hope you can do the math) x = 1.0*10^-13M (so our assumption was OK). I think the calculations are correct but check.

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