you mix a 200.0 ml sample of a solution that is 1.5x10^-3 M in Cu(NO3)2 with a 250.0 ml sample of a solution that is .20 M in NH3. After the solution reaches equilibrium, what concentration of cu2+ remains?
moles of Cu(NO3)2 = 1.5 x 10^-3 M x 0.2 L = 3 x 10^-4 mols
mols of NH3 = 0.2 M x 0.25 L = 0.05 mols
Cu2+ + 4NH3 ---> [Cu(NH3)4]2+
moles of complex formed = 3 x 10^-4 mols
moles of excess NH3 = 0.05 - 1.2 x 10^-3 = 0.0488 mols
[Cu(NH3)4]2+ = 3 x 10^-4/0.45 = 6.67 x 10^-4 M
[NH3] = 0.0488/0.45 = 0.108 M
[Cu(NH3)4]2+ ----> Cu2+ + 4NH3+
let x be the amount of complex dissociated then,
Kf = [Cu(NH3)4]2+/[Cu2+][NH3]^4
1.1 x 10^13 = (6.67 x 10^-4)/(0.108)[Cu2+]
[Cu2+] = 5.61 x 10^-16 M remains
you mix a 200.0 ml sample of a solution that is 1.5x10^-3 M in Cu(NO3)2 with...
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