Pittsburgh Business Daily

How to determine equilibrium values of a system if given concentrations included here?

Given the average Kc of a reaction below is 11.869 copper II + glycine --> copper II - glycine complex what would equilibrium values of copper, glycine, and copper-glycine complex be if 3 mL of 0.05 copper II solution was mixed with 3 mL of 0.12 M glycine? How do I solve this?

Public Comments

1. First of all, you should realize that the final volume of the solution is 6mL. Therefore, for example, the initial concentration of copper II solution is (3)(0.05)/6 . The same calculation process repeated for glycine. Then, let the initial concentration of copper II be [Cu]i , the initial concentration of glycine be [Gly]i Then , the final concentration of copper II will be [Cu]i - X . The final concentration of glycine will be [Gly]i - X . The final concentration of the complex will be X Therefore, we can build up an equation X / ([Cu]i - X)*([Gly]i - X) = equilibrium constant. Then, you can solve about X
2. Okay. Kc in the equilibirum constant and you alway express the equilibirum constant in terms of the equlibirum concentration of the products over the reactants, where each component is raised to its stoichiometric coefficients. Kc = [coperII-glycince complex]/ [copperII] x [glycine] = 11.869 these [] bars mean concentration use an ice table the keep track of the changes in concentration. i=initial, c = change, e = equilibirum i 0.05 0.12 o c -x -x +x e 0.05-x 0.12-x x I hope you can understand that table. read your book if you don't get it. Basically, if you start with copper and glycine and you have no complex. then you form the complex and reduce the copper and glycine by some unknown about the make the complex. we want to find how much copper, glycine, and complex we have at the end. this is our new Kc equation: Kc = (x)/[(0.05-x)(0.12-x)] = 11.869 solve for x 11.869 = x/(0.05-x)(0.12-x) this equation can get tedious to solve and it's going to require the quadratic formula. Here goes: 11.869 = x/ (0.006 - 0.05x - 0.12 x + x^2) if you go through the long equation x should equal about 0.0263. so [complex] = 0.0263 [copper] = 0.05 - 0.0263 = 0.0237 and [glycine] = 0.12 - 0.0263 = 0.0937 check: 0.0263/(0.0237*0.0937) = 11.84 so there is probably a rounding error in my value for x, but this is basically correct. solve the quadratic for x by yourself and be very careful so that you get the exact number, probably slightly higher or lower, so that you get exactly 11.869 when you check. By the way, I don't thin the volume of 3 mL matter since they only ask for concentrations.