In: Chemistry
Consider the coordination of Cu2+ by the tridentate iminodiacetate ligand . When the monoprotonated form, , abbreviated HA , is added to an aqueous solution of Cu2+, the following equilibrium is rapidly established:Cu 2+ + HA- ----------> CuA + H+If a solution prepared by combining 100 mL of 0.2 M Cu2+ and 100 mL of 0.2 M HA- is adjusted to pH 1.60 with NaOH, what are the concentrations of Cu2+, HA- , and CuA in the solution? (The Ka for HA- is 4.70 x 10^-10, and K1 for Cu 2+ + A -2 --------> CuA is 4.26 x 10^10.
[Cu^2+] = 0.2M*100mL /200mL = 0.01 M
[A^2-] = 0.01
[H+] = 10 ^-1.6 = 0.025 M
Two equilibrium will work in the solution :
HA- <===> H+ + A^2- Ka = 4.7*10^-10
Cu^2+ + A2- <==> CuA K1 = 4.26 x 10^10
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Overall : Cu^2+ + HA^- <==> H+ + CuA K = Ka * K1 = 4.7*10^-10 *4.26*10^10 = 20.02
Cu^2+ | HA^2- | CuA | H+ | |
initial | 0.1 | 0.1 | 0 | 0.025 |
change | -x | -x | x | x |
equilibrium | 0.1-x | 0.1-x | x | 0.025-x |
K = [CuA][H+]/[Cu^2+][A^2-]
or, 20.022 = x* (0.025-x)/(0.1-x)(0.1-x)
or, 20.022 = 0.025x-x^2 /0.01 -0.2x -x^2
or, 19.022x^2 + 4.069x-0.2 = 0
or, x = 0.041
[CuA] = 0.041 M
[Cu^2+] = 0.1-0.041 = 0.059 M
[HA^-] = 0.059 M