Consider the coordination of Cu2+ by the tridentate iminodiacetate ligand . When

Question

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.

Explanation / Answer

[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

---------------------------------------------------------------------------

Overall : Cu^2+ + HA^- <==> H+ + CuA      K = Ka * K1 = 4.7*10^-10 *4.26*10^10 = 20.02

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

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

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---