Question

The K_f for the formation of Co(NH₃)₆²⁺ is 7.70 × 10⁴.
The K_f for the formation of Co(en)₃²⁺ is 8.70 × 10¹³.

Suppose a 500.0 mL solution contains 1.000 millimoles of Co(NO₃)₂, 200.0 millimoles of NH₃, and 200.0 millimoles of ethylenediamine. What is the concentration of Co²⁺(aq ) in the solution?

Answer 1

Given data,

[Co2+] ini = 1mM/0.5 L = 2mM = 0.002 M

[NH3] ini = 200 mM/0.5 L = 400 mM = 0.4 M

[en] ini = 200 mM/0.5 L = 400 mM = 0.4 M

Let [Co(en)3]2+ = X and [Co(NH3)6]2+ = Y

As the reaction progresses the change in concentration of [Co2+] is 0.002-X-Y.

Kf for the formation of Co(en)32+= 8.70 × 1013 = X/(0.002-X-Y) x (0.4)3

Kf for the formation of Co(NH3)62+= 7.70 × 104 = Y/(0.002-X-Y) x (0.4)6

X + Y = (8.70 × 1013 x (0.4)3 + 7.70 × 104 x (0.4)6) x (0.002-X-Y)

           = (8.7 x 10 13 x 64 x 10-3 + 7.7 x 104 x 4096 x 10-6) x (0.002-X-Y)

(X+Y)/ (0.002-X-Y) = (0.5568 x 1013 + 0.03154 x 104)

Let 0.002-x-y =a

Therefore X + Y = 0.002-a

(0.002-a) = (0.5568 x 1013 + 0.03154 x 104) x a

0.002/a -1 = (0.5568 x 1013 + 0.03154 x 104)

0.002/a = 1 + (0.5568 x 1013 + 0.03154 x 104)

a = 0.002 /1 + (0.5568 x 1013 + 0.03154 x 104) = 3.592 x 10-16

The final concentration is 3.592 x 10-16 M

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