Determinación de la constante de formación de un compuesto de coordinación.

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DISCUSIÓN

Experiencia 1:

mL KSCN x M KSCN = M FeSCN2+ x 10 mL

Solución 2 → 0,2 mL x 0,002 M = [FeSCN]2+ x 10 mL

4 x 10-5 M = [FeSCN]2+

ε = A/C = 0,13754/4 x 10-5 = 3438.5

Solución 3 → 0,4 mL x 0,002 M = [FeSCN]2+ x 10 mL

8 x 10-5 M = [FeSCN]2+

ε = A/C = 0,28221/ 8 x 10-5 = 3527.6

Solución 4 → 0,6 mL x 0,002 M = [FeSCN]2+ x 10 mL

1,2 x 10-4 M = [FeSCN]2+

ε = A/C = 0,34542/ 1,2 x10-4 = 2878.5

Solución 5 → 0,8 mL x 0,002 M = [FeSCN]2+ x 10 mL

1,6 x 10-4 M = [FeSCN]2+

ε = A/C = 0,59103/1,6 x 10-4 = 3694

Solución 6 → 1,0 mL x 0,002 M = [FeSCN]2+ x 10 mL

2 x 10-4 M = [FeSCN]2+

ε = A/C = 0,62261/2 x 10-4 = 3113

Solución 7 → 1,2 mL x 0,002 M = [FeSCN]2+ x 10 mL

2,4 x 10-4 M = [FeSCN]2+

ε = A/C = 0,763/2,4 x 10-4 = 3179

ε = 3305.1

Experiencia 2:

[FeSCN]2+ = Acorregida/ ε[pic 7]

[Fe3+]eq = [Fe3+]inicial - [FeSCN]2+

[SCN-]eq = [SCN-]inicial - [FeSCN]2+

K = [FeSCN]2+/([Fe3+]eq · [SCN-]eq)

Solución 2 → [FeSCN]2+ = 0.51506/3305.1

[FeSCN]2+ = 1,558 x 10-4 M

[Fe3+]eq = 0,05 M – 1,558 x 10-4 M

[Fe3+]eq = 4,98 x 10-2 M

[SCN-]eq = 2 x 10-4 M – 1,558 x 10-4 M

[SCN-]eq = 4,4 x 10-5 M

K = 1,558 x 10-4/(4,98 x 10-2 · 4,4 x 10-5)

K = 71.1

Solución 3 → [FeSCN]2+ = 1.05082/3305.1

[FeSCN]2+ = 3,79 x 10-4 M

[Fe3+]eq = 0,05 M – 3,79 x 10-4 M

[Fe3+]eq = 4,96 x 10-2 M

[SCN-]eq = 4 x 10-4 M – 3.79 x 10-4 M

[SCN-]eq = 2.1 x 10-5 M

K = 3,79 x 10-4/(4,96 x 10-2 · 2,1 x 10-5)

K = 363.86

Solución 4 → [FeSCN]2+ = 1.66059/3305.1

[FeSCN]2+ = 5.02 x 10-4 M

[Fe3+]eq = 0,05 M – 5.02 x 10-4 M

[Fe3+]eq = 4,95 x 10-2 M

[SCN-]eq = 6 x 10-4 M – 5.02 x 10-4 M

[SCN-]eq = 9.8 x 10-5 M

K = 5,02 x 10-4/(4,95 x 10-2 · 9,8 x 10-5)

K = 103.48

Solución 5 → [FeSCN]2+ = 2.10253/3305.1

[FeSCN]2+ = 6,36 x 10-4 M

[Fe3+]eq = 0,05 M – 6,36 x 10-4 M

[Fe3+]eq = 4,94 x 10-2 M

[SCN-]eq = 8 x 10-4 M – 6,36 x 10-4 M

[SCN-]eq = 1,64 x 10-4 M

K = 6,36 x 10-4/(4,94 x 10-2 · 1,64 x 10-4)

K = 78.5

Solución 6 → [FeSCN]2+ = 2.51835/3305.1

[FeSCN]2+ = 7.62 x 10-4 M

[Fe3+]eq = 0,05 M – 7.62 x 10-4 M

[Fe3+]eq = 4,92 x 10-2 M

[SCN-]eq = 1 x 10-3 M – 7.62 x 10-4 M

[SCN-]eq = 2.38 x 10-4 M

K = 7,62 x 10-4/(4,92 x 10-2 · 2,38 x 10-4)

K = 65.07

[pic 8]

K = 136.402

CONCLUSIONES

REFERENCIAS

- Guía de laboratorio N04 de Química Inorgánica

- Chemical Equilibrium: Finding a Constant, Kc (pdf)

- http://www.uv.es/~baeza/forcom.html

- https://es.scribd.com/doc/69051884/7/Constante-de-Formacion-Kf