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Topic: Complex Ion Formation, Zn(OH)4^2- problem  (Read 18607 times)

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Offline Jinal613

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Complex Ion Formation, Zn(OH)4^2- problem
« on: April 03, 2009, 07:44:22 PM »
The formation constant Kf for the complex ion Zn(OH)4^2- is 2.8x10^15. What is the concentration of zinc ion, Zn^2+, in a solution that is initially 0.50 M in Zn(OH)4^2-?


This problem is really really confusing me. I know how to do complex ion formation problems but this one just had me stumped. Can you show me how to do it, step by step?? Thanks!

Offline Borek

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #1 on: April 04, 2009, 05:22:55 AM »
Treat it - at least initially - as a simple dissociation problem. Write Kf and use ICE table.

However, once you calculate this way, you have to check if OH- from the water autodissociation doesn't play substantial role in the final euilibrium.
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Offline Jinal613

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #2 on: April 04, 2009, 10:40:58 PM »
So would it be...

     Zn(OH)42- <-> Zn2+ + 4OH-
I      0.50                0       0
C       -x                 x       4x
E     0.5-x               x       4x

Dissociation constant, Kd = 1/Kf = 1/(2.8x1015) ?
or would you keep the value of Kf and solve it from there, without changing it into Kd?
or am I just approaching this all wrong..? Like I said, I am really very confused.

Offline Borek

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #3 on: April 05, 2009, 04:55:01 AM »
You started with dissociation equation, so your ICE table is for dissociation - use Kd.
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Offline Jinal613

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #4 on: April 05, 2009, 12:22:25 PM »
Hm alright so then,

Kd = 3.57x10-16 = (0.50-x)/(x)(4x)4 because it would be the reciprocal.

Now the -x would be neglected so the problem would then be...

3.57x10-16 = (0.50)/(256x5)
           x5 = 5.47x1012
           x = 3.53x102 = [Zn2+]

But that doesnt seem right because that is a HUGE number for a concentration. Am I forgetting something?

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #5 on: April 05, 2009, 04:18:55 PM »
(0.50-x)/(x)(4x)4

Looks like Kf to me, complex over metal and ligands.
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Offline Jinal613

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Re: Complex Ion Formation, Zn(OH)4^2- problem
« Reply #6 on: April 05, 2009, 08:55:25 PM »
Ohhhh okay, so I'm not supposed to switch products over reactants to reactants over products. The reciprocal is only to be done on the Kf value.

In that case, that is what I was doing wrong.

Kd = 3.57x10-16 = (x)(4x)4/(0.50-x)
                    x = 2.3x10-4 M
                      = [Zn2+]

Wow that was a rather simple problem. I feel kind of dumb for not getting it earlier, lol.
Thanks for all the assistance! I appreciate the guidance, and not just showing me the way and the answer.
« Last Edit: April 05, 2009, 09:25:18 PM by Jinal613 »

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