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Topic: Calculation of free cadmium concentration.  (Read 5784 times)

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Offline Fmeub

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Calculation of free cadmium concentration.
« on: January 17, 2008, 05:28:45 AM »
This is another question out of the exam paper for which I don't have a memo.

Question:
The stepwise protonation constants of a ligand L are, as log K, 10.5 and 4.3 at 25 degrees Celsius and ionic strength of 0.1 M. Cadmium forms an ML2 complex and its overall formation constant, as log B is 18.8.

Calculate the free cadmium concentration at the following conditions: the total ligand concentration [LT] = 0.02 M, the total metal ion concentration [MT] = 1.0 x 10-4 M, pH = 8.1. Assume that ML2 is fully formed (system is at equilibrium) and is the only metal containing species in the solution.



My final answer is [Cd2+] or [M] = 1.998 x 10-19 M.

The value of alpha 2 calculated as 3.96 x 10-3 M and thus the value of [L] as 7.93 x 10-5 M.

Can I assume that if it was said that ML2 is the only metal containing species that [MT] = [ML2] = 1.0 x 10-4 M ?
The final answer of [M] then follows from [ML2] / [M][L]2 = 6.31 x 1018.

Is there a mistake in my reasoning somewhere?

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Offline Borek

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Re: Calculation of free cadmium concentration.
« Reply #1 on: January 17, 2008, 05:50:46 AM »
Reasoning looks OK (I am not sure what you mean by alpha 2 - but your estimate of L concentration is correct). Unfortunately, final result is off. Check your math.
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Offline Fmeub

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Re: Calculation of free cadmium concentration.
« Reply #2 on: January 17, 2008, 06:27:13 AM »
Yes, it seems I made a calculator-calculation error.
Does 2.52 x 10-15 M sound better for the final [M] concentration?

The alpha2 value is [L] / [LT]. I don't know how to get a greek 'a' on the computer.
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Offline Borek

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Re: Calculation of free cadmium concentration.
« Reply #3 on: January 17, 2008, 06:59:15 AM »
Quote from: Fmeub on January 17, 2008, 06:27:13 AM
Yes, it seems I made a calculator-calculation error.
Does 2.52 x 10-15 M sound better for the final [M] concentration?

Much better now  :)

Quote
The alpha2 value is [L] / [LT].

That's what I supposed. pKa values are very well separated here so it is safe to assume that L2- concentration is governed by pKa2 only, at the same time for pH 8.1 in this case you may safely assume all acid is in the form of HL-. That's how I did it, but systematic approach is more general.

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I don't know how to get a greek 'a' on the computer.

Start->Programs->Accessories->System tools->Character table->α

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Offline Fmeub

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Re: Calculation of free cadmium concentration.
« Reply #4 on: January 17, 2008, 07:30:20 AM »
Wow! I've never seen so many freaky characters huddled together in all my life! Thanks for your *delete me*
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