[Fmeub]

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 [L_T] = 0.02 M, the total metal ion concentration [M_T] = 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 [M_T] = [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?

[Borek]

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.

[Fmeub]

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] / [L_T]. I don't know how to get a greek 'a' on the computer.

[Borek]

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 

The alpha₂ value is [L] / [L_T].

That's what I supposed. pKa values are very well separated here so it is safe to assume that L^2- 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.

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

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

[Fmeub]

Wow! I've never seen so many freaky characters huddled together in all my life! Thanks for your *delete me*