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Topic: Association constant calculation in 3:1 complexes  (Read 1406 times)

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Offline Sunitaa

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Association constant calculation in 3:1 complexes
« on: July 05, 2021, 05:39:13 PM »
I have a question about calculating the association constant related to 3:1 host-guest complexes. I am using the UV-vis titration method, I know the stoichiometry from the molar fraction method which I mentioned earlier is 3 to 1. Below are the formulas that I used to calculate the concentration of complex and the association constant (H=left Host, G=left Guest, H0=initial Host, G0=initial Guest, C=Complex):

aH+bG↽−−⇀C
K=[C]/[H]a[G]b
[H]0=[H]+a[C]
[G]0=[G]+b[C]

I had some issues with understanding how to use these equations.

What exactly is the definition of [G0] (the initial or total concentration of guests) practically? Is it the concentration of prepared metal ion solution? Or do I have to calculate it from the first mole of ionic solution added into the host solution? Or is it the moles of guests divided by the volume of the host solution? Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

By eliminating [G0] from the formula, the concentration of complex became negative, which doesn't make sense as the stoichiometry of the host is 3 so it makes a huge effect on the value of the association constant and makes it too high value.

Offline mjc123

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Re: Association constant calculation in 3:1 complexes
« Reply #1 on: July 06, 2021, 09:08:18 AM »
No, you can't eliminate it. The guests are not consumed (in the sense of destroyed), but merely incorporated into the complex. [G]0 is the total concentration of G in the solution, whether as free G molecules or in the complex.

What that means practically depends on how the experiment is performed, which you have left vague. You say it's a titration, so you are adding something to something, and concentrations will be changing.
For example, suppose you have an initial volume V0 of a solution of H of initial concentration [H]i, and titrate it with a solution of G of concentration [G]i. When you have added V1 of the G solution, the total volume V2 = V0 + V1 and
[H]0 = [H]i*V0/V2
[G]0 = [G]i*V1/V2

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