2
$\begingroup$

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 ($\ce{H}$=Host, $\ce{G}$=Guest, $\ce{C}$=Complex):

$$\ce{aH + bG<=>C}$$ $$K=\frac{[\ce{C}]}{[\ce{H}]^a[\ce{G}]^b}$$ $$[\ce{H}]_0=[\ce{H}]+a[\ce{C}]$$ $$[\ce{G}]_0=[\ce{G}]+b[\ce{C}]$$

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

  1. What exactly is the definition of $\ce{[G0]}$ (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 guest 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?

  2. By eliminating $\ce{[G0]}$ 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.

$\endgroup$

1 Answer 1

2
$\begingroup$

[OP] Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

No. At equilibrium, all species are present. If one of the reactants is depleted, the reaction went to completion, distinct from reaching equilibrium.

Sometimes, you can neglect a species in a sum or a difference, but only in that sense. For the calculation of the equilibrium constant, you need to know all the equilibrium concentrations, and saying that one of them is very little does not help.

What exactly is the definition of [G0] (the initial or total concentration of guests) practically?

It is both the initial concentration of guest (assuming there is no complex at the time) and the total concentration of guest. If you are mixing two components of known concentration, you have to account for the dilution that goes along with mixing them.

Is it the concentration of prepared metal ion solution?

No, because when you add the other component, you are diluting the metal ion solution.

Or do I have to calculate it from the first mole of ionic solution added into the host solution?

I don't understand that idea.

Or is it the moles of guest divided by the volume of the host solution?

No, it is not. That ratio would tell you whether host or guest is in stoichiometric excess.

Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

No, not from the equilibrium constant expression. If the equilibrium lies far on the side of the complex, you could estimate the concentration of complex at equilibrium from the initial concentration of the limiting reactant. You could also estimate the concentration of reactant in excess remaining at equilibrium. However, you need to know the concentration of the limiting reactant at equilibrium to figure out the equilibrium constant.

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.

That's good critical thinking. If you get a nonsense result in a calculation, you either made a mistake in setting up the calculation, or made a numerical error. In this case, it is a mistake in setting up the calculation.

I am using the UV-vis titration method

If you are monitoring the concentration of the complex, you have to try to find conditions where your titration curve is not a straight line (i.e. the complex forms stoichiometrically and the reaction looks like it goes to completion). You can accomplish that by lowering the concentrations, or by changing the temperature in the direction that makes the complex less stable. On the other hand, if you are measuring the "free" host (i.e. the host gives a spectroscopic signal while the guest and the complex do not), you would use the guest in excess (so you can easily calculate its equilibrium concentration) and would try to detect how much (very little) of the host is not complexed at equilibrium.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.