# Equations Used in Isothermal Titration Calorimetry

I have been attempting to understand and derive the formulas used for ITC which can be found here starting on page 310. I am particularly interested in the single set of identical sites. To first briefly summarize, my concern is their treatment for the stoichiometry. The equation they provide (not derived) for the equilibrium on page 312 is $$K=\frac{\theta}{(1-\theta)[X]}\tag1$$ where $\theta$ is the fraction of sites occupied by the ligand $X$ and $[X]$ is the concentration of free unbound ligand at equilibrium.

However, when I attempt to derive the equation, I obtain something slightly different. $$K=\frac{\theta}{(1-\theta)[X]^n}\tag2$$ where $n$ is the stoichiometry of the binding.

My Work

In the notation of the document, let $X$ be the ligand and $M$ be the macromolecule which $X$ binds to. Then the binding event can be written as the following chemical reaction. $$\ce{nX + M <=> MX_n}\tag3$$ Let $X_t$ and $M_t$ be the initial concentrations of the ligand and macromolecule respectively. Define $x$ to be the amount of M that reacts with $X$. Upon reaching equilibrium, the concentrations will now be \begin{align} [X]&=X_t-nx \\ [M]&=M_t-x\tag4 \\ [MX_n]&=x\end{align}

By definition of the equilibrium constant, $$K=\frac{[MX_n]}{[M][X]^n}\tag5$$ Substituting in our equilibrium concentrations gives $$K=\frac{x}{(M_t-x)[X]^n}\tag6$$ Now allow $x=\theta M_t$. This is just the definition of the fraction of occupied sites because the amount of complex made is equal to the fraction of sites times the concentration of the initial amount of macromolecule. So we thus now have $$K=\frac{\theta M_t}{(M_t-\theta M_t)[X]^n}=\frac{\theta}{(1-\theta)[X]^n}\tag7$$ where $[X]=X_t-n\theta M_t$ as they have defined in the document. Equation $(7)$ is actually a rearrangement of the Hill Equation.

My General Question: It appears their treatment is exactly the same except they leave out the power of $n$ in the equilibrium constant expression without reason. What is the justification in ignoring the power in the equilibrium constant?

My analysis using the Origin-built ITC software tends to fail consistently when $n$ is not near unity. Could this be due to the gross oversimplification of the equilibrium? I can understand why they would leave out the $n$ mathematically because it is difficult to solve the equation. Though if the company knew how to solve equations numerically this would not prove to be as great as an issue.

• It's not oversimplification so it's OK. Do you understand what it'd mean on molecular level if order was greater then 1? – Mithoron Sep 8 '17 at 21:36
• @Mithoron It would mean that more than one binds simultaneously which is statistically not likely for an elementary reaction. I think I understand it now. Thanks. – MasterYoda Sep 8 '17 at 21:45
• Yeah, but you may have some point if reaction is more complicated then simple binding. Maybe binding of dimer... Usually such assumption should work, though. – Mithoron Sep 8 '17 at 22:16