1
$\begingroup$

Reactants A, B, and C form the complexes AB and BC according to:

$$ AB \rightleftharpoons A + B, \quad BC\rightleftharpoons B + C $$

Each reaction has a dissociation constant, $K_{dA}$ and $K_{dC}$, respectively, which are defined as:

$$ K_{dA} = \frac{[A][B]}{[AB]}, \quad K_{dC} = \frac{[C][B]}{[BC]} $$

Given starting concentrations $[A]_0, [B]_0, [C]_0$, how does one calculate the equilibrium concentrations of the five species in mixture?

I have tried a couple of different approaches to a solution. First off, I identified the three additional equations

$$ [A] = [A]_0- [AB] $$ $$ [B] = [B]_0 - [AB] - [BC] $$ $$ [C] = [C]_0- [BC] $$

Putting these additional expressions for [A], [B], and [C] into the equations for the dissociation constants ultimately yields two second-degree polynomials in [AB] and [BC] with a cross-term [AB][BC] which I have not figured out how to solve analytically.

$$ 0 = [AB]^2 - (K_{dA} + [A]_0 +[B]_0)[AB] + [A]_0[B]_0 + [AB][BC] - [A]_0[BC]$$ $$ 0 = [BC]^2 - (K_{dB} + [B]_0 +[C]_0)[BC] + [B]_0[C]_0 + [AB][BC] - [C]_0[AB]$$

The other attempt I made was to solve the system of five equations for $K_{dA}$, $K_{dC}$, [A], [B], and [C] numerically using root-finding algorithms. This works okay, but oftentimes the algorithm finds roots with negative concentrations which are obviously not desirable.

I have seen many solutions online for the classic case with one reaction where a so-called ICE-table is used, but I have found no extension of this for the case I have presented. Is there a way to solve this problem analytically?

$\endgroup$
2
  • 1
    $\begingroup$ It is often handy to make variable name substitutions by 1-letter symbols that are more convenient for algebraic manipulations. // Exact problem descriptions usually lead to a set of nonlinear equations, leading to a cubic or higher polynomial equation. It is painful or not possible to solve analytically. It can be solved numerically, or the problem can be often simplified, applying strong inequalities. E.g. famous pH=1/2.(pKa - log c) for weak acid solutions uses assumptions [H+] >> [OH-] and c >> [H+] $\endgroup$
    – Poutnik
    Feb 8, 2023 at 13:17
  • $\begingroup$ See also chemistry.stackexchange.com/a/171096/35806 $\endgroup$
    – Poutnik
    Feb 8, 2023 at 13:53

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.