Derivation of the Distribution of Acid Species
It is often necessary to determine the distribution of acids and conjugate bases as a function of pH from first principles. This can most easily be accomplished by determining the relative fractions, the a, in particular forms. To do this, we start with the mass balance and equilibrium equations for the system, and then solve for the a as functions of the equilibrium constants and [H3O+].
As an example, consider a 'generic' tetra-protic acid (four protons) H4A. In solution, this acid will be distributed in the form of five different species; the acid and the four conjugate bases.
The mass balance equation for this acid is
where FA is the formal concentration (formula weights per liter of solution) of the acid. In addition to the chemical expressions, the acid equilibria are expressed by the four equilibrium equations
One can also find the formation constant, or ß equations. These are found by taking products of the equilibrium equations
Notice that the Ka1 and ß equations give concentrations of the conjugate bases in terms of only the Ka's, [H4A], and [H3O+]. Using these equations, one then solves for the anion concentrations, and substitutes the resulting equations into the mass balance equation. The mass balance equation becomes
One then multiplies both sides of the equation by [H3O+]4 to yield
Rearrangement gives the first a, the fractional amount of acid in the form of H4A
Notice that this equation gives [H4A] in terms of known quantities, e.g., [H3O+], FA, and the equilibrium constants. We have used a convention such that the a subscript signifies the number of protons lost from the most protonated form of the acid. Thus a0 signifies that no protons are lost. a1 would be the fraction of the first conjugate base, e.g., one proton lost, a2 signifies two protons lost, etc.
To obtain the other a, one uses the K and b equilibrium equations to obtain the conjugate base concentration in terms of [H4A], [H3O+], and the appropriate constants, and then substitutes into the a0 equation. For example, a1 is found using the Ka1 equilibrium equation. Solving the Ka1 equilibrium equation for [H4A], one obtains
Substitution into the a0 equation and rearranging, one gets
Similarly, solving the ba2 equation for [H4A] yields
which when substituted into the a0 equation yields
Using the same procedure, i.e., solving the b equations for [H4A] followed by substitution into the a0 equation, gives the remaining a equations
In general, for an n-proton acid, one finds that the " expression all have the same denominator
and the ak, where ak is the fractional amount of the k'th ion, will be of the form
This page was created by Professor Stephen Bialkowski, Utah State University.
August 03, 2004