Introduction to Buffers

Introduction to Buffers (lig)

Most reversible biomolecular reactions involve weak interactions between the biomolecule itself and its specific ligand(s). Such systems effectively behave like the standard pH buffers. Thus, the quantitative and graphical aspects of simple and complex pH buffers will be developed, starting here with acetic acid, a monoprotic (monovalent) buffer representing the simplest of buffer systems.

Buffers effectively "control" the pH or free H⁺ ion concentration, [H⁺], in a solution by adsorbing or releasing weakly bound H⁺ either as the free acid (H⁺ ion) is added to the system or free H⁺ ion is removed from the system by the addition of base or free hydroxyl ion, OH^-.

For weak dissociation/association reactions, the concentrations of products and reactants at equilibrium are all related to each other by a fixed constant called an equilibrium constant.

For acetic acid, the dissociation of is defined by the following equilibrium reaction and equilibrium dissociation constant, Kd-cooh.

One-step dissociationof Acetic Acid

CH₃-COOH

- H⁺

+ H⁺

CH₃-COO^-

Kd-cooh =[ H⁺] * [CH₃-COO^-] / [CH₃-COOH]
Kd-cooh =10^-4.76

Now consider the following question. What is the the H⁺ ion concentration, [H⁺], at equilibrium for a 0.1 M solution of acetic acid in pure water?

Since no other acid or base has been introduced into this system, the following conditions and approximations hold.

[ H⁺] = [CH₃-COO^-] (each dissociated proton generates an acetate ion by mass action)
[CH₃-COOH] = 0.1 M - [ H⁺] ~ 0.1 M since acidic acid is weakly dissociating system.
Kd-cooh = 10^-4.76 = [ H⁺] * [ H⁺] / (0.1) = [ H⁺]² / (0.1)

Solve for [ H⁺] using conventional algebra:

[ H⁺] = (0.1 * 10^-4.76)^1/2 = (10^-1 * 10^-4.76)^1/2 = (10^-1-4.76)^1/2 (10^-5.76)^1/2
[ H⁺] = 10^-2.88 = 10^-0.88-2 = 10^-0.88 * 10^-2= 1.32 * 10^-2
[ H⁺] = 0.0132

Typically, H⁺ ion concentrations are not measured directly but they are measured as pH values. In this case, it makes more sense to solve this problem with logarithmic algebra applying the following definitions:

pH = - log [ H⁺] and pKd = - log Kd.

Starting with the equation, Kd-cooh = [ H⁺] * [ H⁺] / (0.1), solve for [ H⁺] using logarithmic algebra by taking the log value of both sides of this equation:

log (Kd-cooh) = log { [ H⁺]² / (0.1) } = log (10^-4.76) = - 4.76
= log { [ H⁺]² / (0.1) } = log [ H⁺]² - { log (10^-1) }= 2 log [ H⁺] + 1
2 log [ H⁺] = - 4.76 - 1 = - 5.76
log [ H⁺] = - 5.76 / 2 = - 2.88
[ H⁺] = 10 ^{log [}^H+^] = 10 ^-^pH = 10^-2.88 = 0.0132

These equations can easily be generalized for any equilibrium solution of acetic acid at total concentration, C_o as follows:

pH = ( pKd - log C_o) / 2
[ H⁺] = 10 ^-^pH = 10 ⁽^pKd^{- log Co) / 2} = 10 ^{( 4.76 - log Co) / 2}

This is a useful equation because it can be used to quickly predict pH and [ H⁺] over a range of initial acetic acid concentrations, C_o.

equals [ H⁺] * [CH₃-COO^-] / [CH₃-COOH] or the product of the H⁺ ion concentration, [H⁺], and the acetate concentration, [CH₃-COO^-], divided by the concentration of acetic acid, [CH₃-COOH], at equilibrium.

Continue the acid-base titration analysis of acetic acid/acetate by opening the Excel spreadsheet and following the directions found there. You will find four types of saturation graphs showing

Ya, the saturation (association) fraction, plotted against [H⁺] -- Rectangular Hyperbolic Plot;
Ya plotted against pH ( i.e., -log [H⁺] ) -- pH Saturation Plot;
Ya plotted against plotted against [H⁺] on a log axis -- Log Saturation Plot;
log (Yd / Ya) plotted against plotted against pH ( i.e., - log [H⁺] ) -- Henderson-Hasselbalch Plot.

Acid-Base Titration of Acetic Acid

Monovalent Ligand Binding

Quit Excel before moving to the next exercise.

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