General Chemistry/Buffer Systems

Buffer Systems

Introduction

Buffer systems are systems in which there is a significant (and nearly equivalent) amount of a weak acid and its conjugate base -- or a weak base and its conjugate acid -- present in solution. This coupling provides a resistance to change in the solution's pH. The resistance can be overcome by providing a substantial quantity of H⁺ or OH^- ion to the solution.

Consider an arbitrary weak acid, HA, and its conjugate base, A^-, in equilibrium:

$\text{HA}+{\text{H}}_2\text{O}\iff {\text{H}}_3{\text{O}}^{+}+{\text{A}}^{-}$

$\text{HA}>\text{0},{\text{A}}^{-}>\text{0},\text{HA}\approx {\text{A}}^{-}$

This system will prove resistant to an attempt to alter its pH via addition of any Brønsted-Lowry strong acid or strong base.

A Weak Acid and its Conjugate Base

Once again let is consider an arbitrary weak acid, HA, which is present in a solution. If we introduce a salt of the acid's conjugate base, say NaA (which will provide the A^- ion), we now have a buffer solution. Also, let us say that we have .30 mols of HA and .25 mols of A^-. For reference, below is the chemical equation with the concentrations of the respective molecules.

$\text{HA}+{\text{H}}_2\text{O}\iff {\text{A}}^{-}+{\text{H}}_3{\text{O}}^{+}$

Problem Sets

1) Suppose we have 100ml of a .5M solution of Acetic Acid (CH₃COOH). To this solution we add 10.0g of Sodium Acetate (NaCH₃COO). Would you consider this a buffer solution? If so, how much HCl would we need to add to shift the pH .03 units? (Assume the change in volume attributed to the addition of the Sodium Acetate is negligible). The K_a(CH₃COOH) = 1.7 * 10^-5

Solutions

1) Yes, this is a buffer solution because there is a weak acid and its conjugate base present in the solution.
Molar Mass_(CH3COOH):$\frac{60.05g}{mol}$

Molar mass_(NaCH3COO):$\frac{82.03g}{mol}$

The chemical equation for the decomposition of NaCH₃COO is:

    ${\text{NaCH}}_3\text{OO}\Rightarrow {\text{Na}}^{+}+{\text{CH}}_3{\text{OO}}^{-}$

The equilibrium reaction is:

    ${\text{CH}}_3\text{OOH}+{\text{H}}_2\text{O}\iff {\text{CH}}_3{\text{OO}}^{-}+{\text{H}}_3{\text{O}}^{+}$

The total chemical reaction including spectator ions is:

    ${\text{CH}}_3\text{OOH}+{\text{H}}_2\text{O}+{\text{NaCH}}_3\text{OO}\iff {\text{CH}}_3{\text{OO}}^{-}+{\text{H}}_3{\text{O}}^{+}+{\text{Na}}^{+}$

$[{\text{CH}}_3{\text{OO}}^{-}]=\frac{10g\times \frac{1mol}{82.03g}}{.1L}=\frac{1.29mol}{L}$

$[{\text{CH}}_3\text{OOH}]=\frac{.5mol}{L}$