9.9: Buffers

Last updated
Save as PDF

Page ID: 84557

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Buffer solutions, which are of enormous importance in controlling pH in various processes, can be understood in terms of acid/base equilibrium. A buffer is created in a solution which contains both a weak acid and its conjugate base. This creates to absorb excess H⁺ or supply H⁺ to replace what is lost due to neutralization. The calculation of the pH of a buffer is straightforward using an ICE table approach.

Example \(\PageIndex{1}\):

What is the pH of a solution that is 0.150 M in KF and 0.250 M in HF?

Solution

The reaction of interest is

\[HF \rightleftharpoons H^+ + F^- \nonumber \]

Let’s use an ICE table!

	\(HF\)	\(H^+\)	\(F^-\)
Initial	0.250 M	0	0.150 M
Change	-x	+x	+x
Equilibrium	0.250 M - x	x	0.150 M + x

\[ K_a = \dfrac{[H^+][F^-]}{[HF]} \nonumber \]

\[ 10^{-3.17} M = \dfrac{x(0.150 \,M + x)}{0.250 \,M - x} \nonumber \]

This expression results in a quadratic relationship, leading to two values of \(x\) that will make it true. Rejecting the negative root, the remaining root of the equation indicates

\[[ H^+]= 0.00111\,M \nonumber \]

So the pH is given by

\[ pH = -log_{10} (0.00111) = 2.95 \nonumber \]

For buffers made from acids with sufficiently large values of pK_a the buffer problem can be simplified since the concentration of the acid and its conjugate base will be determined by their pre-equilibrium values. In these cases, the pH can be calculated using the Henderson-Hasselbalch approximation.

If one considers the expression for \(K_a\)

\[ K_a = \dfrac{[H^+][A^-]}{[HA]} = [H^+]\dfrac{[H^-]}{[HA]} \nonumber \]

Taking the log of both sides and multiplying by -1 yields

\[ pK_a= pH - \log_{10} \dfrac{[A^-]}{[HA]} \nonumber \]

An rearrangement produces the form of the Henderson-Hasselbalch approxmimation.

\[ pH= pK_a - \log_{10} \dfrac{[A^-]}{[HA]} \nonumber \]

It should be noted that this approximation will fail if:

the \(pk_a\) is too small,
the concentrations \([A^-]\) is too small, or
\([HA]\) is too small,

since the equilibrium concentration will deviate wildly from the pre-equilibrium values under these conditions.