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9.9: Buffers

Last updated

Jul 19, 2021
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- 9.8: Acid-Base Equilibria
- 9.10: Solubility of Ionic Compounds

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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.

Example 9.9.1\PageIndex{1}:

Solution

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Example $\PageIndex{1}$ :