Buffer Capacity & Henderson-Hasselbalch Examples Cheat Sheet

This cheat sheet covers how buffers resist changes in pH and how the Henderson-Hasselbalch equation connects pH to the amounts of weak acid and conjugate base. Students need this reference because buffer problems often combine equilibrium ideas, logarithms, mole ratios, and stoichiometry. It is especially useful for titration questions, biology-related pH systems, and AP or honors chemistry review.

The key formula is $\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$ , where $\mathrm{HA}$ is the weak acid and $\mathrm{A}^-$ is its conjugate base. A buffer works best when $[\mathrm{A}^-]\approx[\mathrm{HA}]$ , so $\mathrm{pH}\approx\mathrm{p}K_a$ . Buffer capacity increases when the total concentration of $\mathrm{HA}$ and $\mathrm{A}^-$ increases, and it is greatest when the acid-base amounts are similar.

Key Facts

The Henderson-Hasselbalch equation is $\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$ .
For a base buffer pair, the related form is $\mathrm{pOH}=\mathrm{p}K_b+\log\left(\frac{[\mathrm{BH}^+]}{[\mathrm{B}] }\right)$ .
The relationship between acid strength and $K_a$ is $\mathrm{p}K_a=-\log(K_a)$ .
A buffer is most effective when $\mathrm{pH}=\mathrm{p}K_a$ , which occurs when $[\mathrm{A}^-]=[\mathrm{HA}]$ .
A useful buffer range is usually $\mathrm{pH}=\mathrm{p}K_a\pm1$ , because the ratio $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}$ stays between about $0.1$ and $10$ .
When strong acid is added, it reacts mainly as $\mathrm{H}^+ + \mathrm{A}^- \rightarrow \mathrm{HA}$ before the Henderson-Hasselbalch equation is used.
When strong base is added, it reacts mainly as $\mathrm{OH}^- + \mathrm{HA} \rightarrow \mathrm{A}^- + \mathrm{H_2O}$ before the Henderson-Hasselbalch equation is used.
Buffer capacity increases as the total buffer concentration $[\mathrm{HA}]+[\mathrm{A}^-]$ increases, even if the ratio $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}$ stays the same.

Vocabulary

Buffer: A solution that resists large changes in $\mathrm{pH}$ when small amounts of strong acid or strong base are added.
Weak acid: An acid such as $\mathrm{HA}$ that only partially ionizes in water and establishes an equilibrium with $\mathrm{A}^-$ and $\mathrm{H}^+$ .
Conjugate base: The particle $\mathrm{A}^-$ formed when the weak acid $\mathrm{HA}$ loses one proton.
$\mathrm{p}K_a$: A logarithmic measure of acid strength defined by $\mathrm{p}K_a=-\log(K_a)$ .
Buffer capacity: The amount of strong acid or strong base a buffer can neutralize before its $\mathrm{pH}$ changes significantly.
Henderson-Hasselbalch equation: An equation that estimates buffer $\mathrm{pH}$ using $\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$ .

Common Mistakes to Avoid

Using concentrations before neutralization, which is wrong when strong acid or strong base has been added. First update the moles of $\mathrm{HA}$ and $\mathrm{A}^-$ using the reaction stoichiometry.
Flipping the ratio in the Henderson-Hasselbalch equation, which gives the wrong sign for the logarithm. The acid form goes in the denominator as $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}$ .
Assuming dilution changes buffer $\mathrm{pH}$ , which is usually wrong if both $[\mathrm{HA}]$ and $[\mathrm{A}^-]$ are diluted by the same factor. The ratio $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}$ stays the same.
Choosing a buffer with $\mathrm{p}K_a$ far from the target $\mathrm{pH}$ , which makes the buffer weak in that range. A good buffer usually has $\mathrm{p}K_a$ within about $1$ unit of the desired $\mathrm{pH}$ .
Confusing buffer capacity with buffer $\mathrm{pH}$ , which ignores total amount. Two buffers can have the same $\mathrm{pH}$ but different capacities if one has larger $[\mathrm{HA}]+[\mathrm{A}^-]$ .

Practice Questions

1 A buffer contains $0.200\,\mathrm{mol}$ of $\mathrm{CH_3COOH}$ and $0.300\,\mathrm{mol}$ of $\mathrm{CH_3COO^-}$ in $1.00\,\mathrm{L}$ . If $\mathrm{p}K_a=4.76$ , find the $\mathrm{pH}$ .
2 A buffer has $0.500\,\mathrm{mol}$ of $\mathrm{HA}$ and $0.500\,\mathrm{mol}$ of $\mathrm{A}^-$ . After $0.100\,\mathrm{mol}$ of $\mathrm{NaOH}$ is added, what ratio $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}$ should be used in the Henderson-Hasselbalch equation?
3 A chemist needs a buffer at $\mathrm{pH}=7.40$ . Which acid is the better choice, one with $\mathrm{p}K_a=7.21$ or one with $\mathrm{p}K_a=4.75$ ? Explain briefly.
4 Two buffers have the same ratio $\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}=1$ , but Buffer $1$ has $0.10\,\mathrm{M}$ total buffer concentration and Buffer $2$ has $1.00\,\mathrm{M}$ total buffer concentration. Which has greater buffer capacity, and why?

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Buffer Capacity & Henderson-Hasselbalch Examples Cheat Sheet

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