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Biochemistry · Acids, bases, and buffers

Henderson-Hasselbalch practice problems, with answers and a calculator

The Henderson-Hasselbalch equation links a buffer's pH to its pKa and the ratio of base form to acid form. The one big idea: every pH unit away from the pKa is a factor of 10 in that ratio. Learn that and most college biochemistry buffer questions take seconds.

Already know the basics? Jump to the 8 practice questions or the worked examples.

The equation and when you can use it

pH = pKa + log([A⁻]/[HA])

HA is the acid form (the one holding the proton) and A⁻ is the conjugate base. The equation is just the Ka expression, rearranged and put on a log scale. In biochemistry the "acid" is often not neutral: in a phosphate buffer, HA is H₂PO₄⁻ and A⁻ is HPO₄²⁻. What matters is "protonated form" over "deprotonated form."

It gives a good answer when:

  1. It is a real buffer. Both the weak acid and its conjugate base are present in meaningful amounts. A strong acid alone, or a weak acid with no base added, is a different problem.
  2. The ratio stays between about 0.1 and 10. That keeps the pH within 1 unit of the pKa, which is the useful buffer range.
  3. The solution is not very dilute. The equation assumes the acid barely ionizes on its own, so the amounts you put in are the amounts present. At very low concentrations that shortcut breaks down.

You can plug in concentrations or moles. Both forms are in the same volume, so the volume cancels in the ratio.

The ratio intuition: read pH without a calculator

Two facts carry most exam questions:

  1. pH = pKa when the ratio is 1:1. log(1) = 0. Half the molecules are protonated and half are not.
  2. Each pH unit is a factor of 10 in the ratio. One unit above the pKa, base outnumbers acid 10 to 1. Two units above, 100 to 1. One unit below, acid wins 10 to 1.
Ratio and percent deprotonated at different distances from the pKa
pH − pKa[A⁻]/[HA]% deprotonated (A⁻)% protonated (HA)
−21/100≈ 1%≈ 99%
−11/10≈ 9%≈ 91%
0150%50%
+0.3≈ 2≈ 67%≈ 33%
+110≈ 91%≈ 9%
+2100≈ 99%≈ 1%

To get the fraction protonated from the ratio r = [A⁻]/[HA], use fraction HA = 1/(1 + r), and fraction A⁻ = r/(1 + r). A handy shortcut: pH below pKa means mostly protonated, and pH above pKa means mostly deprotonated.

Buffers you meet in biochemistry

Phosphate. The H₂PO₄⁻/HPO₄²⁻ pair is the main buffer inside cells. The LibreTexts biochemistry text gives its pKa as 7.21; you will also see values near 6.8 in some books, because the effective pKa shifts with ionic strength and temperature. Use whatever value your problem gives you.

Bicarbonate in blood. Blood holds about 20 parts bicarbonate (HCO₃⁻) to 1 part carbonic acid. Physiology courses usually use an apparent pKa of 6.1 at body temperature, and 6.1 + log(20) = 7.40, the normal blood pH. Textbooks report different pKa values here (OpenStax Chemistry 2e uses 6.4, and one LibreTexts text uses an effective 6.3), so again, use the number in the problem. A 20:1 ratio is outside the textbook 0.1 to 10 range. The system still works because your lungs remove CO₂ and your kidneys adjust bicarbonate.

Histidine. The imidazole side chain has a pKa of about 6.0 (6.00 in the OpenStax Organic Chemistry table). That is close to physiological pH, which is why histidine so often acts as an acid or a base in enzyme active sites.

Making a buffer. Pick a weak acid whose pKa is within 1 unit of the pH you want, ideally close to it. Then use the equation to get the ratio, and split your total concentration in that ratio. Worked example 3 below does exactly this.

Worked examples

Example 1: pH of a phosphate buffer

A buffer contains 0.15 M Na₂HPO₄ and 0.10 M NaH₂PO₄. The pKa of H₂PO₄⁻ is 7.21. What is the pH?

  1. Identify the forms. H₂PO₄⁻ holds the extra proton, so it is HA. HPO₄²⁻ is A⁻.
  2. Ratio: [A⁻]/[HA] = 0.15/0.10 = 1.5.
  3. log(1.5) = 0.176.
  4. pH = 7.21 + 0.176 = 7.386.
  5. Sanity check: more base than acid, so the pH should sit a little above the pKa. It does.

Answer: pH ≈ 7.39

Example 2: How much histidine is protonated at pH 7.4?

A histidine side chain has a pKa of 6.0. What fraction of side chains carry the extra proton (the protonated, positively charged form) at pH 7.4?

  1. pH − pKa = 7.4 − 6.0 = 1.4.
  2. Ratio [A⁻]/[HA] = 101.4 = 25.1. The neutral (deprotonated) form wins by about 25 to 1.
  3. Fraction protonated = 1/(1 + 25.1) = 0.038.
  4. Sanity check: pH is above the pKa, so the protonated form should be the minority. It is.

Answer: about 3.8% of side chains are protonated at pH 7.4.

Example 3: Making a pH 7.40 phosphate buffer

You need 1.0 L of a 0.100 M phosphate buffer at pH 7.40 (pKa 7.21). How much of each form do you need?

  1. Find the ratio: log([A⁻]/[HA]) = 7.40 − 7.21 = 0.19, so [A⁻]/[HA] = 100.19 = 1.549.
  2. Set up the total: [A⁻] + [HA] = 0.100 M, and [A⁻] = 1.549 × [HA].
  3. Substitute: 2.549 × [HA] = 0.100 M, so [HA] = 0.0392 M.
  4. [A⁻] = 0.100 − 0.0392 = 0.0608 M.
  5. In 1.0 L, that is 0.0608 mol HPO₄²⁻ (for example as Na₂HPO₄) and 0.0392 mol H₂PO₄⁻ (as NaH₂PO₄).

Answer: about 0.061 M HPO₄²⁻ and 0.039 M H₂PO₄⁻.

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Henderson-Hasselbalch calculator

Pick what you want to find, enter the pKa and the other value, then press Calculate. The ratio is always base form over acid form, [A⁻]/[HA].

Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 1A solution contains 0.20 M acetic acid and 0.20 M sodium acetate. Acetic acid has a pKa of 4.74. What is the pH?

Show solution

[A⁻]/[HA] = 0.20/0.20 = 1. pH = 4.74 + log(1) = 4.74 + 0 = 4.74. Whenever the two forms are equal, pH = pKa.

Question 2A buffer is at a pH that is 2 units above the pKa of its weak acid. What is the ratio [A⁻]/[HA]?

Show solution

pH − pKa = log([A⁻]/[HA]) = 2. Undo the log: [A⁻]/[HA] = 10² = 100. About 99% of the acid is deprotonated.

Question 3Blood holds about 20 times as much HCO₃⁻ as carbonic acid. Using an apparent pKa of 6.1, what pH does the Henderson-Hasselbalch equation predict?

Show solution

HCO₃⁻ is A⁻ and carbonic acid is HA, so the ratio is 20. log(20) = 1.301. pH = 6.1 + 1.301 = 7.40.

Question 4A histidine side chain (pKa 6.0) is in a solution at pH 5.0. About what percent of the side chains are protonated?

Show solution

log([A⁻]/[HA]) = 5.0 − 6.0 = −1, so the ratio is 0.1. Fraction protonated = 1/(1 + 0.1) = 1/1.1 = 0.909, about 91%.

Question 5You need a buffer at pH 7.4. Which weak acid is the best choice?

Show solution

Choose the pKa closest to the target pH. 7.21 is 0.19 units from 7.4; 4.74 and 9.25 are both more than 1.8 units away, outside the useful range of pKa ± 1.

Question 6A phosphate buffer (pKa 7.21) has a pH of 6.91. What is the ratio [HPO₄²⁻]/[H₂PO₄⁻]?

Show solution

log([A⁻]/[HA]) = 6.91 − 7.21 = −0.30. [A⁻]/[HA] = 10−0.30 = 0.501. So there is about 1 HPO₄²⁻ for every 2 H₂PO₄⁻.

Question 7A 1.00 L buffer contains 0.100 mol acetic acid and 0.100 mol acetate (pKa 4.74). You add 0.010 mol NaOH. What is the new pH?

Show solution

OH⁻ reacts with HA: HA goes from 0.100 to 0.090 mol and A⁻ goes from 0.100 to 0.110 mol. pH = 4.74 + log(0.110/0.090) = 4.74 + 0.087 = 4.83. Moles work fine because both forms share the same volume.

Question 8A lysine side chain amine has a pKa of about 10.5. At pH 7.5, which statement is true?

Show solution

log([A⁻]/[HA]) = 7.5 − 10.5 = −3, so the ratio is 10−3 = 0.001. Fraction protonated = 1/(1 + 0.001) = 0.999, about 99.9%. That is why lysine counts as a positively charged amino acid at physiological pH.

Answer key

  1. B, 4.74: a 1:1 ratio means pH = pKa.
  2. D, 100: each pH unit is a factor of 10.
  3. A, 7.40: 6.1 + log(20).
  4. C, about 91%: one unit below the pKa, acid wins 10 to 1.
  5. B, phosphate: pKa closest to the target pH.
  6. B, about 1:2: 10−0.30 ≈ 0.50.
  7. C, 4.83: update moles first, then use the equation.
  8. D, about 99.9% protonated: 3 units below the pKa.

Frequently asked questions

Is the Henderson-Hasselbalch equation on the MCAT?

Buffers and acid-base equilibria appear in the AAMC content outline for the chemical and biological foundations section, and Henderson-Hasselbalch reasoning shows up in passages about buffers, blood pH, and amino acid charge. You are not given a calculator, so the ratio rules (pH = pKa at 1:1, a factor of 10 per pH unit) are the part worth memorizing.

Can I use moles instead of concentrations?

Yes. The acid and base forms are in the same volume, so the volume cancels in the ratio. That is handy after adding strong acid or base: update the moles, then plug the mole ratio in.

When does the Henderson-Hasselbalch equation fail?

It fails when there is no real buffer, such as a strong acid alone or a weak acid with no conjugate base. It also gets less accurate when the ratio is far outside 0.1 to 10 or the solution is very dilute, because the acid's own ionization then changes the concentrations noticeably.

Why do textbooks list different pKa values for phosphate and bicarbonate?

The apparent pKa depends on temperature, ionic strength, and, for bicarbonate, whether dissolved CO₂ is counted with carbonic acid. Values near 6.8 to 7.2 for phosphate and 6.1 to 6.4 for bicarbonate are all common. On an exam, use the value given.

Sources

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