Henderson-Hasselbalch Practice: pH, pKa, Ratio, Concentrations

If you're practicing for a Henderson–Hasselbalch practice exam and your question gives you pKa and the [A⁻]/[HA] ratio, the calculation is one line: pH = pKa + log([A⁻]/[HA]). The most common mistake here isn't the math—it's plugging the ratio in upside down. Conjugate base goes on top, weak acid on bottom. Flip them and your answer is wrong by 2 × log(ratio) pH units, which on a buffer problem can easily be a full point gone.

Try this: pKa = 4.76, [A⁻]/[HA] = 3.0. pH = 4.76 + log(3.0) = 4.76 + 0.477 = 5.24. If you accidentally used [HA]/[A⁻] = 3.0 instead, you'd get pH = 4.76 + log(0.333) = 4.76 − 0.477 = 4.28. That's nearly a full pH unit off in the wrong direction.

The equation also tells you something intuitive: when [A⁻] > [HA], pH > pKa (more base form means more basic). When [HA] > [A⁻], pH < pKa (more acid form means more acidic). At equal concentrations, pH = pKa exactly. If your answer contradicts these rules, you've made an error somewhere.

The reverse problem: given pH and pKa, find the ratio. Rearrange HH: [A⁻]/[HA] = 10^(pH − pKa). If pH = 5.50 and pKa = 4.76, then [A⁻]/[HA] = 10^(0.74) = 5.50. That means for every mole of acetic acid, you have 5.50 moles of acetate.

The most frequent error is computing the exponent with the wrong sign. It's pH − pKa, not pKa − pH. If pH > pKa, the exponent is positive and the ratio is greater than 1 (more conjugate base). If pH < pKa, the exponent is negative and the ratio is less than 1 (more acid). Getting the sign wrong gives you the reciprocal of the right answer.

The Henderson–Hasselbalch equation has three unknowns (pH, pKa, ratio), so given any two you can find the third. Three forms of the same equation:

The third form is useful for finding pKa from experimental data—measure the pH of a buffer where you know the ratio. If you dissolved 0.020 mol acid and 0.050 mol conjugate base in 1 L and measured pH = 5.16, then pKa = 5.16 − log(0.050/0.020) = 5.16 − log(2.5) = 5.16 − 0.40 = 4.76.

Notice that volumes cancel in the ratio. Since [A⁻] = n(A⁻)/V and [HA] = n(HA)/V, the volume V cancels: [A⁻]/[HA] = n(A⁻)/n(HA). You can use moles directly without converting to molarity. This shortcut saves time and eliminates a common source of arithmetic errors.

HH depends on the ratio of conjugate base to acid, not their absolute concentrations. A buffer with 0.5 M A⁻ and 0.5 M HA has the same pH as one with 0.01 M A⁻ and 0.01 M HA (same ratio = 1, same log(1) = 0). The pH is identical: both equal pKa.

Where concentration matters is buffer capacity. The 0.5 M buffer can absorb 50× more acid or base than the 0.01 M buffer before the ratio shifts enough to change pH significantly. Concentration determines capacity, ratio determines pH. Conflating the two is a trap.

Another trap: diluting a buffer doesn't change its pH (the ratio stays the same). But it does reduce capacity. If you take 100 mL of a 0.5 M acetate buffer and dilute to 1 L, the pH stays the same but it can now absorb only 1/10 the amount of strong acid it could before. Dilution lowers capacity without changing pH—until you push the buffer so far that the 5% approximation fails or water's autoionization becomes significant.

Exam-style problem: A buffer contains 0.30 mol CH₃COOH and 0.45 mol CH₃COONa in 2.0 L. pKa = 4.76. (a) What is the pH? (b) What is the pH after adding 0.05 mol NaOH?

The pH shifted only 0.12 units despite adding 0.05 mol of strong base. That's the buffer working. Notice we used moles directly (not concentrations) because volume cancels. If you divided by 2.0 L first, you'd get [A⁻]/[HA] = 0.25/0.125 = 2.00—same ratio, same answer, more arithmetic.