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Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives a buffer's pH from the ratio of conjugate base to weak acid. When the ratio is 1, the log is 0 and pH = pKa; more conjugate base raises pH above pKa, more weak acid lowers it.

UNIT 8 TOPIC 8.9 • HENDERSON-HASSELBALCH EQUATION HENDERSON-HASSELBALCH The acid-base ratio [A⁻]/[HA] shifts pH around pKa. pH = pKa + log([A⁻]/[HA]) where pKa = − log Ka [A⁻]/[HA] = 0.10 log 0.1 = −1 HA A⁻ more HA (acid) pH = pKa − 1 [A⁻]/[HA] = 1 log 1 = 0 HA A⁻ equal amounts pH = pKa [A⁻]/[HA] = 10 log 10 = +1 HA A⁻ more A⁻ (base) pH = pKa + 1 CED ANCHOR Valid only for buffers — an appreciable amount of BOTH the weak acid (HA) and its conjugate base (A⁻) must be present. AP Chemistry · Unit 8 · Acids and Bases
The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), gives a buffer's pH from the ratio of conjugate base to weak acid. When the ratio is 1, pH = pKa; more conjugate base raises the pH above pKa, more acid lowers it.
Henderson-Hasselbalch Ratio · Open the sandbox →

The trap misapplies the equation — inverting the ratio, using it outside the buffer region, or confusing which component goes on top. The conjugate base [A⁻] is the numerator; get the ratio right and the pH follows.

The work

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Lesson
Henderson-Hasselbalch ›

pH = pKa + log([A⁻]/[HA]) gives a buffer's pH from its component ratio. The lesson applies the equation correctly, then closes with a ten-scenario check.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items spanning the Topic 8.9 misconception: the Henderson-Hasselbalch equation misapplied — the ratio inverted or the equation used improperly.

Not started · 10 items · ~15 min
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions
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