8.9 Henderson-Hasselbalch Equation

The Henderson–Hasselbalch equation connects a weak acid’s pKa to the pH of a buffer made from that acid and its conjugate base: pH = pKa + log([A-]/[HA]) Use it when you have a buffer (a weak acid HA and its conjugate base A-, often as the salt) and you know their concentrations. It tells you the buffer pH from the ratio [A-]/[HA]. It’s most accurate when both species are present in comparable amounts (within about a factor of 10) and the acid is weak. Useful quick rules: when [A-] = [HA], pH = pKa; buffer works best within ≈ ±1 pH unit of the pKa. On the AP exam, you don’t need to derive the equation and you won’t be asked to compute pH changes after adding acid/base to a buffer (that specific calculation is excluded). For a focused study guide on this exact topic see Fiveable’s Henderson–Hasselbalch study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW). For unit review and lots of practice questions, check the Unit 8 page (https://library.fiveable.me/ap-chemistry/unit-8) and Fiveable’s practice set (https://library.fiveable.me/practice/ap-chemistry).

Use pH = pKa + log([A−]/[HA]). Steps: (1) identify the conjugate acid (HA) and base (A−); (2) find the acid’s pKa (from Ka or a table); (3) use the buffer concentrations (molarities of A− and HA) as the ratio; (4) plug into the Henderson–Hasselbalch eqn and calculate. Quick example: acetic acid (pKa = 4.76). If [A−] = 0.10 M (acetate) and [HA] = 0.15 M (acetic acid), pH = 4.76 + log(0.10/0.15) = 4.76 + log(0.667) ≈ 4.76 − 0.176 = 4.58. Remember AP specifics: you’ll be asked to identify pH from pKa and the [A−]/[HA] ratio (CED 8.9.A), not to derive the equation or compute pH change after additions (those are excluded). Buffers resist small additions of acid or base because the ratio stays nearly constant. For more examples and review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and unit resources (https://library.fiveable.me/ap-chemistry/unit-8). For extra practice, use Fiveable’s practice problems (https://library.fiveable.me/practice/ap-chemistry).

In the Henderson–Hasselbalch equation pH = pKa + log([A-]/[HA]), HA is the weak acid (the protonated form) and A- is its conjugate base (the deprotonated form). Think of HA ⇌ H+ + A-: HA donates H+; A- can accept H+. The ratio [A-]/[HA] tells you whether the buffer is more basic (ratio > 1 → pH > pKa) or more acidic (ratio < 1 → pH < pKa). For example, an acetic acid/acetate buffer: HA = CH3COOH, A- = CH3COO-. The CED emphasizes that HH is just the Ka equilibrium rearranged and that small additions of acid or base don’t change the [A-]/[HA] ratio much, so the pH hardly changes (buffer behavior). You don’t need to derive the equation for the AP exam, but you should be able to identify pH from the identity and concentrations of the conjugate pair (CED 8.9.A). For a quick review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and unit overview (https://library.fiveable.me/ap-chemistry/unit-8). For extra practice questions, check Fiveable’s AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

The Henderson–Hasselbalch equation comes from the Ka expression for a weak acid: when you rearrange Ka = [H+][A–]/[HA] you get pH = pKa + log([A–]/[HA]). That algebra only gives a useful, accurate number when both members of the conjugate pair (HA and A–) are present in appreciable amounts—i.e., a buffer. In a “regular” solution of a weak acid without its conjugate base, [A–] is produced only by the acid’s small dissociation (often ≪[HA]), so the approximation that [A–] and [HA] equal their initial concentrations breaks down. Also the HH form assumes the change in concentrations from dissociation or small additions is negligible (buffer capacity), which isn’t true for a lone acid. For AP Chem, use HH to identify buffer pH from the pKa and [A–]/[HA] ratio (CED 8.9.A); don’t use it for pure-acid pH unless you do a full equilibrium (ICE) calculation. For a quick study guide, see Fiveable’s Topic 8.9 page (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW). For extra practice, try the AP practice problems (https://library.fiveable.me/practice/ap-chemistry).

pKa is a property of the acid; pH is the actual acidity of the solution. In the Henderson–Hasselbalch equation pH = pKa + log([A-]/[HA]), pKa = −log(Ka) is a constant that describes how easily the acid donates a proton (intrinsic to that acid). pH is the solution’s hydrogen-ion activity (what you measure). The term log([A-]/[HA]) is what shifts pH away from pKa: if [A-] = [HA], log = 0 and pH = pKa. Changing the ratio [A-]/[HA] (concentrations of conjugate base and acid) changes pH; adding small amounts of acid or base to a buffer usually won’t change that ratio much, so pH stays nearly constant (buffering). Useful rules: buffer effective range ≈ pKa ± 1; pKa is fixed for a given acid, pH depends on concentration and conditions. For AP review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8). For extra practice, try the AP practice set (https://library.fiveable.me/practice/ap-chemistry).

Use [A-] on top and [HA] on bottom—that is, conjugate base over conjugate acid. The formula from the CED is pH = pKa + log([A-]/[HA]). To choose which is A- vs HA, write the acid dissociation: HA ⇌ H+ + A-. The species that loses H+ is HA (put it in the denominator); the species that results after losing H+ is A− (put it in the numerator). Examples: acetic acid/acetate → pH = pKa + log([CH3COO−]/[CH3COOH]); ammonium/ammonia → pH = pKa(NH4+) + log([NH3]/[NH4+]). For buffer problems on the AP, use the concentrations of the conjugate pair present (initial or effective concentrations—the Henderson-Hasselbalch form assumes the equilibrium shift is small). For a quick review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and practice buffers at Fiveable (https://library.fiveable.me/practice/ap-chemistry).

A buffer is made of a weak acid (HA) and its conjugate base (A–). When you add a little strong acid (H+), the A– reacts to form HA; when you add a little base (OH–), the HA reacts to form A–. Those neutralizing reactions change the absolute amounts but hardly change the ratio [A–]/[HA]. Since pH = pKa + log([A–]/[HA]) (Henderson–Hasselbalch), a nearly constant ratio means pH hardly moves. That’s buffer capacity: the larger the amounts of HA and A–, the more H+ or OH– it can absorb with minimal pH change. The buffering range is typically pKa ±1. Note: AP’s CED states you should understand this idea qualitatively and that calculating the exact pH change after adding acid/base won’t be tested (exclusion). For a quick review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and more Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8). For practice, try the AP problems at (https://library.fiveable.me/practice/ap-chemistry).

Use the Henderson–Hasselbalch equation and solve for pKa: pH = pKa + log([A–]/[HA]) → pKa = pH − log([A–]/[HA]). So if you’re given pH and the concentrations of the conjugate base [A–] and acid [HA], just plug them in. Example: pH = 4.76, [A–] = 0.050 M, [HA] = 0.10 M → ratio = 0.50, log(0.50) = −0.301, so pKa = 4.76 − (−0.301) = 5.06. Notes for AP: this is exactly what Topic 8.9.A expects—identify pH from pKa and concentration ratio (and vice versa). You don’t need to derive the equation on the exam (that’s excluded), just use it correctly. For more examples and a quick study guide, see the Henderson–Hasselbalch study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW). For extra practice, try AP-style problems on Fiveable (https://library.fiveable.me/practice/ap-chemistry).

Good question—the log in the Henderson–Hasselbalch equation makes the chemistry easy to read and use. Ka is usually a very small number, so chemists take its negative log to get pKa (a convenient number). Algebraically, when you start from the equilibrium expression for a weak acid, you end up with pH = pKa + log([A-]/[HA]). The log converts the ratio [A-]/[HA] (which multiplies into Ka) into an additive term, so pH depends on pKa plus how much base vs. acid you have. Because log is base-10, a change of 1 in that log term means a tenfold change in the ratio—so the equation shows directly how big a shift in concentrations is needed to change pH. That’s why buffers resist pH change: small additions change [A-]/[HA] only a little, so pH barely moves. Remember: the AP CED gives pH = pKa + log([A-]/[HA]) and says you should be able to identify buffer pH from identity and concentrations (deriving the equation isn’t tested). For a quick review, check the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW). For more practice, see the Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8) and AP practice problems (https://library.fiveable.me/practice/ap-chemistry).

A buffer is made from a weak acid and its conjugate base (or a weak base and its conjugate acid). To identify the pair, look for two species that differ by one H+: - If you see HA and A– (same formula except A– has one fewer H), that’s the acid/conjugate base pair (e.g., CH3COOH / CH3COO–). - If you see B and BH+, that’s the base/conjugate acid pair (e.g., NH3 / NH4+). For buffer pH use Henderson–Hasselbalch: pH = pKa + log([A–]/[HA]). So pick the weak acid (HA) and its conjugate base (A–) and plug their concentrations. If you’re given a salt (like NaA), that’s the source of A–. Remember AP focus: you’ll be asked to identify pH from identity and concentrations of the conjugate pair (CED 8.9.A), not to derive the equation or compute pH-changes after additions. For a quick review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and more unit resources (https://library.fiveable.me/ap-chemistry/unit-8). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).

If [A–] = [HA], the ratio [A–]/[HA] = 1 so log(1) = 0. Plugging into the Henderson–Hasselbalch equation gives pH = pKa. In practice that means a buffer made from equal concentrations of a weak acid and its conjugate base has a pH equal to the acid’s pKa—this is the midpoint of the buffer’s effective range and is also the half-equivalence point in a titration. Small additions of acid or base change the ratio only a little, so the pH stays close to pKa (buffering behavior described in the CED, EQN: pH = pKa + log([A–]/[HA])). The AP exam won’t ask you to derive the equation, but you should be able to identify pH from pKa and the [A–]/[HA] ratio. For a focused review, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and try practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Buffers resist pH changes because they contain a conjugate acid-base pair (HA and A−) that can neutralize added H+ or OH−. The Henderson–Hasselbalch equation, pH = pKa + log([A−]/[HA]), makes this clear: pH depends on the ratio [A−]/[HA], not the absolute amounts. If you add a small amount of acid (H+), A− converts to HA, changing numerator and denominator only a little, so the log term (and thus pH) changes very little. Likewise, adding base converts some HA to A−. When [A−] = [HA], pH = pKa—the buffer is most effective near the pKa (buffering range ≈ pKa ± 1). Buffer capacity (how much acid/base it can absorb) increases with higher concentrations of both components. Note: the AP CED says you don’t need to derive Henderson–Hasselbalch or compute pH change after additions (see Topic 8.9). For a concise study guide, check the Fiveable topic page (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Use the Henderson–Hasselbalch equation: pH = pKa + log([A–]/[HA]). To compare buffers: - Identify the weak acid’s pKa (from table or problem). - Compare the ratio [A–]/[HA]. Larger ratio → higher pH; smaller ratio → lower pH. - If [A–]/[HA] = 1 → pH = pKa. - If [A–]/[HA] = 10 → pH = pKa + 1. - If [A–]/[HA] = 0.1 → pH = pKa – 1. - If two buffers use the same acid (same pKa), the one with the larger [A–]/[HA] has the higher pH. If pKa differs, pKa shifts the baseline: a lower pKa → lower pH for the same ratio. Remember buffering is effective near pKa (about ±1 pH unit). The AP CED focuses on identifying buffer pH from pKa and the conjugate ratio (8.9.A); you won’t be asked to derive the equation. For extra examples and practice, see the Topic 8.9 study guide (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW) and unit review (https://library.fiveable.me/ap-chemistry/unit-8). For lots of practice problems, check (https://library.fiveable.me/practice/ap-chemistry).

Use Henderson-Hasselbalch (HH) when you have a buffer: a weak acid (HA) and its conjugate base (A–) both present in appreciable amounts. Then pH ≈ pKa + log([A–]/[HA]) (CED 8.9.A.1). HH is fast and gives pH from the concentration ratio—it relies on the equilibrium being only a small shift (so [HA] and [A–] don’t change much). Good rule: total concentrations >> Ka and neither species is nearly zero. Use the full acid–base equilibrium (Ka with an ICE table) when: - you don’t have both conjugate pair components (just a weak acid or just its salt), - concentrations are very dilute or comparable to Ka, or - one species is present in trace amount so the HH approximation breaks down. Remember: AP won’t ask you to derive HH or compute pH change after adding acid/base to buffers (exclusions), but you should identify buffer situations and use pH = pKa + log([A–]/[HA]). For the topic study guide, see (https://library.fiveable.me/ap-chemistry/unit-8/henderson-hasselbalch-equation/study-guide/9jNGg5JlJAYF5QilwfJW). For extra practice, check Fiveable’s AP Chem problems (https://library.fiveable.me/practice/ap-chemistry).