8.4 Acid-Base Reactions and Buffers

Strong acid + strong base: they react completely: H+(aq) + OH-(aq) → H2O(l). After mixing, only the excess strong reagent determines pH (calculate moles, divide by total volume). No conjugate pairs or buffer form. This is what the CED calls quantitative neutralization (8.4.A.1) and is tested on the exam by asking you to find pH from excess H+ or OH-. Weak acid + strong base: they also react quantitatively, but the weak acid (HA) is only partly ionized, so the product is its conjugate base (A−): HA + OH− ⇌ A− + H2O (8.4.A.2). If HA is in excess you get a buffer (HA/A−) and use the Henderson–Hasselbalch equation (pH = pKa + log([A−]/[HA])). If strong base is in excess, find pH from excess OH−. If equimolar, you must treat A− hydrolysis (A− + H2O ⇌ HA + OH−) to get a slightly basic pH. Know Ka/pKa, buffer capacity, and common-ion effects for AP problems. For a focused study and examples, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

You have a buffer when the solution contains a significant amount of a weak acid and its conjugate base (or a weak base and its conjugate acid) at the same time so they can neutralize added OH– or H3O+. Practically, that happens after reaction if: - you mix a weak acid (HA) with a strong base and HA is in excess → leftover HA + produced A– form a buffer (CED 8.4.A.2); or - you mix a weak base (B) with a strong acid and B is in excess → leftover B + produced HB+ form a buffer (CED 8.4.A.3). If the strong reagent is in excess, no buffer forms—pH is set by the excess H3O+ or OH–. If they’re exactly equimolar you don’t get a full buffer; you get mostly the conjugate species and must treat hydrolysis equilibria (slightly acidic/basic pH). When a buffer exists you can use the Henderson–Hasselbalch equation to find pH (see CED and Topic 8.4 study guide: https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh). For extra practice, try the AP-style problems at (https://library.fiveable.me/practice/ap-chemistry).

Use the Henderson–Hasselbalch (H–H) equation when you have a buffer—that means a conjugate acid/base pair both present in significant amounts (a weak acid HA and its conjugate base A−, or a weak base B and its conjugate acid HB+). The equation pH = pKa + log([A−]/[HA]) lets you find pH quickly from the concentrations of the major species without doing an ICE table. When to use it in AP Chem (CED 8.4.A.2–3): - After mixing a weak acid with a strong base when the weak acid is in excess (buffer formed). - After mixing a weak base with a strong acid when the weak base is in excess. - At the half-equivalence point in a titration (where [HA] = [A−], pH = pKa). Don’t use H–H if one component is negligible (strong acid/base excess) or if you need high precision when concentrations are extremely dilute. For guided examples and AP-style practice, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and the unit page (https://library.fiveable.me/ap-chemistry/unit-8). For more practice problems, use Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).

If you mix equal moles of a weak acid HA and a strong base OH−, the OH− completely deprotonates HA to give its conjugate base A−. At equivalence (equimolar), you end up with a solution of A− (no HA left), so it’s not a buffer—it’s a basic salt solution (CED 8.4.A.2). The pH is determined by hydrolysis of A−: A− + H2O ⇌ HA + OH− with Kb = Kw/Ka. If the concentration of A− after mixing is c, then [OH−] ≈ sqrt(Kb·c) = sqrt((Kw/Ka)·c) (use ICE to justify). Then pOH = −log[OH−], pH = 14 − pOH. For example, smaller Ka (weaker acid) → larger Kb → more OH− → higher pH. Remember Henderson-Hasselbalch applies when both HA and A− are present (a buffer), not at this equimolar point. For more review on Topic 8.4, see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Because H+ (from a strong acid) and OH− (from a strong base) form H2O, the equilibrium lies essentially all the way to products—the reaction is quantitative. Water is a very, very weak acid/base: its autoionization constant Kw = [H+][OH−] ≈ 1.0×10^−14 (25 °C). That tiny value means the reverse reaction H2O → H+ + OH− is vanishingly small under normal conditions, so H+ + OH− → H2O is treated as effectively irreversible (CED 8.4.A.1). Other acid–base pairs (weak acids or weak bases) produce conjugates with appreciable tendencies to re-donate or re-accept protons, so the forward and reverse reactions have comparable rates and an equilibrium with a finite K (see 8.4.A.2–8.4.A.4). In those cases you must use Ka, Kb, or the Henderson–Hasselbalch equation to find pH and species concentrations. For more practice and review of these CED ideas, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and the AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

“Quantitatively” means you track actual amounts (moles/concentrations) and use stoichiometry + equilibrium to find exact concentrations and pH after reaction—not just saying “they neutralize.” For AP Chem Topic 8.4 that means: - For strong acid + strong base: moles H+ react 1:1 with OH− to form H2O; calculate moles of limiting reagent, find moles of excess, divide by total volume to get [H+] or [OH−], then pH (CED 8.4.A.1). - For weak acid + strong base (or weak base + strong acid): do mole ICE/limiting-reagent work. If both conjugate pair remain (buffer), use Henderson–Hasselbalch to get pH (CED 8.4.A.2–3). If one is in large excess, use excess moles to find [H3O+] or [OH−]. - For weak acid + weak base: set up equilibrium with Ka or Kb and solve (CED 8.4.A.4). On the exam you’ll be asked to calculate pH from concentrations, identify excess reagent, or set up equilibrium—so practice mole tables and H–H problems (Fiveable Topic 8.4 study guide: https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh; practice problems: https://library.fiveable.me/practice/ap-chemistry).

Short answer: do a mole (stoichiometry) check, find the excess H+ or OH−, convert to concentration using total volume, then calculate pH (or pOH). This is exactly what EK 8.4.A.1–A.3 says for strong/weak mixes. Steps you can follow: 1. Write the neutralization: H+ + OH− → H2O. 2. Calculate moles of acid and base (M × L). Subtract the smaller from the larger to get moles of excess reagent. 3. Divide excess moles by total solution volume to get [H+] if acid is in excess, or [OH−] if base is in excess. 4. pH = −log[H+]. If you have [OH−], pOH = −log[OH−] and pH = 14 − pOH (or use Kw at other T if needed). 5. If a weak acid/base is in excess, treat the mixture as a buffer and use Henderson–Hasselbalch (or, if strong reagent is in excess, use the excess moles method above). For equimolar weak+strong conjugate cases use the appropriate equilibrium (A− + H2O ⇌ HA + OH− or HB+ + H2O ⇌ B + H3O+). This aligns with AP CED Topic 8.4 guidance—see the Topic 8.4 study guide for examples (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh). For lots of practice problems, check Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).

A- and HB+ are just the conjugate base and conjugate acid of a weak acid/base pair, but they behave oppositely in water: - A- acting as a base: A- + H2O ⇌ HA + OH-. Here A- accepts a proton from water (Bronsted–Lowry base), producing OH−. This equilibrium is governed by Kb (or by Ka of HA via Kb = Kw/Ka). If A− is present with HA (buffer) you use the Henderson–Hasselbalch equation to find pH (Topic 8.4 and 8.9 in the CED). - HB+ acting as an acid: HB+ + H2O ⇌ B + H3O+. Here HB+ donates a proton to water (Bronsted–Lowry acid), producing H3O+. The equilibrium is governed by Ka for HB+ (or by Kb of B). Which direction matters depends on what's in excess: when A− and HA are equimolar the solution is slightly basic (A− hydrolyzes); when HB+ and B are equimolar the solution is slightly acidic (HB+ hydrolyzes). For more examples and practice, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and hundreds of practice problems (https://library.fiveable.me/practice/ap-chemistry).

If both reactants are weak (HA + B), they don’t fully donate or accept protons—their tendency to transfer H+ is limited. So the proton transfer HA + B ⇌ A− + HB+ stops before going to completion because the forward and reverse rates become equal. The position of that equilibrium depends on the acid/base strengths (Ka of HA and Kb of B)—you can combine those to get the equilibrium constant for HA + B ⇌ A− + HB+ (see CED 8.4.A.4). If HA is much stronger than HB+ (or B much stronger than A−), the equilibrium lies to the right; if not, it lies to the left. That’s why you can’t assume “complete neutralization” like with strong acids/bases (CED 8.4.A.1–A.3). For more on predicting pH and using Ka/Kb and Henderson–Hasselbalch when buffers form, check the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8). Practice problems on Fiveable will help you apply Ka/Kb to specific mixes (https://library.fiveable.me/practice/ap-chemistry).

Use total volume whenever you need a molarity (concentration) for an equilibrium or pH calculation. Quick rules tied to AP CED 8.4.A: - Strong acid + strong base with excess reagent: find moles of excess H3O+ or OH–, then divide by total volume to get [H3O+] or [OH–] → pH or pOH. (You must use total volume.) - Weak acid + strong base (or weak base + strong acid): do the reaction with moles first to find what's left and what's formed. If you end up with a buffer (both HA and A– present), use Henderson-Hasselbalch with concentrations—the dilution cancels, so you can use moles ratio (no need to compute total volume if both species were diluted equally). (CED 8.4.A.2–3) - If strong reagent is in excess or at equivalence with only conjugate present, convert final moles to concentration using total volume and then use Ka/Kb equilibrium or ICE tables. For practice, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and try problems from Fiveable’s practice set (https://library.fiveable.me/practice/ap-chemistry).

Short answer: compare moles of strong species and whether acids/bases are weak or strong. Steps to predict pH (use CED 8.4.A rules): 1. Strong acid + strong base → they neutralize: find moles of H+ and OH−, subtract to get excess, divide by total volume to get [H3O+] or [OH−], then pH or pOH (8.4.A.1). 2. Weak acid (HA) + strong base → do stoichiometry. If HA in excess → buffer (use Henderson–Hasselbalch with pKa and [HA]/[A−]) (8.4.A.2). If strong base in excess → treat excess OH− (calculate pOH). If equimolar → solve base hydrolysis of A− (slightly basic). 3. Weak base + strong acid → analogous (8.4.A.3): buffer if base excess; excess H3O+ if strong acid excess; equimolar gives slightly acidic salt hydrolysis. 4. Weak acid + weak base → set up equilibrium using Ka/Kb or net K (8.4.A.4). For worked examples and AP-style practice, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and thousands of practice problems (https://library.fiveable.me/practice/ap-chemistry).

“Slightly acidic” or “slightly basic” at the equivalence point means the pH is not 7 (neutral) because the salt formed from the titration hydrolyzes in water. For a weak acid titrated with a strong base (CED 8.4.A.2), at the equivalence point all HA → A−. The A− is a weak base: A− + H2O ⇌ HA + OH−, so you get a small [OH−] and pH > 7 (slightly basic). For a weak base titrated with a strong acid (CED 8.4.A.3), the conjugate acid HB+ hydrolyzes: HB+ + H2O ⇌ B + H3O+, producing a small [H3O+] and pH < 7 (slightly acidic). “Slightly” means the pH shift is modest because the hydrolysis equilibrium lies mostly left—you can estimate pH by using Kb = Kw/Ka (or Ka = Kw/Kb) and solving an equilibrium for [OH−] or [H3O+]. For worked examples and AP-style practice, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Saying “reaction is complete” ignores what’s left in solution—and for AP Chem that leftover (the excess reagent) controls the pH. The CED explicitly tells you: when strong acid + strong base react quantitatively, pH is set by the concentration of the excess H+ or OH− (8.4.A.1). For weak acid + strong base (or vice versa) you either: (a) get a buffer if the weak species is in excess and must use Henderson–Hasselbalch (8.4.A.2–3), or (b) calculate pH from the moles of excess strong acid/base divided by total volume if the strong reagent is in excess. In short: reaction stoichiometry tells you what’s consumed; excess concentrations tell you what remains and therefore the solution’s pH—exactly what the exam asks you to calculate. For more practice and step-by-step examples, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and extra problems at (https://library.fiveable.me/practice/ap-chemistry).

Titration curves are just pH tracking the chemical reactions the CED describes. Key links to spot on a curve: - Strong acid + strong base: H+ + OH- → H2O. The curve is very steep at the equivalence point, and pH ≈ 7 there (because neither conjugate weak species affects pH). - Weak acid + strong base: HA + OH- → A- + H2O. Before the equivalence point you get a buffer region (mixture of HA and A-)—pH is given by Henderson–Hasselbalch; at the half-equivalence point pH = pKa. The equivalence point is >7 because A- hydrolyzes to produce OH-. - Weak base + strong acid: analogous; half-equivalence gives pOH = pKb (or pH = pKa of conjugate), equivalence pH < 7. - Weak acid + weak base: curve is shallow; position depends on Ka and Kb (salt hydrolysis/equilibrium). Also note buffer capacity: the flatter the buffer region, the more moles of conjugate pair you can add without big pH change. For AP exam work, be ready to: identify equivalence/half-equivalence points, use H–H, calculate excess reagent pH from remaining [H+] or [OH–], and explain why pH shifts differ for strong vs weak species. Review Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Write the reaction HA + B ⇌ A– + HB+. The equilibrium constant K for that proton-transfer is K = [A–][HB+]/[HA][B]. You can get K from acid dissociation constants: K = Ka(HA) / Ka(HB+). (Derivation: Ka(HA) = [H+][A–]/[HA] and Ka(HB+) = [H+][B]/[HB+]; dividing cancels [H+] and gives the expression above.) In pKa form: log K = pKa(HB+) – pKa(HA), so K = 10^[pKa(HB+) – pKa(HA)]. So if pKa(HA) < pKa(HB+) (i.e., HA is the stronger acid), K > 1 and products are favored; if pKa(HA) > pKa(HB+), reactants are favored. This is exactly what Topic 8.4 (CED 8.4.A.4) expects—use Ka/pKa to predict equilibrium position. For a quick refresher, see the Topic 8.4 study guide (https://library.fiveable.me/ap-chemistry/unit-8/acid-base-reactions-buffers/study-guide/aXiB6ONME0VEX1JR9Kwh) and practice problems (https://library.fiveable.me/practice/ap-chemistry).