AP Chemistry: Acid?Base Equilibria (Ka, Kb, Kw, pH, pOH, buffers)

AP Chemistry – Acid?Base Equilibria (Ka, Kb, Kw, pH, pOH, buffers)

AP Chemistry Study Guide: Acid-Base Equilibria (Ka, Kb, Kw, pH, pOH, Buffers)

What This Is

Acid-base equilibria explain how acids and bases behave in water, including their strength, pH, and how they resist pH changes (buffers). This topic is essential for the AP Chemistry exam—expect multiple-choice questions (MCQs) and free-response questions (FRQs) on calculating pH, interpreting titration curves, and designing buffers. Real-world example: Antacids (like Tums) neutralize stomach acid (HCl) using weak bases (CaCO?), demonstrating acid-base reactions in action.

Key Terms & Concepts

• Arrhenius Acid/Base: An acid produces H? in water; a base produces OH?.

• Example: HCl (acid)-H? + Cl?; NaOH (base)-Na? + OH?.

• Brønsted-Lowry Acid/Base: An acid donates H?; a base accepts H?.

• Example: NH? (base) + H?O (acid)-NH + OH?.

• Ka (Acid Dissociation Constant): Ka = [H?][A?] / [HA]

• Measures acid strength (larger Ka = stronger acid).

• Example: Acetic acid (CH?COOH) has Ka = 1.8 × 10.

• Kb (Base Dissociation Constant): Kb = [BH?][OH?] / [B]

• Measures base strength (larger Kb = stronger base).

• Example: Ammonia (NH?) has Kb = 1.8 × 10.

• Kw (Ionization Constant of Water): Kw = [H?][OH?] = 1.0 × 10?¹? at 25°C

• Relates H? and OH? concentrations in any aqueous solution.

• pH and pOH: pH = -log[H?] pOH = -log[OH?] pH + pOH = 14 (at 25°C).

• Example: If [H?] = 1 × 10?³ M, pH = 3.

• Strong vs. Weak Acids/Bases:

• Strong acids/bases dissociate 100% in water (e.g., HCl, NaOH).

• Weak acids/bases partially dissociate (e.g., CH?COOH, NH?).

• Buffer Solution: A mixture of a weak acid + its conjugate base (or weak base + conjugate acid) that resists pH changes.

• Example: CH?COOH + CH?COO? (acetic acid + acetate ion).

• Henderson-Hasselbalch Equation: pH = pKa + log([A?]/[HA])

• Used for buffer calculations (only valid when [HA] and [A?] are large).

• Titration Curve: A graph of pH vs. volume of titrant added (used to find equivalence point).

• Key points: Half-equivalence point (pH = pKa), equivalence point (pH depends on salt formed).

Step-by-Step / Process Flow

1. Calculating pH of a Strong Acid/Base

1. Write dissociation equation (e.g., HCl-H? + Cl?).
2. Determine [H?] or [OH?] (since strong acids/bases dissociate 100%, [H?] = initial acid concentration).
3. Calculate pH or pOH using pH = -log[H?] or pOH = -log[OH?].
4. Convert pOH to pH (if needed) using pH + pOH = 14.

Example: What is the pH of 0.01 M HCl? - [H?] = 0.01 M-pH = -log(0.01) = 2.

2. Calculating pH of a Weak Acid/Base

1. Write dissociation equation (e.g., CH?COOH-H? + CH?COO?).
2. Set up ICE table (Initial, Change, Equilibrium).
3. Use Ka or Kb expression to solve for [H?] or [OH?].
4. Calculate pH (may require quadratic formula if Ka is large).

Example: What is the pH of 0.1 M CH?COOH (Ka = 1.8 × 10)? - ICE table-[H?] = ?(Ka × [HA]) = ?(1.8 × 10 × 0.1)-1.34 × 10?³ M. - pH = -log(1.34 × 10?³)-2.87.

3. Designing a Buffer

1. Choose a weak acid/base pair (e.g., CH?COOH/CH?COO?).
2. Use Henderson-Hasselbalch equation to find ratio of [A?]/[HA].
3. Adjust concentrations to get desired pH (pH-pKa if [A?] = [HA]).

Example: How much NaCH?COO must be added to 0.1 M CH?COOH to make a pH 5 buffer (pKa = 4.74)? - pH = pKa + log([A?]/[HA])-5 = 4.74 + log([A?]/0.1). - log([A?]/0.1) = 0.26-[A?]/0.1 = 10?·²?-1.82-[A?]-0.182 M.

4. Interpreting a Titration Curve

1. Identify equivalence point (steepest part of curve).
2. Determine pH at equivalence point (strong acid + strong base-pH = 7; weak acid + strong base-pH > 7).
3. Find half-equivalence point (pH = pKa of weak acid/base).

Example: Titration of 50 mL 0.1 M CH?COOH with 0.1 M NaOH. - Equivalence point: pH > 7 (basic salt formed). - Half-equivalence point: pH = pKa = 4.74.

Common Mistakes

• Mistake: Assuming all acids/bases are strong. Correction: Only HCl, HBr, HI, HNO?, H?SO?, HClO? (acids) and Group 1/2 hydroxides (bases) are strong. Everything else is weak!

• Mistake: Forgetting that Kw = [H?][OH?] = 1 × 10?¹? applies to all aqueous solutions. Correction: Even in acidic/basic solutions, [H?] and [OH?] are related by Kw.

• Mistake: Using Henderson-Hasselbalch for strong acids/bases. Correction: Henderson-Hasselbalch only works for buffers (weak acid + conjugate base).

• Mistake: Ignoring dilution when mixing solutions. Correction: Always recalculate concentrations after mixing (e.g., adding water to a buffer changes [HA] and [A?]).

• Mistake: Confusing Ka and Kb for conjugate pairs. Correction: For a conjugate acid-base pair, Ka × Kb = Kw (e.g., Ka of CH?COOH × Kb of CH?COO? = 1 × 10?¹?).

AP Exam Insights

• MCQ Traps:
• Trick: Giving a weak acid’s Ka and asking for pH of its salt (e.g., pH of NaCH?COO). Solution: The salt’s pH depends on the conjugate base’s Kb (Kb = Kw/Ka).

• Trick: Asking for pH of a diluted strong acid (e.g., 1 × 10 M HCl). Solution: At very low concentrations, water’s autoionization dominates (pH-6.98, not 8!).

• FRQ Trends:

• Buffer design: Expect a question like, “How would you prepare a pH 4.7 buffer using acetic acid and sodium acetate?”
• Titration curves: You may need to sketch a curve and label key points (equivalence, half-equivalence).

• Ka/Kb calculations: Often paired with ICE tables or percent dissociation.

• Key Distinctions:

• Strong vs. weak acids/bases: Strong = 100% dissociation; weak = partial dissociation.
• pH vs. pKa: pH measures solution acidity; pKa measures acid strength.
• Buffer capacity: A buffer works best when pH-pKa and concentrations of HA/A? are high and equal.

Quick Check Questions

1. Multiple Choice

What is the pH of a 0.05 M solution of NH? (Kb = 1.8 × 10)? (A) 3.0 (B) 5.3 (C) 8.7 (D) 11.0

Answer: (D) 11.0 Explanation: NH? is a weak base-[OH?] = ?(Kb × [NH?])-9.5 × 10 M-pOH-3.0-pH = 14 - 3 = 11.0.

2. Short FRQ

A student titrates 25.0 mL of 0.10 M HCOOH (Ka = 1.8 × 10) with 0.10 M NaOH. (a) What is the pH at the half-equivalence point? (b) What is the pH at the equivalence point?

Answers: (a) pH = pKa = 3.74 (at half-equivalence, [HCOOH] = [HCOO?]). (b) pH > 7 (basic salt formed; exact pH requires Kb calculation for HCOO?).

Last-Minute Cram Sheet

1. Kw = [H?][OH?] = 1 × 10?¹? at 25°C.
2. pH = -log[H?]; pOH = -log[OH?]; pH + pOH = 14.
3. Strong acids/bases dissociate 100%; weak acids/bases use Ka/Kb.
4. Henderson-Hasselbalch: pH = pKa + log([A?]/[HA]). Only for buffers!
5. Ka × Kb = Kw for conjugate acid-base pairs.
6. Buffer capacity is highest when pH-pKa and [HA] = [A?].
7. Equivalence point pH: strong acid + strong base = 7; weak acid + strong base > 7.
8. Half-equivalence point pH = pKa of weak acid/base.
9. Dilution changes [H?] but not Ka/Kb. Don’t assume pH stays the same!
10. Autoionization of water matters at very low [H?]/[OH?] (e.g., 1 × 10 M HCl).