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Unit 8 · 11–15% of the exam

Acids and Bases

Unit 8 applies equilibrium to proton transfer. You will calculate pH for strong and weak acids and bases, read titration curves, and explain how buffers resist pH change. It also links acid strength to molecular structure, and it is one of the most heavily tested units on the exam.

Titration simulator

Drip base into acid, plot the pH curve live, and find the equivalence and half-equivalence points.

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Topics in this unit

  1. 8.1

    Introduction to Acids and Bases

    pH = −log[H3O+], and in water at 25 °C, Kw = [H3O+][OH−] = 1.0 × 10−14, so pH + pOH = 14.

  2. 8.2

    pH and pOH of Strong Acids and Bases

    Strong acids and bases ionize completely, so [H3O+] or [OH−] comes straight from the concentration and stoichiometry.

  3. 8.3

    Weak Acid and Base Equilibria

    Weak acids ionize only partially; use Ka (or Kb) with an ICE table, often [H3O+] ≈ √(Ka × C).

  4. 8.4

    Acid-Base Reactions and Buffers

    Reactions between acids and bases go essentially to completion when one is strong; mixing a weak acid with its conjugate base makes a buffer.

  5. 8.5

    Acid-Base Titrations

    Titration curves show the buffer region, the half-equivalence point (pH = pKa for a weak acid) and the equivalence point, whose pH depends on the species present.

  6. 8.6

    Molecular Structure of Acids and Bases

    Acid strength increases when the conjugate base is more stable: more electronegative atoms, more O atoms (oxyacids) or resonance delocalization of the negative charge.

  7. 8.7

    pH and pKa

    When pH < pKa the protonated form (HA) predominates; when pH > pKa the deprotonated form (A−) predominates.

  8. 8.8

    Properties of Buffers

    A buffer resists pH change because its weak acid neutralizes added OH− and its conjugate base neutralizes added H3O+.

  9. 8.9

    Henderson-Hasselbalch Equation

    pH = pKa + log([A−]/[HA]) gives the pH of a buffer from the ratio of base to acid.

  10. 8.10

    Buffer Capacity

    A buffer's capacity grows with the amounts of HA and A− present; it resists both added acid and added base best when [HA] ≈ [A−].

Must know

  • pH = −log[H3O+]; pOH = −log[OH−]; pH + pOH = 14.00 at 25 °C
  • Kw = [H3O+][OH−] = 1.0 × 10−14 at 25 °C, and Ka × Kb = Kw for a conjugate pair
  • Weak acid: Ka = [H3O+][A−]/[HA]; [H3O+] ≈ √(Ka × C) when ionization is small
  • pKa = −log Ka; a smaller pKa means a stronger acid
  • Henderson–Hasselbalch: pH = pKa + log([A−]/[HA])
  • Half-equivalence point of a weak acid titration: pH = pKa
  • Equivalence pH: strong acid + strong base = 7; weak acid + strong base > 7; weak base + strong acid < 7
  • Strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4; strong bases: Group 1 hydroxides and Ca(OH)2, Sr(OH)2, Ba(OH)2

Common mix-ups

  • A strong acid and a concentrated acid mean the same thing.

    Strength is how completely an acid ionizes; concentration is how much is dissolved. Dilute HCl is still strong, and concentrated acetic acid is still weak.

  • The equivalence point of every titration is at pH 7.

    Only strong acid–strong base titrations are neutral at equivalence; a weak acid titrated with NaOH has a basic equivalence point because A− is a base.

  • A buffer keeps the pH completely constant.

    A buffer only minimizes pH change, and only until one of its components is used up (its capacity is exceeded).

Practice questions

AP-style multiple choice. Every answer comes with the reasoning, including why the tempting wrong answer is wrong.

1 / 6

What is the pH of 0.0050 M Ba(OH)2(aq) at 25 °C?

Free-response practice

25.0 mL of 0.100 M CH3COOH (Ka = 1.8 × 10−5) is titrated with 0.100 M NaOH. (a) Calculate the initial pH. (b) What volume of NaOH is needed to reach the equivalence point? (c) What is the pH after 12.5 mL of NaOH has been added? (d) Calculate the pH at the equivalence point.

Flashcards

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See it in real life

Everyday reactions that use Unit 8 ideas, animated atom by atom.