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Study Guide: How to Solve: Acids and Bases Problems
Source: https://www.fatskills.com/k-12-assessment-tests/chapter/how-to-solve-acids-and-bases-problems

How to Solve: Acids and Bases Problems

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~7 min read

How to Solve: Acids and Bases Problems

For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script


Introduction

"Master acids and bases, and you’ll predict chemical reactions, calculate pH in seconds, and crush every titration question on your exam—no more guessing!


What You Need To Know First

Before diving into acids and bases, ensure you understand: 1. Mole concept & stoichiometry – Calculating moles, concentrations, and balanced equations. 2. Logarithms (base 10) – Used in pH calculations (e.g., pH = -log[H⁺]). 3. Equilibrium basics – Weak acids/bases don’t fully dissociate; they reach equilibrium.

If any of these are shaky, review them first—this guide assumes you’re solid on them.


Key Vocabulary

Term Plain-English Definition Quick Example
Acid A substance that donates H⁺ ions (protons) in solution. HCl → H⁺ + Cl⁻
Base A substance that accepts H⁺ ions or donates OH⁻ ions in solution. NaOH → Na⁺ + OH⁻
pH A measure of how acidic or basic a solution is (0-14 scale). pH 2 = strong acid; pH 7 = neutral; pH 12 = strong base
Strong acid/base Fully dissociates in water (100% ions). HCl, H₂SO₄, NaOH, KOH
Weak acid/base Partially dissociates in water (equilibrium). CH₃COOH (acetic acid), NH₃ (ammonia)
Titration A lab technique to find an unknown concentration by reacting it with a known solution. Using NaOH to find the concentration of HCl.

Formulas To Know

Formula Variables Memorise?
pH = -log[H⁺] [H⁺] = concentration of H⁺ ions (mol/L) MEMORISE THIS
[H⁺] = 10⁻ᵖᴴ pH = pH value MEMORISE THIS
pOH = -log[OH⁻] [OH⁻] = concentration of OH⁻ ions (mol/L) MEMORISE THIS
[OH⁻] = 10⁻ᵖᴼᴴ pOH = pOH value MEMORISE THIS
pH + pOH = 14 At 25°C (room temperature) MEMORISE THIS
Kₐ = [H⁺][A⁻] / [HA] Kₐ = acid dissociation constant, [HA] = weak acid, [A⁻] = conjugate base Given on exam sheet
K_b = [BH⁺][OH⁻] / [B] K_b = base dissociation constant, [B] = weak base, [BH⁺] = conjugate acid Given on exam sheet
Kₐ × K_b = K_w = 1 × 10⁻¹⁴ K_w = ionisation constant of water (at 25°C) Given on exam sheet
C₁V₁ = C₂V₂ C = concentration (mol/L), V = volume (L or mL) – used in dilutions/titrations MEMORISE THIS

Step-by-Step Method

How to Solve Any Acid/Base Problem

Follow these steps in order for every question:

  1. Identify the type of problem
  2. Is it a pH calculation? (Strong/weak acid/base)
  3. Is it a titration? (Finding unknown concentration)
  4. Is it a buffer solution? (Weak acid + conjugate base)
  5. Is it a Kₐ/K_b calculation? (Dissociation constants)

  6. Write the balanced equation

  7. For strong acids/bases: Full dissociation (e.g., HCl → H⁺ + Cl⁻).
  8. For weak acids/bases: Equilibrium (e.g., CH₃COOH ⇌ H⁺ + CH₃COO⁻).

  9. List knowns and unknowns

  10. Write down given values (concentrations, volumes, pH, Kₐ, etc.).
  11. Circle what you need to find.

  12. Choose the right formula

  13. Strong acid/base? Use pH = -log[H⁺] or pOH = -log[OH⁻].
  14. Weak acid/base? Use Kₐ or K_b and ICE tables (Initial, Change, Equilibrium).
  15. Titration? Use C₁V₁ = C₂V₂ (moles of acid = moles of base at equivalence point).

  16. Solve step-by-step

  17. Show all working (examiners give marks for method, not just answers).
  18. Check units (mol/L for concentration, L or mL for volume).

  19. Check your answer

  20. Does the pH make sense? (Acid = pH < 7, Base = pH > 7)
  21. Are units correct?
  22. Did you round too early? (Keep 3-4 sig figs until the end.)

Worked Example (Using the Steps Above)

Question: What is the pH of a 0.050 mol/L solution of HCl?

Solution:

  1. Identify the problem:
  2. Strong acid pH calculation.

  3. Write the balanced equation:

  4. HCl → H⁺ + Cl⁻ (fully dissociates).

  5. List knowns and unknowns:

  6. [HCl] = 0.050 mol/L
  7. [H⁺] = ? (same as [HCl] because strong acid)
  8. pH = ?

  9. Choose the right formula:

  10. pH = -log[H⁺]

  11. Solve step-by-step:

  12. [H⁺] = 0.050 mol/L (since HCl fully dissociates)
  13. pH = -log(0.050)
  14. pH = -(-1.30) = 1.30

  15. Check your answer:

  16. HCl is a strong acid → pH < 7 (correct).
  17. Units are correct (no units for pH).

Answer: pH = 1.30


Worked Examples

Example 1 – Basic: pH of a Strong Base

Question: What is the pH of a 0.020 mol/L NaOH solution?

Solution:

  1. Identify: Strong base pH calculation.
  2. Equation: NaOH → Na⁺ + OH⁻ (fully dissociates).
  3. Knowns:
  4. [NaOH] = 0.020 mol/L
  5. [OH⁻] = 0.020 mol/L
  6. pH = ?
  7. Formula:
  8. pOH = -log[OH⁻]
  9. pH + pOH = 14
  10. Solve:
  11. pOH = -log(0.020) = 1.70
  12. pH = 14 - 1.70 = 12.30
  13. Check:
  14. NaOH is a strong base → pH > 7 (correct).

What we did and why: - Strong bases fully dissociate, so [OH⁻] = [NaOH]. - We used pOH first because the question gave [OH⁻], then converted to pH.


Example 2 – Medium: Weak Acid pH (Using Kₐ)

Question: Calculate the pH of a 0.10 mol/L acetic acid (CH₃COOH) solution. Kₐ = 1.8 × 10⁻⁵.

Solution:

  1. Identify: Weak acid pH calculation (requires Kₐ).
  2. Equation: CH₃COOH ⇌ H⁺ + CH₃COO⁻
  3. Knowns:
  4. [CH₃COOH] = 0.10 mol/L
  5. Kₐ = 1.8 × 10⁻⁵
  6. [H⁺] = ? (not equal to [CH₃COOH] because weak acid)
  7. Formula:
  8. Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
  9. Use ICE table:
    | Species | Initial | Change | Equilibrium |
    |---------------|---------|--------|-------------|
    | CH₃COOH | 0.10 | -x | 0.10 - x |
    | H⁺ | 0 | +x | x |
    | CH₃COO⁻ | 0 | +x | x |
  10. Solve:
  11. Kₐ = x² / (0.10 - x) ≈ x² / 0.10 (since x is small)
  12. 1.8 × 10⁻⁵ = x² / 0.10
  13. x² = 1.8 × 10⁻⁶
  14. x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/L = [H⁺]
  15. pH = -log(1.34 × 10⁻³) = 2.87
  16. Check:
  17. Weak acid → pH < 7 but not too low (correct, since 0.10 mol/L strong acid would be pH 1).

What we did and why: - Weak acids don’t fully dissociate, so we used Kₐ and an ICE table. - We assumed x was small (0.10 - x ≈ 0.10) because Kₐ is small (check: x = 1.34 × 10⁻³ << 0.10).


Example 3 – Exam Style: Titration Calculation

Question: 25.0 mL of HCl is titrated with 0.100 mol/L NaOH. The equivalence point is reached after adding 30.0 mL of NaOH. What is the concentration of the HCl solution?

Solution:

  1. Identify: Titration (finding unknown concentration).
  2. Equation: HCl + NaOH → NaCl + H₂O (1:1 mole ratio).
  3. Knowns:
  4. V_HCl = 25.0 mL = 0.0250 L
  5. V_NaOH = 30.0 mL = 0.0300 L
  6. C_NaOH = 0.100 mol/L
  7. C_HCl = ?
  8. Formula:
  9. Moles of NaOH = C × V = 0.100 × 0.0300 = 0.00300 mol
  10. At equivalence point: moles of HCl = moles of NaOH
  11. C_HCl × V_HCl = moles of HCl
  12. C_HCl × 0.0250 = 0.00300
  13. Solve:
  14. C_HCl = 0.00300 / 0.0250 = 0.120 mol/L
  15. Check:
  16. Units are mol/L (correct).
  17. Makes sense: more NaOH volume needed → HCl was less concentrated than NaOH.

What we did and why: - Titrations use C₁V₁ = C₂V₂ (or moles = moles at equivalence). - We converted mL to L because concentration is in mol/L. - The 1:1 ratio means moles of acid = moles of base at equivalence.


Common Mistakes

Mistake Why It Happens Correct Approach
Forgetting strong vs. weak Assuming all acids/bases fully dissociate. Strong = full dissociation (HCl, NaOH). Weak = partial (CH₃COOH, NH₃).
Using [H⁺] = [acid] for weak acids Treating weak acids like strong acids. Use Kₐ and ICE table for weak acids.
Ignoring units (mL vs. L) Mixing mL and L in calculations. Convert all volumes to litres before using in formulas.
Rounding too early Rounding intermediate steps (e.g., pH = -log(0.05) = 1.3, not 1). Keep 3-4 sig figs until the final answer.
Misapplying pH + pOH = 14 Using it for non-25°C solutions or strong acids/bases only. Only applies at 25°C and for aqueous solutions.

Exam Traps

Trap How to Spot It How to Avoid It
"Diprotic" or "triprotic" acids Question mentions H₂SO₄, H₃PO₄, or "diprotic acid." Remember: First proton fully dissociates (e.g., H₂SO₄ → H⁺ + HSO₄⁻), second is weak.
Buffer solutions in disguise Question gives a weak acid + its salt (e.g., CH₃COOH + CH₃COONa). Use the Henderson-Hasselbalch equation: pH = pKₐ + log([A⁻]/[HA]).
Tricky units (e.g., % dissociation) Asks for "% dissociation" instead of pH. % dissociation = ([H⁺] / [initial acid]) × 100.

1-Minute Recap

"Alright, let’s lock this in—here’s what you must remember for acids and bases:

  1. Strong vs. weak:
  2. Strong acids/bases fully dissociate (HCl, NaOH).
  3. Weak acids/bases partially dissociate (CH₃COOH, NH₃)—use Kₐ or K_b.

  4. pH and pOH:

  5. pH = -log[H⁺], pOH = -log[OH⁻], and pH + pOH = 14 (at 25°C).
  6. If you know [H⁺], use pH = -log[H⁺]. If you know pH, use [H⁺] = 10⁻ᵖᴴ.

  7. Titrations:

  8. At equivalence point, moles of acid = moles of base.
  9. Use C₁V₁ = C₂V₂ (but convert mL to L first!).

  10. Weak acids/bases:

  11. Set up an ICE table, assume x is small, and solve for [H⁺] or [OH⁻].
  12. If Kₐ is given, use it—don’t assume [H⁺] = [acid]!

  13. Common traps:

  14. Diprotic acids (H₂SO₄) have two protons—first is strong, second is weak.
  15. Buffers hide in questions with weak acids + their salts—use Henderson-Hasselbalch.
  16. Always check units—mL vs. L can ruin your answer.

Now go crush those problems—you’ve got this!




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