The central idea
An equilibrium constant relates equilibrium activities; in many school-level dilute-solution calculations, concentrations provide the working approximation. For a weak monoprotic acid with initial concentration C and dissociation x, Ka = x²/(C − x). Only replace C − x by C when x is small relative to C.
A worked example
If C = 0.0100 M and Ka = 10⁻⁶, estimate x ≈ √(KaC) = 10⁻⁴ M. The dissociation is about 1%, supporting the approximation. For a buffer, first react any added strong acid or base stoichiometrically, then apply pH ≈ pKa + log([base]/[acid]).
A common mistake
The acid/base amounts change when a strong reagent is added. Using the old ratio gives the wrong buffer pH.
Try the idea
These original practice questions also appear in the foundation mock. Attempt them before opening the explanation.
For A ⇌ B, Kc = 4.0. Initially a vessel contains 1.0 mol A and no B. If volume is constant, the equilibrium amount of B is:
Answer: D
Let x mol A react. Then n(B) = x and n(A) = 1 − x. The common volume cancels in Kc = x/(1 − x) = 4. Solving gives x = 0.80 mol.
For N₂O₄(g) ⇌ 2NO₂(g), one mole of pure N₂O₄ is allowed to dissociate. The equilibrium total pressure is P and the degree of dissociation is α. Using partial pressures, Kp is:
Answer: B
At equilibrium the amounts are 1 − α and 2α, totalling 1 + α. Hence p(NO₂) = 2αP/(1 + α) and p(N₂O₄) = (1 − α)P/(1 + α). Substitute into p(NO₂)²/p(N₂O₄) to get 4α²P/(1 − α²). This uses the pressure-based school convention.
A buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻. After adding 0.010 mol HCl, what is its approximate pH? Use pKa = 4.76 and log₁₀(9/11) = −0.087.
Answer: A
Strong acid first reacts with acetate: H⁺ + CH₃COO⁻ → CH₃COOH. The new amounts are 0.090 mol base and 0.110 mol acid. Henderson–Hasselbalch gives pH = 4.76 + log(0.090/0.110) = 4.673.
At 25 °C, a 0.0100 M solution of a weak monoprotic acid has Ka = 1.00 × 10⁻⁶. Its approximate pH is:
Answer: C
For small dissociation, [H⁺] ≈ √(KaC) = √(10⁻⁶ × 10⁻²) = 10⁻⁴ M, giving pH ≈ 4.00. The estimated dissociation is 1%, so the approximation is reasonable.
For AgCl, Ksp = 1.0 × 10⁻¹⁰. Ignoring complex formation and activity effects, its approximate molar solubility in 0.010 M NaCl is:
Answer: B
Write Ksp = [Ag⁺][Cl⁻]. The NaCl supplies approximately 0.010 M chloride, far more than dissolving AgCl contributes. Thus s = 10⁻¹⁰/10⁻² = 10⁻⁸ M. The common ion suppresses solubility.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at equilibrium, which statements are correct for ideal gases?
Answer: B, C
Compression raises the reacting partial pressures and favours the side with fewer gas molecules, here NH₃. An inert gas at fixed V and T does not change nᵢRT/V for any reacting species. A catalyst changes the speed of reaching equilibrium, not Kp. Equilibrium is dynamic: both reaction rates remain equal and non-zero.
For a weak-acid/conjugate-base buffer in the range where Henderson–Hasselbalch applies, which statements are correct?
Answer: A, B, C, D
pH = pKa + log([base]/[acid]), so equal concentrations give pH ≈ pKa. Added H⁺ consumes conjugate base. Buffer capacity is finite. Moderate dilution preserves the ratio and approximately the pH, but reduces capacity; at extreme dilution the approximation can fail. All four statements are correct, so all four must be selected.
What to do next
Explain your answer aloud, then find a related question in an official paper. If you can only follow a worked solution but cannot start a fresh question, revisit the central idea and try again without the notes.
Apply it to an official paper