Ch 1 · Acids and bases · AS & A Level Chemistry 9701

Topic 1 of 2 in Equilibria — AS & A Level Chemistry 9701 — 68 lessons.

Introduction

Building on Brønsted–Lowry theory, A2 acid–base chemistry treats acids and bases as equilibria, using K_a and K_w to calculate pH quantitatively, explaining how buffer solutions and blood pH regulation work, and extending equilibrium ideas to the solubility of sparingly soluble salts through K_sp.

Key ideas — Conjugate acid–base pairs

In any Brønsted–Lowry acid–base reaction, an acid donates a proton to form its conjugate base, and a base accepts a proton to form its conjugate acid; the two species in each pair differ by exactly one H^+. In CH_3COOH + H_2O leftharpoons CH_3COO^- + H_3O^+, CH_3COOH and CH_3COO^- are one conjugate pair, and H_2O and H_3O^+ are the other.

🧑‍🏫 Teacher's explanation

Think of a Brønsted-Lowry acid-base reaction like a game of hot potato. The 'proton' (H^+) is the potato. An acid is a player that throws the potato away; once it has thrown it, it becomes a 'conjugate base'—it is the version of the player that no longer has the potato. A base is a player that catches the potato; once it catches it, it becomes a 'conjugate acid'—the version of the player holding the potato.

📖 Definitions

🔍 Worked example — Easy

In the reaction NH_3 + H_2O ightleftharpoons NH_4^+ + OH^-, identify the conjugate acid and the conjugate base.

🔍 Worked example — Standard

Identify the conjugate acid-base pairs in the following equilibrium: HCO_3^- + H_2O ightleftharpoons CO_3^2- + H_3O^+

🔍 Worked example — Exam-style

Given the reaction HF + H_2O ightleftharpoons F^- + H_3O^+, identify the conjugate acid-base pairs and explain why F^- is the conjugate base of HF.

⚠️ Common errors

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🎯 Exam question

ceNH_3(aq) + H_2O(l) <-> NH_4^+(aq) + OH^-(aq) (a) Identify the conjugate acid produced in this reaction. [1] (b) Identify the conjugate base produced in this reaction. [1] (c) Explain, in terms of proton transfer, why NH_4^+ is the conjugate acid of NH_3. [2] [4 marks]

Key ideas — Defining pH, Ka, pKa and Kw

pH = -log_10[H^+] K_w = [H^+][OH^-] = 1.00 × 10^-14\ mathrmmol^2\ dm^-6\ at\ 298\ K

🧑‍🏫 Teacher's explanation

Think of pH as a 'magnifying glass' for acidity. Because the concentration of hydrogen ions in water is often a tiny, awkward number like 0.0000001, we use a logarithmic scale to turn those messy decimals into simple numbers like 7. The smaller the number, the more 'aggressive' the acid is.

📖 Definitions

🔍 Worked example — Easy

Calculate the pH of a 0.050 mol dm^-3 solution of HCl.

🔍 Worked example — Standard

A weak acid HA has a K_a = 1.8 × 10^-5. Calculate the pK_a of this acid.

🔍 Worked example — Exam-style

Calculate the pH of a 0.10 mol dm^-3 solution of a weak acid HA where K_a = 2.0 × 10^-4. Assume [H^+] ≈ √(K_a[HA]).

⚠️ Common errors

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🎯 Exam question

Key ideas — pH of strong acids, strong alkalis and weak acids

For a strong acid, full dissociation gives $[\mathrm{H^+}] = $ acid concentration directly. For a strong alkali, $[\mathrm{OH^-}] = $ alkali concentration, and [H^+] = K_w / [OH^-]. For a weak acid, assuming [H^+] ≈ [A^-] and that dissociation is small enough that [HA]_eqm ≈ [HA]_initial, K_a = [H^+]^2/[HA], so [H^+] = √(K_a[HA]). For example, 0.100\ mathrmmol\ dm^-3 ethanoic acid (K_a = 1.74 × 10^-5\ mathrmmol\ dm^-3) gives [H^+] = sqrt1.74 × 10^-6 = 1.32 × 10^-3\ mathrmmol\ dm^-3, so pH = 2.88.

🧑‍🏫 Teacher's explanation

Think of acid dissociation like a crowd of people in a room. In a 'strong' acid, every single person (molecule) immediately jumps out of their seat to become an ion; the concentration of ions is exactly the same as the concentration of the original molecules. In a 'weak' acid, only a tiny fraction of people decide to stand up, while most stay seated.

📖 Definitions

🔍 Worked example — Easy

Calculate the pH of 0.20 mol dm^-3 HCl.

🔍 Worked example — Standard

Calculate the pH of 0.050 mol dm^-3 NaOH.

🔍 Worked example — Exam-style

Calculate the pH of 0.10 mol dm^-3 ethanoic acid (CH_3COOH) given K_a = 1.75 × 10^-5 mol dm^-3.

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🎯 Exam question

Key ideas — Buffer solutions: composition, action and uses

An acidic buffer solution is made from a weak acid and a soluble salt of its conjugate base (e.g. CH_3COOH and CH_3COONa), giving a large reservoir of both HA and A^-. It resists pH change because added H^+ is removed by the conjugate base, CH_3COO^- + H^+ arrow CH_3COOH, while added OH^- is neutralised by the weak acid, CH_3COOH + OH^- arrow CH_3COO^- + H_2O; in each case the ratio [A^-]/[HA] changes only slightly, so pH is only slightly altered rather than fixed. Buffers control pH in biochemical processes and in industry; in blood, the carbonic acid–hydrogencarbonate buffer (H_2CO_3/HCO_3^-) keeps plasma pH between 7.35 and 7.45, removing excess H^+ via HCO_3^-(aq) + H^+(aq) arrow H_2CO_3(aq) and excess OH^- via H_2CO_3(aq) + OH^-(aq) arrow HCO_3^-(aq) + H_2O(l).

🧑‍🏫 Teacher's explanation

Imagine you are at a large party. If one person enters or leaves, the 'vibe' (the pH) doesn't change much because there are so many people to absorb the change. A buffer solution works exactly like this. It is a 'reservoir' of two components that can neutralize both acids and bases.

📖 Definitions

🔍 Worked example — Easy

Calculate the pH of a buffer solution containing 0.10 mol dm^-3 ethanoic acid (CH_3 extCOOH) and 0.10 mol dm^-3 sodium ethanoate (CH_3 extCOONa). The K_a for ethanoic acid is 1.8 × 10^-5.

🔍 Worked example — Standard

A buffer solution is prepared by mixing 0.50 mol dm^-3 ammonia (NH_3) and 0.50 mol dm^-3 ammonium chloride (NH_4Cl). Calculate the pH if K_b for NH_3 is 1.8 × 10^-5.

🔍 Worked example — Exam-style

A buffer solution contains 0.20 mol dm^-3 CH_3 extCOOH and 0.05 mol dm^-3 CH_3 extCOONa. Calculate the pH. (K_a = 1.8 × 10^-5)

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🎯 Exam question

Key ideas — Calculating buffer pH

Rearranging K_a = [H^+][A^-]/[HA] for a buffer gives [H^+] = K_a × [HA]/[A^-], or in logarithmic (Henderson–Hasselbalch) form:

🧑‍🏫 Teacher's explanation

Imagine a buffer solution is like a sponge for acidity. If you pour a little acid into a glass of water, the pH jumps wildly. But if you pour that acid into a sponge, the sponge soaks up the extra H^+ ions, and the pH barely moves. In chemistry, a buffer is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid). The weak acid provides a 'reservoir' of H^+ ions, while the conjugate base acts as a 'sink' to absorb any extra H^+ ions added to the system.

📖 Definitions

🔍 Worked example — Easy

Calculate the pH of a buffer solution containing 0.10 mol dm^-3 of ethanoic acid (CH_3COOH) and 0.20 mol dm^-3 of sodium ethanoate (CH_3COONa). The K_a for ethanoic acid is 1.8 × 10^-5.

🔍 Worked example — Standard

A buffer solution is prepared by mixing 50.0 cm^3 of 0.200 mol dm^-3 ammonia (NH_3) with 50.0 cm^3 of 0.100 mol dm^-3 ammonium chloride (NH_4Cl). Calculate the pH. (K_b for NH_3 = 1.8 × 10^-5)

🔍 Worked example — Exam-style

A buffer contains 0.50 mol of HCN and 1.50 mol of KCN in a total volume of 2.00 dm^3. Calculate the pH. (K_a for HCN = 4.9 × 10^-10)

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🎯 Exam question

Key ideas — Solubility product, Ksp, and the common ion effect

The solubility product, K_sp, is the equilibrium constant for the dissolving of a sparingly soluble ionic compound, expressed as the product of the molar ion concentrations, each raised to the power of its coefficient in the dissolving equation, e.g. for AgCl(s) leftharpoons Ag^+(aq) + Cl^-(aq), K_sp = [Ag^+][Cl^-]. The common ion effect occurs when a solution already contains one of the compound's ions (from a second, soluble source); by Le Chatelier's principle, this increases that ion's concentration, so the equilibrium concentration of the other ion — and hence the solubility of the sparingly soluble compound — must decrease to keep K_sp constant.

🧑‍🏫 Teacher's explanation

Imagine a crowded dance floor where people are trying to enter from a hallway. The 'solubility product' is like the maximum number of people allowed on the floor at once before the hallway gets too crowded and people stop entering. If the floor is full, the system is at equilibrium.

Ch 1 · Acids and bases · AS & A Level Chemistry 9701

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