Chapter 1: Acids, Bases and Salts (Advanced)

Why this matters

(I once asked a class, "Why doesn't your blood become acidic when you drink a glass of orange juice?" Most students shrugged. Then I said, "Because of buffers — a chemical system that resists changes in pH. And the same principles that keep your blood safe are used in medicine, agriculture, and industry." That is why we study acid-base chemistry at this level. It is not just about litmus paper. It is about understanding the chemistry that keeps you alive, that grows our food, and that runs our industries. In this chapter, we go deeper than we did before. We will look at the theories of acids and bases, calculate pH, understand buffers, interpret titration curves, and explore salt hydrolysis. Let us begin.) In your earlier studies, you learned that acids turn litmus red, taste sour, and react with metals. You learned that bases turn litmus blue, feel slippery, and neutralise acids. That is a good start — but it is only the surface. At the advanced level, we ask deeper questions. What really makes something an acid? How can we measure acidity precisely? Why do some salt solutions turn litmus red while others turn it blue? How does the body keep its blood pH within a tiny range? These are the questions that this chapter answers. We will explore the Brønsted-Lowry theory, which defines acids and bases in terms of proton transfer. We will learn to calculate pH and pOH for strong and weak acids and bases. We will study buffers — the chemical systems that resist pH changes. We will interpret titration curves. And we will see how salt hydrolysis explains why some salts are acidic, some are basic, and some are neutral. By the end, you will have a deep understanding of acid-base chemistry — one that you can apply to medicine, agriculture, environmental science, and industry. Let us begin.

Learning objectives

• Describe the Brønsted-Lowry theory of acids and bases.
• Define conjugate acid-base pairs.
• Distinguish between strong and weak acids and bases, and explain the difference in terms of dissociation.
• Calculate pH and pOH for strong and weak acids and bases.
• Describe buffers and explain how they resist changes in pH.
• Interpret titration curves for strong-strong, strong-weak, and weak-weak acid-base reactions.
• Describe salt hydrolysis and explain why salt solutions can be acidic, basic, or neutral.
• Appreciate the importance of acid-base chemistry in medicine, agriculture, and industry.

Concept build-up

1.1 The Brønsted-Lowry Theory

The Brønsted-Lowry theory defines acids and bases in terms of proton (H⁺) transfer:

• An acid is a proton (H⁺) donor.
• A base is a proton (H⁺) acceptor.

This is a more general definition than the Arrhenius theory (which defined acids as producing H⁺ in water and bases as producing OH⁻ in water). The Brønsted-Lowry theory works for reactions that do not involve water, and it explains why some substances that do not contain OH⁻ can still act as bases.

Example: In the reaction between hydrochloric acid and ammonia:

HCl + NH₃ → NH₄⁺ + Cl⁻

• HCl donates a proton (H⁺) to NH₃ — so HCl is the acid.
• NH₃ accepts a proton from HCl — so NH₃ is the base.

(The Brønsted-Lowry theory is elegant because it focuses on the proton. A proton is just a hydrogen atom that has lost its electron. When an acid donates a proton, it becomes a base. When a base accepts a proton, it becomes an acid. This give-and-take is the heart of all acid-base chemistry.)

1.2 Conjugate Acid-Base Pairs

When an acid donates a proton, it becomes a conjugate base. When a base accepts a proton, it becomes a conjugate acid. Together, they form a conjugate acid-base pair.

Example: In the reaction CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺:

• CH₃COOH (acid) donates a proton to become CH₃COO⁻ (conjugate base).
• H₂O (base) accepts a proton to become H₃O⁺ (conjugate acid).

Important: The stronger the acid, the weaker its conjugate base. This is because a strong acid gives up its proton easily, so its conjugate base holds onto protons weakly.

1.3 Strong and Weak Acids and Bases

Strong acids dissociate completely in water, releasing all their H⁺ ions. Weak acids dissociate only partially, establishing an equilibrium.

Important distinction: "Strong" refers to how completely the acid or base dissociates, not to its concentration. A strong acid can be dilute, and a weak acid can be concentrated.

1.4 The Dissociation Constant (Ka and Kb)

For a weak acid HA, the dissociation equilibrium is:

HA ⇌ H⁺ + A⁻

The acid dissociation constant (Ka) is given by:

Ka = [H⁺][A⁻] / [HA]

The larger the Ka, the stronger the acid. For a weak base B, the base dissociation constant (Kb) is given by:

Kb = [BH⁺][OH⁻] / [B]

The larger the Kb, the stronger the base.

Relationship between Ka and Kb for a conjugate pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C.

pKa and pKb: pKa = −log₁₀(Ka); pKb = −log₁₀(Kb). The smaller the pKa, the stronger the acid.

1.5 pH and pOH

pH is a measure of the concentration of hydrogen ions in a solution. It is defined as:

pH = −log₁₀[H⁺]

pOH is defined similarly:

pOH = −log₁₀[OH⁻]

At 25°C, the relationship between pH and pOH is:

pH + pOH = 14

This comes from the ionisation of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C, so pKw = 14.

(The pH scale is logarithmic, which means each step represents a tenfold change in H⁺ concentration. A solution of pH 3 is ten times more acidic than one of pH 4, and a hundred times more acidic than one of pH 5. This is why small changes in pH matter so much — in your blood, in a lake, in a swimming pool.)

1.6 Calculating pH for Strong Acids and Bases

Strong acids: Since they dissociate completely, [H⁺] equals the concentration of the acid.

Example: Calculate the pH of 0.01 M HCl.

Solution: HCl is a strong acid, so [H⁺] = 0.01 M = 1.0 × 10⁻² M. pH = −log(1.0 × 10⁻²) = 2.

Strong bases: Since they dissociate completely, [OH⁻] equals the concentration of the base (multiplied by the number of OH⁻ ions per formula unit).

Example: Calculate the pH of 0.01 M NaOH.

Solution: NaOH is a strong base, so [OH⁻] = 0.01 M = 1.0 × 10⁻² M. pOH = −log(1.0 × 10⁻²) = 2. pH = 14 − 2 = 12.

1.7 Calculating pH for Weak Acids and Bases

For weak acids, we use the acid dissociation constant (Ka). Since weak acids dissociate only partially, [H⁺] is not equal to the concentration of the acid.

Approximation: For a weak acid HA with concentration C and dissociation constant Ka:

[H⁺] = √(Ka × C)

Example: Calculate the pH of 0.1 M ethanoic acid (Ka = 1.8 × 10⁻⁵).

Solution: [H⁺] = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M. pH = −log(1.34 × 10⁻³) = 2.87.

Note: The approximation is valid when the degree of dissociation is small (less than about 5%).

1.8 Buffers

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are essential in many chemical and biological systems.

How buffers work: A buffer consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). The weak acid neutralises added base, and the conjugate base neutralises added acid.

Example: The ethanoic acid/ethanoate buffer system:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

• If acid (H⁺) is added, the ethanoate ions (CH₃COO⁻) react with it to form ethanoic acid, removing the added H⁺.
• If base (OH⁻) is added, the ethanoic acid reacts with it to form ethanoate ions and water, removing the added OH⁻.

Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA])

This equation allows us to calculate the pH of a buffer solution.

Biological Buffers

(Your blood pH is kept between 7.35 and 7.45 — a very narrow range. If it goes outside this range, you can become seriously ill. The carbonic acid/bicarbonate buffer system keeps it stable. Every breath you take is part of this buffer system. It is one of the most beautiful examples of chemistry keeping you alive.)

1.9 Titration Curves

A titration curve is a graph of pH against the volume of titrant added during a titration. The shape of the curve depends on the strengths of the acid and base involved.

Strong Acid — Strong Base

• Initial pH is very low (strong acid).
• pH changes gradually at first, then sharply near the equivalence point.
• Equivalence point is at pH 7.
• The curve is steep at the equivalence point.

Strong Acid — Weak Base

• Initial pH is very low (strong acid).
• Equivalence point is below pH 7 (acidic) because the salt formed is acidic.
• The curve is less steep at the equivalence point than for a strong-strong titration.

Weak Acid — Strong Base

• Initial pH is higher than for a strong acid (weak acid).
• Equivalence point is above pH 7 (basic) because the salt formed is basic.
• The curve shows a buffer region before the equivalence point.

Weak Acid — Weak Base

• Initial pH is moderate.
• Equivalence point is near pH 7, but the curve is very gradual — no sharp change.
• Not suitable for accurate titration because the end point is not sharp.

1.10 Salt Hydrolysis

Salt hydrolysis is the reaction of ions from a salt with water, leading to a change in pH. Salts can be acidic, basic, or neutral depending on the strengths of the acid and base from which they were formed.

Explanation:

• NH₄Cl: The NH₄⁺ ion is the conjugate acid of the weak base NH₃, so it donates a proton to water, producing H₃O⁺ and making the solution acidic.
• CH₃COONa: The CH₃COO⁻ ion is the conjugate base of the weak acid CH₃COOH, so it accepts a proton from water, producing OH⁻ and making the solution basic.
• NaCl: Neither Na⁺ nor Cl⁻ reacts with water, so the solution is neutral.

1.11 Applications of Acid-Base Chemistry

Worked examples

Example 1: pH of a strong acid

Question: Calculate the pH of 0.001 M HCl.

Solution: HCl is a strong acid, so [H⁺] = 0.001 M = 1.0 × 10⁻³ M. pH = −log(1.0 × 10⁻³) = 3.

Example 2: pH of a strong base

Question: Calculate the pH of 0.005 M NaOH.

Solution: [OH⁻] = 0.005 M = 5.0 × 10⁻³ M. pOH = −log(5.0 × 10⁻³) = 2.3. pH = 14 − 2.3 = 11.7.

Example 3: pH of a weak acid

Question: Calculate the pH of 0.1 M ethanoic acid (Ka = 1.8 × 10⁻⁵).

Solution: [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M. pH = −log(1.34 × 10⁻³) = 2.87.

Example 4: Buffer pH

Question: Calculate the pH of a buffer solution containing 0.1 M ethanoic acid and 0.1 M sodium ethanoate (Ka = 1.8 × 10⁻⁵).

Solution: Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). pKa = −log(1.8 × 10⁻⁵) = 4.74. pH = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74.

Common mistakes

• Confusing strong and concentrated. Strong refers to complete dissociation; concentrated refers to the amount dissolved.
• Forgetting that pH is a logarithmic scale. A change of 1 pH unit is a tenfold change in [H⁺].
• Assuming all salt solutions are neutral. Salt hydrolysis can make solutions acidic or basic.
• Mixing up conjugate acid-base pairs. The conjugate base of an acid is what remains after it loses a proton.
• Forgetting that buffers work by neutralising added acid or base. They do not prevent pH changes entirely — they just minimise them.
• Using the wrong formula for pH of weak acids. For weak acids, [H⁺] is not equal to the acid concentration.
• Assuming titration curves are always sharp. Weak acid–weak base titrations have very gradual curves.
• Forgetting the relationship pH + pOH = 14. This applies at 25°C.

Checkpoint questions

Structured exam-style questions

Challenge questions

"Acid-base chemistry is the chemistry of balance — the balance of protons, the balance of pH, and the balance that keeps you alive. Now you understand it at a deeper level."