Chemistry (9701): Comprehensive Study Notes on Acids and Bases

Hello future Chemists! This chapter takes you deep into the world of acids, bases, and the delicate balance known as equilibrium. Understanding this topic is vital—it underpins everything from industrial processes (like the contact process) to biological systems (like how your blood stays at a safe pH). Don't worry if the calculations seem tough; we'll break down the concepts step-by-step!

1. The Foundation: Brønsted-Lowry Theory and Strength (AS Level Review)

1.1 Defining Acids and Bases (The Proton Game)

The most useful definition for A-Level Chemistry is the Brønsted-Lowry Theory.

  • Acid: A species that is a proton donor. (It gives away an \(H^+\) ion).
  • Base: A species that is a proton acceptor. (It takes an \(H^+\) ion).
The Conjugate Pair Concept

When an acid donates a proton, it forms a conjugate base. When a base accepts a proton, it forms a conjugate acid. These pairs are connected by the loss or gain of a single \(H^+\) ion. (Syllabus 7.2.1, 7.2.2)

Analogy: Think of a "proton-passing" team. The acid passes the proton, and the species left behind (the conjugate base) is ready to grab it back.

Example Reaction: Dissociation of Ethanoic Acid (a weak acid):
\(CH_3COOH \text{ (acid)} + H_2O \text{ (base)} \rightleftharpoons CH_3COO^- \text{ (conjugate base)} + H_3O^+ \text{ (conjugate acid)}\)

Quick Review: Conjugate acid-base pairs differ only by one \(H^+\).

1.2 Strong vs. Weak Acids and Bases

This is one of the most common confusion points! Strength is about dissociation; Concentration is about how many moles are dissolved in the volume. (Syllabus 7.2.4, 7.2.6)

Strong Acids and Bases

A strong acid or strong base is one that fully dissociates (ionises) in aqueous solution.

  • Strong Acid: \(HCl(aq) \longrightarrow H^+(aq) + Cl^-(aq)\) (The reaction arrow is one-way, indicating 100% dissociation).
  • Common Strong Acids: Hydrochloric acid (\(HCl\)), Sulfuric acid (\(H_2SO_4\)), Nitric acid (\(HNO_3\)). (Syllabus 7.2.1)
  • Common Strong Alkalis: Sodium hydroxide (\(NaOH\)), Potassium hydroxide (\(KOH\)). (Syllabus 7.2.2)
Weak Acids and Bases

A weak acid or weak base is one that partially dissociates in aqueous solution. An equilibrium is established.

  • Weak Acid: \(CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)\) (The reversible arrows indicate partial dissociation).
  • Common Weak Acid: Ethanoic acid (\(CH_3COOH\)). (Syllabus 7.2.1)
  • Common Weak Base: Ammonia (\(NH_3\)). (Syllabus 7.2.2)

Did you know? Even a highly concentrated solution of a weak acid will still have a higher pH than a dilute solution of a strong acid because the strong acid produces vastly more \(H^+\) ions due to full dissociation.

1.3 The pH Scale and the Ionic Product of Water (\(K_w\))

The pH Scale (Syllabus 7.2.5)

pH is a measure of the hydrogen ion concentration, $[H^+]$, in a solution.

The mathematical definition is: $$pH = -\log_{10}[H^+]$$

  • Acidic solutions have \(pH < 7\)
  • Alkaline solutions have \(pH > 7\)
  • Neutral solutions have \(pH = 7\) (at $25^{\circ}C$ or 298 K)
The Ionic Product of Water, \(K_w\)

Pure water undergoes self-ionisation: $$\text{2}H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)$$
or simply: $$\text{H}_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$$

The equilibrium expression for this is the ionic product of water, \(K_w\): $$K_w = [H^+][OH^-]$$

At the standard temperature of 298 K ($25^{\circ}C$): $$K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$$

Why is $K_w$ important? In any aqueous solution (acidic, neutral, or alkaline), $K_w$ is constant at a fixed temperature. This allows you to calculate $[H^+]$ if you know $[OH^-]$ (and thus calculate the pH of strong alkalis). (Syllabus 25.1.3)

Trick for strong bases:
1. Find $[OH^-]$ (it equals the concentration of the strong base).
2. Calculate $[H^+]$ using $$[H^+] = \frac{K_w}{[OH^-]}$$
3. Calculate pH using $$pH = -\log_{10}[H^+]$$

Key Takeaway for Section 1: Strong means full dissociation; weak means partial dissociation (equilibrium). The Brønsted-Lowry theory defines acids as proton donors. $K_w$ links $[H^+]$ and $[OH^-]$ in all aqueous solutions.

2. Quantifying Weak Acids: \(K_a\) and \(pK_a\) (A Level)

Because weak acids set up an equilibrium, we need a special equilibrium constant to describe exactly how "weak" they are.

2.1 The Acid Dissociation Constant (\(K_a\))

For a general weak acid, \(HA\), the dissociation equilibrium is: $$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$

The expression for the equilibrium constant, $K_c$, is modified and called the acid dissociation constant, \(K_a\): (Syllabus 25.1.3) $$K_a = \frac{[H^+][A^-]}{[HA]}$$

What does $K_a$ tell you?

  • The larger the \(K_a\) value, the greater the concentration of products ($H^+$ and $A^-$) at equilibrium. This means the acid is stronger.
  • The smaller the \(K_a\) value, the less it dissociates, and the acid is weaker.

2.2 The \(pK_a\) Term

Just as pH simplifies $10^{-x}$ concentrations, $pK_a$ simplifies very small $K_a$ values: $$pK_a = -\log_{10} K_a$$

This is often easier to compare. (Syllabus 25.1.3)

  • A small \(pK_a\) (more negative log) means a stronger acid.
  • A large \(pK_a\) means a weaker acid.

Memory Aid: Think "Small p, Stronger acid."

Note: You must be able to define $pH$, $K_a$, and $pK_a$ mathematically, and use them in calculations. (Syllabus 25.1.3) (The base dissociation constant, $K_b$, and the relationship $K_w = K_a \times K_b$ are not assessed).

2.3 Calculating the pH of a Weak Acid

Calculating the pH of a weak acid requires using the $K_a$ expression and making two simplifying assumptions. (Syllabus 25.1.4(c))

For the equilibrium: \(HA \rightleftharpoons H^+ + A^-\)

Step 1: The assumption of equal concentration.
If the acid is the only source of $H^+$, then $[H^+] = [A^-]$.
Substituting this into the $K_a$ expression: $$K_a = \frac{[H^+][A^-]}{[HA]} \implies K_a = \frac{[H^+]^2}{[HA]}$$

Step 2: The assumption of negligible dissociation.
Since the acid is weak, almost all of the $HA$ remains undissociated. Therefore, the equilibrium concentration of $HA$ is approximately equal to its initial concentration, $[HA]_{\text{initial}}$.
The equation becomes: $$K_a \approx \frac{[H^+]^2}{[HA]_{\text{initial}}}$$

Step 3: Calculate $[H^+]$ and pH.
$$[H^+] = \sqrt{K_a \times [HA]_{\text{initial}}}$$
Then, calculate the pH: \(pH = -\log_{10}[H^+]\).

Key Takeaway for Section 2: $K_a$ quantifies acid strength. A higher $K_a$ (lower $pK_a$) means a stronger acid. For weak acid pH calculation, we assume $[H^+]=[A^-]$ and $[HA]$ remains essentially constant.

3. Buffer Solutions (A Level)

Imagine you need to keep a chemical reaction or a biological process (like your blood pH) perfectly stable, even when small amounts of acid or alkali are added. This is the job of a buffer solution.

3.1 Definition and Composition

A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. (Syllabus 25.1.5(a))

A buffer must contain both the weak acid and its conjugate base in significant concentrations (or the weak base and its conjugate acid).

  • Acidic Buffer (pH < 7): Made from a weak acid (e.g., \(CH_3COOH\)) and a salt of that acid (e.g., \(CH_3COONa\)) which provides the conjugate base (\(CH_3COO^-\)).
  • Alkaline Buffer (pH > 7): Made from a weak base (e.g., \(NH_3\)) and a salt of that base (e.g., \(NH_4Cl\)) which provides the conjugate acid (\(NH_4^+\)).

3.2 The Mechanism: How Buffers Control pH

The key is that the buffer contains a 'proton sponge' (the conjugate base) and a supply of protons (the weak acid) to counteract any added stress. (Syllabus 25.1.5(c))

Consider the Ethanoic Acid/Sodium Ethanoate buffer: $$CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)$$

The weak acid (\(CH_3COOH\)) is the reservoir for $H^+$. The conjugate base (\(CH_3COO^-\)) is the scavenger for $H^+$.

Action when strong acid ($H^+$) is added:

The added \(H^+\) is mopped up by the conjugate base: $$H^+(aq) + CH_3COO^-(aq) \longrightarrow CH_3COOH(aq)$$
The ratio of acid to salt barely changes, so the pH is maintained.

Action when strong alkali ($OH^-$) is added:

The added \(OH^-\) reacts with the weak acid: $$OH^-(aq) + CH_3COOH(aq) \longrightarrow H_2O(l) + CH_3COO^-(aq)$$
The $OH^-$ is neutralized, preventing the pH from soaring.

3.3 Buffer Calculations

Buffer pH calculations use the same $K_a$ expression, but now both $[HA]$ and $[A^-]$ are high (because the salt fully dissociates). (Syllabus 25.1.6)

$$K_a = \frac{[H^+][A^-]}{[HA]}$$
Rearranging to find $[H^+]$: $$[H^+] = K_a \times \frac{[HA]}{[A^-]}$$

Since the concentration of the weak acid $[HA]$ and the conjugate base $[A^-]$ are known (from the amounts used to make the buffer), you can calculate $[H^+]$ directly and then the pH.

Common Mistake Alert: When calculating the pH of a buffer, you must use the initial concentrations of the weak acid and the salt, as the equilibrium shift caused by the buffer components is negligible.

3.4 Uses of Buffers (Syllabus 25.1.5(d))

Buffers are essential in systems where pH stability is critical.

The most important biological example is the bicarbonate buffer system (\(HCO_3^-\)) in blood.

When cellular respiration produces $CO_2$, it dissolves in blood to form carbonic acid (\(H_2CO_3\)), decreasing the pH. The bicarbonate ion acts as the conjugate base, immediately neutralising the added acid to keep the blood pH constant (around 7.4).

$$H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)$$

Key Takeaway for Section 3: Buffers contain a weak acid/conjugate base pair (or weak base/conjugate acid pair). They work by having the conjugate component neutralise the added stress ($H^+$ or $OH^-$) immediately. The blood buffer is a crucial example.

4. Acid-Base Neutralisation and Titration Curves

4.1 Neutralisation Reactions

Neutralisation is simply the reaction between $H^+(aq)$ and $OH^-(aq)$ to form water. (Syllabus 7.2.7, 7.2.8)

The net ionic equation for all strong acid-strong base reactions is: $$\text{H}^+(aq) + OH^-(aq) \longrightarrow H_2O(l)$$

A salt is always formed in a neutralisation reaction (e.g., $NaCl$, $KNO_3$).

4.2 Sketching and Interpreting pH Titration Curves

Titration curves plot the pH change of a solution as a titrant (acid or base) is added. You must be able to sketch the four main types of curves. (Syllabus 7.2.9)

The equivalence point is the point where the moles of acid equal the moles of base.


1. Strong Acid vs. Strong Base (e.g., \(HCl\) vs. \(NaOH\))

  • Starting pH is very low (e.g., 1).
  • Equivalence point is exactly at pH 7.
  • The vertical region (steepest part of the curve) is long, spanning approximately pH 3 to pH 11.


2. Weak Acid vs. Strong Base (e.g., \(CH_3COOH\) vs. \(NaOH\))

  • Starting pH is higher (e.g., 3-5).
  • A buffer region exists early on (where the slope is minimal).
  • Equivalence point is alkaline (\(pH > 7\)), because the salt formed (\(CH_3COONa\)) hydrolyses in water to produce \(OH^-\) ions.
  • The vertical region is shorter (approx. pH 7 to pH 11).


3. Strong Acid vs. Weak Base (e.g., \(HCl\) vs. \(NH_3\))

  • Starting pH is high (e.g., 11).
  • Equivalence point is acidic (\(pH < 7\)), because the salt formed (\(NH_4Cl\)) hydrolyses in water to produce \(H^+\) ions.
  • The vertical region is shorter (approx. pH 3 to pH 7).


4. Weak Acid vs. Weak Base (e.g., \(CH_3COOH\) vs. \(NH_3\))

  • Starting pH is high, but the initial acid pH is also high.
  • Crucial Point: There is no sharp vertical jump, making it impossible to determine the equivalence point accurately using simple indicators. This titration is generally not performed.

4.3 Indicator Selection (Syllabus 7.2.10)

An indicator is a weak acid or base whose colour changes over a narrow pH range.

For an indicator to be suitable for a titration, its $\text{pH}$ range must fall entirely within the steep vertical region of the titration curve.

  • Strong Acid/Strong Base: Many indicators work (e.g., Phenolphthalein (8.3–10.0) or Methyl Orange (3.1–4.4)).
  • Weak Acid/Strong Base: Requires an indicator that changes colour on the alkaline side (e.g., Phenolphthalein). Methyl orange would be unsuitable.
  • Strong Acid/Weak Base: Requires an indicator that changes colour on the acidic side (e.g., Methyl Orange). Phenolphthalein would be unsuitable.
Key Takeaway for Section 4: The $\text{pH}$ at the equivalence point depends on the strength of the reactants (Strong/Strong = pH 7; Weak Acid/Strong Base = pH > 7; Strong Acid/Weak Base = pH < 7). The indicator range must match the steep part of the curve.

5. Solubility Equilibria: The Solubility Product, \(K_{sp}\) (A Level)

While some salts are very soluble, others dissolve so minimally that we need a specific equilibrium constant to describe their solubility. This is the solubility product, \(K_{sp}\). (Syllabus 25.1.7)

5.1 Defining the Solubility Product (\(K_{sp}\))

$K_{sp}$ applies to salts that are sparingly soluble (or effectively insoluble). When such a salt is placed in water, an equilibrium is set up between the undissolved solid and the aqueous ions.

For a generic sparingly soluble salt, \(X_aY_b\), the equilibrium is: $$X_aY_b(s) \rightleftharpoons aX^{b+}(aq) + bY^{a-}(aq)$$

The solubility product, \(K_{sp}\), is the equilibrium constant for this dissociation. Since the solid concentration is constant, it is excluded: (Syllabus 25.1.8) $$K_{sp} = [X^{b+}]^a [Y^{a-}]^b$$

Example: Silver Chloride (\(AgCl\)) $$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$$ $$K_{sp} = [Ag^+][Cl^-]$$

Example: Calcium Hydroxide (\(Ca(OH)_2\)) $$Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$$ $$K_{sp} = [Ca^{2+}][OH^-]^2$$

Calculations: You must be able to calculate $K_{sp}$ from ion concentrations or vice versa. (Syllabus 25.1.9)

If the solubility of $AgCl$ is $S$ mol $\text{dm}^{-3}$, then $[Ag^+]=S$ and $[Cl^-]=S$. Thus, $K_{sp} = S^2$.

5.2 The Common Ion Effect

The Common Ion Effect is a specific application of Le Chatelier’s principle to solubility equilibria. (Syllabus 25.1.10)

How it works:

If you add an ion that is already present in the solubility equilibrium (the "common ion"), the equilibrium shifts to the left, decreasing the solubility of the salt.

Consider the equilibrium for silver chloride: $$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$$

If you add aqueous sodium chloride (which supplies excess $\mathbf{Cl^-}$ ions), the concentration of $\mathbf{Cl^-}$ increases.

According to Le Chatelier, the system reduces the concentration of $Cl^-$ by shifting the equilibrium to the left, causing more $AgCl$ to precipitate (solidify).

Result: The solubility of $AgCl$ is lower in a solution containing a common ion ($Cl^-$) than it is in pure water. (Syllabus 25.1.10(a))

Practical Application: This effect is used in chemical analysis. For example, when testing for halide ions, silver nitrate is added to precipitate $AgCl$. To ensure maximum precipitation, a large excess of the precipitating agent is often used, relying on the common ion effect.

Key Takeaway for Section 5: $K_{sp}$ is the equilibrium constant for dissolving sparingly soluble salts. Adding an ion already present in the equilibrium (the common ion effect) forces the equilibrium left, drastically reducing the solubility of the salt.