Acid-Base Equilibria: Notes, Solved Examples & Exam Questions | Grade 12 Chemistry Unit 1

Welcome, dear student! In this unit, we will learn about acid-base equilibria. This is one of the most important topics in Grade 12 Chemistry. Take your time, read each section carefully, and try the practice questions. Let’s begin!

1.1 Acid-Base Concepts

1.1.1 Arrhenius Concept of Acids and Bases

The Swedish chemist Svante Arrhenius gave the first successful definition of acids and bases. His idea was simple but powerful:

For example, when hydrogen chloride dissolves in water:

HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)

HCl increases H₃O⁺ concentration, so it is an Arrhenius acid.

When sodium hydroxide dissolves in water:

NaOH(s) + H₂O(l) → Na⁺(aq) + OH⁻(aq)

NaOH increases OH⁻ concentration, so it is an Arrhenius base.

Strong acids (HCl, HNO₃, HClO₄, H₂SO₄, HI, HBr) ionize completely. Strong bases (Group IA and Group IIA hydroxides, except Be) also ionize completely.

But can you see the problem with Arrhenius’s idea? Think about it — what about ammonia (NH₃)? It acts as a base but it does NOT contain OH⁻! This is one of the limitations of the Arrhenius concept.

Practice Question 1: Classify NaCl(aq) as an Arrhenius acid, Arrhenius base, or neither. Explain your answer.

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Practice Question 2: State two limitations of the Arrhenius concept of acids and bases.

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1.1.2 Brønsted-Lowry Concept of Acids and Bases

To solve the problems of Arrhenius’s theory, Brønsted and Lowry (in 1923) proposed a broader definition. Have you noticed that in acid-base reactions, something is being transferred? Let’s see:

This is a wonderful definition because it is NOT limited to water! Let’s look at ammonia reacting with water:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Here, NH₃ accepts a proton from H₂O. So NH₃ is a Brønsted-Lowry base, and H₂O is a Brønsted-Lowry acid. See? NH₃ doesn’t need OH⁻ to be a base!

Conjugate Acid-Base Pairs

Now here is a very important idea. In every Brønsted-Lowry reaction, acids and bases come in pairs. Look at this reaction:

CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)

When CH₃COOH donates a proton, it becomes CH₃COO⁻. These two form a conjugate acid-base pair. Similarly, H₂O (which accepted a proton to become H₃O⁺) and H₃O⁺ form another conjugate pair.

Worked Example 1

Question: Identify the Brønsted-Lowry acids, bases, and conjugate pairs in: NH₃ + H₂PO₄⁻ ⇌ NH₄⁺ + HPO₄²⁻

Solution: Look for proton donors and acceptors.

• H₂PO₄⁻ donates a proton to become HPO₄²⁻ → Acid (and HPO₄²⁻ is its conjugate base)
• NH₃ accepts a proton to become NH₄⁺ → Base (and NH₄⁺ is its conjugate acid)
• Conjugate pair 1: H₂PO₄⁻ / HPO₄²⁻
• Conjugate pair 2: NH₃ / NH₄⁺

Worked Example 2

Question: Identify acids, bases, and conjugate pairs in: HCl + H₂PO₄⁻ ⇌ Cl⁻ + H₃PO₄

Solution:

• HCl donates a proton → Acid; Cl⁻ is its conjugate base
• H₂PO₄⁻ accepts a proton → Base; H₃PO₄ is its conjugate acid
• Conjugate pair 1: HCl / Cl⁻
• Conjugate pair 2: H₂PO₄⁻ / H₃PO₄

Strengths of Conjugate Acid-Base Pairs

Here is a very important rule — please remember it well for your exam:

For example: HCl is a strong acid, so Cl⁻ is a very weak base. CH₃COOH is a weak acid, so CH₃COO⁻ is a relatively stronger base (compared to Cl⁻).

Practice Question 3: Identify the Brønsted-Lowry acids, bases, and conjugate pairs in: HClO₂ + H₂O ⇌ ClO₂⁻ + H₃O⁺

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Practice Question 4: In the reaction OCl⁻ + H₂O ⇌ HOCl + OH⁻, identify the acid, base, and both conjugate pairs.

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Amphiprotic Species and Auto-ionization

Can a substance be BOTH an acid and a base? Yes! Such substances are called amphiprotic species. They can either donate or accept a proton depending on what they react with.

Water is the most important amphiprotic species:

• With NH₃ (a base): H₂O donates a proton → acts as an acid
• With CH₃COOH (an acid): H₂O accepts a proton → acts as a base

Another example: HCO₃⁻

• With OH⁻: HCO₃⁻ donates H⁺ → acts as an acid
• With HF: HCO₃⁻ accepts H⁺ → acts as a base

Auto-ionization (Self-ionization) is when two identical molecules react to produce ions. Water auto-ionizes:

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

One water molecule acts as an acid, the other as a base!

Ammonia also undergoes auto-ionization: 2NH₃(l) ⇌ NH₄⁺ + NH₂⁻

Practice Question 5: Write equations to show the amphiprotic behavior of H₂PO₄⁻.

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1.1.3 Lewis Concept of Acids and Bases

The Brønsted-Lowry concept focuses on proton transfer. But what about reactions where no proton is involved at all? G.N. Lewis proposed an even broader definition:

Think about it this way: a Lewis acid has an empty orbital (it is “electron-hungry”), and a Lewis base has a lone pair of electrons to share.

Worked Example 3

Question: Identify the Lewis acid and Lewis base in: BF₃ + NH₃ → F₃B—NH₃

Solution: Boron in BF₃ has only 6 valence electrons — it needs 2 more. So BF₃ accepts an electron pair → Lewis acid. Nitrogen in NH₃ has a lone pair → it donates an electron pair → Lewis base.

Worked Example 4

Question: Identify Lewis acid and base in: CaO + CO₂ → CaCO₃

Solution: The O²⁻ in CaO donates an electron pair → Lewis base (CaO). CO₂ accepts the electron pair (carbon is electron-deficient) → Lewis acid.

Worked Example 5

Question: Identify Lewis acid and base in: BeCl₂ + 2Cl⁻ → BeCl₄²⁻

Solution: Be in BeCl₂ has only 4 electrons around it (electron-deficient) → Lewis acid. Cl⁻ ions have lone pairs to donate → Lewis bases.

Practice Question 6: Identify the Lewis acid and base in: H₂O + SO₃ → H₂SO₄

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Practice Question 7: Why is BF₃ a Lewis acid but NOT a Brønsted-Lowry acid?

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1.2 Ionic Equilibria of Weak Acids and Bases

1.2.1 Ionization of Water

Did you know that even pure water conducts electricity — just a very tiny amount? This is because water undergoes self-ionization (auto-ionization):

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

The equilibrium constant for this reaction is called the ion-product constant for water (K_w):

We often write [H⁺] instead of [H₃O⁺], so:

In pure water: [H⁺] = [OH⁻] = x, so x² = 1.0 × 10⁻¹⁴, therefore x = 1.0 × 10⁻⁷ M.

This gives us three important cases for any aqueous solution at 25°C:

• Neutral: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M
• Acidic: [H⁺] > [OH⁻] (i.e., [H⁺] > 10⁻⁷ M)
• Basic: [OH⁻] > [H⁺] (i.e., [OH⁻] > 10⁻⁷ M)

But no matter what is dissolved in water, the product [H⁺][OH⁻] always equals 1.0 × 10⁻¹⁴ at 25°C. This is extremely useful!

Worked Example 6

Question: A solution has [H₃O⁺] = 3.0 × 10⁻⁴ M at 25°C. Calculate [OH⁻]. Is the solution acidic, basic, or neutral?

Solution:

Since [H₃O⁺] > [OH⁻], the solution is acidic.

Practice Question 8: Calculate [OH⁻] in a solution where [H⁺] = 2.0 × 10⁻⁵ M at 25°C. Is the solution acidic or basic?

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Practice Question 9: In a solution, [OH⁻] = 6.7 × 10⁻² M. Calculate [H⁺] at 25°C and state if it is acidic, basic, or neutral.

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The pH Scale

Working with very small numbers like 10⁻⁷ is not convenient. Scientists use the pH scale instead:

Key pH values to remember:

• pH = 7 → Neutral
• pH < 7 → Acidic (lower pH = stronger acid)
• pH > 7 → Basic (higher pH = stronger base)

To find [H⁺] from pH: $[H^+] = 10^{-pH}$

To find [OH⁻] from pOH: $[OH^-] = 10^{-pOH}$

Worked Example 7

Question: Calculate the pH and pOH of a juice solution in which [H₃O⁺] = 5.0 × 10⁻³ M.

Solution:

Worked Example 8

Question: Human blood has pH = 7.40. Calculate [H⁺] and [OH⁻].

Solution:

Practice Question 10: An antacid solution has pH = 9.18 at 25°C. Calculate [H⁺], [OH⁻], and pOH.

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Practice Question 11: Calculate the pH of 0.0063 M HNO₃ at 25°C. (HNO₃ is a strong acid.)

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1.2.2 Measures of the Strength of Acids and Bases

How do we measure exactly how strong or weak an acid is? We have several tools: concentration of H⁺, pH, percent ionization, and acid dissociation constant (K_a).

Acid Dissociation Constant (K_a)

For a weak acid HA dissolving in water:

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

K_a tells us how much the acid ionizes. A larger K_a means a stronger acid. A smaller K_a means a weaker acid.

Percent Ionization

For weak acids, percent ionization is small. Strong acids ionize nearly 100%.

Worked Example 9

Question: A 0.250 M butyric acid solution has pH = 2.72. Determine K_a.

Solution (Step-by-step):

Step 1: Find [H₃O⁺] from pH:

Step 2: Set up the ICE table:

Step 3: Substitute into K_a expression:

Worked Example 10

Question: Calculate the pH of a 0.50 M HF solution. K_a = 6.8 × 10⁻⁴.

Solution (Step-by-step):

Step 1: ICE table:

Step 2: Write K_a expression:

Step 3: Use approximation (since HF is weak, x is small compared to 0.50):

Step 4: Check approximation: (0.018 / 0.50) × 100% = 3.6% < 5% ✓

Step 5: Calculate pH:

Worked Example 11

Question: Calculate the percent ionization of acetic acid in a 0.100 M solution. K_a = 1.8 × 10⁻⁵.

Solution:

Practice Question 12: Calculate the pH of a 0.10 M solution of acetic acid. K_a = 1.8 × 10⁻⁵.

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Practice Question 13: Calculate the percent ionization of a 0.10 M acetic acid solution that has pH = 2.89.

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Base Dissociation Constant (K_b)

For weak bases, we use K_b. The process is very similar to K_a, but we calculate [OH⁻] first instead of [H⁺].

For a weak base B:

B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq)

Worked Example 12

Question: Calculate [H₃O⁺] in a 0.20 M NH₃ solution. K_b = 1.8 × 10⁻⁵.

Solution (Step-by-step):

Step 1: ICE table:

Step 2: K_b expression:

Step 3: Solve for x:

Step 4: Find [H₃O⁺]:

Practice Question 14: For a 0.040 M NH₃ solution, calculate [OH⁻], pOH, and pH. K_b = 1.8 × 10⁻⁵.

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1.3 Common Ion Effect and Buffer Solutions

1.3.1 The Common Ion Effect

Do you remember Le Chatelier’s principle from Grade 11? It says that if you disturb a system at equilibrium, the system shifts to counteract the change. The common ion effect is a direct application of this principle.

Let’s understand with an example. Consider acetic acid in water:

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)

Now, if we add HCl (a strong acid) to this solution, HCl provides H⁺ ions. This H⁺ is already on the product side! According to Le Chatelier’s principle, the equilibrium shifts to the left, meaning less acetic acid ionizes. The degree of ionization decreases.

Similarly, if we add sodium acetate (CH₃COONa), it provides CH₃COO⁻ ions, which also shifts the equilibrium to the left.

Worked Example 13

Question: Determine [H₃O⁺] and [CH₃COO⁻] in a solution that is 0.10 M in both CH₃COOH and HCl. K_a = 1.8 × 10⁻⁵.

Solution:

HCl ionizes completely: [H⁺] from HCl = 0.10 M. The Cl⁻ is a spectator ion.

Since x is very small: (0.10 + x) ≈ 0.10 and (0.10 − x) ≈ 0.10

So [CH₃COO⁻] = 1.8 × 10⁻⁵ M and [H₃O⁺] = 0.10 + 1.8 × 10⁻⁵ ≈ 0.10 M

Notice: Compare this with pure 0.10 M CH₃COOH (where [H⁺] ≈ 1.34 × 10⁻³ M). Adding the common ion H⁺ dramatically reduced the ionization!

Practice Question 15: A solution is 0.15 M in CH₃COOH and 0.15 M in CH₃COONa. Calculate [H⁺] and pH. K_a = 1.8 × 10⁻⁵.

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1.3.2 Buffer Solutions

Have you ever wondered why our blood stays at pH 7.40 even when we eat acidic or basic food? The answer is buffer solutions!

A buffer consists of two components:

• A weak acid and its conjugate base (e.g., CH₃COOH / CH₃COO⁻)
• OR a weak base and its conjugate acid (e.g., NH₃ / NH₄⁺)

How does a buffer work? Let’s take the CH₃COOH / CH₃COONa buffer:

• If you add a strong acid (H⁺): the CH₃COO⁻ (base) neutralizes it: CH₃COO⁻ + H⁺ → CH₃COOH
• If you add a strong base (OH⁻): the CH₃COOH (acid) neutralizes it: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

In both cases, the pH changes only very slightly!

Henderson-Hasselbalch Equation

This is the most important formula for buffer calculations:

For a base buffer: pH = pK_a + log([base] / [conjugate acid]), where pK_a is for the conjugate acid.

Worked Example 14

Question: Calculate the pH of a buffer containing 0.10 M CH₃COOH and 0.10 M CH₃COONa. K_a = 1.8 × 10⁻⁵.

Solution:

Notice: When [acid] = [base], pH = pK_a. This is a very useful fact!

Worked Example 15

Question: What is the pH of a buffer made with 0.20 M NH₃ and 0.30 M NH₄Cl? K_b for NH₃ = 1.8 × 10⁻⁵.

Solution:

First find K_a for NH₄⁺ (the conjugate acid):

Practice Question 16: A buffer contains 0.15 M HCOOH and 0.25 M HCOONa. K_a = 1.8 × 10⁻⁴. Calculate the pH.

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Practice Question 17: Explain how the CH₃COOH / CH₃COONa buffer resists pH change when a small amount of NaOH is added.

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1.4 Hydrolysis of Salts

When a salt dissolves in water, sometimes the solution is NOT neutral. Why? Because the ions from the salt can react with water — this is called hydrolysis.

Have you noticed that NaCl solution is neutral, but NH₄Cl solution is acidic, and CH₃COONa solution is basic? Let’s understand why!

Type 1: Salt of Strong Acid + Strong Base

Example: NaCl (from HCl + NaOH)

Neither Na⁺ nor Cl⁻ reacts with water. The solution is neutral (pH = 7).

Type 2: Salt of Weak Acid + Strong Base

Example: CH₃COONa (from CH₃COOH + NaOH)

The CH₃COO⁻ ion reacts with water:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

This produces OH⁻, so the solution is basic (pH > 7).

Type 3: Salt of Strong Acid + Weak Base

Example: NH₄Cl (from HCl + NH₃)

The NH₄⁺ ion reacts with water:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

This produces H₃O⁺, so the solution is acidic (pH < 7).

Type 4: Salt of Weak Acid + Weak Base

Example: NH₄CH₃COO (from CH₃COOH + NH₃)

Both ions hydrolyze. The pH depends on the relative values of K_a and K_b:

• If K_a > K_b: solution is acidic
• If K_b > K_a: solution is basic
• If K_a = K_b: solution is neutral

Practice Question 18: Predict whether each solution will be acidic, basic, or neutral: (a) NaNO₃ (b) KCN (c) NH₄NO₃ (d) NaF

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1.5 Acid-Base Indicators and Titrations

1.5.1 Acid-Base Indicators

An acid-base indicator is a weak acid or weak base whose color changes depending on the pH of the solution. Indicators have a color change range (usually about 2 pH units wide).

How to choose an indicator: The indicator’s pH range should overlap with the equivalence point pH of the titration.

1.5.2 Equivalents of Acids and Bases

The equivalent of an acid or base is a measure that considers how many H⁺ or OH⁻ ions one mole of the substance can provide:

where n = number of H⁺ (for acids) or OH⁻ (for bases) per formula unit.

Examples: For HCl, n = 1; For H₂SO₄, n = 2; For Ca(OH)₂, n = 2.

1.5.3 Acid-Base Titrations

A titration is a laboratory technique to determine the unknown concentration of an acid or base by reacting it with a standard solution of known concentration.

Key terms:

• Titrant: The solution of known concentration in the burette.
• Analyte: The solution of unknown concentration in the flask.
• Equivalence point: The point where moles of H⁺ = moles of OH⁻.
• End point: The point where the indicator changes color (should be close to equivalence point).

At the equivalence point:

• Strong acid – Strong base titration: pH = 7 (use bromothymol blue)
• Strong acid – Weak base titration: pH < 7 (use methyl orange or methyl red)
• Weak acid – Strong base titration: pH > 7 (use phenolphthalein)

Worked Example 16

Question: 25.0 mL of HCl solution is titrated with 0.10 M NaOH. The equivalence point is reached at 20.0 mL of NaOH. Calculate the concentration of HCl.

Solution:

At equivalence point: moles of H⁺ = moles of OH⁻

Worked Example 17

Question: Calculate the number of equivalents in 500 mL of 0.2 M H₂SO₄.

Solution:

H₂SO₄ has n = 2 (it provides 2 H⁺ per molecule).

Practice Question 19: 30.0 mL of 0.15 M CH₃COOH is titrated with 0.10 M NaOH. What volume of NaOH is needed to reach the equivalence point? What indicator would you use?

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Practice Question 20: Calculate the number of equivalents in 200 mL of 0.5 M Ca(OH)₂.

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