This MCQ module is based on: Ionic Equilibrium
TOPIC 27 OF 28
Ionic Equilibrium
🎓 Class 11
Chemistry
CBSE
Theory
Ch 6 – Equilibrium
⏱ ~14 min
🌐 Language: [gtranslate]
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Ionic Equilibria, pH, Buffer and Solubility Product
6.13 Ionic Equilibrium in Solution
Many salts, acids and bases ionise in water. The equilibrium between the un-ionised molecules (or solid) and ions in solution is called ionic equilibrium. Strong electrolytes (NaCl, HCl, NaOH) dissociate ~100%; weak electrolytes (CH₃COOH, NH₃) dissociate only partially.
6.14 Acids and Bases — Three Theories
| Theory | Acid | Base | Example |
|---|---|---|---|
| Arrhenius (1887) | Releases H⁺ in water | Releases OH⁻ in water | HCl, NaOH |
| Brønsted–Lowry (1923) | Proton (H⁺) donor | Proton (H⁺) acceptor | NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ |
| Lewis (1923) | Electron-pair acceptor | Electron-pair donor | BF₃ + NH₃ → F₃B–NH₃ |
6.14.1 Conjugate Acid–Base Pairs (Brønsted)
For HA + B ⇌ A⁻ + BH⁺, the pairs (HA, A⁻) and (B, BH⁺) differ by one proton — they are conjugate pairs. A strong acid has a weak conjugate base and vice versa.
6.15 Ionisation of Water — Kw
Water itself dissociates slightly:
\[2\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)\] \[K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at 298 K}\]In pure water, [H₃O⁺] = [OH⁻] = √Kw = 10⁻⁷ M.
6.16 The pH Scale
pH = −log₁₀[H⁺]; pOH = −log₁₀[OH⁻]; pH + pOH = 14 at 298 K.
Acidic: pH < 7 | Neutral: pH = 7 | Basic: pH > 7.
Acidic: pH < 7 | Neutral: pH = 7 | Basic: pH > 7.
6.17 Strength of Acids — Ka and Kb
For a weak acid HA: HA ⇌ H⁺ + A⁻; Ka = [H⁺][A⁻]/[HA]. pKa = −log Ka.
For a weak base B: B + H₂O ⇌ BH⁺ + OH⁻; Kb = [BH⁺][OH⁻]/[B]; pKb = −log Kb.
For a conjugate acid–base pair: Ka × Kb = Kw ⇒ pKa + pKb = 14 (at 298 K).
| Weak acid | Ka | pKa |
|---|---|---|
| HF | 6.8 × 10⁻⁴ | 3.17 |
| CH₃COOH | 1.8 × 10⁻⁵ | 4.74 |
| HCN | 4.9 × 10⁻¹⁰ | 9.31 |
| NH₄⁺ | 5.6 × 10⁻¹⁰ | 9.25 |
| H₂CO₃ (Ka1) | 4.3 × 10⁻⁷ | 6.37 |
6.17.1 Degree of Ionisation (α) and Ostwald's Dilution Law
For a weak acid HA of concentration c:
\[K_a = \frac{c\alpha^2}{1-\alpha} \quad\overset{\alpha\ll 1}{\approx}\quad c\alpha^2\] \[\Rightarrow \alpha = \sqrt{K_a/c}\]So dilution increases the degree of ionisation but decreases [H⁺].
6.18 Hydrolysis of Salts
| Salt of | Solution pH | Example |
|---|---|---|
| Strong acid + strong base | Neutral (7) | NaCl, KNO₃ |
| Strong acid + weak base | Acidic (< 7) | NH₄Cl, FeCl₃ |
| Weak acid + strong base | Basic (> 7) | CH₃COONa, NaCN |
| Weak acid + weak base | Depends on Ka vs Kb | NH₄CN (basic if Kb > Ka) |
6.19 Buffer Solutions
Buffer: A solution that resists pH change on small additions of acid or base. Two main types:
Acidic buffer: weak acid + its salt (e.g., CH₃COOH + CH₃COONa)
Basic buffer: weak base + its salt (e.g., NH₃ + NH₄Cl)
Acidic buffer: weak acid + its salt (e.g., CH₃COOH + CH₃COONa)
Basic buffer: weak base + its salt (e.g., NH₃ + NH₄Cl)
The Henderson–Hasselbalch equation gives buffer pH:
\[\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}\]Maximum buffer capacity: pH = pKa (when [salt] = [acid]).
6.20 Common Ion Effect
Adding a salt with a common ion to a weak acid/base solution suppresses ionisation (Le Chatelier).
Example: Add NH₄Cl to NH₄OH solution → [NH₄⁺] rises → equilibrium shifts left → less OH⁻ → solution less basic.
6.21 Solubility Product (Ksp)
For a sparingly soluble salt MX(s) ⇌ M⁺(aq) + X⁻(aq):
\[K_{sp} = [\text{M}^+][\text{X}^-]\]
For M_aX_b: K_sp = [M^(b+)]^a [X^(a−)]^b. Smaller Ksp = less soluble salt.
| Salt | Ksp at 25 °C |
|---|---|
| AgCl | 1.8 × 10⁻¹⁰ |
| AgBr | 5.4 × 10⁻¹³ |
| AgI | 8.5 × 10⁻¹⁷ |
| BaSO₄ | 1.1 × 10⁻¹⁰ |
| CaF₂ | 3.4 × 10⁻¹¹ |
| PbCl₂ | 1.7 × 10⁻⁵ |
6.21.1 Ionic Product (Q_sp) and Precipitation
| Comparison | Result |
|---|---|
| Q_sp < K_sp | Unsaturated; more solute can dissolve |
| Q_sp = K_sp | Saturated; equilibrium |
| Q_sp > K_sp | Supersaturated; precipitate forms |
🎯 Interactive: Buffer pH Designer (Henderson–Hasselbalch)
Pick a weak acid and a salt:acid ratio. The simulator gives the buffer pH.
pH = pKa + log([salt]/[acid]) = 4.74
Equimolar mixture: maximum buffer capacity at pH = pKa.
🧪 Activity 6.4 — Common-Ion Effect
Setup: Take three test tubes, each with 5 mL of saturated NH₃ solution and 2 drops of phenolphthalein (pink). Add: (a) nothing, (b) a few drops of NH₄Cl solution, (c) a few drops of dilute HCl.
Predict: What happens to the pink colour in each tube?
(a) Pink stays — NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ unchanged.
(b) Pink fades — adding NH₄⁺ shifts equilibrium LEFT (common-ion effect), [OH⁻] drops, less basic.
(c) Pink disappears completely — HCl removes OH⁻ outright by neutralisation.
Lesson: The common-ion effect is just Le Chatelier's principle applied to ionic equilibrium.
Worked Example 6.8: pH of a Weak Acid
Calculate the pH and degree of ionisation of 0.10 M CH₃COOH; Ka = 1.8 × 10⁻⁵.
α = √(Ka/c) = √(1.8 × 10⁻⁵ / 0.10) = √(1.8 × 10⁻⁴) = 0.0134 (1.34%).
[H⁺] = cα = 0.10 × 0.0134 = 1.34 × 10⁻³ M.
pH = −log(1.34 × 10⁻³) = 2.87.
[H⁺] = cα = 0.10 × 0.0134 = 1.34 × 10⁻³ M.
pH = −log(1.34 × 10⁻³) = 2.87.
Worked Example 6.9: Buffer pH
A buffer is made of 0.20 M CH₃COOH and 0.30 M CH₃COONa. Find pH. (pKa = 4.74)
pH = pKa + log([salt]/[acid]) = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92.
Worked Example 6.10: Solubility from Ksp
Calculate the molar solubility of AgCl in pure water; Ksp = 1.8 × 10⁻¹⁰.
Let s = molar solubility. AgCl(s) ⇌ Ag⁺ + Cl⁻. [Ag⁺] = [Cl⁻] = s.
Ksp = s² → s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M.
That is ~1.9 mg of AgCl per litre — almost insoluble.
Ksp = s² → s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M.
That is ~1.9 mg of AgCl per litre — almost insoluble.
🎯 Competency-Based Questions
Q1. According to Brønsted–Lowry, an acid is a: L1 Remember
Answer: (b) Proton donor.
Q2. The pH of pure water at 50 °C (where Kw = 5.5 × 10⁻¹⁴) is: L3 Apply
Answer: [H⁺] = √Kw = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ M. pH = −log(2.35 × 10⁻⁷) = 6.63. Note: water is still NEUTRAL ([H⁺] = [OH⁻]) but pH < 7 because Kw is greater at higher T.
Q3. Why is NH₄Cl solution acidic? L2 Understand
Answer: NH₄Cl is the salt of a strong acid (HCl) and a weak base (NH₃). The cation NH₄⁺ undergoes hydrolysis: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, releasing H₃O⁺ → solution is acidic (pH < 7).
Q4. Calculate the pH of 0.05 M NaOH. L3 Apply
Answer: NaOH is a strong base; complete dissociation. [OH⁻] = 0.05 M → pOH = −log(0.05) = 1.30 → pH = 14 − 1.30 = 12.70.
Q5. HOT (Analyse): 0.10 mol HCl is added to 1 L of buffer (0.10 M CH₃COOH + 0.10 M CH₃COONa). Find the new pH (pKa = 4.74). L4 Analyse
Answer: Initial: pH = 4.74 (equimolar buffer).
HCl converts acetate (CH₃COO⁻) to acetic acid: 0.10 mol salt becomes 0 mol salt and acid becomes 0.20 mol.
Wait — only 0.10 mol HCl is added but only 0.10 mol salt available — all consumed.
After: [acid] = 0.20 M, [salt] = 0 → no longer a buffer; this is just a weak acid solution. Better: add LESS HCl.
Practical demo: add only 0.02 mol HCl to the buffer. Then [acid] = 0.12 M, [salt] = 0.08 M → pH = 4.74 + log(0.08/0.12) = 4.74 − 0.18 = 4.56 (small change!). Compare with 0.02 mol HCl in pure water → pH would be ~1.7. Buffer drastically reduces pH change.
HCl converts acetate (CH₃COO⁻) to acetic acid: 0.10 mol salt becomes 0 mol salt and acid becomes 0.20 mol.
Wait — only 0.10 mol HCl is added but only 0.10 mol salt available — all consumed.
After: [acid] = 0.20 M, [salt] = 0 → no longer a buffer; this is just a weak acid solution. Better: add LESS HCl.
Practical demo: add only 0.02 mol HCl to the buffer. Then [acid] = 0.12 M, [salt] = 0.08 M → pH = 4.74 + log(0.08/0.12) = 4.74 − 0.18 = 4.56 (small change!). Compare with 0.02 mol HCl in pure water → pH would be ~1.7. Buffer drastically reduces pH change.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: A buffer solution resists drastic changes in pH on addition of small amounts of acid or base.
R: A buffer contains comparable amounts of a weak acid and its conjugate base.
Answer: (A). Both true; R explains A. The conjugate base neutralises added acid; the weak acid neutralises added base.
A: AgCl is more soluble in pure water than in 0.1 M NaCl.
R: Common ion (Cl⁻) suppresses dissolution of AgCl.
Answer: (A). Both true; R explains A. By Le Chatelier, increasing [Cl⁻] shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ backward, reducing solubility.
A: The pH of a salt of a strong acid and strong base is 7.
R: Such salts undergo extensive hydrolysis.
Answer: (C). A is TRUE — at 298 K, NaCl, KNO₃ etc. give pH 7. R is FALSE — these salts do NOT undergo significant hydrolysis precisely because their parent acid and base are both strong.
Frequently Asked Questions — Ionic Equilibria, pH, Buffer and Solubility Product
What are the three theories of acids and bases?
NCERT Class 11 Chemistry Chapter 6 covers three theories of acids and bases: (1) Arrhenius — acids produce H⁺ in water (HCl), bases produce OH⁻ (NaOH); limited to aqueous solutions; (2) Brønsted-Lowry — acids are proton donors, bases are proton acceptors; works in any solvent (NH₃ + HCl → NH₄⁺ + Cl⁻ even in gas phase); conjugate acid-base pairs are introduced; (3) Lewis — acids are electron-pair acceptors (BF₃, H⁺, Cu²⁺), bases are electron-pair donors (NH₃, OH⁻, H₂O); broadest definition, covers reactions without protons such as BF₃ + NH₃ → F₃B-NH₃.
What is pH and how is it calculated for strong and weak acids?
pH = −log₁₀[H⁺] expresses hydrogen ion concentration on a logarithmic scale. Pure water at 25°C has [H⁺] = 10⁻⁷ M, so pH = 7 (neutral). For a strong acid (HCl, HNO₃, H₂SO₄), [H⁺] = molar concentration, so 0.1 M HCl has pH = 1. For a weak acid (CH₃COOH, K_a = 1.8 × 10⁻⁵), [H⁺] = √(K_a × C) for dilute solutions, so 0.1 M acetic acid has [H⁺] = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ M, pH ≈ 2.87. NCERT Class 11 Chemistry Chapter 6 also defines pOH = −log[OH⁻] and pH + pOH = 14 at 25°C.
What is a buffer solution and how does it work?
A buffer solution resists changes in pH when a small amount of acid or base is added. Two types: (1) acidic buffer — weak acid + its conjugate base salt (CH₃COOH + CH₃COONa); (2) basic buffer — weak base + its conjugate acid salt (NH₃ + NH₄Cl). The Henderson-Hasselbalch equation gives pH = pK_a + log([salt]/[acid]) for acidic buffers. NCERT Class 11 Chemistry Chapter 6 explains that added H⁺ reacts with the conjugate base and added OH⁻ reacts with the weak acid, both regenerating buffer components and keeping pH nearly constant. Maximum buffer capacity is when [salt] = [acid] (pH = pK_a).
What is the common ion effect?
The common ion effect is the suppression of ionisation of a weak electrolyte when a strong electrolyte providing a common ion is added. Example from NCERT Class 11 Chemistry Chapter 6: adding NaCH₃COO (acetate) to acetic acid (CH₃COOH) suppresses ionisation of the acid because Le Chatelier's principle shifts CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left, lowering [H⁺] and raising pH. The common ion effect is the basis for buffer action. Another example: adding NH₄Cl to NH₃ suppresses ammonia ionisation. The effect is used in qualitative analysis to precipitate selected ions.
What is solubility product (Ksp) and how is it used?
Solubility product (K_sp) is the equilibrium constant for the dissolution of a sparingly soluble salt: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), K_sp = [Ag⁺][Cl⁻]. K_sp depends only on temperature and indicates how soluble a salt is — small K_sp means low solubility. NCERT Class 11 Chemistry Chapter 6 uses K_sp to calculate molar solubility, predict precipitation (compare ionic product Q with K_sp) and explain selective precipitation. Example: K_sp(AgCl) = 1.8 × 10⁻¹⁰, so solubility s = √K_sp = 1.34 × 10⁻⁵ M. The common ion effect reduces solubility further.
What is salt hydrolysis and how does it affect pH?
Salt hydrolysis is the reaction of a salt with water producing acidic or basic solution, depending on which parent acid and base formed the salt. Four cases in NCERT Class 11 Chemistry Chapter 6: (1) salt of strong acid + strong base (NaCl) — no hydrolysis, pH = 7; (2) salt of strong acid + weak base (NH₄Cl) — cation hydrolyses, pH < 7; (3) salt of weak acid + strong base (CH₃COONa) — anion hydrolyses, pH > 7; (4) salt of weak acid + weak base (CH₃COONH₄) — both ions hydrolyse, pH depends on K_a and K_b values. Hydrolysis is the reverse of neutralisation.
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