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Ionic Equilibria

Ch. 3Std 12

Easy Overview

Ever used a pH strip in biology class and wondered how it works? Or why your stomach can digest food but your mouth cannot? The answer lies in ionic equilibria — the tug-of-war between molecules that want to stay together and those that want to split into ions. This chapter gives you the math to calculate exactly how acidic or basic a solution is, how to make solutions that resist pH change (buffers), and how to predict whether a salt will dissolve or precipitate. When you dissolve an electrolyte in water, it behaves in one of two ways. Strong electrolytes like HCl, NaOH, and NaCl dissociate COMPLETELY — every molecule falls apart into ions. There is no equilibrium here; the reaction goes to completion. Weak electrolytes like acetic acid (CH₃COOH) or ammonia (NH₃) only partially dissociate, setting up a dynamic equilibrium between the undissociated molecule and its ions. This is where the interesting chemistry — and the calculations — begin. For a weak acid HA ⇌ H⁺ + A⁻, we define K_a = [H⁺][A⁻]/[HA]. This acid dissociation constant tells you how strong the acid is — larger K_a means more dissociation. Similarly for weak bases, K_b = [BH⁺][OH⁻]/[B]. Water itself ionizes: H₂O ⇌ H⁺ + OH⁻, and K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. This is the rock-bottom foundation of acid-base chemistry. A beautiful relationship: K_a × K_b = K_w for any conjugate acid-base pair. The pH scale makes tiny hydrogen ion concentrations manageable. pH = -log[H⁺]. A change of one pH unit means a 10-fold change in [H⁺]. Then come buffers — solutions that resist pH change like a stubborn mule. A buffer has either a weak acid plus its salt (CH₃COOH + CH₃COONa) or a weak base plus its salt (NH₃ + NH₄Cl). The Henderson-Hasselbalch equation (pH = pK_a + log[salt]/[acid]) governs buffer pH. Blood's buffer system (H₂CO₃/HCO₃⁻) keeps your blood pH at 7.35-7.45 — if it deviates outside this range, you are in serious medical trouble. Solubility product K_sp tells you when a salt will precipitate — essential for qualitative analysis. Understanding these principles connects classroom chemistry to medicine, biology, and everyday life.

Strong vs weak electrolytes

An electrolyte dissociates in water to produce ions that conduct electricity. Strong electrolytes: 100% dissociation — no equilibrium. Examples: strong acids (HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄), strong bases (NaOH, KOH, Ca(OH)₂, Ba(OH)₂), and most soluble salts (NaCl, KBr, NH₄NO₃). Their solutions conduct electricity well. Weak electrolytes: partial dissociation — equilibrium between ions and undissociated molecules. Examples: weak acids (CH₃COOH, H₂CO₃, H₃PO₄, HF), weak bases (NH₃, C₆H₅NH₂), and some salts (HgCl₂). Their solutions conduct poorly. The calculation approach differs completely. For strong acids: pH = -log C. For weak acids: you must use K_a. Common mistake: confusing concentration with strength — 0.1 M acetic acid is still a weak acid. Strength is about the tendency to dissociate (K_a), not the actual concentration. HCl is strong at ANY concentration.

Degree of ionization and Ostwald's dilution law

Degree of ionization (α) = fraction of electrolyte molecules that dissociate. For a weak electrolyte AB ⇌ A⁺ + B⁻ with initial concentration C: at equilibrium, [AB] = C(1 - α), [A⁺] = Cα, [B⁻] = Cα. K = (Cα × Cα)/[C(1 - α)] = Cα²/(1 - α). For weak electrolytes, α is very small (1 - α ≈ 1): K = Cα², so α = √(K/C). This is Ostwald's dilution law — as C decreases (dilution), α increases. At infinite dilution, α → 1. For 0.1 M acetic acid (K_a = 1.8 × 10⁻⁵): α = √(1.8 × 10⁻⁵/0.1) = 0.0134 or 1.34%. This means only 13 out of 1000 molecules are dissociated. Factors affecting α: (1) Nature of electrolyte (K value). (2) Concentration — dilution increases α. (3) Temperature — ionization generally increases with temperature (endothermic). (4) Common ion effect — adding an ion present in equilibrium suppresses ionization. Ostwald's law fails for strong electrolytes because they are already completely dissociated.

Ionization of water and K_w

Water autoionizes: H₂O ⇌ H⁺ + OH⁻. The equilibrium constant K = [H⁺][OH⁻]/[H₂O]. Since [H₂O] ≈ 55.5 M and essentially constant, K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. K_w increases with temperature — at 50°C it is about 5.5 × 10⁻¹⁴, at 100°C about 5.1 × 10⁻¹³. In pure water at 25°C, [H⁺] = [OH⁻] = 10⁻⁷ M. Important exam nuance: at higher temperatures, the pH of pure water is less than 7 (because K_w is larger), but it is still NEUTRAL because [H⁺] = [OH⁻]. Acidity depends on the relative concentrations of H⁺ and OH⁻, not on the absolute pH value. K_w is the master key relating [H⁺] and [OH⁻] in any aqueous solution. For 0.1 M HCl: [H⁺] = 0.1 M, so [OH⁻] = 10⁻¹⁴/0.1 = 10⁻¹³ M. This tiny concentration is negligible but theoretically present.

pH, pOH, and the logarithmic scale

pH = -log₁₀[H⁺]. This converts tiny concentrations into manageable numbers. For pure water at 25°C: [H⁺] = 10⁻⁷, pH = 7. pOH = -log[OH⁻], and pH + pOH = 14 at 25°C (from K_w). For strong acids: pH = -log C (if C ≥ 10⁻⁶ M). For weak acids (HA): [H⁺] = √(K_a × C) (approximation), and pH = -½ log K_a - ½ log C = ½ pK_a - ½ log C. For weak bases (B): [OH⁻] = √(K_b × C), pOH = -½ log K_b - ½ log C, and pH = 14 - pOH. Example: 0.1 M acetic acid (K_a = 1.8 × 10⁻⁵): [H⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M, pH = -log(1.34 × 10⁻³) = 2.87. For 0.1 M NaOH (strong base): [OH⁻] = 0.1 M, pOH = 1, pH = 13. Watch out: for very dilute strong acids (< 10⁻⁶ M), the contribution from water autoionization becomes significant. A 10⁻⁸ M HCl solution does NOT have pH = 8 (which would be basic) — you must solve the full charge balance equation.

K_a, K_b, and the conjugate seesaw

For a weak acid HA: K_a = [H⁺][A⁻]/[HA]. For its conjugate base A⁻: K_b = [HA][OH⁻]/[A⁻]. Multiply: K_a × K_b = [H⁺][OH⁻] = K_w = 1.0 × 10⁻¹⁴. So K_a × K_b = K_w and pK_a + pK_b = 14. This is the conjugate seesaw — a strong acid (small pK_a) has a weak conjugate base (large pK_b). HCl (pK_a ≈ -7) has Cl⁻ with no measurable basicity (pK_b ≈ 21). Acetic acid (K_a = 1.8 × 10⁻⁵, pK_a = 4.76) has acetate: K_b = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰, pK_b = 9.24. NH₃ (K_b = 1.8 × 10⁻⁵) has NH₄⁺: K_a = 5.56 × 10⁻¹⁰. This relationship means you never need separate K_a and K_b tables — given one, you calculate the other. Classic exam question: 'K_a for acetic acid is 1.8 × 10⁻⁵. Find K_b for acetate ion.' Answer: K_b = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.

Salt hydrolysis — why salt solutions can be acidic or basic

Not all salt solutions are neutral. When a salt dissolves, its ions may react with water (hydrolysis), changing pH. Four cases: (1) Salt of strong acid + strong base (NaCl, KNO₃): neither ion hydrolyzes → neutral, pH = 7. (2) Salt of weak acid + strong base (CH₃COONa, Na₂CO₃): the anion hydrolyzes: A⁻ + H₂O ⇌ HA + OH⁻ → basic, pH > 7. (3) Salt of strong acid + weak base (NH₄Cl, AlCl₃): the cation hydrolyzes: BH⁺ + H₂O ⇌ B + H₃O⁺ → acidic, pH < 7. (4) Salt of weak acid + weak base (CH₃COONH₄): both ions hydrolyze — pH depends on relative K_a and K_b values. For CH₃COONa 0.1 M (K_a = 1.8 × 10⁻⁵): pH = 7 + ½pK_a + ½log C = 7 + 2.38 + (-0.5) = 8.88. For NH₄Cl 0.1 M (K_b = 1.8 × 10⁻⁵): pH = 7 - ½pK_b - ½log C = 7 - 2.38 - (-0.5) = 5.12. Common mistake: assuming all salt solutions are neutral. Always check the parent acid and base.

Buffer solutions — the pH guardians

A buffer resists pH change when small amounts of acid or base are added. Two types: acidic buffer (weak acid + its salt with strong base) and basic buffer (weak base + its salt with strong acid). How it works: take CH₃COOH + CH₃COONa. Add H⁺ → CH₃COO⁻ + H⁺ → CH₃COOH (H⁺ consumed). Add OH⁻ → CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (OH⁻ consumed). The buffer contains both a reservoir of the weak acid (to neutralize added base) and its conjugate base (to neutralize added acid). Buffer capacity is the amount of acid/base the buffer can neutralize before pH changes significantly. It depends on absolute concentrations of components and the ratio [salt]/[acid]. Maximum buffer capacity occurs when [salt] = [acid]. The effective pH range is pK_a ± 1. Beyond this range, the buffer cannot handle large additions. Real-life: blood buffer (H₂CO₃/HCO₃⁻, pK_a = 6.35) maintains blood pH at 7.35-7.45 — the ratio [HCO₃⁻]/[H₂CO₃] ≈ 20:1, controlled by lungs and kidneys.

Henderson-Hasselbalch equation — the buffer formula

This is the most important equation in buffer chemistry. For an acidic buffer (HA + A⁻): pH = pK_a + log([A⁻]/[HA]) = pK_a + log([salt]/[acid]). Derivation: from K_a = [H⁺][A⁻]/[HA], rearrange: [H⁺] = K_a × [HA]/[A⁻]. Take -log: pH = pK_a - log([HA]/[A⁻]) = pK_a + log([A⁻]/[HA]). For a basic buffer (B + BH⁺): pOH = pK_b + log([BH⁺]/[B]) and pH = 14 - pOH. When [salt] = [acid], pH = pK_a — the buffer is most effective. Example: prepare a buffer with pH 5.0 using acetic acid (pK_a = 4.76). 5.0 = 4.76 + log([salt]/[acid]), so log([salt]/[acid]) = 0.24, [salt]/[acid] = 10⁰·²⁴ = 1.74. You need 1.74 moles of sodium acetate for every mole of acetic acid. Classic problem: 'Calculate pH of a buffer containing 0.1 M CH₃COOH and 0.2 M CH₃COONa (K_a = 1.8 × 10⁻⁵).' pK_a = -log(1.8 × 10⁻⁵) = 4.74. pH = 4.74 + log(0.2/0.1) = 4.74 + 0.30 = 5.04.

Common ion effect

The common ion effect is a direct application of Le Chatelier's principle. When you add an ion already present in a dissociation equilibrium, the equilibrium shifts to SUPPRESS dissociation. Example: CH₃COOH ⇌ H⁺ + CH₃COO⁻. Add CH₃COONa (provides CH₃COO⁻) → equilibrium shifts left → [H⁺] decreases → pH increases. This is how buffers work. Another example: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Add NH₄Cl (provides NH₄⁺) → equilibrium shifts left → [OH⁻] decreases → less basic. In qualitative analysis: when you add HCl to a mixture of Group I cations (Ag⁺, Pb²⁺, Hg₂²⁺), the Cl⁻ from HCl precipitates these ions as chlorides. The excess Cl⁻ also supplies a common ion, suppressing dissociation of other metal chlorides — this is why CuCl₂, FeCl₃, etc. stay dissolved. For Group III analysis: NH₄Cl + NH₄OH together — NH₄⁺ (common ion) suppresses NH₄OH dissociation, keeping [OH⁻] low enough that only Fe³⁺, Al³⁺, Cr³⁺ precipitate as hydroxides. Exam tip: the common ion must be identical to an ion in the equilibrium, not just any ion.

Solubility product K_sp — predicting precipitation

For a sparingly soluble salt A_xB_y(s) ⇌ xA^(y⁺)(aq) + yB^(x⁻)(aq): K_sp = [A^(y⁺)]ˣ × [B^(x⁻)]ʸ at saturation (constant temperature). Examples: AgCl(s) ⇌ Ag⁺ + Cl⁻, K_sp = 1.8 × 10⁻¹⁰ M². PbI₂(s) ⇌ Pb²⁺ + 2I⁻, K_sp = 7.1 × 10⁻⁹ M³. Fe(OH)₃: K_sp = 1.1 × 10⁻³⁶ M⁴. Smaller K_sp = less soluble, but you cannot directly compare K_sp for different stoichiometries — convert to solubility (S in mol/L) first. The ion product Q is calculated like K_sp but from actual concentrations at any moment. Compare Q and K_sp: Q = K_sp → saturated (equilibrium); Q < K_sp → unsaturated (more can dissolve); Q > K_sp → supersaturated → precipitation occurs until Q = K_sp. Example: mix 50 mL 0.01 M AgNO₃ + 50 mL 0.02 M NaCl. After mixing: [Ag⁺] = 0.005 M, [Cl⁻] = 0.01 M. Q = [Ag⁺][Cl⁻] = 5 × 10⁻⁵ > 1.8 × 10⁻¹⁰ → AgCl precipitates. This principle is the basis of all qualitative analysis separations.

Solubility and K_sp — the math

The relationship between molar solubility (S) and K_sp depends on stoichiometry. AB type (1:1, like AgCl): K_sp = S², S = √K_sp. Example: K_sp of AgCl = 1.8 × 10⁻¹⁰, S = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M. AB₂ or A₂B type (PbI₂, Ag₂CrO₄): PbI₂ → Pb²⁺ + 2I⁻. If S mol/L dissolves: [Pb²⁺] = S, [I⁻] = 2S. K_sp = S × (2S)² = 4S³. S = ³√(K_sp/4). For PbI₂ (K_sp = 7.1 × 10⁻⁹): S = ³√(1.775 × 10⁻⁹) = 1.21 × 10⁻³ M. AB₃ type (Fe(OH)₃): K_sp = 27S⁴. General formula: for AₓB_y, K_sp = xˣ × yʸ × S^(x+y). Common mistake: forgetting the stoichiometric coefficients. For PbI₂, [I⁻] = 2S, not S. If you use just S, your answer is off by a factor of 4. Exam favorite: 'Calculate solubility from K_sp' and 'Calculate K_sp from solubility' — practice both directions.

Common ion effect on solubility

Adding a common ion DECREASES the solubility of a sparingly soluble salt. Le Chatelier's principle: AgCl(s) ⇌ Ag⁺ + Cl⁻. Add NaCl (supplies Cl⁻) → equilibrium shifts left → more AgCl precipitates. Numerical example: Solubility of AgCl in water = 1.34 × 10⁻⁵ M. Solubility in 0.01 M NaCl: Let S' be the new solubility. [Ag⁺] = S', [Cl⁻] = 0.01 + S' ≈ 0.01 (since S' is tiny). K_sp = S' × 0.01 → S' = K_sp/0.01 = 1.8 × 10⁻¹⁰/0.01 = 1.8 × 10⁻⁸ M. About 750 times less soluble! This is why AgCl is much less soluble in brine than in pure water. In gravimetric analysis, a common ion is added to ensure complete precipitation and minimize losses. In qualitative analysis, excess HCl is added to Group I to ensure complete precipitation of AgCl, PbCl₂, and Hg₂Cl₂. Board exam favorite: 'Calculate solubility of BaSO₄ in 0.01 M Na₂SO₄ (K_sp = 1.1 × 10⁻¹⁰).'

Ionic equilibria in biological systems

Your body is a walking ionic equilibrium experiment. Blood pH is maintained at 7.35-7.45 by the bicarbonate buffer: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. Acidosis (pH < 7.35): body breathes faster to exhale CO₂, shifting equilibrium left. Alkalosis (pH > 7.45): breathing slows to retain CO₂. The kidneys provide slower but more powerful regulation by excreting H⁺ or HCO₃⁻. Cells use the phosphate buffer (H₂PO₄⁻/HPO₄²⁻) internally (pH ≈ 6.9). Stomach acid (HCl, pH ≈ 1.5-2) activates digestive enzymes (pepsin) and kills bacteria. The stomach lining is protected by mucus and HCO₃⁻ secretion. The small intestine has pH ≈ 7-8 where pancreatic enzymes work best. Each protein has an isoelectric point (pI) — the pH at which it has no net charge and is least soluble. Below pI, the protein is positively charged; above pI, it is negatively charged. This affects protein structure, separation (isoelectric focusing), and function. Medical IV fluids, cell culture media, and pharmaceutical formulations all use carefully designed buffer systems. Without ionic equilibria, life as we know it would not exist.

Key Points

  • •Strong electrolytes: 100% dissociation; Weak: partial dissociation (equilibrium)
  • •Ostwald's dilution law: α = √(K/C) for weak electrolytes — diluted = more dissociated
  • •K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C; increases with temperature
  • •pH = -log[H⁺]; pOH = -log[OH⁻]; pH + pOH = 14 at 25°C
  • •For weak acids: [H⁺] = √(K_a × C); pH = ½pK_a - ½log C
  • •K_a × K_b = K_w; pK_a + pK_b = 14 — conjugate seesaw relationship
  • •Salt hydrolysis: strong + strong → neutral; weak + strong → basic; strong + weak → acidic
  • •Buffer = weak acid + its salt (acidic) OR weak base + its salt (basic)
  • •Henderson-Hasselbalch: pH = pK_a + log([salt]/[acid]); maximum capacity at ratio = 1
  • •Effective buffer range: pK_a ± 1
  • •Common ion effect: adding common ion → suppresses dissociation (Le Chatelier)
  • •K_sp = solubility product; Q = ion product; Q > K_sp → precipitation
  • •AB type: K_sp = S²; AB₂: K_sp = 4S³; AB₃: K_sp = 27S⁴
  • •Common ion decreases solubility of sparingly soluble salts
  • •Blood buffer: H₂CO₃/HCO₃⁻, pH 7.35-7.45, controlled by lungs and kidneys
  • •Qualitative analysis uses K_sp and common ion effect for selective precipitation

Practice Questions

  • Calculate the pH of 0.001 M HCl and 0.01 M NaOH. What is [OH⁻] in the HCl solution?
  • Define buffer. Derive the Henderson-Hasselbalch equation. Calculate pH of a buffer with 0.1 M CH₃COOH and 0.2 M CH₃COONa (K_a = 1.8 × 10⁻⁵).
  • The solubility of BaSO₄ in water is 1.05 × 10⁻⁵ M. Calculate K_sp. Also find solubility in 0.01 M Na₂SO₄.
  • Explain common ion effect with examples. How is it used in qualitative analysis for Group II precipitation?
  • Distinguish strong and weak electrolytes. The degree of ionization of 0.1 M CH₃COOH is 1.34%. Calculate K_a.
  • Calculate pH of 0.1 M CH₃COOH and 0.1 M CH₃COONa separately (K_a = 1.8 × 10⁻⁵). Why is the salt pH higher?
  • What is salt hydrolysis? Explain why CH₃COONa is basic, NH₄Cl is acidic, and NaCl is neutral.
  • If you mix 25 mL 0.001 M AgNO₃ with 75 mL 0.002 M NaCl, will AgCl precipitate? (K_sp = 1.8 × 10⁻¹⁰)