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Chemistry — Class 12 — L12: Ionic Equilibrium

NIOS Code 313 · Module 5 · Chemical Dynamics

Notes extracted from NIOS Chemistry Course (313), Lesson 12 — Ionic Equilibrium (313_Chemistry_Eng_Lesson12.pdf). Content covers sections 12.1–12.8.
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Overview — Module 5: Ionic Equilibrium

In the first lesson of Module 5 you studied chemical equilibrium and Le Chatelier's principle. This lesson extends those ideas to equilibria involving ions — especially acid-base equilibria and solubility equilibria of sparingly soluble salts. Acid-base chemistry is central to living systems (blood pH ~7.4, saliva ~6.8), agriculture, and industrial processes. Buffer solutions maintain pH; the solubility product governs precipitation — including the calcium phosphate in bones and teeth.

You will learn strong and weak electrolytes, degree of ionization and Ostwald's dilution law, Arrhenius/Brønsted-Lowry/Lewis acid-base concepts, Ka and Kb, auto-ionization of water and Kw, the pH scale, common ion effect, buffer solutions and the Henderson-Hasselbalch equation, salt hydrolysis, solubility product Ksp, and applications in qualitative analysis.

Section 1: Strong and Weak Electrolytes (12.1)

Electrolytes are compounds that produce ions in aqueous solution and conduct electricity. Strong electrolytes (NaCl, KCl, HCl, NaOH) ionize almost completely — shown with a single arrow. Weak electrolytes (CH₃COOH, NH₄OH, C₆H₅NH₂) ionize partially; a dynamic ionic equilibrium exists between unionized molecules and ions (reversible arrows).

Strong vs Weak Electrolytes Strong NaCl → Na⁺ + Cl⁻ ~100% ionized Weak CH₃COOH ⇌ H₃O⁺ + CH₃COO⁻ Partial ionization, equilibrium Degree of ionization α = fraction ionized
Strong electrolytes: complete dissociation. Weak: equilibrium between molecules and ions.

The degree of ionization α is the fraction of electrolyte present as ions. For AB ⇌ An+ + Bn− with initial concentration c:

K = cα²/(1−α)  ≈  cα²  |  α = √(K/c)
Ostwald's Dilution Law — α increases with dilution (↓c) at constant T · Example: NH₄OH 0.001 M, K=1.8×10⁻⁵ → α=0.134

At equilibrium: concentrations are c(1−α) for AB, cα for each ion. For weak electrolytes α ≪ 1, so (1−α) ≈ 1 and K ≈ cα². Qualitatively, dilution promotes ionization — more water molecules available to solvate ions.

The ionization constant K is a characteristic property of the electrolyte at a given temperature. Strong electrolytes have very large effective K values (complete dissociation), while weak electrolytes have small K values reflecting partial ionization. Conductivity measurements and colligative properties (from L7) can also reveal the extent of dissociation — electrolytes with i > 1 show abnormal colligative behavior when dissociation occurs.

Dynamic equilibrium means the rates of forward (ionization) and reverse (recombination) processes are equal. Adding solvent (dilution) decreases c and shifts equilibrium toward more ionized form per Ostwald's law — this is why weak acid conductivity increases on dilution even though total ion count per unit volume may change in a complex way.

Section 2: Acid-Base Concepts (12.2–12.3)

12.2.1 Arrhenius Concept

Acid: produces H⁺ (H₃O⁺) in water — HA → H⁺ + A⁻. Base: produces OH⁻ — MOH → M⁺ + OH⁻. Limitations: aqueous only; cannot explain acidic AlCl₃ or basic NH₃/Na₂CO₃ (no OH⁻ in formula).

12.2.2 Brønsted-Lowry Concept

Acid: proton (H⁺) donor. Base: proton acceptor. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ — NH₃ accepts proton (base), H₂O donates (acid). Conjugate acid-base pairs differ by one H⁺: NH₃/NH₄⁺, H₂O/OH⁻, CH₃COOH/CH₃COO⁻. In any acid-base reaction, the product on the left is the conjugate base of the reactant acid, and vice versa.

12.2.3 Lewis Concept

Acid: electron pair acceptor (AlCl₃, BF₃, Fe³⁺). Base: electron pair donor (NH₃). AlCl₃ + NH₃ → Cl₃Al←NH₃ (coordinate bond). Explains acidity without H⁺. All Brønsted acids are also Lewis acids (H⁺ accepts electron pair from base), but Lewis definition is broader — CO₂ + H₂O → H₂CO₃ involves Lewis acid-base interaction at carbon. Fajan's rules and polarizing power connect to Lewis acidity of small, highly charged cations that hydrolyze water.

Relative strength (12.3): Strong acids (HCl) ionize completely; weak acids (CH₃COOH) partially. Strong bases (NaOH) fully dissociate; weak bases (NH₄OH) establish equilibrium. Larger Ka → stronger acid; larger Kb → stronger base.

In Brønsted equilibria, the stronger acid has the weaker conjugate base. HCl is completely ionized because Cl⁻ is a weaker base than H₂O — it cannot accept the proton back. HF is only partially ionized because F⁻ is a stronger base than H₂O and competes for the proton. This inverse relationship between acid strength and conjugate base strength is fundamental to predicting reaction direction: acids react with bases whose conjugate acids are weaker.

Amphiprotic species can act as both acid and base — HCO₃⁻ is basic toward HF but acidic toward CN⁻; H₂O is acidic toward NH₃ and basic toward HCl. Neutralization in Arrhenius terms (H⁺ + OH⁻ → H₂O) is proton transfer in Brønsted terms. The Lewis concept completes the picture for reactions like AlCl₃ + NH₃ where no proton transfers but a coordinate bond forms.

Section 3: Quantitative Strengths — Ka, Kb, α (12.4)

Weak acid: HA + H₂O ⇌ H₃O⁺ + A⁻

Ka = [H₃O⁺][A⁻]/[HA]  |  Kb = [B⁺][OH⁻]/[BOH]
CH₃COOH Ka=1.8×10⁻⁵ · HCN Ka=4.9×10⁻¹⁰ · Larger K → stronger · Polyprotic acids: Ka1 > Ka2 (H₂SO₄)

Degree of dissociation: α = √(K/c) for weak acids/bases when α is small. Example 12.2: For weak base with Kb and concentration c, same Ostwald form applies. Percent dissociation = α × 100.

Polyprotic acids (H₂SO₄, H₃PO₄) ionize stepwise; each step has its own Ka. First ionization is always strongest because removing H⁺ from a neutral molecule is easier than from a negatively charged ion.

For a conjugate acid-base pair: Ka × Kb = Kw. A weak acid with Ka = 1.8×10⁻⁵ has conjugate base CH₃COO⁻ with Kb = Kw/Ka = 5.6×10⁻¹⁰. Example 12.1 asks for Ka expression for acetic acid: Ka = [H₃O⁺][CH₃COO⁻]/[CH₃COOH]. Comparing Ka values ranks acid strength: acetic acid (1.8×10⁻⁵) is much stronger than hydrocyanic acid (4.9×10⁻¹⁰).

Percent dissociation = α × 100%. As concentration increases, α decreases (Ostwald) even though absolute ion concentration may increase. Exam problems often give K and c and ask for α, [H₃O⁺], or pH — always write the equilibrium table with initial and equilibrium concentrations first.

Section 4: Auto-Ionization, pH and Common Ion Effect (12.5)

Water self-ionizes: 2H₂O ⇌ H₃O⁺ + OH⁻. Ionic product: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. In pure water: [H₃O⁺] = [OH⁻] = 10⁻⁷ M (neutral).

pH Scale at 25°C 7 neutral Acidic pH<7 Basic pH>7 0 14 pH + pOH = 14
pH = −log[H₃O⁺]. Neutral = 7; acidic < 7; basic > 7.
pH = −log[H₃O⁺]  |  pOH = −log[OH⁻]  |  pH + pOH = 14
Examples: 0.01 M HCl → pH=2 · 0.01 M NaOH → pH=12 · Rain pH=5 → [H₃O⁺]=10⁻⁵ M

Example 12.3: In 0.01 M HCl, [H₃O⁺] ≈ 0.01 M (strong acid dominates); [OH⁻] = Kw/0.01 = 10⁻¹² M. Example 12.7: 0.1 M CH₃COOH, α=0.0134 → [H₃O⁺]=0.00134 → pH=2.87.

The p notation extends beyond pH: pOH = −log[OH⁻], pKa = −log Ka, pKw = 14. Taking logs of Kw = [H₃O⁺][OH⁻] gives pKw = pH + pOH. Strongly acidic solutions can have pH < 0; strongly alkaline can exceed 14 — but the 0–14 range covers most laboratory and biological systems. Sorensen's pH scale (1909) replaced awkward powers of ten with manageable numbers.

Self-ionization equilibrium applies in every aqueous solution — even strong acid solutions contain trace OH⁻ from water, and strong base solutions contain trace H₃O⁺. The dominant species sets pH; the minor species comes from Kw. In 0.01 M HCl, water's contribution to [H₃O⁺] (10⁻¹²) is negligible compared to 0.01 — validating the assumption that [H₃O⁺] equals strong acid concentration.

12.5.3 Common Ion Effect

Adding a salt with a common ion suppresses dissociation of weak acid/base (Le Chatelier). CH₃COOH + CH₃COONa: acetate ion shifts equilibrium left — α decreases. Example 12.8: 0.1 M CH₃COOH + 0.1 M CH₃COONa → [H₃O⁺]=1.85×10⁻⁵, pH=4.73, α=1.85×10⁻⁴.

Section 5: Buffer Solutions (12.6)

Buffers resist pH change when small amounts of acid or base are added. Two types: (i) weak acid + salt of conjugate base (CH₃COOH + CH₃COONa — acidic buffer, pH < 7); (ii) weak base + salt of conjugate acid (NH₄OH + NH₄Cl — basic buffer, pH > 7).

Buffer Action — CH₃COOH / CH₃COONa Add H⁺ (strong acid) H⁺ + CH₃COO⁻ → CH₃COOH Add OH⁻ (strong base) OH⁻ + CH₃COOH → CH₃COO⁻ Acid reserve (HA) and base reserve (A⁻) consume added H⁺ or OH⁻ Blood, enzymes, industrial processes need stable pH
Added acid reacts with base reserve; added base reacts with acid reserve — [H₃O⁺] nearly unchanged.
pH = pKa + log([Salt]/[Acid])  |  pOH = pKb + log([Salt]/[Base])
Henderson-Hasselbalch · Ex 12.9: 0.1M/0.1M acetic/acetate → pH=pKa=4.73 · Ex 12.10: NH₄OH/NH₄Cl → pOH=8.25, pH=5.75

When [Salt]=[Acid], pH = pKa. Biological fluids use carbonate, phosphate, and protein buffers to maintain pH within narrow limits essential for enzyme activity and oxygen transport.

Example 12.10: 0.1 M NH₄OH + 0.01 M NH₄Cl gives pOH = 9.25 + log(0.01/0.1) = 8.25, hence pH = 5.75. Buffer capacity depends on total concentrations of acid and base reserves — higher concentrations absorb more added acid/base before pH shifts significantly. Buffer action fails if large amounts of strong acid or base are added (reserves exhausted) or if components are too dilute.

The common ion effect and buffer action are linked: a buffer is essentially a weak electrolyte solution with high common-ion concentration from added salt. CH₃COOH alone has pH ~2.87 (0.1 M); adding CH₃COONa raises pH toward pKa and stabilizes it against further change.

Section 6: Salt Hydrolysis (12.7)

Some salts give acidic or basic solutions through hydrolysis (reaction with water):

  • SA + SB (NaCl): no hydrolysis → neutral
  • SA + WB (NH₄Cl): NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ → acidic
  • WA + SB (CH₃COONa): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ → basic
  • WA + WB (CH₃COONH₄): both ions hydrolyze — nature depends on relative Ka and Kb

Cations of weak bases and anions of weak acids hydrolyze; ions from strong acids/bases do not. This explains why Na₂CO₃ is basic despite no OH⁻ in the formula — CO₃²⁻ hydrolyzes.

For WA + WB salts like CH₃COONH₄, both NH₄⁺ (weak base cation) and CH₃COO⁻ (weak acid anion) hydrolyze. If Ka of the acid part equals Kb of the base part, the solution is approximately neutral. If Ka > Kb, solution is slightly acidic; if Kb > Ka, slightly basic. Hydrolysis constant Kh = Kw/K for the conjugate — weaker conjugate means stronger hydrolysis.

Practical applications: aqueous FeCl₃ is acidic (Fe³⁺ hydrolyzes — Lewis acid behavior); baking soda (NaHCO₃) is weakly basic; laundry detergents often contain basic phosphates. Understanding salt hydrolysis explains why pH indicators behave differently in solutions of different salts.

Section 7: Solubility Equilibrium (12.8)

For sparingly soluble salts like AgCl: AgCl(s) ⇌ Ag⁺ + Cl⁻ (heterogeneous equilibrium). Solubility product: Ksp = [Ag⁺][Cl⁻]. Solid activity = 1 by convention.

Ksp and Solubility (s) AB: K_sp = s² AB₂: K_sp = 4s³ A₂B: K_sp = 4s³ A_xB_y: K_sp = x^x · y^y · s^(x+y) Ex 12.11: CaSO₄ s=4.9×10⁻³ → K_sp=2.4×10⁻⁵ Ex 12.12: AgI K_sp=8.5×10⁻¹⁷ → s=9.2×10⁻⁹ Common ion: AgI in 0.1M AgNO₃ → s=8.5×10⁻¹⁶
Solubility s relates to Ksp by stoichiometry. Common ion drastically lowers solubility.
Ksp = [Ay+]x[Bx−]y = xxyys(x+y)
Common ion effect: Adding AgNO₃ to AgI solution → [Ag⁺] high → s drops · Qualitative analysis uses selective precipitation via Ksp

12.8.3 Qualitative analysis: Cation group separation uses controlled [S²⁻] via H₂S dissociation — acidic medium precipitates Group II sulphides (low [S²⁻]); alkaline medium precipitates Group IV (higher [S²⁻]). Ksp determines whether precipitation occurs when ion product exceeds Ksp.

Ion product Q compared to Ksp: if Q < Ksp, unsaturated (no precipitate); Q = Ksp, saturated; Q > Ksp, supersaturated — precipitation occurs until Q returns to Ksp. Ca₃(PO₄)₂ in bones has very small Ksp — slightly soluble but biologically essential. Fluoridation shifts equilibrium by common ion effect (F⁻) modifying tooth enamel solubility.

Example 12.13 dramatically shows common ion suppression: AgI solubility drops from 9.2×10⁻⁹ M in water to 8.5×10⁻¹⁶ M in 0.1 M AgNO₃ — a factor of ~10⁷. This principle is used in gravimetric analysis (precipitating ions completely) and in controlling water hardness removal.

Exam Connections and Chapter Summary

Terminal exercises cover: α and K calculations; pH of strong/weak acids and bases; Henderson equation for buffers; Ksp and solubility conversions; common ion effect on solubility. Intext 12.1 tests conjugate pairs and Lewis acids. Intext 12.2: HF Ka expression, glycine pH, lime juice pH.

Key skills: write Ka, Kb, Ksp expressions; use Kw to find [H₃O⁺] or [OH⁻]; apply Henderson-Hasselbalch; predict salt solution pH from hydrolysis type; calculate s from Ksp and vice versa; explain how buffers and common ions work via Le Chatelier.

Ionic equilibrium connects the abstract equilibrium constant to measurable pH, buffer design, water chemistry, bone mineral solubility, and laboratory qualitative analysis — one of the most practically important chapters in NIOS Class 12 Chemistry.

Intext 12.3 practice: benzoic acid/sodium benzoate buffer with pKa=4.2 — apply Henderson directly. Ag₂SO₄ Ksp from [SO₄²⁻]=2.5×10⁻² requires writing correct dissolution stoichiometry (2 Ag⁺ per formula unit). Always state whether assumptions (α << 1, [common ion] dominates) are valid and check them after calculation.

Module 5 links equilibrium (Lesson 1) with ionic systems (this lesson) and will extend to electrochemistry (L13) where electrode potentials relate to Gibbs energy and equilibrium constants. Mastering K, Ka, Kb, Kw, and Ksp as specific applications of the law of mass action prepares you for the unified treatment of chemical thermodynamics and kinetics across the syllabus.

MCQ Quiz — L12 Ionic Equilibrium

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Flashcards — L12

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Golden Rules — L12 Ionic Equilibrium

Most exam-important points from this chapter:

Electrolytes & α

Strong = full ionization. Weak = equilibrium with K and α. Ostwald: α = √(K/c); dilute → more ionization.

Acid-base concepts

Arrhenius (H⁺/OH⁻ in water), Brønsted (proton transfer, conjugate pairs), Lewis (e⁻ pair acceptor/donor). Larger K_a = stronger acid.

pH & K_w

K_w = 10⁻¹⁴ at 298 K. pH = −log[H₃O⁺]. pH + pOH = 14. Strong acid: [H₃O⁺] = c. Neutral: both ions 10⁻⁷ M.

Buffers & common ion

Buffer = weak electrolyte + conjugate salt. Henderson: pH = pK_a + log([Salt]/[Acid]). Common ion suppresses α (Le Chatelier).

K_sp & hydrolysis

K_sp(AB)=s²; K_sp(AB₂)=4s³. Common ion lowers solubility. Salt: SA+SB neutral, SA+WB acidic, WA+SB basic.

K = cα²/(1−α)
α = √(K/c) (Ostwald)
Ka = [H₃O⁺][A⁻]/[HA]
Kb = [B⁺][OH⁻]/[BOH]
Kw = 1.0×10⁻¹⁴
pH = −log[H₃O⁺]
pH + pOH = 14
pH = pKa + log[Salt]/[Acid]
Ksp = [A⁺][B⁻] (AB type)

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry 313, Module 5 — Ionic Equilibrium (sections 12.1–12.8).

Weak Electrolyte — K = cα²/(1−α) ≈ cα²

Ostwald's Dilution Law: α = √(K/c) at constant T

Where

• α = degree of ionization (fraction ionized)
• c = initial concentration (mol dm⁻³)
• K = ionization/dissociation constant

When to use

Calculate α for weak electrolytes; Example: NH₄OH 0.001 M, K = 1.8×10⁻⁵ → α = 0.134.

Memory aid: "Dilute more → α grows" — α inversely related to √c.

❌ Using (1−α) ≈ 1 when α is large.
✓ Approximation valid only for weak electrolytes (α << 1).

Worked Examples

Exam

Q: α for 0.001 M NH₄OH if K = 1.8×10⁻⁵?

α = √(1.8×10⁻⁵/0.001)

Answer: α = 0.134 (13.4%)

Weak Acid — Ka = [H₃O⁺][A⁻]/[HA]

Weak base: Kb = [B⁺][OH⁻]/[BOH]

Strength

Larger Ka → stronger acid (CH₃COOH Ka = 1.8×10⁻⁵; HCN Ka = 4.9×10⁻¹⁰)

Relation

Ka·Kb = Kw for conjugate acid-base pair

Intermediate

Q: pH of 0.1 M CH₃COOH, α = 0.0134?

[H₃O⁺] = cα = 0.00134; pH = −log(0.00134)

Answer: pH = 2.87 (Example 12.7)

Ionic Product of Water — Kw = [H₃O⁺][OH⁻] = 1.0×10⁻¹⁴ (298 K)

Pure/neutral water: [H₃O⁺] = [OH⁻] = 1.0×10⁻⁷ mol dm⁻³

Solution types

Acidic: [H₃O⁺] > [OH⁻] · Basic: [H₃O⁺] < [OH⁻] · Neutral: equal

Basic

Q: pH of 0.01 M HCl?

[H₃O⁺] = 0.01; pH = 2.0

Answer: pH = 2.0 (Example 12.4)

Intermediate

Q: pH of 0.010 M NaOH?

[OH⁻]=10⁻²; [H₃O⁺]=10⁻¹²; pH=12

Answer: pH = 12 (Example 12.5)

pH Scale — pH = −log[H₃O⁺]  |  pOH = −log[OH⁻]

Relation: pH + pOH = pKw = 14 at 298 K

Ranges

Neutral pH = 7 · Acidic pH < 7 · Basic pH > 7 · Usual range 0–14

Henderson-Hasselbalch — Acidic Buffer

pH = pKa + log([Salt]/[Acid])

Basic buffer: pOH = pKb + log([Salt]/[Base])

Buffer types

Acidic: weak acid + salt (CH₃COOH + CH₃COONa) · Basic: weak base + salt (NH₄OH + NH₄Cl)

Exam

Q: pH of 0.1 M CH₃COOH + 0.1 M CH₃COONa (pKa=4.73)?

pH = 4.73 + log(0.1/0.1) = 4.73

Answer: pH = 4.73 (Example 12.9)

Solubility Product — Ksp

AB: Ksp = s²  |  AB₂: Ksp = 4s³  |  A₂B: Ksp = 4s³

AxBy: Ksp = xx·yy·s(x+y)

Common ion effect

Adding common ion shifts equilibrium left → solubility decreases (AgI in AgNO₃).

Intermediate

Q: Ksp for CaSO₄ if s = 4.9×10⁻³ M?

Ksp = s²

Answer: 2.4×10⁻⁵ mol² dm⁻⁶ (Example 12.11)

Exam

Q: Solubility of AgI if Ksp = 8.5×10⁻¹⁷?

s = √Ksp

Answer: 9.2×10⁻⁹ mol dm⁻³ (Example 12.12)

Section 2: Detailed Definitions

DEFINITION: Strong Electrolyte
Almost completely ionized in solution (NaCl, HCl, NaOH). Single arrow in equation.

DEFINITION: Weak Electrolyte
Partially ionized; dynamic equilibrium between molecules and ions (CH₃COOH, NH₄OH).

DEFINITION: Conjugate Acid-Base Pair
Differ by one H⁺. HA/A⁻ or BH⁺/B. In any equilibrium, acid₁/base₂ and base₁/acid₂ are conjugate pairs.

DEFINITION: Buffer Solution
Resists pH change on adding small amounts of acid or base. Contains conjugate acid-base pair at high concentration.

DEFINITION: Salt Hydrolysis
Reaction of salt ions with water. SA+SB → neutral; SA+WB → acidic; WA+SB → basic; WA+WB → depends on relative Ka/Kb.

Section 3: Diagrams & Visuals

Ionic Equilibrium Map — L12 Weak acid K_a K_w, pH Henderson K_sp Common ion → suppress dissociation / solubility Arrhenius · Brønsted-Lowry · Lewis acid-base concepts Salt hydrolysis: SA+WB acidic · WA+SB basic · SA+SB neutral pH 7 neutral · <7 acid · >7 base · pH+pOH=14

Weak electrolytes → pH → buffers → solubility equilibria

K_sp vs SOLUBILITY (s) ═══════════════════════════════════════ AB type: K_sp = s² AB₂ type: K_sp = 4s³ A₂B type: K_sp = 4s³ A_xB_y: K_sp = x^x · y^y · s^(x+y) Common ion: s decreases (Le Chatelier) ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Strong vs weak electrolyte?

Strong: almost complete ionization (NaCl, HCl). Weak: partial ionization with equilibrium (CH₃COOH). Weak electrolytes have K and α.

Q2: Arrhenius vs Brønsted-Lowry acid?

Arrhenius: produces H⁺ in water. Brønsted-Lowry: proton donor (works in non-aqueous media). NH₃ is Brønsted base (accepts H⁺) but not Arrhenius base (no OH⁻).

Q3: What is conjugate pair in NH₃ + H₂O → NH₄⁺ + OH⁻?

NH₃/NH₄⁺ and H₂O/OH⁻. Acid donates proton; base accepts. Reverse reaction has reverse roles.

Q4: Why is CH₃COONa solution basic?

Weak acid + strong base salt. CH₃COO⁻ hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ — generates OH⁻.

Q5: Common ion effect on weak acid?

Adding salt with common anion (CH₃COONa to CH₃COOH) shifts ionization left — α decreases, pH increases slightly from pure weak acid.

Q6: How do buffers work?

Added H⁺ reacts with base reserve (A⁻); added OH⁻ reacts with acid reserve (HA). [H₃O⁺] changes minimally — pH stays nearly constant.

Q7: pH when [Salt]=[Acid] in buffer?

pH = pKa + log(1) = pKa. Equal concentrations give pH equal to pKa of weak acid.

Q8: Why is NH₄Cl solution acidic?

Strong acid + weak base salt. NH₄⁺ hydrolyzes: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ — generates H₃O⁺.

Q9: Ksp for AgCl expression?

AgCl(s) ⇌ Ag⁺ + Cl⁻; Ksp = [Ag⁺][Cl⁻]. Solid concentration taken as 1 (heterogeneous equilibrium).

Q10: Common ion effect on AgI solubility?

In 0.1 M AgNO₃, [Ag⁺] is high → solubility drops from 9.2×10⁻⁹ to 8.5×10⁻¹⁶ mol dm⁻³ (Example 12.13).

Q11: Lewis acid example?

Electron pair acceptor — AlCl₃ + NH₃ → Cl₃Al←NH₃. BF₃, Fe³⁺ also Lewis acids. Explains acidity without H⁺.

Q12: Exam — pH of rain water pH=5?

[H₃O⁺] = 10⁻⁵ mol dm⁻³ (Example 12.6). Acid rain has pH below 5.6 (normal CO₂ equilibrium).

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • pH 7: neutral at 25°C
  • pH + pOH: always 14
  • Buffer: weak + conjugate salt
  • K_sp AB: s = √K_sp
  • Ostwald: dilute → ionize more

Exam Tips

  • Strong acid/base: [H₃O⁺] or [OH⁻] = concentration
  • Weak acid: use Ka or given α
  • Henderson: log ratio of salt to acid/base
  • K_sp: write dissolution equation first
  • Common ion: add external ion concentration

Section 7: Connections & Relationships

Builds on: L7 solutions (concentration units), chemical equilibrium (Le Chatelier, K).
Leads to: L13 electrochemistry, qualitative analysis, biochemistry (blood buffers pH 7.4).
Related: van't Hoff factor (L7); ΔG and K (equilibrium module).

Section 8: Complete Quick Reference

• K = cα²/(1−α) ≈ cα² · α = √(K/c)

• Ka, Kb · Kw = 10⁻¹⁴ · pH = −log[H₃O⁺] · pH + pOH = 14

• pH = pKa + log([Salt]/[Acid]) · pOH = pKb + log([Salt]/[Base])

• Ksp(AB) = s² · Ksp(AB₂) = 4s³ · s = (Ksp/xxyy)1/(x+y)

Acid-base concepts: Arrhenius · Brønsted-Lowry · Lewis

Remember: ✓ Common ion suppresses ✓ Buffers resist pH change ✓ SA+SB salt neutral ✓ Ksp heterogeneous

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L12 — Ionic Equilibrium only. Use Model Answer for marking points; Explanation for concept clarity.

L12 — Ionic Equilibrium

4 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section B — Short / Long answer (from papers)

PYQ1. Derive the unit for Ksp of the salt of AB type. Explain, why the salt solution of a strong acid and a weak base like NH4Cl is acidic in nature. AB àH$ma Ho$ bdU Ho$ {bE Ksp Ho$ ‘mÌH$

2 marks · Q32 · 313/MAY/205A

Model Answer

Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.

Explanation

Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q32 · 2 mark(s) · L12.

PYQ2. Derive the unit for Ksp of the salt of AB type. Explain, why the salt solution of a strong acid and a weak base like NH4Cl is acidic in nature. AB àH$ma Ho$ bdU Ho$ {bE Ksp Ho$ ‘mÌH$

2 marks · Q33 · 313/MAY/205B

Model Answer

Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.

Explanation

Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q33 · 2 mark(s) · L12.

PYQ3. Derive the unit for Ksp of the salt of AB type. Explain, why the salt solution of a strong acid and a weak base like NH4Cl is acidic in nature. AB àH$ma Ho$ bdU Ho$ {bE Ksp Ho$ ‘mÌH$

2 marks · Q33 · 313/MAY/205C

Model Answer

Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.

Explanation

Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q33 · 2 mark(s) · L12.

PYQ4. What is meant by ionic product constant of water? Write its mathematical expression

2 marks · Q33 · 313/TUS/105A

Model Answer

Ionic product of water Kw is the product of molar concentrations of H⁺ and OH⁻ in pure water (or aqueous solution) at a given temperature: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.

Explanation

At 25 °C, pure water has [H⁺] = [OH⁻] = 10⁻⁷ M. Kw rises with temperature. Linked to pH scale (L12).

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q33 · 2 mark(s) · L12.

Problem Solving — L12 Ionic Equilibrium

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6pH

Draw a pH scale sketch. Calculate pH of 0.010 M HCl (strong acid, complete dissociation).

pH = −log[H⁺]
pH + pOH = 14 (25°C)

Pencil sketch (labelled)

pH scale 0 7 14 acidic neutral basic pH = −log[H⁺] · pOH = −log[OH⁻] pH + pOH = 14 (25°C)
Pencil sketch: pH scale 0–14 labelled

Solution — step by step with formulas

  1. [H⁺] = 0.010 = 10⁻² M.
  2. pH = 2.0.

Final answer: pH = 2.0

Formulas used in this problem

pH = −log[H⁺]
pH + pOH = 14 (25°C)

Textbook formal language

pH is the negative logarithm of hydrogen ion concentration (activity ideally).

Working formulas: pH = −log[H⁺]; pH + pOH = 14 (25°C). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

0.01 M strong acid → [H⁺]=0.01 → pH 2 (acidic).

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — pH definition

At 25°C, pH + pOH = 14 for aqueous solutions.

Linked to chapter notes (L12). Remember: pH = −log[H⁺]; pH + pOH = 14 (25°C). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write pH = −log[H⁺]; pH + pOH = 14 (25°C) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Ka

Write K_a expression for weak acid HA ⇌ H⁺ + A⁻.

K_a = [H⁺][A⁻]/[HA]

Solution — step by step with formulas

  1. K_a = [H⁺][A⁻]/[HA].

Final answer: K_a = [H⁺][A⁻]/[HA]

Formulas used in this problem

K_a = [H⁺][A⁻]/[HA]

Textbook formal language

Acid dissociation constant measures acid strength; larger K_a ⇒ stronger acid.

Working formulas: K_a = [H⁺][A⁻]/[HA]. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

How much the acid splits into ions at equilibrium.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Weak acid equilibrium

Often use ICE tables for weak acid pH problems.

Linked to chapter notes (L12). Remember: K_a = [H⁺][A⁻]/[HA]. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write K_a = [H⁺][A⁻]/[HA] before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Buffer

Draw/label components of an acidic buffer. State Henderson–Hasselbalch equation.

pH = pK_a + log([salt]/[acid]) (Henderson–Hasselbalch)

Pencil sketch (labelled)

Acidic buffer weak acid HA + salt A⁻ (conjugate base) pH = pKₐ + log([salt]/[acid])
Pencil sketch: buffer components labelled

Solution — step by step with formulas

  1. Weak acid + salt of conjugate base.
  2. pH = pK_a + log([A⁻]/[HA]).

Final answer: pH = pK_a + log([salt]/[acid])

Formulas used in this problem

pH = pK_a + log([salt]/[acid]) (Henderson–Hasselbalch)

Textbook formal language

Buffers resist pH change on small addition of acid/base via common ion.

Working formulas: pH = pK_a + log([salt]/[acid]) (Henderson–Hasselbalch). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Mix weak acid with its salt—like a shock absorber for pH.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Buffer solution

Maximum buffer capacity near pH = pK_a.

Linked to chapter notes (L12). Remember: pH = pK_a + log([salt]/[acid]) (Henderson–Hasselbalch). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write pH = pK_a + log([salt]/[acid]) (Henderson–Hasselbalch) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Kw

In pure water at 25°C, [H⁺] = ?

K_w = [H⁺][OH⁻] = 10⁻¹⁴ (25°C)

Solution — step by step with formulas

  1. [H⁺] = [OH⁻] = 10⁻⁷ M; pH = 7.

Final answer: [H⁺] = 1.0 × 10⁻⁷ M

Formulas used in this problem

K_w = [H⁺][OH⁻] = 10⁻¹⁴ (25°C)

Textbook formal language

Autoionisation of water: H₂O ⇌ H⁺ + OH⁻ with K_w = 10⁻¹⁴ at 25°C.

Working formulas: K_w = [H⁺][OH⁻] = 10⁻¹⁴ (25°C). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Pure water is neutral: equal H⁺ and OH⁻ at 10⁻⁷ each.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Ionic product of water

K_w increases with temperature.

Linked to chapter notes (L12). Remember: K_w = [H⁺][OH⁻] = 10⁻¹⁴ (25°C). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write K_w = [H⁺][OH⁻] = 10⁻¹⁴ (25°C) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Solubility

For sparingly soluble AB(s) ⇌ A⁺ + B⁻, write K_sp.

K_sp = [A⁺][B⁻] for AB(s)

Solution — step by step with formulas

  1. K_sp = [A⁺][B⁻].

Final answer: K_sp = [A⁺][B⁻]

Formulas used in this problem

K_sp = [A⁺][B⁻] for AB(s)

Textbook formal language

Solubility product is the equilibrium constant for dissolution of a sparingly soluble salt.

Working formulas: K_sp = [A⁺][B⁻] for AB(s). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

How many ions can sit dissolved before solid starts precipitating.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Solubility product

Common ion effect reduces solubility.

Linked to chapter notes (L12). Remember: K_sp = [A⁺][B⁻] for AB(s). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write K_sp = [A⁺][B⁻] for AB(s) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Hydrolysis

Is aqueous CH₃COONa acidic, basic or neutral? Why?

Salt of weak acid + strong base → basic solution

Solution — step by step with formulas

  1. Basic: CH₃COO⁻ hydrolyses to give OH⁻.

Final answer: Basic (acetate hydrolysis)

Formulas used in this problem

Salt of weak acid + strong base → basic solution

Textbook formal language

Anions of weak acids hydrolyse producing OH⁻; cations of weak bases produce H⁺.

Working formulas: Salt of weak acid + strong base → basic solution. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Acetate steals H⁺ from water leaving OH⁻—solution turns basic.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Salt hydrolysis idea

Classify salts by parent acid/base strength.

Linked to chapter notes (L12). Remember: Salt of weak acid + strong base → basic solution. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Salt of weak acid + strong base → basic solution before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.