Is it correct to say: Applying Ostwald's dilution law to a strong electrolyte such as HCl.?

The law is for weak electrolytes only. A strong acid is taken as completely dissociated: [H₃O⁺] = c × number of H⁺ released.

Is it correct to say: Writing Ksp of Ag₂CrO₄ or CaF₂ as S³ instead of 4S³.?

The ion present twice has concentration 2S, and it is squared: (2S)² × S = 4S³.

Is it correct to say: Inverting the ratio in the Henderson–Hasselbalch equation.?

For an acidic buffer it is log([salt]/[acid]). More salt raises the pH above pKa.

Is it correct to say: Using pH + pOH = 14 at every temperature.?

pKw = 14 only at 298 K. Kw increases with temperature, so neutral water has pH below 7 when hot.

Is it correct to say: Writing the pH of 10⁻⁸ M HCl as 8.?

An acid cannot have pH above 7. At such low concentration the H₃O⁺ from water cannot be ignored; the pH is slightly below 7 (about 6.98).

Is it correct to say: Using the original concentrations of two solutions in the ionic product after mixing them.?

Mixing changes the volume. For equal volumes each concentration is halved before calculating IP.

📄 Formula sheet (PDF)

Ionic Equilibria

Ostwald's dilution law, Ka and Kb, ionic product of water, pH and pOH, hydrolysis of salts, buffer solutions and solubility product — Maharashtra HSC Chemistry Ch 3

📐 16 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Derive Ostwald's dilution law (Ka = α²c/(1 − α) ≈ α²c) step by step: write the equilibrium, the concentrations at equilibrium, then the approximation for a weak electrolyte.
→In pH numericals, write [H₃O⁺] or [OH⁻] in the form m × 10⁻ⁿ first, then pH = n − log m. This avoids many log errors.
→For a strong base like Ba(OH)₂ or a strong acid like H₂SO₄, multiply the concentration by the number of OH⁻ or H⁺ ions released before taking the log.
→Learn the Henderson–Hasselbalch equation in both forms: pH = pKa + log([salt]/[acid]) for acidic buffers and pOH = pKb + log([salt]/[base]) for basic buffers.
→For Ksp numericals, write the dissociation equation and the ion concentrations in terms of S (for example [Ag⁺] = 2S for Ag₂CrO₄) before substituting.
→To predict whether a salt solution is acidic, basic or neutral, identify the parent acid and base and decide which of them is weak.

📐 Formulas(16)

Degree of Dissociation

α = \frac{moles of electrolyte dissociated}{total moles of electrolyte dissolved}

Ostwald's Dilution Law (weak acid)★ Board fav

K_{a} = \frac{α ^{2} c}{1 - α} \approx α^{2} c \Rightarrow α = \frac{K _{a}}{c}

Symbol	Meaning
$K_{a}$	Acid dissociation (ionisation) constant
$α$	Degree of dissociation
$c$	Initial molar concentration of the acid (mol L⁻¹)

Ostwald's Dilution Law (weak base)

K_{b} = \frac{α ^{2} c}{1 - α} \approx α^{2} c \Rightarrow α = \frac{K _{b}}{c}

Hydronium Ion Concentration of a Weak Acid★ Board fav

[H_{3} O^{+}] = α c = K_{a} c

Hydroxide Ion Concentration of a Weak Base

[OH^{-}] = α c = K_{b} c

Ionic Product of Water

K_{w} = [H_{3} O^{+}] [OH^{-}] = 1.0 \times 1 0^{- 14} at 298 K

pH and pOH★ Board fav

pH = - lo g_{10} [H_{3} O^{+}], pOH = - lo g_{10} [OH^{-}]

Relation between pH and pOH

pH + pOH = p K_{w} = 14 (298 K)

pKa and pKb

p K_{a} = - lo g_{10} K_{a}, p K_{b} = - lo g_{10} K_{b}

Hydrolysis of Salts — Nature of the Solution

NaCl: pH = 7, CH_{3} COONa: pH > 7, NH_{4} Cl: pH < 7

Salt of a Weak Acid and a Weak Base

K_{a} > K_{b} : acidic; K_{a} < K_{b} : basic; K_{a} = K_{b} : neutral

Henderson–Hasselbalch Equation (acidic buffer)★ Board fav

pH = p K_{a} + lo g_{10} \frac{[ salt ]}{[ acid ]}

Henderson–Hasselbalch Equation (basic buffer)

pOH = p K_{b} + lo g_{10} \frac{[ salt ]}{[ base ]}

Solubility Product

A_{x} B_{y} (s) ⇌ x A^{y +} (a q) + y B^{x -} (a q), K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}

Relation between Ksp and Molar Solubility★ Board fav

K_{s p} = x^{x} y^{y} S^{x + y}

Symbol	Meaning
$K_{s p}$	Solubility product
$S$	Molar solubility (mol L⁻¹)
$x, y$	Numbers of cations and anions in the formula AxBy

Condition for Precipitation

IP > K_{s p} : precipitate; IP = K_{s p} : saturated; IP < K_{s p} : unsaturated

✏️ Solved Examples

1Solved Exampleeasy3 steps

Calculate the pH of 0.002 M Ba(OH)₂ solution, assuming complete dissociation, at 298 K.

Each Ba(OH)₂ gives two OH⁻ ions.

[OH^{-}] = 2 \times 0.002 = 4 \times 1 0^{- 3} M

2Solved Exampleboard3 steps

Calculate the degree of dissociation and the pH of 0.1 M acetic acid. (Ka = 1.8 × 10⁻⁵)

Acetic acid is weak, so use Ostwald's dilution law with α ≪ 1.

α = \frac{K _{a}}{c} = \frac{1.8 \times 1 0 ^{- 5}}{0.1} = 1.8 \times 1 0^{- 4} = 1.342 \times 1 0^{- 2}

3Solved Exampleboard3 steps

Calculate the pH of a buffer solution containing 0.20 M CH₃COOH and 0.15 M CH₃COONa. (Ka of CH₃COOH = 1.8 × 10⁻⁵)

Find pKa.

p K_{a} = - lo g_{10} (1.8 \times 1 0^{- 5}) = 5 - 0.255 = 4.745

4Solved ExampleHOTS4 steps

The solubility of Ag₂CrO₄ at 298 K is 0.0216 g L⁻¹. Calculate its solubility product. (Molar mass of Ag₂CrO₄ = 331.7 g mol⁻¹)

Convert solubility to mol L⁻¹.

S = \frac{0.0216}{331.7} = 6.51 \times 1 0^{- 5} mol L^{- 1}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Applying Ostwald's dilution law to a strong electrolyte such as HCl.
✓The law is for weak electrolytes only. A strong acid is taken as completely dissociated: [H₃O⁺] = c × number of H⁺ released.
2
Writing Ksp of Ag₂CrO₄ or CaF₂ as S³ instead of 4S³.
✓The ion present twice has concentration 2S, and it is squared: (2S)² × S = 4S³.
3
Inverting the ratio in the Henderson–Hasselbalch equation.
✓For an acidic buffer it is log([salt]/[acid]). More salt raises the pH above pKa.
4
Using pH + pOH = 14 at every temperature.
✓pKw = 14 only at 298 K. Kw increases with temperature, so neutral water has pH below 7 when hot.
5
Writing the pH of 10⁻⁸ M HCl as 8.
✓An acid cannot have pH above 7. At such low concentration the H₃O⁺ from water cannot be ignored; the pH is slightly below 7 (about 6.98).
6
Using the original concentrations of two solutions in the ionic product after mixing them.
✓Mixing changes the volume. For equal volumes each concentration is halved before calculating IP.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
The hydroxide ion concentration of a solution is 2.5 × 10⁻⁴ M at 298 K. Find its pH.
Q2
Calculate the degree of dissociation and pH of 0.05 M NH₄OH. (Kb = 1.8 × 10⁻⁵)
Q3
Ksp of AgCl at 298 K is 1.8 × 10⁻¹⁰. Find its molar solubility in pure water and in 0.01 M NaCl.
Q4
The molar solubility of CaF₂ is 2.0 × 10⁻⁴ mol L⁻¹. Calculate its Ksp.
Q5
What ratio [CH₃COONa]/[CH₃COOH] gives a buffer of pH 5.00? (Ka = 1.8 × 10⁻⁵)
Q6
Equal volumes of 2.0 × 10⁻⁴ M AgNO₃ and 2.0 × 10⁻⁴ M NaCl are mixed. Will AgCl precipitate? (Ksp of AgCl = 1.8 × 10⁻¹⁰)

📝 Notes

Ionic Equilibria — Maharashtra HSC Overview

Chapter 3 of the Maharashtra State Board Std XII Chemistry textbook applies the law of chemical equilibrium to ions in water. The chapter has four connected parts: weak electrolytes, the pH scale, hydrolysis and buffers, and sparingly soluble salts.

Weak electrolytes to pH

For a weak acid, Ostwald's dilution law gives $\alpha = \sqrt{K_a/c}$ , and $[\text{H}_3\text{O}^+] = \alpha c = \sqrt{K_a c}$ . A weak base works the same way with $K_b$ and $[\text{OH}^-]$ . From there, $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$ and $\text{pH} + \text{pOH} = 14$ at 298 K. Strong acids and bases skip Ostwald's law: take the concentration directly, multiplied by the number of H⁺ or OH⁻ ions released.

Salts and buffers

When a salt dissolves, the ion from the weak parent hydrolyses. The weak acid's anion makes the solution basic, and the weak base's cation makes it acidic. When both parents are weak, compare $K_a$ and $K_b$ . A buffer is a weak acid with its salt (or a weak base with its salt). The Henderson–Hasselbalch equation gives its pH from $\text{p}K_a$ and the salt-to-acid ratio, so equal concentrations give $\text{pH} = \text{p}K_a$ .

Solubility product

For a sparingly soluble salt $\text{A}_x\text{B}_y$ , write each ion's concentration in terms of $S$ and substitute into $K_{sp}$ . This gives the general result $K_{sp} = x^x y^y S^{x+y}$ . To predict precipitation, compare the ionic product with $K_{sp}$ . Adding a common ion raises the ionic product, so it reduces solubility.