Why does neutral water have a pH of 6.5 at 60°C?

The autoionization of water, H2O(l) H+(aq) + OH-(aq), is an endothermic process (ΔH > 0). According to Le Chatelier's principle, increasing temperature shifts the equilibrium forward, increasing the ionic product Kw. At 60°C, Kw ≈ 10^(-13), so [H+] = [OH-] = √(10^(-13)) = 10^(-6.5) M. Thus pH = 6.5. However, since [H+] = [OH-], the water is strictly NEUTRAL despite pH < 7.

What is Ostwald's Dilution Law and when does it fail?

Ostwald's dilution law states that for a weak binary electrolyte HA H+ + A-, the degree of dissociation is α = √(Ka / c). As concentration c decreases (infinite dilution), α approaches 1. The law fails for strong electrolytes because strong electrolytes are nearly 100% ionized at all moderate concentrations, and their deviations are governed by interionic Debye-Huckel forces rather than incomplete dissociation.

§ 6.2NCERT Class 11 · Chemistry · Chapter 6

Acids, Bases, Autoionization of Water (K_w), pH Scale & Weak Electrolytes

Q: What is Ostwald's Dilution Law and when does it fail?

Ostwald's dilution law states that for a weak binary electrolyte HA H+ + A-, the degree of dissociation is α = √(Ka / c). As concentration c decreases (infinite dilution), α approaches 1. The law fails for strong electrolytes because strong electrolytes are nearly 100% ionized at all moderate concentrations, and their deviations are governed by interionic Debye-Huckel forces rather than incomplete dissociation.

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 8 min read

Ionic equilibrium governs chemical transformations occurring in aqueous solutions. From biological enzyme kinetics to industrial acid-base titrations, understanding the quantitative behavior of hydronium ions, dissociation constants (K_a, K_b), and the logarithmic pH scale forms the analytical backbone of chemistry.

1. Classical and Modern Concepts of Acids and Bases

Theory / Model	Definition of Acid	Definition of Base	Limitations / Key Examples
Arrhenius Theory	Furnishes H⁺ (or H₃O⁺) in aqueous solution (e.g., HCl, HNO₃)	Furnishes OH⁻ in aqueous solution (e.g., NaOH, KOH)	Restricted to aqueous solvents; cannot explain basicity of NH₃ or acidity of AlCl₃.
Brønsted-Lowry Theory	Proton (H⁺) Donor	Proton (H⁺) Acceptor	Introduces Conjugate Acid-Base Pairs differing by exactly one proton (H⁺).
Lewis Electronic Theory	Electron-pair Acceptor (electrophile, vacant orbitals, e.g., BF₃, AlCl₃, Fe³⁺)	Electron-pair Donor (nucleophile, lone pair, e.g., :NH₃, H₂O:, F⁻)	Most universal concept; operates in gas phase and non-aqueous media without proton transfer.

Conjugate Acid-Base Pairs Rule

In any Brønsted acid-base reaction: HA + B ⇔ BH⁺ + A⁻. The species A⁻ is the conjugate base of HA, and BH⁺ is the conjugate acid of B. Fundamental principle:

Strength Relationship: A strong acid has a very weak conjugate base (e.g., HCl is strong, Cl⁻ has negligible basicity). Conversely, a weak acid has a relatively strong conjugate base (e.g., CH₃COOH is weak, CH₃COO⁻ hydrolyzes readily).

2. Autoionization of Water & The Ionic Product (K_w)

Pure liquid water undergoes self-ionization to a tiny extent:

H₂O(l) + H₂O(l) ⇔ H₃O⁺(aq) + OH⁻(aq) (ΔH > 0, Endothermic)

The equilibrium constant expression is:

Ionic Product of Water (K_w):
K_w = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴ (at 298 K / 25°C)
Taking negative logarithm: pK_w = pH + pOH = 14.00 (at 298 K).

Critical Pitfall — Temperature Variation of Kw:
Because water autoionization is endothermic, increasing temperature increases K_w (e.g., at 373 K / 100°C, K_w ≈ 5.5 × 10⁻¹³, so pK_w ≈ 12.26). At 100°C, neutral water has [H⁺] = [OH⁻] ≈ 7.4 × 10⁻⁷ M, meaning pH of neutral water at 100°C is about 6.1, NOT 7.0! Neutrality means [H⁺] = [OH⁻], not pH = 7.

pH Scale & Ionic Concentration Spectrum (at 298 K)

Logarithmic scale mapping hydronium and hydroxide concentrations

[H+] (M)   10^0   10^-2   10^-4   10^-6   10^-7   10^-8   10^-10  10^-12  10^-14
pH          0      2       4       6       7       8       10      12      14
           |------- STRONGLY ACIDIC ------| NEUTRAL |------- STRONGLY BASIC ------|
pOH        14     12      10       8       7       6        4       2       0
[OH-] (M) 10^-14 10^-12  10^-10  10^-8   10^-7   10^-6   10^-4   10^-2   10^0
--------------------------------------------------------------------------------
Key Formulation: pH = -log10 [H3O+]    |    pOH = -log10 [OH-]    |    pH + pOH = pKw

Extreme Dilution Alert: For an acid solution of 10⁻⁸ M HCl, the pH is NOT 8 (an acid cannot have alkaline pH!). Contribution of H⁺ from water autoionization (~10⁻⁷ M) must be accounted for: [H⁺]_total = 10⁻⁸ + x, where x(10⁻⁸ + x) = 10⁻¹⁴ ⇒ [H⁺] ≈ 1.05 × 10⁻⁷ M ⇒ pH ≈ 6.98.

3. Weak Monoprotic Acids & Ostwald's Dilution Law

Weak acids dissociate partially in aqueous media:

HA(aq) ⇔ H⁺(aq) + A⁻(aq)

Let initial analytical concentration be c and degree of dissociation be α (0 < α < 1):

Equilibrium concentrations: [HA] = c(1 − α), [H⁺] = cα, [A⁻] = cα
Acid Ionization Constant: K_a = (cα · cα) / [c(1 − α)] = cα² / (1 − α)

Ostwald's Dilution Law Approximation:
When α ≤ 0.05 (less than 5% dissociation), 1 − α ≈ 1. Therefore:
α = √(K_a / c)
[H⁺] = c α = √(K_a · c)
pH = ½ [pK_a − log c]

Weak Bases and K_b

Similarly, for weak base BOH(aq) ⇔ B⁺(aq) + OH⁻(aq):

K_b = cα² / (1 − α) ≈ cα² ⇒ [OH⁻] = √(K_b · c)
pOH = ½ [pK_b − log c] and pH = 14 − pOH

4. Conjugate Pair Relationship & Polyprotic Acids

For any conjugate acid-base pair (e.g., NH₄⁺ and NH₃, or CH₃COOH and CH₃COO⁻):

K_a · K_b = K_w ⇒ pK_a + pK_b = pK_w = 14.00 (at 298 K)

Ionization of Polyprotic Acids

Acids containing more than one ionizable proton dissociate stepwise:

Step 1: H₂A ⇔ H⁺ + HA⁻ (K_a1)
Step 2: HA⁻ ⇔ H⁺ + A²⁻ (K_a2)

Due to electrostatic attraction, it is vastly more difficult to remove a positively charged proton from a negatively charged anion (HA⁻) than from an electrically neutral molecule (H₂A). Hence:

K_a1 ≫ K_a2 ≫ K_a3
For dibasic acids like H₂S or H₂CO₃ where K_a1 ≫ K_a2:

The [H⁺] and pH are almost entirely determined by the first ionization step: [H⁺] ≈ √(K_a1 · c).
The concentration of the divalent anion [A²⁻] is approximately equal to K_a2, completely independent of the starting acid concentration c!

5. Solved Practice Problems (JEE Main & NEET)

Example 1: Calculate the pH of a 0.05 M solution of acetic acid (CH3COOH). Given Ka = 1.8 × 10⁻⁵. Check whether the Ostwald approximation is valid.

Correct Answer: pH = 3.02 (α = 1.9%, approximation valid)

Step-by-step Solution:
1. Test α = √(K_a / c) = √[(1.8 × 10⁻⁵) / 0.05] = √(3.6 × 10⁻⁴) = 0.019 (1.9%).
2. Since α = 1.9% < 5%, the approximation (1 − α ≈ 1) is completely valid.
3. [H⁺] = cα = 0.05 × 0.019 = 9.5 × 10⁻⁴ M.
4. pH = −log(9.5 × 10⁻⁴) = 4 − log(9.5) = 4 − 0.978 = 3.02.

Example 2: What is the pH of a 10⁻⁸ M aqueous solution of NaOH at 25°C?

Correct Answer: pH = 7.02

Step-by-step Solution:
1. In extremely dilute basic solutions, OH⁻ from autoionization of water cannot be ignored.
2. Let [OH⁻] from water = x.
Total [OH⁻] = 10⁻⁸ + x, and [H⁺] = x.
3. K_w = [H⁺][OH⁻] = x(10⁻⁸ + x) = 10⁻¹⁴.
4. x² + 10⁻⁸ x − 10⁻¹⁴ = 0 ⇒ x ≈ 0.95 × 10⁻⁷ M.
5. Total [OH⁻] = 1.05 × 10⁻⁷ M ⇒ pOH = −log(1.05 × 10⁻⁷) ≈ 6.98.
6. pH = 14 − pOH = 14 − 6.98 = 7.02 (slightly alkaline, as required).

6. Frequently Asked Questions (FAQs)

1. Can a substance be both a Brønsted acid and a Brønsted base?

Yes! Such substances are called amphiprotic (or amphoteric). Common examples include water (H₂O acts as an acid by donating H⁺ to form OH⁻, and as a base by accepting H⁺ to form H₃O⁺) and bicarbonate (HCO₃⁻ forms H₂CO₃ upon protonation and CO₃²⁻ upon deprotonation).

2. Why is [S²⁻] in a saturated H2S solution approximately equal to Ka2?

For H₂S(aq) ⇔ 2H⁺(aq) + S²⁻(aq), the overall dissociation is governed by K_overall = K_a1 · K_a2 = [H⁺]² [S²⁻] / [H₂S]. Because K_a1 ≫ K_a2, essentially all H⁺ comes from the first step where [H⁺] ≈ [HS⁻]. From the second step equilibrium: K_a2 = [H⁺][S²⁻] / [HS⁻]. Canceling [H⁺] ≈ [HS⁻] leaves [S²⁻] ≈ K_a2.

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