Why is the pH of a salt of weak acid and weak base independent of concentration?

In a salt of weak acid and weak base (like CH3COONH4), both the cation and anion hydrolyze simultaneously: CH3COO- + NH4+ + H2O CH3COOH + NH4OH. The degree of hydrolysis h = √(Kw / (Ka * Kb)) does not contain the concentration term c in the denominator. Consequently, [H+] = √(Kw * Ka / Kb) and pH = 7 + (1/2) pKa - (1/2) pKb are entirely independent of the salt concentration.

Does diluting a buffer solution change its pH?

No. According to the Henderson-Hasselbalch equation, pH = pKa + log([Conjugate Base] / [Acid]). Dilution decreases both the concentration of conjugate base and weak acid by the exact same volume factor, leaving the ratio [Base]/[Acid] unchanged. Hence, dilution does NOT alter buffer pH, although it reduces total buffer capacity.

§ 6.3NCERT Class 11 · Chemistry · Chapter 6

Hydrolysis of Salts & Buffer Solutions: Complete Mathematical Formulations

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 7 min read

When salts dissolve in water, their constituent ions interact with solvent water molecules. If an ion originates from a weak acid or weak base, it acts as a conjugate Brønsted partner, perturbing the neutral water equilibrium [H⁺] = [OH⁻] — a phenomenon known as salt hydrolysis. In contrast, buffer solutions leverage conjugate equilibrium pairs to resist drastic pH fluctuations.

1. The Four Categories of Salt Hydrolysis

Salt hydrolysis is the reverse of neutralization: Salt + Water ⇔ Acid + Base. Depending on the nature of the parent acid and base, salts fall into four distinct classes:

Salt Classification	Hydrolyzing Ion	Nature of Solution	Hydrolysis Constant (K_h)	Degree of Hydrolysis (h)	Solution pH Formulation (at 298 K)
1. Strong Acid + Strong Base (NaCl, KNO₃, Na₂SO₄)	Neither cation nor anion hydrolyzes	Neutral [H⁺] = [OH⁻]	K_h = 0	h = 0	pH = 7.00
2. Weak Acid + Strong Base (CH₃COONa, KCN, Na₂CO₃)	Anion hydrolyzes (Anionic Hydrolysis)	Basic / Alkaline [OH⁻] > [H⁺]	K_h = K_w / K_a	h = √[K_w / (K_a c)]	pH = 7 + ½ pK_a + ½ log c
3. Strong Acid + Weak Base (NH₄Cl, CuSO₄, FeCl₃)	Cation hydrolyzes (Cationic Hydrolysis)	Acidic [H⁺] > [OH⁻]	K_h = K_w / K_b	h = √[K_w / (K_b c)]	pH = 7 − ½ pK_b − ½ log c
4. Weak Acid + Weak Base (CH₃COONH₄, NH₄CN)	Both cation & anion hydrolyze simultaneously	Depends on K_a vs K_b (Neutral if K_a = K_b)	K_h = K_w / (K_a K_b)	h = √[K_w / (K_a K_b)] (independent of c)	pH = 7 + ½ pK_a − ½ pK_b (independent of c)

Mechanism of Buffer Action: Neutralizing Added Acid and Base

Dynamic conjugate reservoir absorbing external H+ and OH- disturbances

ACIDIC BUFFER SYSTEM: CH3COOH (Weak Acid Reservoir) + CH3COO- (Conjugate Base Reservoir)
=======================================================================================
Case A: Small Amount of Strong Acid (H+) Added
        H+(added) + CH3COO-(reservoir) → CH3COOH(aq)
        [Strong acid converted into weak, un-ionized CH3COOH. Minimal drop in pH!]

Case B: Small Amount of Strong Base (OH-) Added
        OH-(added) + CH3COOH(reservoir) → CH3COO-(aq) + H2O(l)
        [Strong base converted into weak conjugate base and water. Minimal rise in pH!]
---------------------------------------------------------------------------------------
Henderson-Hasselbalch Equation:
pH = pKa + log10 ( [Conjugate Base] / [Acid] ) = pKa + log10 ( [Salt] / [Acid] )

Buffer Capacity Peak: The resistance of a buffer to pH change is maximal when [Conjugate Base] = [Acid], which occurs when pH = pK_a. The effective buffering range is typically restricted to pH = pK_a ± 1.

2. Buffer Solutions & Henderson-Hasselbalch Formulations

A buffer solution resists significant changes in its hydrogen ion concentration upon minor addition of strong acid, strong base, or upon moderate dilution.

1. Acidic Buffers

Consists of an equimolar or near-equimolar mixture of a weak acid and its salt with a strong base (e.g., CH₃COOH + CH₃COONa, or HCN + NaCN):

Henderson-Hasselbalch Equation for Acidic Buffer:
pH = pK_a + log₁₀ ([Conjugate Base] / [Weak Acid]) = pK_a + log₁₀ (n_salt / n_acid)
Notice that volume cancels out: the ratio can be directly computed using moles (n_salt / n_acid).

2. Basic Buffers

Consists of a weak base and its salt with a strong acid (e.g., NH₄OH + NH₄Cl):

Henderson-Hasselbalch Equation for Basic Buffer:
pOH = pK_b + log₁₀ ([Conjugate Acid] / [Weak Base]) = pK_b + log₁₀ (n_salt / n_base)
pH = 14.00 − pOH (at 298 K)

3. Buffer Capacity and Buffer Index

The buffer capacity (β) quantitatively measures the effectiveness of a buffer solution in resisting pH variations:

β = db / d(pH) = −da / d(pH)
Where db is the number of moles of strong base added per liter of buffer to produce an increase in pH of d(pH), and da is the moles of strong acid added to produce a decrease in pH of d(pH).

JEE Trap — Addition of Strong Reagent to Buffer:
When x moles of strong acid HCl are added to an acidic buffer containing a moles of HA and s moles of NaA:

Moles of weak acid become (a + x)
Moles of conjugate salt become (s − x)
New pH = pK_a + log[(s − x) / (a + x)]

Never attempt to calculate pH by simply computing [H⁺] directly from the added HCl without stoichiometric neutralization!

4. Solved Numerical Problems (JEE Main / NEET)

Example 1: Calculate the pH of a 0.1 M aqueous solution of sodium acetate (CH3COONa). Given Ka for acetic acid = 1.8 × 10⁻⁵. (log 1.8 ≈ 0.255, log 3 ≈ 0.477)

Correct Answer: pH = 8.87

Step-by-step Solution:
1. Identify salt type: CH₃COONa is a salt of Weak Acid + Strong Base ⇒ Alkaline solution (pH > 7).
2. Calculate pK_a: pK_a = −log(1.8 × 10⁻⁵) = 5 − 0.255 = 4.745.
3. Here c = 0.1 M ⇒ log c = log(10⁻¹) = −1.
4. Apply hydrolysis formula: pH = 7 + ½ pK_a + ½ log c
pH = 7 + ½ (4.745) + ½ (−1) = 7 + 2.3725 − 0.50 = 8.87.

Example 2: An acidic buffer is prepared by mixing 50 mL of 0.2 M CH3COOH and 50 mL of 0.1 M CH3COONa. What is the pH of the mixture? What will be the new pH if 5 mL of 0.1 M HCl is added to it? (pKa of CH3COOH = 4.74)

Correct Answer: Initial pH = 4.44; New pH after adding HCl = 4.37

Step-by-step Solution:
1. Initial millimoles:
Acid (CH₃COOH) = 50 × 0.2 = 10.0 mmol.
Salt (CH₃COONa) = 50 × 0.1 = 5.0 mmol.
Initial pH = pK_a + log(salt / acid) = 4.74 + log(5.0 / 10.0) = 4.74 + log(0.5) = 4.74 − 0.301 = 4.44.
2. Millimoles of HCl added = 5 mL × 0.1 M = 0.5 mmol of H⁺.
3. Neutralization stoichiometry:
CH₃COO⁻ + H⁺ → CH₃COOH
New millimoles of salt = 5.0 − 0.5 = 4.5 mmol.
New millimoles of acid = 10.0 + 0.5 = 10.5 mmol.
4. New pH = 4.74 + log(4.5 / 10.5) = 4.74 + log(3 / 7) = 4.74 + (0.477 − 0.845) = 4.74 − 0.368 = 4.37.
(Notice the buffering action: pH dropped by merely 0.07 units!).

5. Frequently Asked Questions (FAQs)

1. Can an equimolar mixture of NH4Cl and NaOH act as a buffer?

No. Complete neutralization occurs: NH₄Cl + NaOH → NH₃ + NaCl + H₂O. Since both react completely in a 1:1 mole ratio, there is no unreacted NH₄Cl left to form a conjugate buffer pair with the produced NH₃. However, if NH₄Cl is in excess (e.g., 2 moles NH₄Cl + 1 mole NaOH), 1 mole of NH₄Cl remains alongside 1 mole of generated NH₃, successfully producing a basic buffer!

2. What happens to the pH of ammonium acetate (CH3COONH4) solution when its temperature increases?

For CH₃COONH₄, pH = ½ pK_w + ½ pK_a − ½ pK_b. Since for acetic acid and ammonium hydroxide pK_a ≈ pK_b ≈ 4.74, pH simplifies to ½ pK_w. As temperature rises, K_w increases and pK_w decreases, so the pH of the neutral solution decreases slightly below 7.00.

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