Salt Hydrolysis — Why Some Salts Are Acidic or Basic

Most people assume all salts give neutral solutions. NaCl (table salt) does — its aqueous solution has pH = 7. But CH₃COONa (sodium acetate) gives a basic solution, and NH₄Cl (ammonium chloride) gives an acidic solution. Why? The answer is salt hydrolysis — the reaction of a salt’s ions with water that shifts the solution’s pH from neutral.

Understanding salt hydrolysis requires a firm grip on acid-base equilibria. Once you understand the mechanism, you’ll never need to memorise whether a given salt is acidic or basic — you can derive it from the parent acid and base.

Key Terms & Definitions

Salt hydrolysis: The reaction of a salt with water, where one or both of the ions react with H₂O, producing either H₃O⁺ (acidic) or OH⁻ (basic).

Parent acid and base: Every salt is formed by the neutralisation of an acid and a base. The nature of that acid and base determines the salt’s behaviour in water.

Strong acid/strong base: Completely dissociated in water. Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄. Strong bases: NaOH, KOH, Ca(OH)₂, Ba(OH)₂.

Weak acid/weak base: Partially dissociated. Weak acids: CH₃COOH, H₂CO₃, H₂S, HCN, HF. Weak bases: NH₃, amines.

Ka: Acid dissociation constant — measures strength of an acid. Higher Ka = stronger acid.

Kb: Base dissociation constant — measures strength of a base.

Kw: Water autoionisation constant = 1.0 × 10⁻¹⁴ at 25°C. Relationship: $K_w = K_a \times K_b$ (for conjugate acid-base pairs).

Hydrolysis constant (Kh): The equilibrium constant for the hydrolysis reaction. $K_h = K_w/K_a$ (for anion hydrolysis) or $K_h = K_w/K_b$ (for cation hydrolysis).

The Four Types of Salts

Based on the strength of the parent acid and base, salts fall into four categories:

Type 1: Strong Acid + Strong Base → Neutral Salt

Examples: NaCl, KNO₃, Na₂SO₄, KBr, NaClO₄

Behaviour: Neither ion reacts with water. pH = 7.

Reasoning: Cl⁻ is the conjugate base of HCl (strong acid) — so Cl⁻ has essentially no tendency to accept protons from water. Na⁺ is the conjugate acid of NaOH (strong base) — so Na⁺ has essentially no tendency to donate protons to water. No hydrolysis → neutral solution.

\text{pH} = 7 \quad (\text{at 25°C})

Neither cation nor anion hydrolyses. The salt is a spectator in water.

Type 2: Strong Acid + Weak Base → Acidic Salt

Examples: NH₄Cl, NH₄NO₃, AlCl₃, FeCl₃, CuSO₄

Behaviour: The cation (from the weak base) hydrolyses. pH < 7.

Reasoning for NH₄Cl: The NH₄⁺ ion is the conjugate acid of the weak base NH₃. Since NH₃ is weak, NH₄⁺ is a reasonably strong acid and readily donates H⁺ to water:

\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+

Cl⁻ does not hydrolyse (it’s from strong HCl). Net result: H₃O⁺ produced → acidic solution.

pH formula (for salt of strong acid + weak base):

\text{pH} = 7 - \frac{1}{2}(pK_b + \log C)

where $C$ is the salt concentration and $pK_b = -\log K_b$ of the weak base.

Equivalently: $\text{pH} = \frac{1}{2}(pK_w - pK_b - \log C) = \frac{1}{2}(14 - pK_b - \log C)$

\text{pH} = 7 - \frac{1}{2}(pK_b + \log C)

Or using $pK_a$ of the cation (where $pK_a = pK_w - pK_b = 14 - pK_b$ ):

\text{pH} = \frac{1}{2}(pK_a - \log C)

Type 3: Weak Acid + Strong Base → Basic Salt

Examples: CH₃COONa (sodium acetate), Na₂CO₃, NaCN, NaHCO₃, Na₂S

Behaviour: The anion (from the weak acid) hydrolyses. pH > 7.

Reasoning for CH₃COONa: CH₃COO⁻ is the conjugate base of the weak acid CH₃COOH (acetic acid). Since acetic acid is weak, its conjugate base (acetate) is a reasonably strong base and accepts H⁺ from water:

\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-

Na⁺ does not hydrolyse (it’s from strong NaOH). Net result: OH⁻ produced → basic solution.

pH formula (for salt of weak acid + strong base):

\text{pH} = 7 + \frac{1}{2}(pK_a + \log C)

\text{pH} = 7 + \frac{1}{2}(pK_a + \log C)

Where $pK_a$ is that of the parent weak acid.

Type 4: Weak Acid + Weak Base → pH depends on Ka vs Kb

Examples: CH₃COONH₄ (ammonium acetate), NH₄CN, NH₄HCO₃

Behaviour: Both cation and anion hydrolyse. pH depends on relative strength.

Both reactions occur simultaneously:

Cation (NH₄⁺) hydrolyses: $\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$
Anion (CH₃COO⁻) hydrolyses: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$

pH formula:

\text{pH} = 7 + \frac{1}{2}(pK_a - pK_b)

Consequences:

If $K_a = K_b$ (or $pK_a = pK_b$ ): pH = 7 (neutral)
If $K_a > K_b$ (acid is stronger than base): pH < 7 (acidic)
If $K_b > K_a$ (base is stronger than acid): pH > 7 (basic)

Example: For CH₃COONH₄: $K_a(\text{acetic acid}) = 1.8 \times 10^{-5}$ ; $K_b(\text{NH}_3) = 1.8 \times 10^{-5}$ . Since $K_a = K_b$ , pH = 7 (neutral). This is why ammonium acetate solution is neutral despite coming from a weak acid and weak base.

\text{pH} = 7 + \frac{1}{2}(pK_a - pK_b)

Buffer Solutions — The Connection to Hydrolysis

A buffer solution resists changes in pH when small amounts of acid or base are added. Buffers work because of equilibria very similar to hydrolysis:

Acidic buffer: Weak acid + its conjugate base (salt of weak acid with strong base)

Example: CH₃COOH + CH₃COONa
pH calculated by Henderson-Hasselbalch equation: $\text{pH} = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$

Basic buffer: Weak base + its conjugate acid (salt of weak base with strong acid)

Example: NH₃ + NH₄Cl
pH: $\text{pOH} = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$ , then pH = 14 − pOH

\text{pH} = pK_a + \log\frac{[\text{A}^-]}{[\text{HA}]}

Where [A⁻] is the salt (conjugate base) concentration and [HA] is the weak acid concentration.

Maximum buffer capacity is at pH = pKa (equal concentrations of acid and salt).

Degree of Hydrolysis

The degree of hydrolysis ( $h$ ) is the fraction of the salt that has hydrolysed:

For the salt of a strong acid and weak base (e.g., NH₄Cl):

h = \sqrt{\frac{K_w}{K_b \cdot C}} = \sqrt{\frac{K_h}{C}}

where $C$ = salt concentration and $K_h = K_w/K_b$ = hydrolysis constant.

Degree of hydrolysis increases when:

Concentration decreases (more dilute solutions → more hydrolysis)
Temperature increases (water ionisation increases → more hydrolysis)
The weak acid/base is weaker (smaller Ka/Kb → more hydrolysis)

Solved Examples

Example 1 — Type Identification (CBSE Level)

Q: Predict whether aqueous solutions of the following are acidic, basic, or neutral: (a) KNO₃, (b) Na₂CO₃, (c) NH₄Cl, (d) NH₄CN.

Solution: (a) KNO₃: K⁺ from strong base (KOH), NO₃⁻ from strong acid (HNO₃). Neither hydrolyses. Neutral (pH = 7).

(b) Na₂CO₃: Na⁺ from strong base (NaOH), CO₃²⁻ from weak acid (H₂CO₃). Anion hydrolyses: $\text{CO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_3^- + \text{OH}^-$ . Basic (pH > 7). This is why washing soda (Na₂CO₃) solutions feel slippery (like soap) — they’re alkaline.

(d) NH₄CN: NH₄⁺ from weak base (NH₃), CN⁻ from weak acid (HCN). Both hydrolyse. Compare $K_b(\text{NH}_3) = 1.8 \times 10^{-5}$ and $K_a(\text{HCN}) = 6.2 \times 10^{-10}$ . Since $K_b >> K_a$ , the base is stronger → anion hydrolysis dominates → basic (pH > 7).

Example 2 — pH Calculation (JEE Level)

Q: Calculate the pH of 0.1 M CH₃COONa solution. Given: $K_a$ of CH₃COOH = $1.8 \times 10^{-5}$ .

Solution: Salt of weak acid (CH₃COOH, $K_a = 1.8 \times 10^{-5}$ , $pK_a = 4.74$ ) and strong base (NaOH). Solution is basic.

\text{pH} = 7 + \frac{1}{2}(pK_a + \log C) = 7 + \frac{1}{2}(4.74 + \log 0.1)

$= 7 + \frac{1}{2}(4.74 + (-1)) = 7 + \frac{1}{2}(3.74) = 7 + 1.87 = \mathbf{8.87}$

The solution is basic (pH 8.87), as expected for a salt of weak acid + strong base.

Example 3 — Buffer Calculation (JEE Main Level)

Q: A buffer solution is made by mixing 0.2 M CH₃COOH and 0.1 M CH₃COONa. Calculate the pH. ( $K_a = 1.8 \times 10^{-5}$ , $pK_a = 4.74$ )

Solution: Using Henderson-Hasselbalch:

\text{pH} = pK_a + \log\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 4.74 + \log\frac{0.1}{0.2} = 4.74 + \log(0.5)

$= 4.74 + (-0.301) = \mathbf{4.44}$

The buffer has pH 4.44. Since [acid] > [salt], pH < pKa — makes sense.

Example 4 — Degree of Hydrolysis (JEE Advanced Level)

Q: Calculate the degree of hydrolysis and pH of 0.01 M NH₄Cl solution at 25°C. $K_b(\text{NH}_3) = 1.8 \times 10^{-5}$ , $K_w = 1.0 \times 10^{-14}$ .

Solution: Hydrolysis constant: $K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

Degree of hydrolysis: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{5.56 \times 10^{-10}}{0.01}} = \sqrt{5.56 \times 10^{-8}} = 2.36 \times 10^{-4}$

Concentration of H₃O⁺ produced: $[\text{H}^+] = h \times C = 2.36 \times 10^{-4} \times 0.01 = 2.36 \times 10^{-6}$ M

$\text{pH} = -\log(2.36 \times 10^{-6}) = 6 - \log(2.36) = 6 - 0.37 = 5.63$

Alternatively using the formula: $\text{pH} = 7 - \frac{1}{2}(pK_b + \log C) = 7 - \frac{1}{2}(4.74 + (-2)) = 7 - \frac{1}{2}(2.74) = 7 - 1.37 = 5.63$ ✓

Exam-Specific Tips

CBSE Class 11/12: Know all four salt types with examples and whether each gives acidic/basic/neutral solution. Memorise: “strong acid + weak base → acidic” and “weak acid + strong base → basic.” These two rules cover 80% of marks from this chapter. pH formulas appear in Class 12 equilibrium chapter.

JEE Main: The Henderson-Hasselbalch equation, buffer capacity, and pH calculations for all four salt types are direct marks. Know the degree of hydrolysis formula and when hydrolysis increases (dilution and temperature). Multiple choice options in JEE often differ by whether pH is above or below 7 — using the parent acid/base rule never fails.

Common Mistakes to Avoid

Mistake 1 — “All salts have pH = 7”: Only salts from strong acid + strong base are neutral. This misconception is the root of all salt hydrolysis errors. Always ask: what are the parent acid and base? Are they strong or weak?

Mistake 2 — Henderson-Hasselbalch with wrong acid: Always use the pKa of the weak acid in the buffer. Students sometimes use pKa of H₂O or pKb values. The formula uses pKa directly.

Mistake 3 — Confusing log base and log formula signs: In the salt of weak acid + strong base pH formula: pH = 7 + ½(pKa + log C). For salt of strong acid + weak base: pH = 7 − ½(pKb + log C). The sign changes — weak acid salt → add, weak base salt → subtract. Mixing these signs is a very common calculation error.

Mistake 4 — Thinking dilution doesn’t affect pH of a hydrolysed salt: For a hydrolysed salt, the degree of hydrolysis increases on dilution ( $h \propto 1/\sqrt{C}$ ). More hydrolysis means more OH⁻ (for weak acid salts) — so the solution becomes more basic on dilution. For a strong acid/strong base salt (no hydrolysis), dilution just dilutes and pH approaches 7.

Practice Questions

Q1: Arrange the following in order of increasing pH: NaCl, Na₂CO₃, AlCl₃, CH₃COONa.

NaCl = neutral (pH 7). Na₂CO₃ = basic (strong base + weak acid). AlCl₃ = acidic (strong acid + weak base Al(OH)₃). CH₃COONa = basic (strong base + weak acid), but carbonate is a stronger base than acetate.

Order (increasing pH, most acidic to most basic): AlCl₃ < NaCl < CH₃COONa < Na₂CO₃

Justification: AlCl₃ solution is quite acidic (Al³⁺ is a strongly hydrolysing cation); NaCl is neutral; CH₃COONa is weakly basic (pKa of CH₃COOH = 4.74 → pH ≈ 8.87 at 0.1 M); Na₂CO₃ is strongly basic (pKa of HCO₃⁻ ≈ 10.3 for the second ionisation → pH ≈ 11 at 0.1 M).

Q2: A buffer solution is prepared by adding 50 mL of 0.2 M acetic acid to 50 mL of 0.2 M sodium acetate. Calculate the pH. (pKa = 4.74)

After mixing, both concentrations are halved (equal volumes): [CH₃COOH] = 0.1 M, [CH₃COO⁻] = 0.1 M.

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.1/0.1) = 4.74 + log(1) = 4.74 + 0 = 4.74

This makes perfect sense: when [acid] = [salt], pH = pKa. This is the principle of buffer preparation at maximum buffer capacity.

Q3: Explain why blood pH is maintained around 7.4.

Blood pH is maintained between 7.35–7.45 by several buffer systems, the most important being the carbonic acid-bicarbonate buffer:

H₂CO₃ (carbonic acid) ⇌ H⁺ + HCO₃⁻ (bicarbonate)

Normal plasma [HCO₃⁻]/[H₂CO₃] ≈ 20:1. Using Henderson-Hasselbalch with pKa (H₂CO₃) = 6.1: pH = 6.1 + log(20) = 6.1 + 1.3 = 7.4 ✓

If CO₂ (which hydrates to H₂CO₃) builds up (e.g., during shallow breathing), pH falls (acidosis). If CO₂ is lost (hyperventilation), pH rises (alkalosis). The lungs control CO₂ levels; the kidneys control HCO₃⁻ levels — both regulate this buffer equilibrium.

This is a physiological application of the Henderson-Hasselbalch equation and buffer chemistry — demonstrating that salt hydrolysis and buffer theory are not just textbook topics but life-sustaining chemistry.

FAQs

Q: Is hydrolysis the same as decomposition? No. Hydrolysis is a reversible equilibrium reaction with water that shifts the pH of a solution. Decomposition is a permanent chemical breakdown into simpler substances. Hydrolysis of CH₃COO⁻ gives CH₃COOH and OH⁻ — these still exist in equilibrium and the reaction can reverse.

Q: Why does adding a small amount of strong acid to a buffer change the pH only slightly? The added H⁺ reacts with the conjugate base (A⁻) in the buffer: H⁺ + A⁻ → HA. This consumes the H⁺ before it can significantly affect [H⁺]. The [A⁻]/[HA] ratio changes slightly, shifting pH slightly — but the large reservoir of A⁻ means the shift is small. If no buffer were present, the same addition of H⁺ would cause a dramatic pH drop.

Q: Can a salt be both acidic AND basic in different concentrations? For salts of weak acid + weak base (Type 4), the relative hydrolysis of cation vs anion determines pH. The pH formula pH = 7 + ½(pKa − pKb) is concentration-independent (the concentration terms cancel for this type) — so concentration doesn’t change the direction of the pH deviation, only its exact value slightly. For other salt types, pH changes with concentration (as seen in the formulas for Types 2 and 3).