Ionic Equilibrium — pH, Buffers, Solubility Product

What Is Ionic Equilibrium?

When an ionic compound dissolves in water, it dissociates into ions. At some point, the rate of dissociation equals the rate of recombination — the system reaches ionic equilibrium. Understanding this equilibrium is the key to explaining why vinegar is sour, why blood maintains a constant pH, and why certain salts crystallise out of solution.

This is one of the highest-weightage chapters in JEE Main and NEET. The concepts here connect directly to qualitative analysis, buffer action, and solubility — topics that appear in almost every paper.

Key Terms and Definitions

Strong Electrolyte: Completely dissociates in water. HCl, NaOH, NaCl — 100% dissociation. No equilibrium to speak of, because the reaction goes fully to completion.

Weak Electrolyte: Partially dissociates. CH₃COOH, NH₃, HCN — only a small fraction of molecules are ionised at any time. An equilibrium constant (Ka or Kb) describes this.

Degree of Dissociation (α): Fraction of the original substance that has dissociated. If 0.1 mol of acetic acid is placed in solution and 0.001 mol dissociates, α = 0.01 (1%).

Conjugate Acid-Base Pair: A pair of species differing by one proton. CH₃COOH and CH₃COO⁻ are a conjugate pair. The stronger the acid, the weaker its conjugate base.

K_a = \frac{[H^+][A^-]}{[HA]}

For acetic acid: $K_a = 1.8 \times 10^{-5}$ at 25°C

Larger $K_a$ → stronger acid → greater dissociation

pH and pOH — The Scale Explained

pH is not just a number on a scale — it’s the negative logarithm of hydrogen ion concentration. The logarithm means each unit change represents a 10-fold change in acidity.

\text{pH} = -\log_{10}[H^+]

\text{pOH} = -\log_{10}[OH^-]

\text{pH} + \text{pOH} = 14 \quad (\text{at } 25°C)

K_w = [H^+][OH^-] = 1 \times 10^{-14} \quad (\text{at } 25°C)

The neutral pH is 7 only at 25°C. At higher temperatures, $K_w$ increases, so neutral pH drops below 7 — water is still neutral (equal [H⁺] and [OH⁻]), but the pH number is less than 7. This is a classic exam trap.

Calculating pH for Weak Acids

For a weak acid HA at concentration C and dissociation constant Ka:

[H^+] = \sqrt{K_a \times C}

This approximation holds when α < 5% (which means $K_a \ll C$ ). Always check this condition.

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

Buffer Solutions — The Body’s pH Controller

A buffer resists changes in pH when small amounts of acid or base are added. Blood (pH ≈ 7.4) is maintained by the carbonic acid–bicarbonate buffer system.

Acidic buffer: Weak acid + its salt with strong base (e.g., CH₃COOH + CH₃COONa)

Basic buffer: Weak base + its salt with strong acid (e.g., NH₃ + NH₄Cl)

\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}

For basic buffers:

\text{pOH} = \text{p}K_b + \log\frac{[\text{Salt}]}{[\text{Base}]}

Buffer capacity is maximum when [salt] = [acid], i.e., when pH = pKa. At this point, adding acid or base causes the least change in pH.

Solubility Product (Ksp)

When a sparingly soluble salt like AgCl is placed in water, very little dissolves. The equilibrium between the undissolved solid and the dissolved ions is described by the solubility product.

For $A_mB_n \rightleftharpoons mA^{n+} + nB^{m-}$ :

K_{sp} = [A^{n+}]^m [B^{m-}]^n

For AgCl: $K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}$

If solubility = s mol/L: $K_{sp} = s^2 \Rightarrow s = \sqrt{K_{sp}}$

Common Ion Effect: Adding a soluble salt with a common ion reduces the solubility of the sparingly soluble salt. Adding NaCl to a saturated AgCl solution increases [Cl⁻], pushing the equilibrium backward — more AgCl precipitates.

Ionic product (Q) vs Ksp:

If Q < Ksp: solution is unsaturated, more salt can dissolve
If Q = Ksp: solution is at equilibrium (just saturated)
If Q > Ksp: precipitate forms

Hydrolysis of Salts

Not all salts are neutral. When dissolved, they can make the solution acidic or basic depending on which ions hydrolyse.

Salt Type	Example	Solution pH
Strong acid + Strong base	NaCl	7 (neutral)
Weak acid + Strong base	CH₃COONa	> 7 (basic)
Strong acid + Weak base	NH₄Cl	< 7 (acidic)
Weak acid + Weak base	CH₃COONH₄	Depends on Ka vs Kb

CH₃COO⁻ from sodium acetate accepts a proton from water, releasing OH⁻ — making the solution basic. NH₄⁺ from ammonium chloride donates a proton to water, releasing H⁺ — making the solution acidic.

Solved Examples

Example 1 (CBSE Level): Calculate pH of 0.1 M HCl

HCl is a strong acid, so it dissociates completely.

[H^+] = 0.1\ \text{M} = 10^{-1}\ \text{M}

\text{pH} = -\log(10^{-1}) = 1

pH = 1

Example 2 (JEE Main Level): pH of 0.01 M acetic acid (Ka = 1.8 × 10⁻⁵)

[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.01}

= \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4}\ \text{M}

\text{pH} = -\log(4.24 \times 10^{-4}) = 4 - \log 4.24 \approx 4 - 0.627 = 3.37

Check α: $\alpha = \dfrac{4.24 \times 10^{-4}}{0.01} = 0.0424 = 4.24\%$ < 5%, so approximation holds.

Example 3 (JEE Advanced Level): Ksp of Ag₂CrO₄ = 1.12 × 10⁻¹². Find solubility.

Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}

If solubility = s: $[Ag^+] = 2s$ , $[CrO_4^{2-}] = s$

K_{sp} = (2s)^2(s) = 4s^3

4s^3 = 1.12 \times 10^{-12}

s^3 = 2.8 \times 10^{-13}

s = (2.8 \times 10^{-13})^{1/3} \approx 6.54 \times 10^{-5}\ \text{mol/L}

Exam-Specific Tips

JEE Main 2024 had a direct question on comparing buffer capacity at different salt-to-acid ratios. Remember: maximum buffer capacity occurs when pH = pKa (equal concentrations of acid and salt). This appeared in both Shift 1 and Shift 2 of recent years.

NEET frequently asks about the effect of common ion on solubility and about which salt solutions are acidic, basic, or neutral. Salt hydrolysis and buffer pH calculation are guaranteed 2-mark questions.

For competitive exams, always check whether $\alpha < 5\%$ before using the simplified formula $[H^+] = \sqrt{K_a C}$ . If α is large (weak acid, very dilute solution), you must solve the quadratic. Examiners specifically create problems where the approximation fails.

Common Mistakes to Avoid

Mistake 1: Assuming neutral pH is always 7. Neutral means [H⁺] = [OH⁻] — the pH value depends on temperature. At body temperature (37°C), neutral pH is about 6.8.

Mistake 2: Writing Ksp for a salt that includes the solid phase. Ksp is only for the dissolved ions. The concentration of the solid is constant and absorbed into Ksp.

Mistake 3: Forgetting stoichiometry in Ksp expressions. For Ag₂CrO₄, [Ag⁺] = 2s, not s. Many students write Ksp = s × s = s² instead of (2s)² × s = 4s³.

Mistake 4: Confusing Ka and pKa. Larger Ka = smaller pKa = stronger acid. Students who memorise numbers without understanding the relationship often get this backwards.

Mistake 5: In Henderson-Hasselbalch, using moles instead of concentrations when the volume is the same for both components. Volume cancels only if both are in the same solution — always state your assumption.

Practice Questions

Q1. What is the pH of a 0.001 M NaOH solution at 25°C?

[OH⁻] = 0.001 = 10⁻³ M
pOH = 3
pH = 14 – 3 = 11

Q2. A buffer contains 0.1 M CH₃COOH and 0.2 M CH₃COONa. Given pKa = 4.74, find pH.

pH = pKa + log([Salt]/[Acid]) = 4.74 + log(0.2/0.1) = 4.74 + log 2 = 4.74 + 0.30 = 5.04

Q3. Ksp of BaSO₄ = 1.1 × 10⁻¹⁰. What is its solubility in mol/L?

BaSO₄ → Ba²⁺ + SO₄²⁻
Ksp = s²
s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ mol/L

Q4. Why does acetic acid have a lower degree of dissociation in a solution already containing sodium acetate?

Common ion effect. CH₃COO⁻ from CH₃COONa shifts the equilibrium of acetic acid dissociation to the left, suppressing further ionisation. This is an application of Le Chatelier’s principle to ionic equilibrium.

Q5. Is a solution of FeCl₃ acidic, basic, or neutral? Why?

Acidic. FeCl₃ is a salt of strong acid (HCl) and weak base Fe(OH)₃. Fe³⁺ undergoes hydrolysis: Fe³⁺ + 3H₂O ⇌ Fe(OH)₃ + 3H⁺, releasing H⁺ and making the solution acidic.

Q6. What happens to the solubility of AgCl when a few drops of AgNO₃ solution are added?

Solubility decreases. The common ion Ag⁺ from AgNO₃ increases [Ag⁺], pushing the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left. More AgCl precipitates, reducing solubility.

Q7. Calculate pOH of a 0.1 M NH₃ solution. Kb = 1.8 × 10⁻⁵.

[OH⁻] = √(Kb × C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M
pOH = –log(1.34 × 10⁻³) ≈ 2.87
pH = 14 – 2.87 = 11.13

Indicators and Titration Curves

An acid-base indicator is a weak acid (or base) whose conjugate pair has different colours. The indicator changes colour when the pH of the solution matches the indicator’s pKa range.

Indicator	pH Range	Colour Change (Acid → Base)
Methyl orange	3.1 – 4.4	Red → Yellow
Litmus	5.0 – 8.0	Red → Blue
Phenolphthalein	8.2 – 10.0	Colourless → Pink

Choosing the right indicator: The indicator must change colour at the pH of the equivalence point of the titration.

Strong acid + strong base → equivalence at pH 7 → any indicator works
Strong acid + weak base → equivalence at pH < 7 → methyl orange (changes in acidic range)
Weak acid + strong base → equivalence at pH > 7 → phenolphthalein (changes in basic range)

CBSE Class 12 and NEET both ask: “Which indicator is suitable for the titration of CH₃COOH with NaOH?” Answer: phenolphthalein, because the equivalence point is at pH ~8.7 (basic side, due to hydrolysis of sodium acetate). Methyl orange would change colour too early (at pH ~4) — well before the equivalence point.

Titration Curves — What They Tell Us

A titration curve plots pH vs volume of titrant added. The shape reveals:

Initial pH — tells you the strength of the acid/base being titrated
Buffer region — the flat section before equivalence where pH changes slowly (Henderson-Hasselbalch applies here)
Equivalence point — the steep vertical section where pH changes rapidly
Half-equivalence point — where exactly half the acid is neutralised; at this point, $[\text{acid}] = [\text{salt}]$ and pH = pKa (for weak acid titrations)

At half-equivalence in a weak acid–strong base titration:

\text{pH} = \text{p}K_a

This gives a direct experimental method to determine $K_a$ of any weak acid — just titrate it, find the half-equivalence volume, and read off the pH.

Q8. The half-equivalence point in a titration of a weak acid with NaOH occurs at pH = 4.74. What is the acid?

At half-equivalence, pH = pKa. So pKa = 4.74, meaning Ka = $10^{-4.74} = 1.82 \times 10^{-5}$ . This is the Ka of acetic acid (CH₃COOH). The pKa value 4.74 is a signature of acetic acid and should be memorised.

FAQs

Why does the pH of a buffer barely change when acid is added? The added H⁺ ions react with the conjugate base (A⁻) present in the buffer: H⁺ + A⁻ → HA. This converts the added acid into the weak acid form, which barely changes [H⁺]. The system absorbs the perturbation rather than letting it affect pH.

Can Ksp change with temperature? Yes. Ksp is an equilibrium constant and depends on temperature. For most salts, solubility (and therefore Ksp) increases with temperature. For a few exceptions like CaSO₄, solubility decreases with temperature.

What is the difference between solubility and Ksp? Solubility is the mass or moles of salt that dissolve per litre. Ksp is derived from solubility and depends on stoichiometry. Two salts with the same Ksp can have different solubilities if their formulas are different (e.g., 1:1 vs 1:2 stoichiometry).

Why is blood pH maintained so precisely at 7.4? Enzymes in the body have narrow pH optima. Even small deviations (acidosis below 7.35 or alkalosis above 7.45) denature enzymes and disrupt cellular function. The bicarbonate buffer (H₂CO₃/HCO₃⁻) is the main buffer, supplemented by phosphate and protein buffers.

How do you know if a precipitate will form when two solutions are mixed? Calculate the ionic product Q for the mixed solution. If Q > Ksp, a precipitate will form. If Q < Ksp, no precipitate forms. If Q = Ksp, the solution is exactly saturated.