IIT-JAM Ionic Equilibrium — pH, Buffers and Solubility Product

By Aniket Bhardwaj · 15 September 2026 · IIT-JAM Chemistry

Ionic equilibrium is one of the friendliest units in IIT-JAM physical chemistry, because almost every question reduces to four or five equations that you can apply mechanically once you understand them. It is also the unit where careless students lose the most marks: a forgotten square root, a pH taken as pOH, or an approximation used where it is not valid. This guide walks through the whole unit — weak acid and base equilibria, buffers, salt hydrolysis and the solubility product — with every number actually computed so you can check your own working line by line.

The equations the whole unit is built on

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K → pH + pOH = 14

Weak acid HA: K_a = [H⁺][A⁻] / [HA]
Ostwald dilution law: α = √(K_a / C) and [H⁺] = √(K_a · C)

Henderson–Hasselbalch: pH = pK_a + log ( [salt] / [acid] )

Salt of weak acid + strong base: K_b = K_w / K_a, [OH⁻] = √(K_b · C)

Sparingly soluble salt A_xB_y: K_sp = (xs)^x(ys)^y

What each symbol means

K_a, K_b — acid and base dissociation constants. Both are dimensionless in the thermodynamic sense; numerically they are quoted for concentrations in mol L⁻¹.
C — the analytical (total) concentration of the acid you dissolved, not the equilibrium concentration of the undissociated form.
α — degree of dissociation, the fraction of the acid that has ionised. It has no unit and must lie between 0 and 1.
K_sp — solubility product. s is molar solubility in mol L⁻¹; they are related but they are not the same quantity.

The square-root forms of Ostwald's law and the hydrolysis expression are approximations. They assume the acid is weak enough that the amount dissociated is small compared with C, which in practice means α is below about 5%. Always check α at the end; if it comes out large, solve the full quadratic instead.

Worked example 1 — pH and degree of dissociation of a weak acid

Find the pH and α of 0.10 M acetic acid, K_a = 1.8 × 10⁻⁵.

[H⁺] = √(K_a · C) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶)

√1.8 = 1.3416, so [H⁺] = 1.3416 × 10⁻³ mol L⁻¹

pH = −log(1.3416 × 10⁻³) = 3 − log 1.3416 = 3 − 0.1276 = 2.87

α = [H⁺] / C = 1.3416 × 10⁻³ / 0.10 = 0.0134, i.e. 1.34%

Validity check: α = 1.34% is well under 5%, so the approximation was justified. Cross-check by Ostwald's law directly: α = √(K_a/C) = √(1.8 × 10⁻⁵/0.10) = √(1.8 × 10⁻⁴) = 1.342 × 10⁻² — the same answer by a second route.

Worked example 2 — a buffer, and proof that it buffers

1.0 L of solution contains 0.25 mol acetic acid and 0.15 mol sodium acetate. Find the pH. Then find the new pH after adding 0.050 mol NaOH (no volume change).

pK_a = −log(1.8 × 10⁻⁵) = 5 − log 1.8 = 5 − 0.2553 = 4.7447

pH = pK_a + log(0.15 / 0.25) = 4.7447 + log 0.60 = 4.7447 − 0.2218 = 4.52

Adding strong base converts acid into its conjugate base, mole for mole:

acid: 0.25 − 0.050 = 0.20 mol · acetate: 0.15 + 0.050 = 0.20 mol

pH = 4.7447 + log(0.20 / 0.20) = 4.7447 + 0 = 4.74

The point of the question: 0.050 mol of NaOH in 1 L of pure water would give [OH⁻] = 0.050 M, pOH = 1.30, pH = 12.70. In the buffer the pH moved only from 4.52 to 4.74. That contrast — about 0.2 units instead of about 5.7 — is the whole idea of a buffer, and JAM likes to ask it as a comparison rather than as a single number.

Worked example 3 — hydrolysis of a salt

Find the pH of 0.10 M sodium acetate (K_a of acetic acid = 1.8 × 10⁻⁵).

Acetate is the conjugate base of a weak acid, so it hydrolyses:

K_b = K_w / K_a = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.556 × 10⁻¹⁰

[OH⁻] = √(K_b · C) = √(5.556 × 10⁻¹⁰ × 0.10) = √(5.556 × 10⁻¹¹) = 7.454 × 10⁻⁶ mol L⁻¹

pOH = 6 − log 7.454 = 6 − 0.8724 = 5.128

pH = 14 − 5.128 = 8.87

Sensible: the salt of a weak acid and a strong base must be basic, so any answer below 7 would be wrong on inspection alone. Train yourself to make that check before you compute.

Worked example 4 — Ksp and the common ion effect

K_sp(AgCl) = 1.8 × 10⁻¹⁰. Find the molar solubility in pure water, in g L⁻¹, and then in 0.010 M NaCl.

In water, AgCl ⇌ Ag⁺ + Cl⁻ gives s mol L⁻¹ of each ion:

K_sp = s², so s = √(1.8 × 10⁻¹⁰) = 1.3416 × 10⁻⁵ mol L⁻¹

M(AgCl) = 107.868 + 35.45 = 143.318 g mol⁻¹

Solubility = 1.3416 × 10⁻⁵ × 143.318 = 1.92 × 10⁻³ g L⁻¹ ≈ 1.9 mg per litre

In 0.010 M NaCl the chloride is fixed by the added salt, so [Cl⁻] ≈ 0.010 M and

s = K_sp / [Cl⁻] = 1.8 × 10⁻¹⁰ / 0.010 = 1.8 × 10⁻⁸ mol L⁻¹

The solubility falls by a factor of 1.3416 × 10⁻⁵ / 1.8 × 10⁻⁸ ≈ 745. That single line is the common ion effect, quantified.

Common mistakes that cost marks

Forgetting the square root. [H⁺] = K_aC is a very common slip. It is √(K_aC). The units alone give it away: K_aC has units of concentration squared.
Using the approximation when α is large. For a fairly strong weak acid or a very dilute solution, α can exceed 5% and you must solve x² + K_ax − K_aC = 0 properly.
Inverting the Henderson–Hasselbalch ratio. It is log([salt]/[acid]). Written upside down you get a pH on the wrong side of pK_a. Sanity check: more salt than acid must give pH above pK_a.
Confusing s with K_sp. For a 1:1 salt K_sp = s², but for a 1:2 salt such as Mg(OH)₂ it is K_sp = 4s³, and for Ca₃(PO₄)₂ it is 108s⁵. Derive it from the stoichiometry every single time.
Stopping at pOH. In hydrolysis problems the answer wanted is almost always pH. Subtract from 14 before you write the final line.
Assuming K_w = 10⁻¹⁴ at every temperature. It is the value at 298 K. K_w increases with temperature, so the neutral pH of pure water is below 7 at higher temperatures — water is still neutral, just not at pH 7.

How to prepare this unit for JAM

Sub-topic	What you must be able to do without hesitation	Tool to check with
Weak acid / weak base	pH and α from K_a and C; decide whether the approximation holds	pH / pOH calculator
Buffers	Apply Henderson–Hasselbalch; recompute after adding strong acid or base	Buffer calculator
Salt hydrolysis	Convert K_a ⇌ K_b through K_w; predict acidic/basic/neutral by inspection	pH / pOH calculator
Solubility equilibria	Write K_sp from stoichiometry; common ion effect; decide whether a precipitate forms by comparing Q with K_sp	Ksp calculator
Indicators and titration curves	Choose an indicator whose pK_In sits in the steep part of the curve	pH / pOH calculator

For the authoritative syllabus and the current paper pattern, always read the official IIT-JAM notification for your year rather than relying on any coaching summary, including this one — the organising institute rotates and details can change.

Check every buffer answer in seconds. The Henderson–Hasselbalch buffer calculator takes pK_a and the salt-to-acid ratio and returns the pH, so you can verify a full page of practice problems instead of trusting one hand calculation.

Open the Buffer (Henderson–Hasselbalch) Calculator →

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