Class 11 Ionic Equilibrium — the Seven Problem Types

By Aniket Bhardwaj · 26 September 2026 · CBSE / ICSE

The second half of the Class 11 Equilibrium chapter looks enormous, but almost every numerical set in a school or entrance paper is one of just seven problem types. If you can recognise which type a question is within ten seconds, you already know which formula to reach for. This guide names all seven, gives the formula for each, and works one example of each with the arithmetic shown.

First, the definitions the theory questions come from

Theory	Acid is…	Base is…	Example
Arrhenius	Gives H⁺ in water	Gives OH⁻ in water	HCl, NaOH
Brønsted–Lowry	Proton donor	Proton acceptor	NH₃ is a base without any OH⁻
Lewis	Electron-pair acceptor	Electron-pair donor	BF₃ is an acid without any H at all

A conjugate pair differs by exactly one proton: HCl/Cl⁻, H₂O/H₃O⁺, NH₄⁺/NH₃. A strong acid has a weak conjugate base, and the other way round.

The formulas, all in one place

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K
pH = −log[H⁺] · pOH = −log[OH⁻] · pH + pOH = 14 (at 298 K)

Ostwald's dilution law: K_a = cα²/(1 − α); if α ≪ 1, α ≈ √(K_a/c)
Weak acid: [H⁺] = √(K_a · c)

Buffer (Henderson–Hasselbalch): pH = pK_a + log([salt]/[acid])
Basic buffer: pOH = pK_b + log([salt]/[base])

Solubility product: for A_xB_y, K_sp = [A]^x[B]^y

Two cautions that belong with these formulas. First, K_w = 10⁻¹⁴ and pH + pOH = 14 hold only at 298 K; K_w rises with temperature, so neutral water at 60 °C has a pH below 7 and is still neutral. Second, [H⁺] = √(K_a·c) assumes the acid is only slightly dissociated — check afterwards that α is small, and if it is more than about 5 % you must solve the full quadratic instead.

Type 1 — pH of a strong acid or strong base

A strong acid is fully dissociated, so its concentration is the H⁺ concentration. Nothing else is needed.

Q. Find the pH of (a) 0.001 M HCl and (b) 0.01 M NaOH at 298 K.

(a) [H⁺] = 1 × 10⁻³ M
pH = −log(10⁻³) = 3

(b) [OH⁻] = 1 × 10⁻² M
pOH = −log(10⁻²) = 2
pH = 14 − 2 = 12

Type 2 — pH of a weak acid, and the degree of dissociation

Q. Calculate the pH and the degree of dissociation of 0.1 M acetic acid. Take K_a = 1.8 × 10⁻⁵.

Step 1. [H⁺] = √(K_a × c) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶)
√1.8 = 1.3416, so [H⁺] = 1.3416 × 10⁻³ M

Step 2. pH = −log(1.3416 × 10⁻³) = 3 − log 1.3416 = 3 − 0.1276 = 2.87

Step 3. α = [H⁺] ÷ c = 1.3416 × 10⁻³ ÷ 0.1 = 1.3416 × 10⁻² = 1.34 %

Cross-check by a second route: Ostwald's dilution law gives α = √(K_a/c) = √(1.8 × 10⁻⁵ ÷ 0.1) = √(1.8 × 10⁻⁴) = 1.3416 × 10⁻² — the same answer. And since α is only 1.34 %, well under 5 %, the approximation was valid.

Different NCERT printings list K_a for acetic acid as anything between 1.74 × 10⁻⁵ and 1.8 × 10⁻⁵. Use the value in your own table and quote it in your answer; the method is what is being marked.

Type 3 — the common ion effect

Adding an ion that is already part of the equilibrium pushes the equilibrium back, by Le Chatelier's principle. Sodium acetate added to acetic acid supplies acetate ions and suppresses the acid's dissociation.

Q. To 0.1 M acetic acid, sodium acetate is added so that its concentration is also 0.1 M. What is the new pH, and by how much has it changed? (K_a = 1.8 × 10⁻⁵)

pK_a = −log(1.8 × 10⁻⁵) = 5 − log 1.8 = 5 − 0.2553 = 4.74

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

The acid alone was pH 2.87 (Type 2). Adding the common ion has raised the pH by 4.74 − 2.87 = 1.87 units, that is, the hydrogen-ion concentration has fallen by a factor of about 74. This is exactly the common ion effect.

Type 4 — buffer solutions

A buffer resists a change in pH. An acidic buffer is a weak acid with its salt (acetic acid + sodium acetate, useful near pH 4.74); a basic buffer is a weak base with its salt (ammonia + ammonium chloride). Notice from Type 3 that when the salt and acid concentrations are equal, pH = pK_a — which is how you pick the right acid for a buffer at a target pH.

Q. What ratio of sodium acetate to acetic acid is needed for a buffer of pH 5.04? (pK_a = 4.74)

5.04 = 4.74 + log([salt]/[acid])
log([salt]/[acid]) = 5.04 − 4.74 = 0.30
[salt]/[acid] = antilog 0.30 = 2.0

So twice as much salt as acid. Sensible: more of the conjugate base makes the buffer less acidic, and 0.30 is very nearly log 2.

Type 5 — pH of a salt solution (hydrolysis)

Salt from	Example	Nature of solution
Strong acid + strong base	NaCl	Neutral — no hydrolysis
Weak acid + strong base	CH₃COONa	Basic (pH > 7)
Strong acid + weak base	NH₄Cl	Acidic (pH < 7)
Weak acid + weak base	CH₃COONH₄	Depends on which of K_a and K_b is larger

Weak acid + strong base salt: pH = 7 + ½pK_a + ½log c
Strong acid + weak base salt: pH = 7 − ½pK_b − ½log c
(both at 298 K, where ½pK_w = 7)

Q. Find the pH of 0.1 M sodium acetate. (pK_a = 4.74)

pH = 7 + ½(4.74) + ½(log 0.1)
= 7 + 2.37 + ½(−1)
= 7 + 2.37 − 0.50
= 8.87

The answer is above 7, which is what the table predicts for the salt of a weak acid with a strong base — always check your number against the table before writing it down.

Type 6 — solubility product and solubility

Salt type	Example	K_sp in terms of solubility s
AB	AgCl, BaSO₄	s²
AB₂ or A₂B	Ca(OH)₂, Ag₂CrO₄	4s³
A₂B₃	Ca₃(PO₄)₂ type	108s⁵

Derive rather than memorise: for AB₂ the ions are s and 2s, so K_sp = (s)(2s)² = 4s³.

Q. K_sp of AgCl is 1.8 × 10⁻¹⁰. Find its solubility in mol/L and in g/L, and then its solubility in 0.01 M NaCl. (Ag = 107.868, Cl = 35.45)

In pure water. AgCl ⇌ Ag⁺ + Cl⁻, so K_sp = s²
s = √(1.8 × 10⁻¹⁰) = 1.3416 × 10⁻⁵ mol/L

M(AgCl) = 107.868 + 35.45 = 143.318 g/mol
Solubility = 1.3416 × 10⁻⁵ × 143.318 = 1.92 × 10⁻³ g/L

In 0.01 M NaCl. The chloride from the salt swamps the tiny amount from AgCl, so [Cl⁻] ≈ 0.01 M and
s′ = K_sp ÷ [Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 0.01 = 1.8 × 10⁻⁸ mol/L

That is 1.3416 × 10⁻⁵ ÷ 1.8 × 10⁻⁸ ≈ 745 times less soluble than in pure water — the common ion effect again, now acting on a solubility instead of on a pH.

Q. A sparingly soluble hydroxide M(OH)₂ has a solubility of 1.0 × 10⁻² mol/L. Find its K_sp.

M(OH)₂ ⇌ M²⁺ + 2OH⁻, so [M²⁺] = s and [OH⁻] = 2s.
K_sp = (s)(2s)² = 4s³ = 4 × (1.0 × 10⁻²)³ = 4 × 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Type 7 — will a precipitate form?

Work out the ionic product Q using the concentrations actually present after mixing, then compare it with K_sp:

Q < K_sp → unsaturated, no precipitate
Q = K_sp → exactly saturated
Q > K_sp → supersaturated, a precipitate forms

Q. Equal volumes of 2.0 × 10⁻⁴ M AgNO₃ and 2.0 × 10⁻⁴ M NaCl are mixed. Will AgCl precipitate? (K_sp = 1.8 × 10⁻¹⁰)

Step 1 — dilution. Mixing equal volumes halves each concentration:
[Ag⁺] = [Cl⁻] = 1.0 × 10⁻⁴ M

Step 2 — ionic product.
Q = (1.0 × 10⁻⁴) × (1.0 × 10⁻⁴) = 1.0 × 10⁻⁸

Step 3 — compare. 1.0 × 10⁻⁸ is larger than 1.8 × 10⁻¹⁰, so Q > K_sp and a white precipitate of AgCl does form.

Step 1 is the one students skip. Mixing two solutions dilutes both — always recalculate the concentrations before computing Q.

Mistakes that cost marks

Using pH + pOH = 14 at any temperature. It is true at 298 K. K_w changes with temperature and so does the pH of neutral water.
Using [H⁺] = √(K_ac) without checking α. The formula assumes 1 − α ≈ 1. If α exceeds roughly 5 %, solve the quadratic K_a = cα²/(1 − α) properly.
Averaging pH values when two solutions are mixed. pH is a logarithm. Add moles of H⁺, divide by the total volume, and only then take the log.
Applying Henderson–Hasselbalch to a strong acid. It is derived for a weak acid with its conjugate base and means nothing outside a buffer.
Confusing K_sp with solubility. They are different quantities with different units. Two salts of different formula types cannot be ranked by comparing their K_sp values directly.
Dropping the stoichiometric exponent in K_sp. For M(OH)₂ it is (s)(2s)², not (s)(2s).
Forgetting dilution on mixing. Equal volumes halve every concentration before any reaction is considered.
Applying pH = −log(concentration) to very dilute acid. Below about 10⁻⁶ M the H⁺ from water itself matters, and a naive calculation would give an acid a pH above 7, which is impossible.

Where these problem types are examined

Exam	How they appear
CBSE / ICSE Class 11	All seven types, usually with numbers chosen so the approximations hold
Class 12	Return inside electrochemistry (the Nernst equation needs [H⁺]) and in salt analysis
JEE / NEET	Types 3, 4 and 7 in combination, often with a titration wrapped around them
IIT-JAM / CUET-PG / CSIR-NET	The same seven types plus activity coefficients, polyprotic acids and full quadratic treatments

Qualitatively, this is one of the most formula-dense parts of Class 11 chemistry, so the single best use of revision time is sorting past questions into these seven bins rather than reading the chapter again. Once the type is identified, the arithmetic is short.

Check every pH you calculate. The pH / pOH tool converts between [H⁺], [OH⁻], pH and pOH, so you can verify each answer above — for example that 1.3416 × 10⁻³ M really does give pH 2.87 — before you trust your own log working.

Open the pH / pOH Calculator →

Working through Class 11 chemistry and looking for structured practice? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — details at abcchemistry.in. Families in Delhi-NCR who prefer one-to-one home tuition can see delhihometutor.com.