Chapter 5 of 11

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NEET UG Chemistry★ 2–3 Q within Chemistry

Equilibrium and Ionic Equilibrium

A complete NEET treatment of equilibrium — the equilibrium constant Kc/Kp and their relation, the reaction quotient, and Le Chatelier's principle for every stress; then ionic equilibrium in full — Brønsted and Lewis acids and bases, Kw and the pH scale, weak-acid dissociation and Ostwald's law, the common-ion effect, buffers with the Henderson equation, salt hydrolysis and the solubility product. Every constant is derived and every calculation worked.

⏱ 40 min readModerate difficulty✓ Free · exam-pattern aligned

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Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

NEET UG PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Write Kc and Kp, relate them via (RT)^Δn_g and use Q vs K to predict direction
2Apply Le Chatelier's principle to concentration, pressure, temperature and catalyst
3Classify acids and bases by Arrhenius, Brønsted–Lowry and Lewis definitions and identify conjugate pairs
4Calculate pH/pOH for strong and weak acids and bases using Kw, Ka and Ostwald's law
5Compute buffer pH with the Henderson–Hasselbalch equation and explain the common-ion effect
6Predict salt-solution acidity by hydrolysis and calculate solubility from Ksp

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Why this chapter matters in NEET UG

Equilibrium is where chemistry becomes two-directional: reactions settle into a dynamic balance whose position we control in industry and in the body. The block yields a reliable 2–3 NEET questions spanning equilibrium constants, Le Chatelier shifts, pH, buffers and solubility — and it underpins electrochemistry, kinetics and biochemistry. This chapter builds Kc and Kp from the law of mass action, applies Le Chatelier to every stress, then develops ionic equilibrium in full: acid–base definitions, the pH scale, weak-acid dissociation, buffers with the Henderson equation, salt hydrolysis and the solubility product, with every calculation worked.

Before you start — revise these

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States of Matter and Thermodynamics

ΔG° = −RT ln K links thermodynamics to the equilibrium constant.

Equilibrium and Ionic Equilibrium — NEET Chemistry

Equilibrium is where chemistry stops being one-directional. Reactions rarely go to completion — they settle into a dynamic balance in which forward and reverse rates are equal, and the position of that balance is what we control in a lab, a blast furnace or a cell. NEET draws 2–3 questions from this block, spanning equilibrium constants, Le Chatelier shifts, pH calculations, buffers and solubility. This chapter builds the equilibrium constant from first principles, then applies the same logic to acids, bases and sparingly soluble salts. Every relationship is derived and drilled with the exact calculations the exam repeats.

Part A — Chemical Equilibrium

1. Dynamic equilibrium and the equilibrium constant

In a reversible reaction $a A + b B ⇌ c C + d D$ , equilibrium is reached when the forward and reverse rates become equal — concentrations then stop changing, though both reactions continue (hence dynamic). The law of mass action gives a constant ratio:

$K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

$K_{c}$ is constant at a given temperature; it changes only with temperature.
Large $K_{c}$ (≫1) → products favoured; small $K_{c}$ (≪1) → reactants favoured.
Pure solids and pure liquids are omitted (their "concentration" is constant).

For gases, an equivalent constant uses partial pressures, $K_{p}$ , related to $K_{c}$ by:

$K_{p} = K_{c} (R T)^{Δ n_{g}}, Δ n_{g} = (moles gas products) - (moles gas reactants)$

So $K_{p} = K_{c}$ only when $Δ n_{g} = 0$ .

Worked example 1.1. For N₂ + 3H₂ ⇌ 2NH₃, write $K_{c}$ and relate $K_{p}$ to $K_{c}$ . $K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ . Here $Δ n_{g} = 2 - 4 = - 2$ , so $K_{p} = K_{c} (R T)^{- 2}$ .

Reaction quotient $Q$ has the same form as $K$ but with current (non-equilibrium) concentrations. Comparing $Q$ with $K$ tells you the direction of the net reaction:

$Q < K$ → forward reaction proceeds (more product forms).
$Q > K$ → reverse reaction proceeds.
$Q = K$ → at equilibrium.

Worked example 1.2. For a reaction $K_{c} = 10$ . If at some instant $Q = 2$ , which way does it shift? $Q < K$ , so the reaction proceeds forward (toward products) until $Q$ rises to 10.

2. Le Chatelier's principle

Le Chatelier's principle: if a system at equilibrium is disturbed, it shifts to partly counteract the disturbance. Applied to four stresses:

Stress applied	Equilibrium shifts to…
Increase [reactant]	forward (consume added reactant)
Increase [product]	backward
Increase pressure (↓ volume)	side with fewer gas moles
Increase temperature	endothermic direction (absorbs heat)
Add catalyst	no shift (speeds both rates equally)
Add inert gas at constant V	no shift (partial pressures unchanged)

Temperature is the only stress that changes $K$ itself — the others merely move the position along a fixed $K$ . For an exothermic reaction ( $Δ H < 0$ ), heating decreases $K$ (shifts backward); for endothermic, heating increases $K$ .

Worked example 2.1 (Haber process). N₂ + 3H₂ ⇌ 2NH₃, $Δ H < 0$ . How do high pressure and low temperature affect the yield? High pressure favours the side with fewer gas moles (2 < 4) → more NH₃. Low temperature favours the exothermic forward reaction → more NH₃, but slows the rate; industry compromises at ~450 °C with a catalyst.

Worked example 2.2. For an endothermic reaction, what happens to $K$ when temperature is raised? Heating drives an endothermic reaction forward, so $K$ increases — more product at equilibrium.

Part B — Ionic Equilibrium

3. Acids and bases: three definitions

Theory	Acid	Base
Arrhenius	gives H⁺ in water	gives OH⁻ in water
Brønsted–Lowry	proton (H⁺) donor	proton acceptor
Lewis	electron-pair acceptor	electron-pair donor

The Brønsted view introduces conjugate acid–base pairs — an acid becomes its conjugate base on losing H⁺:

$acid H A + base H_{2} O ⇌ conj. base A^{-} + conj. acid H_{3} O^{+}$

A strong acid has a weak conjugate base and vice versa. The Lewis view is broadest — it covers acids with no H at all (BF₃, AlCl₃ accept electron pairs; NH₃, H₂O donate them).

Worked example 3.1. Identify the conjugate base of HCO₃⁻ and its conjugate acid. Losing H⁺: conjugate base is CO₃²⁻. Gaining H⁺: conjugate acid is H₂CO₃. (HCO₃⁻ is amphoteric — it can do both.)

4. Ionic product of water and the pH scale

Water self-ionises: $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ , with

$K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 1 0^{- 14} at 25 °C$

Because $K_{w}$ is constant, $[H^{+}]$ and $[O H^{-}]$ are inversely linked. The pH scale compresses these into logarithms:

$pH = - lo g [H^{+}], pOH = - lo g [O H^{-}], pH + pOH = 14$

Neutral: $[H^{+}] = [O H^{-}] = 1 0^{- 7}$ , pH = 7.
Acidic: pH < 7; basic: pH > 7.
$K_{w}$ rises with temperature, so neutral pH is below 7 in hot water (but $[H^{+}] = [O H^{-}]$ still means neutral).

Worked example 4.1. pH of 0.01 M HCl (a strong acid, fully dissociated)? $[H^{+}] = 0.01 = 1 0^{- 2}$ M, so pH $= - lo g (1 0^{- 2}) = 2$ .

Worked example 4.2. pH of 0.001 M NaOH? $[O H^{-}] = 1 0^{- 3}$ , pOH $= 3$ , so pH $= 14 - 3 = 11$ .

5. Weak acids, Ka and Ostwald's dilution law

A weak acid dissociates only partially. For $H A ⇌ H^{+} + A^{-}$ with initial concentration $C$ and degree of dissociation $α$ :

$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} = \frac{C α ^{2}}{1 - α} \approx C α^{2} (α ≪ 1)$

This is Ostwald's dilution law: for a weak electrolyte, $α \approx K_{a} / C$ — dilution increases the degree of dissociation. And:

$[H^{+}] = K_{a} C, pH = \frac{1}{2} (p K_{a} - lo g C), p K_{a} = - lo g K_{a}$

A smaller $p K_{a}$ (larger $K_{a}$ ) means a stronger acid.

Worked example 5.1. A 0.1 M weak acid has $K_{a} = 1 0^{- 5}$ . Find $[H^{+}]$ , pH and $α$ . $[H^{+}] = K_{a} C = 1 0^{- 5} \times 1 0^{- 1} = 1 0^{- 6} = 1 0^{- 3}$ M → pH = 3. $α = K_{a} / C = 1 0^{- 5} /1 0^{- 1} = 1 0^{- 4} = 1 0^{- 2} = 0.01$ (1% dissociated).

6. Common-ion effect and buffers

Common-ion effect: adding an ion already present in an equilibrium suppresses the dissociation of a weak electrolyte (Le Chatelier). Adding CH₃COONa to acetic acid pushes $C H_{3} C O O H ⇌ H^{+} + C H_{3} C O O^{-}$ backward, lowering $[H^{+}]$ .

This is exactly how a buffer works — a solution that resists pH change on adding small amounts of acid or base. A buffer is a weak acid + its salt (acidic buffer) or a weak base + its salt (basic buffer). Its pH follows the Henderson–Hasselbalch equation:

$pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$

pH = $p K_{a}$ when [salt] = [acid] (maximum buffer capacity).
Blood is buffered near pH 7.4 by the H₂CO₃/HCO₃⁻ system — the biological headline example.

Worked example 6.1. A buffer has 0.2 M acetic acid ( $p K_{a} = 4.74$ ) and 0.2 M sodium acetate. Its pH? $pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]} = 4.74 + lo g \frac{0.2}{0.2} = 4.74 + 0 = 4.74$ .

Worked example 6.2. What ratio [salt]/[acid] gives a buffer of pH 5.74 with the same acid? $5.74 = 4.74 + lo g r \Rightarrow lo g r = 1 \Rightarrow r = 10$ . Ten times as much salt as acid.

7. Salt hydrolysis — the pH of salt solutions

A salt can make its solution acidic, basic or neutral depending on the strength of its parent acid and base:

Salt from…	Example	Solution	Reason
Strong acid + strong base	NaCl	neutral (pH 7)	neither ion hydrolyses
Strong acid + weak base	NH₄Cl	acidic (pH < 7)	cation (NH₄⁺) hydrolyses
Weak acid + strong base	CH₃COONa	basic (pH > 7)	anion (CH₃COO⁻) hydrolyses
Weak acid + weak base	CH₃COONH₄	depends on $K_{a}$ vs $K_{b}$	both hydrolyse

The ion of the weaker parent hydrolyses and dictates the pH.

Worked example 7.1. Is an aqueous solution of ammonium chloride (NH₄Cl) acidic, basic or neutral? NH₄Cl comes from a strong acid (HCl) and a weak base (NH₃). The ammonium ion hydrolyses (NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺), releasing H⁺ → the solution is acidic.

8. Solubility product and its applications

For a sparingly soluble salt $A_{x} B_{y} ⇌ x A^{y +} + y B^{x -}$ , the solubility product is:

$K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$

If solubility is $s$ mol/L, then for a salt like AgCl ( $1 : 1$ ), $K_{s p} = s^{2}$ ; for $A_{2} B$ or $A B_{2}$ ( $1 : 2$ ), $K_{s p} = 4 s^{3}$ ; for $A_{2} B_{3}$ , $K_{s p} = 108 s^{5}$ .

Precipitation: a precipitate forms when the ionic product $Q_{s p}$ exceeds $K_{s p}$ . ( $Q_{s p} < K_{s p}$ : unsaturated; $= K_{s p}$ : saturated.)
Common-ion effect on solubility: adding a common ion lowers solubility (why AgCl is less soluble in NaCl solution than in water).

Worked example 8.1. $K_{s p}$ of AgCl is $1.8 \times 1 0^{- 10}$ . Its molar solubility in water? $K_{s p} = s^{2} \Rightarrow s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}$ mol/L.

Worked example 8.2. For a salt $A B_{2}$ with $K_{s p} = 3.2 \times 1 0^{- 11}$ , find $s$ . $K_{s p} = 4 s^{3} \Rightarrow s^{3} = 8 \times 1 0^{- 12} \Rightarrow s = 2 \times 1 0^{- 4}$ mol/L.

9. Common traps NEET sets here

$K_{p} = K_{c} (R T)^{Δ n_{g}}$ — count only gaseous mole change; they're equal only if $Δ n_{g} = 0$ .
Only temperature changes $K$ — pressure, concentration and catalyst move the position, not $K$ .
Catalyst and inert gas (constant V) cause no shift — a favourite trap in Le Chatelier questions.
$Q$ vs $K$ : $Q < K$ goes forward, $Q > K$ goes backward.
pH + pOH = 14 at 25 °C; for a strong base, find pOH first.
Dilution increases $α$ (Ostwald) but the acid still gets weaker in [H⁺].
Buffer pH = $p K_{a}$ + log(salt/acid) — not log(acid/salt); watch the ratio direction.
Salt hydrolysis: the ion of the weaker parent decides acidity/basicity.
$K_{s p}$ shape: $s^{2}$ (1:1), $4 s^{3}$ (1:2), $108 s^{5}$ (2:3) — don't forget the coefficient.

10. Memory aids

" $Q$ low, go" — $Q < K$ drives the forward reaction.
"Fewer gas moles win under pressure" — the Le Chatelier pressure rule.
"Heat feeds the endothermic side" — temperature shift direction.
"Strong acid, weak conjugate" — the Brønsted pairing.
"pH = ½(p $K_{a}$ − log C)" — weak-acid pH in one line.
"Buffer = salt over acid, log it onto p $K_{a}$ " — Henderson equation.
"Weaker parent's ion hydrolyses" — salt-solution pH.

11. Exam protocol

Write $K_{c}$ (omit solids/liquids); convert to $K_{p}$ with $(R T)^{Δ n_{g}}$ using gaseous moles only.
Compare $Q$ with $K$ to get the direction; remember only temperature changes $K$ .
Apply Le Chatelier: reactant/product for concentration, fewer-gas-moles for pressure, endothermic side for heat; catalyst and inert gas → no shift.
For strong acids/bases, pH directly from concentration; use pH + pOH = 14.
For weak acids, $[H^{+}] = K_{a} C$ , pH = ½(p $K_{a}$ − log C); dilution raises $α$ .
Buffer pH from Henderson $p K_{a} + lo g (salt/acid)$ ; recognise the common-ion effect.
Salt hydrolysis: identify the strong/weak parents; the weaker parent's ion sets the pH.
Solubility: use the correct $K_{s p}$ – $s$ relation and precipitate when $Q_{s p} > K_{s p}$ .

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Equilibrium constant

K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}

Omit pure solids and liquids; K changes only with temperature.

Kp–Kc relation

K_{p} = K_{c} (R T)^{Δ n_{g}}

Δn_g = gaseous product moles − reactant moles; equal only if Δn_g = 0.

Ionic product & pH

K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14}, pH + pOH = 14

At 25 °C; neutral pH = 7, acidic < 7, basic > 7.

Weak-acid pH (Ostwald)

[H^{+}] = K_{a} C, α = K_{a} / C

Dilution increases the degree of dissociation α.

Henderson–Hasselbalch

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

pH = pKa when [salt] = [acid]; the basis of buffer action.

Solubility product

K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}

AgCl: s²; AB₂: 4s³; A₂B₃: 108s⁵; precipitate when Q_sp > K_sp.

NEET UG test series

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Equilibrium and Ionic Equilibrium shows up in every NEET UG paper — sit a timed one and find out where you actually stand.

2–3Qs in paper

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Traps NEET UG sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Thinking a catalyst or an inert gas shifts the equilibrium.

✓ A catalyst speeds forward and reverse reactions equally, so it changes only the rate, not the position or K. Adding an inert gas at constant volume leaves all partial pressures unchanged, so it too causes no shift.

WATCH OUT

✗ Believing pressure or concentration changes the value of K.

✓ Only temperature changes K itself. Concentration, pressure and catalyst move the position of equilibrium along a fixed K. For an exothermic reaction, heating decreases K; for endothermic, heating increases it.

WATCH OUT

✗ Forgetting Δn_g counts only gaseous species in Kp = Kc(RT)^Δn_g.

✓ Δn_g is the change in moles of gas only. Solids and liquids are excluded, and Kp equals Kc only when Δn_g = 0.

WATCH OUT

✗ Using log(acid/salt) in the Henderson equation.

✓ The buffer equation is pH = pKa + log([salt]/[acid]) — salt over acid. Inverting the ratio flips the sign of the log term and gives the wrong pH.

WATCH OUT

✗ Thinking all salts give neutral solutions.

✓ Only salts of a strong acid and strong base (like NaCl) are neutral. The ion of the weaker parent hydrolyses: NH₄Cl (weak base) is acidic, CH₃COONa (weak acid) is basic.

WATCH OUT

✗ Dropping the coefficient in the Ksp–solubility relation.

✓ For a 1:1 salt Ksp = s², but for AB₂ or A₂B it is 4s³ and for A₂B₃ it is 108s⁵. Missing the numerical factor gives a wrong solubility.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Equilibrium and Ionic Equilibrium?

15 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

15 questions~11 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Dynamic equilibrium: forward rate = reverse rate; Kc = products/reactants (omit solids/liquids)
•Kp = Kc(RT)^Δn_g (gaseous moles only); K changes only with temperature
•Q < K forward, Q > K backward, Q = K equilibrium
•Le Chatelier: reactant/product for conc., fewer gas moles for pressure, endothermic side for heat; catalyst & inert gas → no shift
•Acids/bases: Arrhenius (H⁺/OH⁻), Brønsted (proton donor/acceptor), Lewis (e⁻-pair acceptor/donor); conjugate pairs
•Kw = 10⁻¹⁴; pH + pOH = 14; neutral 7, acidic <7, basic >7
•Weak acid: [H⁺] = √(KaC), α = √(Ka/C); dilution raises α; pKa = −log Ka
•Buffer: pH = pKa + log(salt/acid); common-ion effect suppresses dissociation
•Salt hydrolysis: weaker parent's ion sets pH; Ksp shapes s², 4s³, 108s⁵; precipitate when Q_sp > Ksp

NEET UG question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 12

Question style	Marks each	Typical count	What it tests
Equilibrium constant & Le Chatelier	~1 Q
pH, weak acids & buffers	~1 Q
Hydrolysis & solubility product	~1 Q

Prep strategy

Drill Kc/Kp writing and Q-versus-K direction questions
Master the Le Chatelier shift table, especially the catalyst/inert-gas traps
Practise pH for strong and weak acids/bases and buffer (Henderson) calculations
Learn the salt-hydrolysis rules and the Ksp–solubility relations for 1:1, 1:2 and 2:3 salts

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Write Kc (omit solids/liquids); convert to Kp with (RT)^Δn_g using gaseous moles only.
Compare Q with K for direction; remember only temperature changes K.
Apply Le Chatelier: fewer gas moles under pressure, endothermic side on heating; catalyst/inert gas → no shift.
Strong acid/base pH directly; use pH + pOH = 14.
Weak acid [H⁺] = √(KaC), pH = ½(pKa − log C); buffer via Henderson.
Salt hydrolysis by parent strength; solubility from the correct Ksp–s relation, precipitate when Q_sp > Ksp.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Blood pH regulation

The carbonic-acid/bicarbonate buffer holds blood near pH 7.4; small deviations cause acidosis or alkalosis.

Industrial synthesis

Le Chatelier optimisation of temperature and pressure maximises yield in the Haber (ammonia) and Contact (sulphuric acid) processes.

Solubility and kidney stones

The solubility product predicts when salts like calcium oxalate precipitate — the chemistry behind kidney-stone formation.

Antacids and drug absorption

Acid–base equilibria govern stomach pH, antacid action and how drugs ionise and cross membranes.

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE MainEquilibrium constants & ionic equilibrium

JEE AdvancedMulti-step pH & solubility problems

CUET (Science)Equilibrium & acid–base chemistry

State medical/engg CETspH, buffer & Ksp MCQs

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

The equilibrium constant is fixed by the standard free-energy change through ΔG° = −RT ln K, and ΔG° itself depends on temperature. Changing concentration or pressure disturbs the position of equilibrium, and the system shifts to restore the same K — the constant is unchanged. A catalyst speeds both directions equally, so it too leaves K alone. Only altering the temperature changes ΔG° and hence K: heating increases K for an endothermic reaction and decreases it for an exothermic one.

The synthesis N₂ + 3H₂ ⇌ 2NH₃ is exothermic and reduces the number of gas moles from four to two. Le Chatelier says high pressure favours the side with fewer gas moles (more ammonia) and low temperature favours the exothermic forward reaction (more ammonia). But low temperature also makes the reaction impossibly slow, so industry compromises at around 450 °C with an iron catalyst and high pressure (~200 atm), removing ammonia as it forms to keep pulling the equilibrium forward.

In the Brønsted–Lowry view, an acid donates a proton and a base accepts one. When an acid HA loses its proton it becomes A⁻, its conjugate base; when a base gains a proton it becomes the conjugate acid. HA and A⁻ differ by exactly one H⁺ and form a conjugate pair. A strong acid has a weak conjugate base and vice versa. Water is amphoteric — its conjugate base is OH⁻ and its conjugate acid is H₃O⁺.

A buffer contains a weak acid and its conjugate base (or a weak base and its salt) in comparable amounts. When acid is added, the conjugate base neutralises it; when base is added, the weak acid neutralises it. Because both reservoirs are present, the ratio [salt]/[acid] changes only slightly, so the Henderson–Hasselbalch pH = pKa + log([salt]/[acid]) barely moves. Blood is buffered near pH 7.4 by the carbonic-acid/bicarbonate system, which is vital for life.

A salt dissociates into a cation and an anion, and either may react with water (hydrolyse). Only salts of a strong acid and a strong base, like NaCl, are neutral because neither ion hydrolyses. For a salt of a strong acid and weak base (NH₄Cl), the cation hydrolyses to release H⁺, making the solution acidic. For a salt of a weak acid and strong base (CH₃COONa), the anion hydrolyses to release OH⁻, making it basic. The ion derived from the weaker parent controls the pH.