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Equilibrium: one-page cheat sheet

NEET Chemistry · Unit 6 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 3–4 questions · 12–16 marks

1 · Equilibrium constant

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

Reverse

K_{rev} = 1/ K

n

K^{'} = K^{n}

Add equations

K_{net} = K_{1} \times K_{2}

K_{p} = K_{c} (R T)^{Δ n_{g}}

Δ n_{g} = 0

K_{p} = K_{c}

_{2}

+ I

_{2}

, N

_{2}

+ O

_{2}

)

Δ n_{g} > 0

K_{p} > K_{c}

(PCl

_{5}

dissociation)

Δ n_{g} < 0

K_{p} < K_{c}

_{2}

+ 3H

_{2}

)

2 · Reaction quotient

Compare	Result
$Q_{c} < K_{c}$	forward reaction
$Q_{c} = K_{c}$	at equilibrium
$Q_{c} > K_{c}$	backward reaction

Δ G = Δ G^{\circ} + R T ln Q, Δ G^{\circ} = - 2.303 R T lo g K

3 · Le Chatelier's principle

Change	Shift
add reactant / remove product	forward
add product	backward
increase pressure	fewer gas moles
decrease pressure	more gas moles
increase $T$	endothermic direction
decrease $T$	exothermic direction
catalyst	no shift; faster approach
inert gas, constant $V$	no shift
inert gas, constant $P$	more gas moles

Only temperature changes the value of

K

. Concentration, pressure and catalyst do not.

4 · Acids, bases & water

Arrheniusacid gives H

^{+}

, base gives OH

^{-}

Brønsted–Lowryacid donates H

^{+}

, base accepts; conjugate pair differs by one H

^{+}

Lewisacid accepts an electron pair (BF

_{3}

, AlCl

_{3}

); base donates one (NH

_{3}

)

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

$K_{w}$ rises with $T$ , so neutral pH falls below 7 on heating.

5 · Weak electrolytes

K_{a} = \frac{C α ^{2}}{1 - α} \Rightarrow α = \frac{K _{a}}{C}, [H^{+}] = K_{a} C

pH = \frac{1}{2} (p K_{a} - lo g C)

Common ionCH

_{3}

COONa added to CH

_{3}

COOH lowers

[H^{+}]

, raises pH

6 · Hydrolysis & buffers

Salt of	pH at 298 K
weak acid + strong base	$7 + \frac{1}{2} (p K_{a} + lo g C)$
strong acid + weak base	$7 - \frac{1}{2} (p K_{b} + lo g C)$
weak acid + weak base	$7 + \frac{1}{2} (p K_{a} - p K_{b})$

Acidic buffer

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

Basic buffer

pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}

Weak acid + weak base salt: pH does not depend on concentration.

7 · Solubility product

Type	$K_{s p}$	Examples
AB	$s^{2}$	AgCl, BaSO $_{4}$
AB $_{2}$ or A $_{2}$ B	$4 s^{3}$	Mg(OH) $_{2}$ , Ag $_{2}$ CrO $_{4}$
AB $_{3}$	$27 s^{4}$	Fe(OH) $_{3}$ , Al(OH) $_{3}$
A $_{2}$ B $_{3}$	$108 s^{5}$	As $_{2}$ S $_{3}$
A $_{3}$ B $_{4}$	$6912 s^{7}$	Zr $_{3}$ (PO $_{4}$ ) $_{4}$

Precipitateswhen ionic product

>

K_{s p}

Group IIH

_{2}

S in dilute HCl: common H

^{+}

keeps

[S^{2 -}]

low; only CuS, PbS, Bi

_{2}

_{3}

settle

Group IIINH

_{4}

OH with NH

_{4}

Cl: common NH

_{4}^{+}

limits

[OH^{-}]

; Fe(OH)

_{3}

, Al(OH)

_{3}

settle

8 · Quick pH rules

Strong acid

pH = - lo g C

(monoprotic)

Strong base

pOH = - lo g C

, then

pH = 14 - pOH

Acid strengthlarger

K_{a}

means smaller

p K_{a}

means stronger acid

Solubilitya common ion lowers the solubility of a sparingly soluble salt

Seen in NEET 2026

Buffer with equal $[X^{-}]$ and $[HX]$ ; $K_{b} (X^{-}) = 1 0^{- 10}$ . Find pH.

K_{a} = \frac{1 0 ^{- 14}}{1 0 ^{- 10}} = 1 0^{- 4} \Rightarrow pH = p K_{a} + lo g 1 = 4

Seen in NEET 2022

50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid; $p K_{a} = 4.57$ .

pH = 4.57 + lo g \frac{0.10}{0.01} = 5.57

Six traps to remember

Only temperature changes $K$ .
$K_{p} = K_{c} (R T)^{Δ n_{g}}$ : count gas moles only.
$Q > K$ means the reaction runs backward.
A catalyst never shifts the equilibrium; it only speeds up reaching it.
Inert gas at constant volume does nothing; at constant pressure it favours more gas moles.
Henderson equation uses $lo g (salt / acid)$ ; equal amounts give pH $= p K_{a}$ .

Equilibrium: free one-page cheat sheet