Chemistry Chemical and Ionic Equilibrium

Equilibrium Formula Sheet

All key Chemistry formulas for Chemical & Ionic Equilibrium: Kp-Kc, pH, Ka/Kb, Henderson-Hasselbalch, Ksp. JEE Main & Advanced quick revision.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula, constant, and rule from the Equilibrium chapter on one scannable page. Use this for last-minute revision before JEE Main and Advanced.

Chemical Equilibrium

For the general reaction $aA + bB \rightleftharpoons cC + dD$:

$$\boxed{K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}}$$$$\boxed{K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}}$$$$\boxed{K_p = K_c(RT)^{\Delta n_g}}$$

where $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$, $R = 0.0821 \text{ L·atm·mol}^{-1}\text{K}^{-1}$, and $T$ in Kelvin.

QuantityFormula / RuleNotes
Equilibrium constant$K_c = \dfrac{[\text{products}]^{\text{powers}}}{[\text{reactants}]^{\text{powers}}}$Pure solids/liquids excluded
$K_p$ vs $K_c$$K_p = K_c(RT)^{\Delta n_g}$$K_p = K_c$ when $\Delta n_g = 0$
Units of $K$$(\text{mol/L})^{\Delta n}$Unitless if $\Delta n = 0$
Reaction quotient$Q_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$Same form as $K$, any instant
When Kp = Kc
If $\Delta n_g = 0$, then $(RT)^0 = 1$, so $K_p = K_c$. Classic examples: $H_2 + I_2 \rightleftharpoons 2HI$ and $N_2 + O_2 \rightleftharpoons 2NO$ (both $\Delta n_g = 0$).

Predicting Direction with Q

ComparisonDirectionWhat happens
$Q < K$ForwardMore products form
$Q = K$At equilibriumNo net change
$Q > K$BackwardMore reactants form

Magnitude of K

$K$ valueMeaningPosition
$K \gg 1$ ($> 10^3$)Products favouredLies to right (nearly complete)
$K \approx 1$Both favoured equallyMiddle
$K \ll 1$ ($< 10^{-3}$)Reactants favouredLies to left (barely proceeds)

Manipulating Equilibrium Constants

Operation on reactionEffect on K
Reverse reaction$K_{\text{reverse}} = \dfrac{1}{K_{\text{forward}}}$
Multiply equation by $n$$K_{\text{new}} = (K_{\text{old}})^n$
Add reactions (1 + 2)$K_3 = K_1 \times K_2$

Degree of Dissociation (α)

$$\alpha = \frac{\text{amount dissociated}}{\text{initial amount}}, \qquad 0 \leq \alpha \leq 1$$

For $A \rightleftharpoons B + C$ with initial concentration $C_0$:

$$K_c = \frac{C_0\alpha^2}{1-\alpha}$$

When $\alpha \ll 1$ (valid if $K_c/C_0 < 0.0025$):

$$\boxed{\alpha \approx \sqrt{\frac{K_c}{C_0}}}$$
Affects K vs Not
Only temperature changes $K$. Concentration, pressure, and catalyst have no effect on the value of $K$.

Le Chatelier’s Principle

A system at equilibrium shifts to counteract any disturbance.

DisturbanceEquilibrium responseK changes?
Add reactantShifts toward productsNo
Add productShifts toward reactantsNo
Increase pressure ($\Delta n \neq 0$)Toward fewer gas molesNo
Increase pressure ($\Delta n = 0$)No effectNo
Increase temp (endothermic, $\Delta H > 0$)Shifts forwardYes, increases
Increase temp (exothermic, $\Delta H < 0$)Shifts backwardYes, decreases
Add catalystFaster equilibrium, same positionNo
Add inert gas (constant V)No effectNo
Add inert gas (constant P)Toward more gas molesNo
$$\boxed{\Delta n_g = (\text{moles of gas products}) - (\text{moles of gas reactants})}$$
Pressure & Temperature Tricks

Pressure matters only when $\Delta n \neq 0$.

For temperature, treat heat as a reactant (endothermic) or product (exothermic), then apply Le Chatelier as for any species.

Ionic Equilibrium

Electrolytes

  • Strong electrolytes: $\alpha = 1$ (complete ionisation), no equilibrium.
  • Weak electrolytes: $\alpha \ll 1$ (partial ionisation), equilibrium exists.

Ostwald’s Dilution Law

For weak electrolyte $AB \rightleftharpoons A^+ + B^-$ at concentration $C$:

$$K = \frac{C\alpha^2}{1-\alpha}$$

When $\alpha \ll 1$:

$$\boxed{\alpha = \sqrt{\frac{K}{C}}}$$

Dilution increases the degree of ionisation $\alpha$.

Ionisation Constants

QuantityFormulaNotes
Weak acid$K_a = \dfrac{[H^+][A^-]}{[HA]}$$HA \rightleftharpoons H^+ + A^-$
Weak base$K_b = \dfrac{[B^+][OH^-]}{[BOH]}$$BOH \rightleftharpoons B^+ + OH^-$
$[H^+]$ from weak acid$[H^+] = C\alpha = \sqrt{K_a \cdot C}$when $\alpha \ll 1$
$[OH^-]$ from weak base$[OH^-] = \sqrt{K_b \cdot C}$when $\alpha \ll 1$

Conjugate Pair Relation

$$\boxed{K_a \times K_b = K_w = 10^{-14} \text{ (at 25°C)}}$$$$\boxed{pK_a + pK_b = 14}$$
High-Yield Conversion

Given $K_a$ of an acid, the $K_b$ of its conjugate base is $K_b = K_w / K_a$. This appears in nearly every ionic-equilibrium set.

Common Ion Effect

Adding a common ion suppresses ionisation of a weak electrolyte, shifts equilibrium toward the unionised form, and decreases $\alpha$ (Le Chatelier).

Isohydric Solutions

Two weak acids have the same $[H^+]$ when:

$$\sqrt{K_{a1}C_1} = \sqrt{K_{a2}C_2}$$

Acids and Bases

TheoryAcidBase
ArrheniusProduces $H^+$ in waterProduces $OH^-$ in water
Brønsted–LowryProton ($H^+$) donorProton ($H^+$) acceptor
LewisElectron-pair acceptorElectron-pair donor

Conjugate pairs differ by exactly one $H^+$:

$$\underbrace{HA}_{\text{acid 1}} + \underbrace{B}_{\text{base 2}} \rightleftharpoons \underbrace{A^-}_{\text{base 1}} + \underbrace{BH^+}_{\text{acid 2}}$$
  • Strong acid $\leftrightarrow$ weak conjugate base; weak acid $\leftrightarrow$ strong conjugate base.
  • Amphiprotic species (donate and accept $H^+$): $H_2O$, $HCO_3^-$, $HSO_4^-$, $H_2PO_4^-$, $HPO_4^{2-}$, $HS^-$.

Factors Affecting Acid Strength

FactorTrendExample
Electronegativity of atom bonded to HMore EN → stronger acid$HF > H_2O > NH_3 > CH_4$
Atomic size (same group)Larger atom → stronger acid$HI > HBr > HCl > HF$
Oxidation state of central atom (oxoacids)Higher state → stronger acid$HClO_4 > HClO_3 > HClO_2 > HClO$
Resonance stabilisation of conjugate baseMore resonance → stronger acid$CH_3COOH > CH_3OH$
Seven Strong Acids

Memorise these — everything else is weak: $HCl,\ HBr,\ HI,\ HNO_3,\ H_2SO_4,\ HClO_4,\ HClO_3$.

pH and Buffers

Water and the pH Scale

$$\boxed{K_w = [H^+][OH^-] = 10^{-14} \text{ (at 25°C)}}$$$$\boxed{pH = -\log[H^+]} \qquad \boxed{pOH = -\log[OH^-]}$$$$\boxed{pH + pOH = 14 \text{ (at 25°C)}}$$

pH Formulas by Solution Type

Solution typepH formulaKey point
Strong acid (monoprotic)$pH = -\log C$$[H^+] = C_{\text{acid}}$
Strong base$pH = 14 + \log C$$[OH^-] = C_{\text{base}}$
Weak acid$pH = \tfrac{1}{2}(pK_a - \log C)$no salt present
Weak base$pH = 14 - \tfrac{1}{2}(pK_b - \log C)$no salt present
Acidic buffer$pH = pK_a + \log\dfrac{[\text{Salt}]}{[\text{Acid}]}$Henderson–Hasselbalch
Basic buffer$pOH = pK_b + \log\dfrac{[\text{Salt}]}{[\text{Base}]}$, then $pH = 14 - pOH$Henderson–Hasselbalch

For a diprotic strong acid (e.g. $H_2SO_4$): $[H^+] = 2 \times C_{\text{acid}}$.

Henderson–Hasselbalch Equation

$$\boxed{pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = pK_a + \log\frac{[A^-]}{[HA]}}$$$$\boxed{pOH = pK_b + \log\frac{[\text{Salt}]}{[\text{Base}]}}$$

Buffer Capacity and Range

$$\beta = \frac{\text{amount of acid/base added}}{\text{change in pH}}$$
  • Maximum buffer capacity when $[\text{Salt}] = [\text{Acid}]$, i.e. $pH = pK_a$.
  • Effective buffer range: $pH = pK_a \pm 1$ (ratio between 0.1 and 10).
Buffer Reminders

A buffer needs both the weak acid and its conjugate base present.

Use initial concentrations directly in Henderson–Hasselbalch (the common-ion effect keeps ionisation negligible).

Salt Hydrolysis (pH of Salt Solutions)

Salt typepHExample
Strong acid + strong base7$NaCl$
Strong acid + weak base$< 7$$NH_4Cl$
Weak acid + strong base$> 7$$CH_3COONa$
Weak acid + weak baseDepends on $K_a$ vs $K_b$$CH_3COONH_4$

Solubility Product (Ksp)

For a sparingly soluble salt $A_xB_y(s) \rightleftharpoons xA^{y+} + yB^{x-}$:

$$\boxed{K_{sp} = [A^{y+}]^x [B^{x-}]^y}$$

The solid is not included; $K_{sp}$ applies only to saturated solutions and depends on temperature.

Ksp–Solubility Relationships

Salt type$K_{sp}$ expression$K_{sp}$–S relation$S$ from $K_{sp}$Example
AB$[A^+][B^-]$$S^2$$\sqrt{K_{sp}}$$AgCl$, $BaSO_4$
AB₂$[A^{2+}][B^-]^2$$4S^3$$\sqrt[3]{K_{sp}/4}$$CaF_2$, $PbCl_2$
A₂B$[A^+]^2[B^{2-}]$$4S^3$$\sqrt[3]{K_{sp}/4}$$Ag_2S$, $Ag_2CrO_4$
AB₃$[A^{3+}][B^-]^3$$27S^4$$\sqrt[4]{K_{sp}/27}$$AlF_3$
A₂B₃$[A^{3+}]^2[B^{2-}]^3$$108S^5$$\sqrt[5]{K_{sp}/108}$$Al_2(SO_4)_3$

Ionic Product (Q) vs Ksp

ComparisonStateResult
$Q < K_{sp}$UnsaturatedNo precipitation, more can dissolve
$Q = K_{sp}$SaturatedAt equilibrium
$Q > K_{sp}$SupersaturatedPrecipitation occurs

Common Ion Effect on Solubility

For $AgCl$ in a solution already containing the common ion at concentration $c$ (with solubility $s \ll c$):

$$K_{sp} = s \times c \implies s = \frac{K_{sp}}{c}$$

Solubility falls sharply compared with pure water.

Selective Precipitation

The salt requiring the lower counter-ion concentration to reach $Q = K_{sp}$ precipitates first (lower $K_{sp}$, for the same salt type). Minimum ion concentration needed to start precipitation:

$$[\text{counter-ion}] = \left(\frac{K_{sp}}{[\text{given ion}]^{\,p}}\right)^{1/q}$$

where $p$ is the given ion’s coefficient and $q$ is the counter-ion’s coefficient (e.g. for $PbCl_2$ with $Pb^{2+}$ given: $p=1$, $q=2$, so $[Cl^-] = (K_{sp}/[Pb^{2+}])^{1/2}$).

pH and Solubility

For salts with basic anions ($OH^-$, $S^{2-}$, $CO_3^{2-}$): higher pH lowers solubility of metal hydroxides, while lower pH raises solubility of basic salts (acid removes the anion and pulls equilibrium right).

Master Constants & Relations

RelationValue / Form
Ionic product of water$K_w = [H^+][OH^-] = 10^{-14}$ (25°C)
$pH + pOH$$14$ (25°C)
Conjugate pair$K_a \times K_b = K_w$; $pK_a + pK_b = 14$
$K_p$–$K_c$$K_p = K_c(RT)^{\Delta n_g}$
Gas constant$R = 0.0821 \text{ L·atm·mol}^{-1}\text{K}^{-1} = 8.314 \text{ J·mol}^{-1}\text{K}^{-1}$
Max buffer capacityat $pH = pK_a$ ($[\text{Salt}]=[\text{Acid}]$)
Effective buffer range$pK_a \pm 1$