Ionic Equilibrium (11th) formulas

Every formula JEE Advanced expects you to know for Ionic Equilibrium (11th), with the variables spelled out. Free, no sign-up. Read it before a mock, not during one.

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Formulas: 22
Subject: Chemistry
Exam: JEE Advanced
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Ionic Equilibrium (11th), every formula

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Ionic product of water
$K_{w} = [H^{+}] [O H^{-}] = 1 0^{- 14} (2 5^{\circ} C)$
pH
$pH = - lo g [H^{+}]$
pOH
$pOH = - lo g [O H^{-}]$
pH + pOH
$pH + pOH = 14$
Acid dissociation constant
$K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}$
Base dissociation constant
$K_{b} = \frac{[ B H ^{+} ] [ O H ^{-} ]}{[ B ]}$
Ka × Kb
$K_{a} \times K_{b} = K_{w}$
Ostwald's dilution law
$K_{a} = \frac{α ^{2} C}{1 - α} \approx α^{2} C$
Degree of dissociation
$α = \frac{K _{a}}{C}$
[H⁺] of a weak acid
$[H^{+}] = K_{a} C$
pKa
$p K_{a} = - lo g K_{a}$
Henderson equation (acid buffer)
$pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$
Henderson (base buffer)
$pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]}$
Solubility product
$K_{s p} = [A^{+}]^{x} [B^{-}]^{y}$
Ksp of a salt AB
$K_{s p} = S^{2}$
S = molar solubility
Ksp of AB₂
$K_{s p} = 4 S^{3}$
Common-ion effect
$added common ion suppresses dissociation/solubility$
Hydrolysis (salt of weak acid)
$pH = 7 + \frac{1}{2} (p K_{a} + lo g C)$
Degree of hydrolysis
$h = \frac{K _{w}}{K _{a} C}$
Buffer at half-neutralization
$pH = p K_{a}$
Neutral solution
$[H^{+}] = [O H^{-}] = 1 0^{- 7} M$
Precipitation condition
$ionic product > K_{s p} \Rightarrow precipitation$

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Ionic Equilibrium (11th) formulas

Ionic Equilibrium (11th), every formula

Ionic product of water

pH

pOH

pH + pOH

Acid dissociation constant

Base dissociation constant

Ka × Kb

Ostwald's dilution law

Degree of dissociation

[H⁺] of a weak acid

pKa

Henderson equation (acid buffer)

Henderson (base buffer)

Solubility product

Ksp of a salt AB

Ksp of AB₂

Common-ion effect

Hydrolysis (salt of weak acid)

Degree of hydrolysis

Buffer at half-neutralization

Neutral solution

Precipitation condition

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Know the Ionic Equilibrium (11th) formulas. Now use them.

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