Home
Learn
Plan & Focus
Community
Pricing
Play & Compete
Free Resources

Log in Join Free

by JEE Master

Ask. Share. Learn. Grow Together. India's all-in-one community for JEE Main & JEE Advanced aspirants.

Explore

Learn
Plan & Focus
Community
Play & Compete
Free Resources

Resources

JEE Main PYQs
JEE Advanced PYQs
Formula Sheets
NCERT Mapping
Mock Analysis

Support

Help Center
Privacy Policy
Terms of Use
Contact Us

© 2026 PadhoNa. All rights reserved.

Formula Sheets / Chemistry

Chapter 8

Ionic Equilibrium

Revise with flashcards Download this chapter (PDF) All Chemistry formulas (PDF)

Acids, bases and pH Buffers and salt hydrolysis Solubility product

Acids, bases and pH

pH $^{-}] = 1 0^{- 14} (2 5^{\circ}$ H $^{+}],$ $K_{w} = [$ H $+$ OH $^{-}] = 1 0^{- 14} (2 5^{\circ}$ C $),$ pH $+$ pOH $= 14$
Weak acid (Ostwald)
$α = \frac{K _{a}}{C},$ $[$ H $^{+}] = K_{a} C,$ pH $= \frac{1}{2} ($ p $K_{a} - lo g C)$
$K_{a} K_{b} = K_{w}$ (conjugate pair)

Buffers and salt hydrolysis

Henderson
pH $=$ p $K_{a} + lo g \frac{[ salt ]}{[ acid ]};$ pOH $=$ p $K_{b} + lo g \frac{[ salt ]}{[ base ]}$
Weak acid + strong base salt
pH $= 7 + \frac{1}{2} ($ p $K_{a} + lo g C)$
Strong acid + weak base salt
pH $= 7 - \frac{1}{2} ($ p $K_{b} + lo g C)$
Weak acid + weak base salt
pH $= 7 + \frac{1}{2} ($ p $K_{a} -$ p $K_{b})$

Solubility product

$A_{x} B_{y} : K_{s p} = x^{x} y^{y} s^{x + y}$
$A B : K_{s p} = s^{2};$ $A B_{2} : K_{s p} = 4 s^{3};$ $A B_{3} : K_{s p} = 27 s^{4}$
Precipitation
ionic product $> K_{s p} \Rightarrow$ precipitate
Common ion
solubility falls: e.g. $s = \frac{K _{s p}}{[ common ion ]}$ for $A B$

Chemical Equilibrium Redox Reactions