“Solids and liquids sit out”
Writing an equilibrium constant
Pure solids and pure liquids never appear in K. Their concentration does not change, so it is absorbed into the constant.
Chemistry, Class 11 and 12 · Gujarat Board, CBSE, GUJCET, JEE and NEET
Chemical and Ionic Equilibrium formulas
Equilibrium is two chapters treated as one: the equilibrium constant, and what happens in water. Between them they supply pH, buffers and solubility, which are among the most reliably asked numericals in the whole subject.
18 formulas, 4 memory hooks, and the 5 mistakes that cost the most marks.
The equilibrium constant
| What | Formula | Symbols and notes |
|---|---|---|
| In concentrations | Kc = [products]/[reactants], each raised to its coefficient | pure solids and liquids are left out |
| In partial pressures | Kp = Kc(RT)Δn | Δn is moles of gaseous products minus reactants |
| Reaction quotient | Q compared with K | Q < K goes forward, Q > K goes backward, Q = K is equilibrium |
| Reversed reaction | K' = 1/K | and for a reaction multiplied by n, K becomes Kn |
| Le Chatelier's principle | a system at equilibrium opposes any change imposed on it | pressure, concentration or temperature |
Acids, bases and pH
| What | Formula | Symbols and notes |
|---|---|---|
| Ionic product of water | Kw = [H⁺][OH⁻] = 10−14 | at 25 °C |
| pH | pH = −log[H⁺] | 7 is neutral at 25 °C |
| pOH and the relation | pOH = −log[OH⁻], pH + pOH = 14 | at 25 °C |
| Weak acid | [H⁺] = √(KaC) | C is the initial concentration |
| Degree of dissociation | α = √(Ka/C) | Ostwald's dilution law: dilution increases α |
| pK values | pKa = −log Ka | a smaller pKa means a stronger acid |
Buffers and solubility
| What | Formula | Symbols and notes |
|---|---|---|
| Acidic buffer | pH = pKa + log([salt]/[acid]) | the Henderson-Hasselbalch equation |
| Basic buffer | pOH = pKb + log([salt]/[base]) | then pH = 14 − pOH |
| Solubility product, AB | Ksp = s² | for example AgCl |
| Solubility product, AB₂ | Ksp = 4s³ | for example CaF₂ |
| Solubility product, A₂B₃ | Ksp = 108s⁵ | for example Ca₃(PO₄)₂ type salts |
| Precipitation | precipitates when the ionic product exceeds Ksp | otherwise the salt stays dissolved |
| Common ion effect | adding a common ion suppresses dissociation | which is how buffers resist pH change |
Taught by Darshit Bhatt, founder and lead faculty at Darshit Sir Institute. Instrumentation and Control engineer, with a master's degree and a chartered degree, teaching in Paldi, Ahmedabad since 2006.
Know the formula but not when to use it?
Darshit Sir has taught this in Paldi since 2006, in the classroom and live online.
Memory hooks
How to remember this chapter.
“Push it, it pushes back”
Le Chatelier's principle
Add a reactant and the reaction moves forward; raise the pressure and it shifts to the side with fewer gas molecules; heat an exothermic reaction and it goes backwards.
“Buffer is pKa plus the log of the ratio”
Henderson-Hasselbalch
When salt and acid are equal the log is zero, so pH equals pKa. That is the best buffering point, and it is worth remembering on its own.
“s squared, 4s cubed, 108s to the five”
Solubility product expressions
AB gives s², AB₂ gives 4s³, A₂B₃ gives 108s⁵. The coefficients come from the stoichiometry, not from memory alone.
Where marks go
The mistakes that cost the most.
- Including pure solids or liquids in the equilibrium constant expression.
- Forgetting that Kp equals Kc only when Δn is zero.
- Using pH + pOH = 14 at a temperature other than 25 °C, where Kw is different.
- Treating a strong acid with the weak acid formula. A strong acid dissociates completely, so [H⁺] equals its concentration.
- Writing Ksp = s² for every salt. The exponents follow the formula of the salt.
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