Error Analysis — Common Mistakes in Equilibrium — Notes

#	Mistake	Why It Is Wrong	Correct Approach
1	Writing Kc as [products]/[reactants] without using stoichiometric exponents	The law of mass action requires each concentration raised to its coefficient	Kc = [C]^c[D]^d / [A]^a[B]^b
2	Including pure solids/liquids in Kc or Kp expression	Pure solids and pure liquids have constant "concentrations" (activity = 1); they are omitted from K	Only include gaseous and aqueous species
3	Thinking catalyst shifts equilibrium toward products	Catalyst equally accelerates forward and backward reactions; net effect on equilibrium position is zero	Catalyst changes rate, not equilibrium position or K
4	Inert gas at constant V → shifts equilibrium	Partial pressures and concentrations of reactants/products are unchanged at constant volume	No shift at constant V; shift toward more gas moles only at constant P
5	pH of 10^{-8} M HCl = 8	pH 8 is basic; a strong acid cannot produce a basic solution	Must include water autoionization: [ $H^{+}$ ] ≈ $1.05 \times 10^{-7}$ M; pH ≈ 6.98
6	Increase in temperature always increases K	True only for endothermic reactions; for exothermic reactions, increase in T decreases K	Direction of K change depends on sign of $\Delta H$
7	pH = 7 means neutral at all temperatures	Neutral means [ $H^{+}$ ] = [ $OH^{-}$ ]; at higher T, Kw > 10^{-14}, so neutral pH < 7	pH = 7 is neutral only at 25°C
8	Ka of weak acid = Cα for all concentrations	Ka = Cα^{2} (not Cα); α = degree of dissociation	Ka = Cα^{2}; [ $H^{+}$ ] = Cα = √(Ka·C) when α ≪ 1
9	Buffer pH does not depend on Ka	The Henderson-Hasselbalch equation directly uses pKa; Ka determines the intrinsic buffer pH	pH = pKa + log([salt]/[acid])
10	Higher Ksp always means higher solubility	True only for salts of the same formula type (1:1 vs 1:2); e.g., $BaSO_{4}$ (Ksp = $1.1 \times 10^{-10}$ ) vs $Ag_{2}CrO_{4}$ (Ksp = $1.1 \times 10^{-12}$ ): compare s values, not just Ksp	Calculate s from Ksp for each formula type separately