Equilibrium: Chemical and Ionic
Equilibrium: Chemical and Ionic tests reversible reactions, Kc and Kp, Le Chatelier's principle, weak acids and bases, pH, buffers, hydrolysis, and solubility product. The exam expects both equilibrium reasoning and numerical approximations.
Core rules
- For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d/([A]^a[B]^b). Kp = Kc(RT)^Δn, where Δn is gaseous product moles minus reactant moles.
- A catalyst changes the rate of approach but not the equilibrium constant.
- Le Chatelier: increasing pressure favors the side with fewer gas moles; increasing temperature favors the endothermic direction.
- Weak acid: Ka = [H+][A−]/[HA]; pH = −log[H+]. For a weak acid, [H+] ≈ sqrt(Ka × C) when dissociation is small.
- Buffer: pH = pKa + log([base]/[acid]).
- Solubility product: for sparingly soluble salt, Ksp gives molar solubility; a common ion lowers solubility.
Worked example
Find the pH of 0.10 M acetic acid with Ka = 1.8 × 10−5. Let x = [H+]. x^2/(0.10 − x) = 1.8 × 10−5. Since x is small, x^2 ≈ 1.8 × 10−6, so x = 1.34 × 10−3 M. pH = −log(1.34 × 10−3) = 2.87. The approximation is acceptable because x is about 1.3% of 0.10 M.
Traps
Do not use Kp = Kc(RT)^Δn with Δn counted for all species; count gases only. Do not apply the weak-acid square-root formula when dissociation is large. Do not forget that adding a common ion shifts solubility equilibrium left. A buffer fails when the added acid or base exceeds its capacity.