EquilibriumNEET Chemistry · Class 11 · NCERT Chapter 6

Kc is the equilibrium constant written in terms of molar concentrations (mol/L). Kp is written in terms of partial pressures (atm or bar). They are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). For reactions where Δn = 0, Kp = Kc.

Calculate the reaction quotient Q using the current concentrations (same formula as Kc but with current values, not equilibrium values). If Q < K, the reaction goes forward (towards products). If Q > K, the reaction goes backward (towards reactants). If Q = K, the system is already at equilibrium.

It means a system at equilibrium resists change. If you disturb the equilibrium (by adding a reactant, increasing pressure, or changing temperature), the system shifts to partially undo that disturbance. Adding a reactant shifts equilibrium toward products. Increasing pressure shifts toward fewer moles of gas. Increasing temperature shifts toward the endothermic direction.

For a weak acid HA with concentration C and dissociation constant Ka: HA ⇌ H⁺ + A⁻. At equilibrium, [H⁺] = [A⁻] = x. Ka = x²/(C − x). If Ka is small (weak acid), C − x ≈ C, so x = √(Ka × C). Then pH = −log(x). Always check the approximation: it is valid when x/C < 5%.

pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid in the buffer. When [A⁻] = [HA], pH = pKa. Buffers work best when pH is within 1 unit of pKa.

A precipitate forms when the ionic product (IP) of the ions in solution exceeds Ksp. If IP < Ksp, the solution is unsaturated and no precipitate forms. If IP = Ksp, the solution is just saturated. If IP > Ksp, precipitation occurs. The common ion effect lowers solubility by increasing one of the ions already present.