Gaseous Reaction Equilibria and pH

For a reversible reaction $aA + bB \rightleftharpoons cC + dD$, at equilibrium: $$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \qquad K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$ where $K_p = K_c(RT)^{\Delta n}$.

Le Châtelier's principle: a system at equilibrium shifts to counteract any applied change (concentration, pressure, temperature).

Water autoionizes: $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$. The ion product of water: $$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times10^{-14}\;(\text{at }25°\mathrm{C})$$ $$\mathrm{pH} = -\log[H_3O^+]\qquad \mathrm{pOH} = -\log[OH^-]\qquad \mathrm{pH + pOH = 14}$$ Strong acids (HCl, HNO₃, H₂SO₄) dissociate completely → $[H_3O^+] = C_{acid}$. Strong bases (NaOH, KOH) dissociate completely → $[OH^-] = C_{base}$.

Weak acids partially dissociate (equilibrium). The acid dissociation constant: $$K_a = \dfrac{[H_3O^+][A^-]}{[HA]}$$ For weak bases: $K_b = \dfrac{[BH^+][OH^-]}{[B]}$ with $K_a\cdot K_b = K_w$.

The pH calculation uses the approximation (if $C/K_a > 10^3$): $[H_3O^+] \approx \sqrt{K_a\cdot C}$. The exact solution requires solving the quadratic from the mass balance.

The degree of dissociation $\alpha$ is the fraction of acid/base that has dissociated: $$\alpha = \dfrac{[A^-]}{[HA]_0} = \dfrac{K_a}{[H_3O^+] + K_a}$$

Inductive effect: electron-withdrawing groups (e.g. Cl in $\mathrm{CCl_3COOH}$) increase $K_a$ by stabilizing the conjugate base. Leveling effect: the strongest acid that can exist in a solvent is the solvated proton; any stronger acid is leveled to $H_3O^+$.

For polyprotic acids (e.g. $\mathrm{H_3PO_4}$), each dissociation step has its own $K_{a1}, K_{a2}, K_{a3}$.

A buffer solution resists pH change. It consists of a weak acid and its conjugate base (or weak base and its conjugate acid).

The Henderson-Hasselbalch equation: $$\mathrm{pH} = \mathrm{p}K_a + \log\!\left(\dfrac{[A^-]}{[HA]}\right)$$ The buffer capacity is maximum when $[A^-]/[HA] = 1$ (pH = pKₐ).