Buffer Action

To resist pH changes on adding an acid or base, a buffer must contain both an acidic and a basic component that can neutralise the added stress without consuming each other. In a $CH_3COOH$/$CH_3COONa$ buffer, $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq)+H_3O^+(aq)$: if acid is added, the extra $H^+$ is consumed by the conjugate base $CH_3COO^-$ to re-form undissociated acid, so the pH barely drops; if base is added, the extra $OH^-$ is neutralised by $H_3O^+$, which the dissociating acetic acid immediately replenishes to maintain equilibrium, so the pH barely rises. These neutralisation reactions are exactly the ones already met under the common ion effect.

Worked buffer-action calculation: 0.01 mol NaOH added to a 0.8M/0.8M CH3COOH/CH3COONa buffer. One litre of a buffer contains 0.8M $CH_3COOH$ and 0.8M $CH_3COONa$; $K_a$ for $CH_3COOH$ is $1.8\times10^{-5}$. Before addition: since $CH_3COOH$ dissociates only slightly and common-ion suppression makes it even smaller, $[CH_3COOH]\approx0.8$ and $[CH_3COO^-]\approx0.8$ (the salt fully dissociates and its concentration is already $\ge$ the acid's, so $\alpha$ is negligible), giving $[H^+]=K_a\dfrac{[CH_3COOH]}{[CH_3COO^-]}=K_a\dfrac{0.8}{0.8}=K_a$, so $pH=-\log_{10}(1.8\times10^{-5})=5-\log_{10}1.8=5-0.26=4.74$. After adding 0.01 mol NaOH to the 1 litre buffer (negligible volume change): the $OH^-$ consumes 0.01 mol of the acid, giving $[CH_3COOH]=0.8-0.01=0.79$ and $[CH_3COO^-]=0.8+0.01=0.81$; $[H^+]=1.8\times10^{-5}\times\dfrac{0.79}{0.81}=1.76\times10^{-5}$, so $pH=5-\log_{10}1.76=5-0.25=4.75$. The pH shifts only from 4.74 to 4.75 despite adding a strong base -- verifying buffer action. …