Henderson – Hasselbalch Equation

In an acidic buffer, $[H_3O^+]=K_a\dfrac{[acid]_{eq}}{[base]_{eq}}$; since a weak acid dissociates only slightly, and common-ion suppression from the salt weakens it further, the equilibrium acid concentration is close to the acid's initial (unionised) concentration, and the equilibrium conjugate-base concentration is close to the salt's initial concentration -- so $[H_3O^+]=K_a\dfrac{[acid]}{[salt]}$, using the initial concentrations used to prepare the buffer. Taking $-\log_{10}$ of both sides: $-\log_{10}[H_3O^+]=-\log_{10}K_a-\log_{10}\dfrac{[acid]}{[salt]}$, and since $pH=-\log_{10}[H_3O^+]$ and $pK_a=-\log_{10}K_a$, this rearranges to the Henderson–Hasselbalch equation, $pH=pK_a+\log_{10}\dfrac{[salt]}{[acid]}$. The equivalent form for a basic buffer is $pOH=pK_b+\log_{10}\dfrac{[salt]}{[base]}$.

Example 8.6 – pH of an acetate buffer from molar concentrations. Find the pH of a buffer containing 0.20 mol/L sodium acetate and 0.18 mol/L acetic acid, $K_a=1.8\times10^{-5}$. $pK_a=-\log_{10}(1.8\times10^{-5})=5-\log_{10}1.8=5-0.26=4.74$. $pH=pK_a+\log_{10}\dfrac{[salt]}{[acid]}=4.74+\log_{10}\dfrac{0.20}{0.18}=4.74+\log_{10}(10/9)=4.74+(1-0.95)=4.74+0.05=4.79$.

Example 8.7 – pH of a buffer prepared by mass. 6 g of acetic acid and 8.2 g of sodium acetate are made up to 500 mL; $K_a=1.8\times10^{-5}$ (so $pK_a=4.74$, as in Example 8.6). Moles of sodium acetate $=8.2/82=0.1$ mol, so $[salt]=0.1/0.5=0.2$ M. Moles of acetic acid $=6/60=0.1$ mol, so $[acid]=0.1/0.5=0.2$ M. Since $[salt]=[acid]$, $pH=pK_a+\log_{10}(0.2/0.2)=pK_a+\log_{10}1=4.74+0=4.74$. …