Relation between pH and pOH

Adding the defining equations $pH=-\log_{10}[H_3O^+]$ and $pOH=-\log_{10}[OH^-]$ gives $pH+pOH = -\log_{10}([H_3O^+][OH^-]) = -\log_{10}K_w = pK_w$. At $25^\circ C$, $K_w=1\times10^{-14}$, so $pK_w=-\log_{10}(1\times10^{-14})=14$, giving the widely used relation $pH+pOH=14$ at $25^\circ C$. This lets pH and pOH be converted into each other directly once one of them (or $[H_3O^+]$/$[OH^-]$) is known.

Example 8.2 – pH of 0.001M HCl. Calculate the pH of 0.001M HCl. As a strong acid, $HCl$ dissociates completely: $[H_3O^+]=0.001=10^{-3}$ M dominates the auto-ionisation contribution from water (only $10^{-7}$ M), which is negligible by comparison. So $pH=-\log_{10}(10^{-3})=3$. The unit also notes that when the acid/base concentration is below about $10^{-6}$ M, water's own auto-ionisation contribution can no longer be neglected, and must be added: $[H_3O^+]=10^{-7}$ (from water) $+ [H_3O^+]$ (from the acid).

Example 8.3 – pH of 10⁻⁷ M HCl (water's contribution matters). Calculate the pH of $10^{-7}$ M HCl. Ignoring water's own ionisation would give $[H_3O^+]=10^{-7}$ M and pH = 7 -- but $HCl$ is always acidic regardless of concentration, so pH = 7 (neutral) cannot be right. Because the acid concentration ($10^{-7}$ M) is comparable to water's own $H_3O^+$ contribution ($10^{-7}$ M from auto-ionisation), both must be added: $[H_3O^+] = 10^{-7}(\text{from HCl}) + 10^{-7}(\text{from water}) = 2\times10^{-7}$ M. Then $pH = -\log_{10}(2\times10^{-7}) = 7 - \log_{10}2 = 7 - 0.301 = 6.70$. …