Q.x. In NaOH solution $\mathrm{[OH^-]}$ is $2.87 \times 10^{-4}$. Calculate the pH of solution.

Because [H+] and [OH-] in aqueous solutions span an enormous range (roughly 1 M down to about $10^{-14}$ M), it is far more convenient to work on a logarithmic scale. pH is defined as $pH=-\log_{10}[H^+]$, and pOH similarly as $pOH=-\log_{10}[OH^-]$; each single pH unit represents a ten-fold change in [H+]. The two are linked through water's own ionization equilibrium, $H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)$, whose equilibrium constant, the ionic product of water $K_w=[H_3O^+][OH^-]$, equals $1.0\times10^{-14}$ at 298 K (rising at higher temperature, since water's ionization is endothermic). Taking logarithms of $K_w=[H_3O^+][OH^-]$ gives the compact and very widely used relation $pH+pOH=14$ at 298 K. This immediately fixes the three regions of the pH scale: a neutral solution (pure water) has $[H_3O^+]=[OH^-]=1.0\times10^{-7}$ M and pH exactly 7; an acidic sol …