The equilibrium law can be applied to acid–base reactions. Numerical problems can be simplified by making assumptions about the relative concentrations of the species involved.
The use of logarithms is also significant here.

  • $\rm pH = –\log_{10} [H^+]$; $\rm [H^+] = 10^{–pH}$
  • $\rm pOH = –\log_{10}[OH^–]$; $\rm [OH^–] = 10^{–pOH}$
  • $\mathrm K_w = \rm [H^+] [OH^–] = 1.00 \times 10^{–14}$ at $\rm 25°C =$ From the above; $\rm 10^{–pH} \times 10^{–pOH} = 1.00 \times 10^{–14}$ at $\rm 25°C$
    $\rm pH + pOH = 14.00$ at $\rm 25 °C$
  • $\mathrm{pK}_w = –\log_{10}(\mathrm K_w)$; $\mathrm K_w = 10^{−\mathrm{pK}_w}$
  • Taking logs in the equation: $\mathrm K_w = \rm [H^+] [OH^–]$
    $\mathrm{pK}_w = \rm pH + pOH$
  • The value of $\mathrm K_w$, as an equilibrium constant is temperature dependant. $\rm pH$, and $\rm pOH$ are also temperature dependant. The $\rm pH$ of neutral water is $7$ at $\rm 25°C$ but does change with temperature.
  • For strong acids and strong bases, $\rm pH$ can be calculated from the concentration of the acid as all the acid molecules dissociate:
    $\rm HA(aq) \rightarrow H^+(aq) + A^–(aq)$
    $\rm [H^+(aq)] = [HA(aq)]$
    $\rm pH = –\log_{10}[HA(aq)]$
    For weak acids and weak bases, the calculation of $\rm pH$ is more complicated.
    $\rm K_a$ gives a measure of the dissociation of a weak acid.
    $\rm HA(aq) \rightleftharpoons H^+(aq) + A^–(aq)$
    $\rm K_a = \dfrac{[H^+(aq)][A^–(aq)]}{[HA(aq)]}$
    The stronger the acid the higher the value of $\rm K_a$. 
  • $\rm K_b$, is a measure of the strength of a weak base.
    $\rm B(aq) + H_2O(l)\rightleftharpoons BH^+(aq) + OH^–(aq)$
    $\rm K_b = \dfrac{[BH^+][OH^–]}{[B]}$
    The stronger the base the higher the value of $\rm K_b$. 
  • Strong acids and bases have infinite values of $\rm K_a$ and $\rm K_b$.
  • Higher values for $\rm K_a$ and $\rm K_b$ represent stronger acids and bases respectively.
  • The $\rm K_a$ value of an acid and $\rm K_b$ for its conjugate acid are related:
    $\rm HA(aq)\rightleftharpoons H^+(aq) + A^–(aq)$  
    $\rm K_a = \dfrac{[H^+(aq)][A^–(aq)]}{[HA(aq)]}$
    $\rm A^–(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^–(aq)$
    $\rm K_b = \dfrac{[HA(aq)][OH^–(aq)]}{[A^–(aq)]}$
    $\rm K_a \times K_b$ $\rm = \dfrac{[H^+(aq)][A^–(aq)]}{[HA_{(aq)}]} \times \dfrac{[HA(aq)][OH^–(aq)]}{[A^–(aq)]}$ $\rm = [H^+(aq)] [OH^–(aq)]$
    $\rm K_a \times K_b = K_{\mathcal w}$
    Taking logarithms:
    $\rm pK_a = –\log_{10} [K_a]$; $\rm [K_a] = 10^{−pK_a}$
    $\rm pK_b = –\log_{10} [K_b]$; $\rm [K_b] = 10^{−pK_b}$
    $\rm pK_a + pK_b = pK_{\mathcal w}$
  • Stronger acids and bases have lower values for $\rm pK_a$ and $\rm pK_b$.