How to calculate pKa of HCl?

Nov 28, 2014

The ionisation of HCl can be shown as:

$H C l + {H}_{2} O \to {H}_{3} {O}^{+} + C {l}^{-}$

${K}_{a}$ can be calculated as the ratio of the product of the concentrations of the products to that of the reactant, that is,

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[C {l}^{-}\right]}{\left[H C l\right]}$

Since $H C l$ is a strong acid, the value of ${K}_{a}$ turns out to be very large, that is,

${K}_{a} = {10}^{7} / 1 = {10}^{7}$ (approx.)

The value of $p {K}_{a}$ is given by

$p {K}_{a} = - \log {K}_{a}$
$p {K}_{a} = - \left(\log {10}^{7}\right)$
$p {K}_{a} = - \left(7\right) = - 7$

You can learn more about finding the values of ${K}_{a}$ and $p {K}_{a}$ here.