What is the Ka of a strong acid like HCl?

Aug 23, 2016

The approximate value of $p K a$ is $- 7$ for $H C l$.

Explanation:

Hydrochloric acid is a strong acid and it is considered to completely dissociate in water under the following equation:

$H C l \left(a q\right) + {H}_{2} O \left(l\right) \to {H}_{3} {O}^{+} \left(a q\right) + C {l}^{-} \left(a q\right)$

The equilibrium constant ${K}_{a}$ is written as:

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

Since the acid dissociates completely, the concentration of products ($\left[{H}_{3} {O}^{+}\right] \mathmr{and} \left[C {l}^{-}\right]$) is very large and the concentration of the reactant ($\left[H C l\right]$) is very small. Therefore, the value of ${K}_{a}$ is going to be very large (~10^(7)).

Since $p {K}_{a} = - \log {K}_{a}$, therefore, the value of the $p {K}_{a}$ is going to be a negative number.

The approximate value of $p K a$ is $- 7$ for $H C l$.