What is the Ka of a strong acid like HCl?

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What is the Ka of a strong acid like HCl?

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Answer 1 · 2016-08-23T08:50:57

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

Explanation:

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

$H C l (a q) + H_{2} O (l) \to H_{3} O^{+} (a q) + C l^{-} (a q)$ $HCl(aq)+H_2O(l)->H_3O^(+)(aq)+Cl^(-)(aq)$

The equilibrium constant $K_{a}$ $K_a$ is written as:

$K_{a} = \frac{[H_{3} O^{+}] [C l^{-}]}{H C l}$ $K_a=([H_3O^+][Cl^-])/(HCl]$

Since the acid dissociates completely, the concentration of products ( $[H_{3} O^{+}] and [C l^{-}]$ $[H_3O^+] and [Cl^-]$ ) is very large and the concentration of the reactant ( $[H C l]$ $[HCl]$ ) is very small. Therefore, the value of $K_{a}$ $K_a$ is going to be very large ( $~ 10^{7}$ $~10^(7)$ ).

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

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

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What is the Ka of a strong acid like HCl?

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