How does Ka relate to acid strength?

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How does Ka relate to acid strength?

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Answer 1 · 2016-08-01T11:26:36

Strong acids typically possess large $K_{a}$ $K_a$ values.

Explanation:

The generalized protonolysis reaction for an acid in water is as follows:

$H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} + X^{-}$ $HX(aq) + H_2O(l) rightleftharpoons H_3O^(+) + X^-$

And of course $K_{a} = \frac{[H_{3} O^{+}] [X^{-}]}{[H X]}$ $K_a=([H_3O^+][X^-])/([HX])$

For strong acids, i.e. $H X, H_{2} S O_{4} . H C l O_{4}$ $HX, H_2SO_4. HClO_4$ , the equilibrium lies almost quantitatively to the RIGHT. Given this we can look back to the the $K_{a}$ $K_a$ expression and realize that if equilibrium quantities of $[H_{3} O^{+}]$ $[H_3O^+]$ and $[X^{-}]$ $[X^-]$ are large and $[H X]$ $[HX]$ is small, then $K_{a}$ $K_a$ is correspondingly large. And thus large $K_{a}$ $K_a$ values betoken strong acids.

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How does Ka relate to acid strength?

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