Question #69eae

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Question #69eae

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Answer 1 · 2017-05-17T16:43:09

$Dissociate.............$ $"Dissociate............."$

Explanation:

The strength of an acid relates to the extent which the following equilibrium is completed:

$H X (a q) + H_{2} O (l) ⇌ H_{3} O^{+} + X^{-}$ $HX(aq) + H_2O(l) rightleftharpoonsH_3O^(+) + X^-$

For strong acids, $X^{-} = H S O_{4}^{-}, halide (X \neq F), C l O_{4}^{-}$ $X^(-) =HSO_4^-, "halide" (X!=F), ClO_4^-$ , the equilibrium lies STRONGLY to the right. For weaker acids, $H F, H_{2} S O_{3}, H O_{2} C - C H_{3}$ $HF, H_2SO_3, HO_2C-CH_3$ , the equilibrium lies to the left.

And this we can use the equilibrium constant, $K_{a}$ $K_a$ , as a determinant:

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

The given acids, have $K_{a}$ $K_a$ values so large in water that the dissociation is considered to be quantitative.

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Question #69eae

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