a .35 m solution of weak acid, HX, has a h+ of 4.15×10^-3. what is the value of ka for this acid

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03-06-2016
Chemistry

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a .35 m solution of weak acid, HX, has a h+ of 4.15×10^-3. what is the value of ka for this acid

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Hagrid Hagrid · Answer 1 · 2016-06-05T12:16:00+02:00

Hagrid Hagrid

05-06-2016

Ka has the formula:

Ka = [H+] [X-] / [HX]

We have the following given:
[H+] = 4.15x10^-3
[HX] = 0.35

Since the acid dissociates in a ratio of 1 [H+] to 1 X,
[H+] = [X-]

Plugging in the values,
Ka = (4.15x10^-3)(4.15x10-3) / 0.35
Ka = 4.92x10^-5

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