How do you solve for Ka when you only have molarity of the acid and pH?

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How do you solve for Ka when you only have molarity of the acid and pH?

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Answer 1 · 2015-06-01T23:26:29

You start by using the pH of the solution to determine the concentration of the hydronium ions, $H_{3} O^{+}$ $H_3O^(+)$ .

$[H_{3} O^{+}] = 10_{sol}^{- P H_{sol}}$ $[H_3O^(+)] = 10^(-PH_"sol")$

The general dissociation equation for a weak acid looks like this

$H A_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + A_{(a q)}^{-}$ $HA_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + A_((aq))^(-)$

By definition, the acid dissociation constant, $K_{a}$ $K_a$ , will be equal to

$K_{a} = \frac{[H_{3} O^{+}] \cdot [A^{-}]}{[H A]}$ $K_a = ([H_3O^(+)] * [A^(-)])/([HA])$

If you have a $1 : 1$ $1:1$ mole ratio between the acid and the hydronium ions, and between the hydronium ions and the conjugate base, $A^{-}$ $A^(-)$ , then the concentration of the latter will be equal to that of the hydronium ions.

$[A^{-}] = [H_{3} O^{+}]$ $[A^(-)] = [H_3O^(+)]$

Since you know the molarity of the acid, $K_{a}$ $K_a$ will be

$K_{a} = \frac{{[H_{3} O^{+}]}_{3}^{2}}{[H A]}$ $K_a = ([H_3O^(+)]^2)/([HA])$

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How do you solve for Ka when you only have molarity of the acid and pH?

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