What is Kb and Ka in chemistry?

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Answer 1 · 2016-06-27T03:08:59

$K_{b}$ $K_b$ and $K_{a}$ $K_a$ are the base and acid dissociation constants respectively. They are simply equilibrium constants.

For acid dissociation:

$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$ $HA+H_2OrightleftharpoonsH_3O^+ + A^-$

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

For base hydrolysis:

$A^{-} + H_{2} O ⇌ H O^{-} + H A$ $A^(-)+H_2OrightleftharpoonsHO^(-) + HA$

$K_{b} = \frac{[H O^{-}] [H A]}{[A^{-}]}$ $K_b=([HO^-][HA])/([A^-])$ .

As with any mathematical expression, they can be manipulated, and as water itself undergoes self protonolysis, we can write:

$K_axxK_b=([H_3O^+]cancel([A^-]))/cancel([HA])xx([HO^-]cancel([HA]))/cancel([A^-])$

$=$ $[HO^-][H_3O^+]$ $=$ $K_w$ .

Of course, in water, at $298*K$ , $K_w=10^(-14)$ .

And if we take logarithms to the base 10, and mulitply each side by $-1$ .

$14=pK_a + pK_b$

There is a lot of unsolicited info here so use it for reference. The defining equation for $K_a$ is necessary to remember.