# What is Kb and Ka in chemistry?

Jun 27, 2016

${K}_{b}$ and ${K}_{a}$ are the base and acid dissociation constants respectively. They are simply equilibrium constants.

#### Explanation:

For acid dissociation:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

For base hydrolysis:

${A}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H {O}^{-} + H A$

${K}_{b} = \frac{\left[H {O}^{-}\right] \left[H A\right]}{\left[{A}^{-}\right]}$.

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

${K}_{a} \times {K}_{b} = \frac{\left[{H}_{3} {O}^{+}\right] \cancel{\left[{A}^{-}\right]}}{\cancel{\left[H A\right]}} \times \frac{\left[H {O}^{-}\right] \cancel{\left[H A\right]}}{\cancel{\left[{A}^{-}\right]}}$

$=$ $\left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]$ $=$ ${K}_{w}$.

Of course, in water, at $298 \cdot K$, ${K}_{w} = {10}^{- 14}$.

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

$14 = p {K}_{a} + p {K}_{b}$

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