# What are Ka and Kb in acids and bases?

Jun 18, 2018

These are measures of acidity and basicity...

#### Explanation:

And acid in aqueous solution is conceived to undergo a protonolysis reaction...

$H X \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {X}^{-}$

And as for any equilibrium, we can measure and quantify it in the usual way...

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{X}^{-}\right]}{\left[H X \left(a q\right)\right]}$

Note that ${H}_{2} O$ DOES NOT appear in the equilibrium expression because it is present in such high concentration that it is effectively constant..

For strong acids, i.e. $H I$, $H B r$, $H C l$, ${H}_{2} S {O}_{4}$ protonolysis is effectively quantitative: the given equilibrium lies entirely to the right as we face the page, and the acid solution is quantitative in ${H}_{3} {O}^{+}$. For weaker acids, $H F$, ${H}_{3} C - C {O}_{2} H$, the equilibrium lies somewhat to the left...and concentrations of the parent acid remain at equilibrium.

And likewise, we can formalize the performance of a base by an equivalent equilibrium...we use ammonia, because this is a WEAK base in aqueous solution...

$N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + H {O}^{-}$

And ${K}_{b}$ is defined in an equivalent way to ${K}_{a}$...

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]}$, ${K}_{b} \text{(ammonia)} = 1.74 \times {10}^{-} 5$...

Confused yet....?

Well, note that NECESSARILY....for a given acid/conjugate pair, say $N {H}_{4}^{+} \text{/} N {H}_{3}$...

${K}_{a} {K}_{b} = {10}^{-} 14$...or perhaps more usefully...

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