Question

An acid with a p K a of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form of the acid?

Answer 1

Explanation:

According to the Henderson-Hasselbalch equation,

                  pH = $pK_{a}$ + $\frac{log[A^{-}]}{[HA]}$

Given values are pH = 6, $pK_{a}$ = 8

Putting given values into the above equation as follows.

                   6 = 8 + $\frac{log [A^{-}]}{[HA]}$

                   $\frac{log[A^{-}]}{[HA]}$ = -2

                   $\frac{[A^{-}]}{[HA]}$ = antilog -2

                                         = 0.01

But according to the question, we need protonated to deprotonated ratio of $\frac{[HA]}{[A^{-}]}$

            $\frac{[HA]}{[A^{-}]}$ = $\frac{1}{0.01}$

                     $\frac{[HA]}{[A^{-}]}$ = 100

Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is $\frac{100}{1}$.