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

Step-by-step explanation:

According to the Henderson-Hasselbalch equation,

pH =
`pK_(a)` +
`(log[A^(-)])/([HA])`

Given values are pH = 6,
`pK_(a)` = 8

Putting given values into the above equation as follows.

6 = 8 +
`(log [A^(-)])/([HA])`

`(log[A^(-)])/([HA])` = -2

`([A^(-)])/([HA])` = antilog -2

= 0.01

But according to the question, we need protonated to deprotonated ratio of
`([HA])/([A^(-)])`

`([HA])/([A^(-)])` =
`(1)/(0.01)`

`([HA])/([A^(-)])` = 100

Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is
`(100)/(1)`.