Explanation:
According to the Henderson-Hasselbalch equation,
pH =
+ ![\frac{log[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7Blog%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Given values are pH = 6,
= 8
Putting given values into the above equation as follows.
6 = 8 + ![\frac{log [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7Blog%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= -2
= antilog -2
= 0.01
But according to the question, we need protonated to deprotonated ratio of ![\frac{[HA]}{[A^{-}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHA%5D%7D%7B%5BA%5E%7B-%7D%5D%7D)
= 
= 100
Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is
.
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