hey, the answer is D and the clarification of options 1 and 2 is obvious. I just don’t get it about number 3, when we add NAOH the solution gets even more alkaline, well the reaction must undergo the prosses of consumption of CH3OH. why isn’t it right
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Hi!
Adding sodium hydroxide will react with the ethanoic acid which means that less methanol reacts with the ethanoic acid, so 3 is correct. You can think of this as the same thing as taking away ethanoic acid.
Thus CH3OH is increased to compensate low ethanoic acid.
Hope this helps!
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that was pretty helpful, thank you.
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