How to calculate the concentration of a monoprotic organic acid that was extracted into an ethyl ether layer?

Compound A is a monoprotic organic acid. If $25.0~\mathrm{mL}$ of a $0.456~M$ solution of A in water was extracted with $25.0~\mathrm{mL}$ of ethyl ether and it was determined experimentally that $5.45~\mathrm{mL}$ of $0.523~M~\ce{NaOH}$ was required to neutralize the bottom layer from the extraction, what is the concentration of compound A in the ether layer?

My attempt:

$25.0~\mathrm{mL}$ of a $0.456~M$ solution of A in water

This gives the total millimoles of your unknown acid that is split between the two layers.

$5.45~\mathrm{mL}$ of $0.523~M~\ce{NaOH}$ was required to neutralize the bottom layer

That gives the number of millimoles in the bottom layer.
So the number of millimoles in the ether layer must be the difference, and there are $25~\mathrm{mL}$ of ether.

I think I figured the concentration out: $0.523\cdot\frac{0.00545}{0.25} = 0.114~M$

Or do I need to determine the amount in the aqueous layer by subtracting this number? I.e. $0.456 - 0.114 = 0.34~M$

I am left with eithr the answer being $0.34~M$ or $0.114~M$. I think the answer is $0.34~M$, because the $0.114$ applies to the bottom layer which is water and needs to be neutralized by the NaOH solution, therefore the top layer (ether) has the remaining $.34~M$ of compound A.

• Welcome to Chemistry.SE! This appears to be a homework question. Please share your thoughts and attempts towards the solution, otherwise your question may be closed. – ringo Sep 26 '16 at 21:48
• It's great that your problem was solved. Please don't edit like that though, it confuses people who visit the website because they have no clue as to what was happening. – orthocresol Sep 27 '16 at 14:28