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

Question

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

My attempt:

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

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

$\pu{5.45 mL}$ of $\pu{0.523 M}$ $\ce{NaOH}$ 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 $\pu{25 mL}$ of ether.

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

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

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

Answer 1

You arrive at the correct answer, but the way you get there is a bit messy. (And it only works because both phases have the same volume.) You should be a lot clearer in defining what properties you have been given in the question and which ones you need to calculate. Setting up the equation is a lot more important then calculating it in the end, because this does not only show that you know what you are doing, but it also is a way to condense a difficult sentence into a calculator understandable way. \begin{align} V(\mathrm{water}) &= \pu{25.0 mL}\\ c_\mathrm{tot}(\ce{AH}) &= \pu{0.456 mmol/mL}\\ V(\mathrm{ether}) &= \pu{25.0 mL}\\ V(\ce{NaOH}) &= \pu{5.45 mL}\\ c(\ce{NaOH}) &= \pu{0.523 mmol/mL}\\ \color{red}{c_\mathrm{ether}(\ce{AH})} &= ? \end{align}

Then you must always carry the units with the quantity; they belong together. This way you will always have a way to tell if you have made a mistake setting up the equation. If the units don't work out, it cannot be correct.

Since it is a monoprotic acid the following will hold true: \begin{align} \ce{AH &<=>[H2O] A- + H+} & && n_\mathrm{tot}(\ce{AH}) &= c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water}) \end{align}

Now you extracted your solution and split the amount of substance between the water and the ether phase. You can calculate the remaining substance in the water phase from the neutralisation experiment. \begin{align} \ce{AH + NaOH &-> A- + Na+ +H2O} & n_\mathrm{water}(\ce{AH}) &= n(\ce{NaOH})\\ && n_\mathrm{water}(\ce{AH}) &= c(\ce{NaOH}) \cdot V(\ce{NaOH}) \end{align}

The extracted amount of substance of the acid is simply the difference, from that you can easily calculate the concentration in the ether phase. \begin{align} c_\mathrm{ether}(\ce{AH}) &= \frac{n_\mathrm{ether}(\ce{AH})}{V(\mathrm{ether})}\\ c_\mathrm{ether}(\ce{AH}) &= \frac{n_\mathrm{tot}(\ce{AH}) - n_\mathrm{water}(\ce{AH})} {V(\mathrm{ether})}\\ c_\mathrm{ether}(\ce{AH}) &= \frac{c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water}) - c(\ce{NaOH}) \cdot V(\ce{NaOH})} {V(\mathrm{ether})}\\ c_\mathrm{ether}(\ce{AH}) &= \frac{\pu{0.456 mmol/mL} \cdot \pu{25.0 mL} - \pu{0.523 mmol/mL} \cdot \pu{5.45 mL}} {\pu{25.0 mL}}\\ c_\mathrm{ether}(\ce{AH}) &\approx \pu{0.342 mmol/mL} \end{align}