How to calculate the concentration of a monoprotic organic acid that was extracted into an ethyl ether layer?
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})       &amp;= \pu{25.0 mL}\\
c_\mathrm{tot}(\ce{AH}) &amp;= \pu{0.456 mmol/mL}\\
V(\mathrm{ether})       &amp;= \pu{25.0 mL}\\
V(\ce{NaOH})            &amp;= \pu{5.45 mL}\\
c(\ce{NaOH})            &amp;= \pu{0.523 mmol/mL}\\
\color{red}{c_\mathrm{ether}(\ce{AH})} &amp;= ?
\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 &amp;&lt;=&gt;[H2O] A- + H+} &amp; 
&amp;&amp; n_\mathrm{tot}(\ce{AH}) &amp;= 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 &amp;-&gt; A- + Na+ +H2O} &amp; n_\mathrm{water}(\ce{AH}) &amp;= n(\ce{NaOH})\\
&amp;&amp; n_\mathrm{water}(\ce{AH}) &amp;= 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}) &amp;= 
  \frac{n_\mathrm{ether}(\ce{AH})}{V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &amp;= 
  \frac{n_\mathrm{tot}(\ce{AH}) - n_\mathrm{water}(\ce{AH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &amp;= 
  \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}) &amp;= 
  \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}) &amp;\approx \pu{0.342 mmol/mL}
\end{align}