How to calculate partition coefficients? Below is a transcript of this doubtful question:


  A weak monobasic organic acid $\ce{HA}$ is soluble in both water
  and $\ce{CHCl3}$.  $\pu{500.0 cm3}$ of a solution of $\ce{HA}$ in $\ce{CHCl3}$, with a concentration of 
  $\pu{0.057 mol dm-3}$,
  is shaken well with $\pu{500.0 cm3}$ of water and allowed
  to attain equilibrium at $\pu{27°C}$.  An aqueous layer and
  $\ce{CHCl3}$ layer then separate out; the pH of the aqueous layer is
  found to be 3.21 under these conditions.
  
  The dissociation constant of $\ce{HA}$ in water at $\pu{27°C}$
  is $\pu{1E-5 mol dm-3}$.
  
  
  Calculate the partition coefficient at $\pu{27°C}$
  for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.
  In a second experiment, a further $\pu{500.0 cm3}$ portion of the
  same $\ce{HA}$ solution in $\ce{CHCl3}$ in which the concentration
  of $\ce{HA}$ is $\pu{0.057 mol dm-3}$, is shaken well with
  $\pu{500.0 cm3}$ of a $\pu{0.027 mol dm-3}$ aqueous $\ce{NaOH}$
  solution and allowed to reach equilibrium at $\pu{27°C}$.
  Calculate the $\mathrm{pH}$ of the aqueous layer under these conditions. 
  State the assumptions you make, if any, in the above calculations.
  




My solution for Question 1.


Since pH is given, $\ce{[H^+](aq)}$ is determined by


\begin{align}
\mathrm{pH} &amp;=-\log[\ce{H^+~(aq)}]\\
[\ce{H^+}~(\ce{aq})] &amp;= 10^{-3.21}\\
    &amp;= 6.16 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}\\
\end{align} 


By appropriately applying $K_\mathrm{a}$, the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer is $0.038~\mathrm{mol\,  dm^{-3}}$.
Next, $[\ce{HA}~(\ce{aq})]$ present in the organic layer can be determined by subtracting the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer
from the initial concentration of $[\ce{HA}~({\ce{aq}})]$, which results in $0.019~\mathrm{mol\,  dm^{-3}}$.
Finally, I got $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.019~\mathrm{mol\,  dm^{-3}}} = 2$.


I hope that this is correct, but  I am not quite sure.



My incomplete solution for Question 2.


The reaction between $\ce{HA}$ and $\ce{NaOH}$ is


$$\ce{HA}~(\ce{aq}) + \ce{NaOH}~(\ce{aq})  \longrightarrow \ce{NaA}~(\ce{aq}) + \ce{H2O}~(\ce{l})$$


The number of moles of $\ce{NaOH}$ present is $0.0135~\mathrm{mol}$.


I am confused with the concentration of $\ce{HA}$ that I needed to take for this calculation. Initially I have found the concentration of $\ce{HA}$ present in organic layer as $0.019~\mathrm{mol~dm^{-3}}$. Do I need to use that here or $0.057~\mathrm{mol~dm^{-3}}$?

What would be the assumptions that I need to assume to take out this calculation?
Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey. 

I think you're very close to the right solution for part 1, but I'd state it a bit differently.


Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by



  
    $[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$ 
    
    
      The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$
    
  



We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] &gt;&gt; [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.


$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$


Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$



  
    The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.
  




For Question 2.


The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,


$$\ce{HA + NaOH -&gt; Na^+ + A^- + H2O}$$


Assume none of $\ce{NaA}$ migrates into the organic phase. 
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated. 
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic. 
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] &gt;&gt; [OH^-] }$ and $\ce{[Na^+] &gt;&gt; [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.  
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$


\begin{align}
1\times10^{-5} &amp;= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &amp;= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &amp;= 5.11
\end{align}