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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 $\pu{1E-5 mol dm-3}$.

  1. Calculate the partition coefficient at $\pu{27°C}$ for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.
  2. 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.
  3. 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} &=-\log[\ce{H^+~(aq)}]\\ [\ce{H^+}~(\ce{aq})] &= 10^{-3.21}\\ &= 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?

  • $\begingroup$ Next, [HA (aq)] present in the organic layer can be determined by subtracting the [HA (aq)] present in aqueous layer from the initial concentration of [HA (aq)], which results in // Typo the last bit should be from the initial concentration of [HA (CHCl3 )], which results in // Also $K_D = 2.0$ when considering significant figures $\endgroup$
    – MaxW
    Oct 17, 2017 at 19:37
  • $\begingroup$ For part 2 you know that there are 0.019 moles total of (a) HA in CHCl3 (b) HA in H20 (c) A^- in H2O // You also know how HA will partition between CHCL3 and water. // You also know how H+, A- and HA relate in water via Ka // You also know that $\ce{Na+ + H+ = A^- + OH^-}$ so you have enough equations to solve for all the unknowns. $\endgroup$
    – MaxW
    Oct 17, 2017 at 19:48

2 Answers 2


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 $\ce{[A^-] >> [OH^-]}$ 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,

$$\ce{HA + NaOH -> Na^+ + A^- + H2O}$$

  • 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 $\ce{[A^-] >> [OH^-] }$ and $\ce{[Na^+] >> [H^+]}$ 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} &= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\ \ce{[H^+]} &= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\ \mathrm{pH} &= 5.11 \end{align}

  • $\begingroup$ Carboxylic acids in organic solvents normally form dimers, this will make the calculations much harder. I think that it may be impossible to get a good answer to this question with what is provided in the question. $\endgroup$ May 8, 2018 at 5:01

I think the question is a bit clever and may needs deeper analysis, a monobasic organic acid may have carboxylic acid group and amine group, both can exist in aqueous form by dissociation or by hydrogen bond. At the aforementioned equilibrium and the dilute initial concentration, there will be ionic forms in water and molecular forms in aqueous form. The sum HA and ions in that equilibrium is in aqueous form, and could be some molecules in organic phase if their sum is less than the initial moles. So here seems no significant quantity is in $\ce{CHCl3}$ phase. On my point of view.

  • $\begingroup$ You can't just add an amine group for this type of problem. $\endgroup$
    – MaxW
    Oct 17, 2017 at 19:41

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