# How do we calculate the partition coefficient (including titration)

In a separating funnel we add $\pu{25.0 mL}$ of $\pu{8.5 M}$ $\ce{HCl}$ solution. We also add $\pu{25.0 mL}$ of a tributyl phosphate solution. We shake the funnel for 1 minute and then after the two phases have separated we remove the lower phase.

We add in the same separating funnel $\pu{23.0 mL}$ of the same $\pu{8.5 M}$ $\ce{HCl}$ solution and we also add $\pu{2.00 mL}$ of $\pu{0.5 M}$ $\ce{ZnCl2}$ solution. We shake the funnel again and we remove the lower phase. Now the biggest concentration of zinc is in the organic phase.

We add in the separating funnel $\pu{20.0 mL}$ of deionized water. We shake the funnel and now the zinc comes back again the aqueuous phase. After the two phases have separated, we transfer the aqueous phase to a volumetric flask. We repeat the previous step 2 more times and we collect the aqueuous phase in the same volumetric flask. In a conical flask, we add $\pu{25.00 mL}$ of the volumetric flask, we add $\pu{75 mL}$ of deionized water and $\pu{3 mL}$ of a buffer and also drops of an indicator. We titrate this with EDTA solution.

Now we want to calculate the partition coefficient using these formulas; $D_\ce{Zn}$ is the distribution coefficient. $C_\ce{{Zn}_{org}}$ and $C_\ce{{Zn}_{aq}}$ are the concentrations of zinc in the organic and aqueous phases respectively.

$D_\ce{Zn}=\dfrac{C_\ce{{Zn}_{org}}}{C_\ce{{Zn}_{aq}}}$

If $n$ denotes amount, then

$n_\ce{{Zn}_{org}}=\pu{0.0100 mmol//mL}\times V_\mathrm{EDTA}\times 4$

$n_\ce{{Zn}_{total}}=\pu{0.0100 mmol//mL}\times V_\mathrm{{EDTA}_{total}}\times 4$

$n_\ce{{Zn}_{aq}}=n_\ce{{Zn}_{total}}-n_\ce{{Zn}_{org}}$

I get confused about what the difference between the $V_\mathrm{EDTA}$ and the $V_\mathrm{{EDTA}_{total}}$ is.

• Some formatting would certainly help. – Eashaan Godbole Dec 11 '17 at 14:51
• What a hopelessly convoluted way to determine a $D$! Good luck with that... – Gert Dec 11 '17 at 21:49