The partition coefficient (Kpc) is the equilibrium constant that relates the concentration of a solute partitioned between two immiscible solvents at a particular temperature.
Let's unpack it in simple terms. Most substances have different solubilities in different solvents. For example, NaCl has a high solubility in water and a very low solubility in oil. Kpc is the ratio of the solubility of a substance in one solvent to another.
In this case, the Kpc of NaCl would be written as:
\( \text{Kpc} = \frac{[\text{concentration of NaCl in water}]}{[\text{concentration of NaCl in oil}]} \)
Solubility depends on the polarity of the solute and the solvent, following the concept "like dissolves like." Polar molecules have a high solubility in polar solvents like water and a low solubility in non-polar solvents like octane.
Let's try to understand this better by solving a question.
[Question]
100 cm3 of a 0.100 mol dm−3 solution of ammonia in water was shaken with 50 cm3 of an organic solvent and left in a separating funnel for equilibrium to be established. A 20.0 cm3 portion of the aqueous layer was run off and titrated against 0.200 mol dm−3 dilute hydrochloric acid. The end-point was found to be 9.40 cm3 of acid.
What is the partition coefficient of ammonia between these two solvents?
Step 1: Write out the balanced chemical equation
The alkaline ammonia solution is neutralized by dilute hydrochloric acid:
\( \text{NH}_3\text{(aq)} + \text{HCl(aq)} \rightarrow \text{NH}_4\text{Cl(aq)} \)
z=2 as 2 electrons are transferred
Step 2: Calculate moles of HCl used in the titration.
Volume of HCl used = 9.40 cm3
Concentration of HCl = 0.200 mol dm−3
Moles of HCl = \( \frac{\text{Volume}}{1000} \times \text{concentration} \) = \( \frac{9.40}{1000} \times 0.200 = 1.88 \times 10^{-3} \) moles
Step 3: Determine moles of NH3 in the 20.0 cm3 portion.
Since 1 mole of NH3 reacts with 1 mole of HCl, the moles of NH3 in the 20.0 cm3 portion is equal to the moles of HCl used.
Moles of NH3 in 20.0 cm3 portion = 1.88 × 10−3 moles
Step 4: Calculate the moles of NH3 in the 100 cm3 aqueous layer.
Since the titration used a 20.0 cm3 portion of the 100 cm3 aqueous layer, scale up the moles of NH3 to find the total in the aqueous layer.
Moles of NH3 in 100 cm3 aqueous layer = \( \frac{100}{20} \times 1.88 \times 10^{-3} = 9.40 \times 10^{-3} \) mol
Step 5: Calculate the final moles of NH3 in the organic layer
Initial concentration of NH3 (before organic solvent added) = 0.100 mol dm−3
Which translates to \( \frac{100}{1000} \times 0.100 \text{ mol dm}^{-3} = 0.01 \text{ moles} \)
Moles of NH3 in organic layer = Initial moles - Moles in aqueous layer = 0.0100 – 9.40 × 10−3 = 6.00 × 10−4 mol
Step 6: Convert moles to concentrations
Concentration of NH3 in the aqueous layer:
Volume of aqueous layer = 100 cm3
Concentration = \(\frac{\text{Moles} \times 1000}{\text{Volume}} = \frac{9.40 \times 10^{-3} \times 1000}{100} = 0.094 \text{ mol dm}^{-3} \)
Concentration of NH3 in the organic solvent:
Volume of organic solvent = 50 cm3
Step 7: Calculate the partition coefficient (Kpc)
\( K_{pc} = \frac{[NH_3(\text{organic solvent})]}{[NH_3(\text{aq})]} = \frac{0.012}{0.094} = 0.13 \)