Partition coefficients — practice questions

1. Find the mass of X transferred to the ether layer. Initial mass of X in water = 6.0 g Equilibrium mass of X in water = 1.2 g Mass of X in ether = 6.0 g - 1.2 g = 4.8 g
2. Calculate the concentration in each solvent. Concentration is mass/volume. We can use g cm⁻³ as the units will cancel. $[X_{aq}] = \frac{1.2 \text{ g}}{100 \text{ cm}^3} = 0.012 \text{ g cm}^{-3}$ $[X_{ether}] = \frac{4.8 \text{ g}}{20 \text{ cm}^3} = 0.24 \text{ g cm}^{-3}$
3. Calculate $K_{pc}$. The question asks for $K_{pc}$ between ether and water, so ether is the numerator. $K_{pc} = \frac{[X_{ether}]}{[X_{aq}]} = \frac{0.24}{0.012} = 20$

Answer: The partition coefficient, $K_{pc}$, is 20.

1. Define variables. Let 'm' be the mass of iodine extracted into the $CHCl_3$ layer. The mass of iodine remaining in the aqueous layer will be (1.00 - m) g.
2. Set up the $K_{pc}$ expression with concentrations. $[I_2(CHCl_3)] = \frac{m}{20 \text{ cm}^3}$ $[I_2(aq)] = \frac{(1.00 - m)}{100 \text{ cm}^3}$
3. Substitute into the $K_{pc}$ formula and solve for m. $K_{pc} = 130 = \frac{m/20}{(1.00 - m)/100}$ $130 = \frac{m}{20} \times \frac{100}{1.00 - m}$ $130 = \frac{5m}{1.00 - m}$ $130(1.00 - m) = 5m$ $130 - 130m = 5m$ $130 = 135m$ $m = \frac{130}{135} = 0.963$ g (This is the mass extracted)
4. Calculate the mass remaining in the aqueous layer. Mass remaining = Initial mass - mass extracted Mass remaining = 1.00 g - 0.963 g = 0.037 g

Answer: The mass of iodine remaining in the aqueous layer is 0.037 g (to 2 s.f. based on volume).