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Equilibria: Question 10

Syllabus 25.2

Structured A2 7 marks

Iodine, I2\text{I}_2, distributes itself between water and hexane, two immiscible solvents, reaching a partition equilibrium at 298 K298\ \text{K}:

I2(aq)⇌I2(hexane)\text{I}_2\text{(aq)} \rightleftharpoons \text{I}_2\text{(hexane)}

(a) Write the expression for the partition coefficient, KpcK_{pc}, of iodine between hexane and water. [1]

(b) In one experiment, the equilibrium concentration of iodine in the aqueous layer is 2.50×10−4 mol dm−32.50\times10^{-4}\ \text{mol dm}^{-3}, and in the hexane layer is 2.05×10−2 mol dm−32.05\times10^{-2}\ \text{mol dm}^{-3}. Calculate KpcK_{pc}. [2]

(c) In a second experiment at the same temperature, 25.0 cm325.0\ \text{cm}^3 of aqueous iodine solution of concentration 1.00×10−3 mol dm−31.00\times10^{-3}\ \text{mol dm}^{-3} is shaken with 10.0 cm310.0\ \text{cm}^3 of hexane until partition equilibrium is re-established. Given that KpcK_{pc} is unchanged, use conservation of the total amount of iodine to calculate the new equilibrium concentration of iodine in the aqueous layer. [4]

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Worked solution

Part (a): Writing the Kpc expression

The partition coefficient of iodine between hexane and water is defined as the ratio of its equilibrium concentration in hexane to its equilibrium concentration in water:

Kpc=[I2(hexane)][I2(aq)]K_{pc}=\frac{[\text{I}_2\text{(hexane)}]}{[\text{I}_2\text{(aq)}]}

Part (b): Calculating Kpc

Substituting the given equilibrium concentrations: Kpc=2.05×10−22.50×10−4=82.0K_{pc}=\frac{2.05\times10^{-2}}{2.50\times10^{-4}}=82.0

(Check: recomputing independently, 2.05÷2.50=0.8202.05\div2.50=0.820, and 0.820×10(−2)−(−4)=0.820×102=82.00.820\times10^{(-2)-(-4)}=0.820\times10^{2}=82.0. Consistent.)

Part (c): New equilibrium concentration after adding hexane

Total amount of iodine. Before shaking with hexane, all the iodine is in the aqueous layer: n(I2)total=1.00×10−3×25.01000=2.50×10−5 moln(\text{I}_2)_{total}=1.00\times10^{-3}\times\frac{25.0}{1000}=2.50\times10^{-5}\ \text{mol}

Setting up the balance. Let the new equilibrium concentration in the aqueous layer be caqc_{aq}. Since Kpc=82.0K_{pc}=82.0 is unchanged, the equilibrium concentration in hexane is chex=82.0 caqc_{hex}=82.0\,c_{aq}. The total amount of iodine is conserved between the two layers, of volumes Vaq=25.0 cm3=0.0250 dm3V_{aq}=25.0\ \text{cm}^3=0.0250\ \text{dm}^3 and Vhex=10.0 cm3=0.0100 dm3V_{hex}=10.0\ \text{cm}^3=0.0100\ \text{dm}^3:

n(I2)total=caqVaq+chexVhex=caq(0.0250)+82.0 caq(0.0100)n(\text{I}_2)_{total}=c_{aq}V_{aq}+c_{hex}V_{hex}=c_{aq}(0.0250)+82.0\,c_{aq}(0.0100)

2.50×10−5=caq(0.0250+0.820)=caq(0.845)2.50\times10^{-5}=c_{aq}\left(0.0250+0.820\right)=c_{aq}(0.845)

Solving for caqc_{aq}: caq=2.50×10−50.845=2.959×10−5 mol dm−3 ⇒ [I2(aq)]=2.96×10−5 mol dm−3 (3 s.f.)c_{aq}=\frac{2.50\times10^{-5}}{0.845}=2.959\times10^{-5}\ \text{mol dm}^{-3}\ \Rightarrow\ [\text{I}_2\text{(aq)}]=2.96\times10^{-5}\ \text{mol dm}^{-3}\ \text{(3 s.f.)}

(Check: chex=82.0×2.959×10−5=2.426×10−3 mol dm−3c_{hex}=82.0\times2.959\times10^{-5}=2.426\times10^{-3}\ \text{mol dm}^{-3}. Moles: naq=2.959×10−5×0.0250=7.40×10−7 moln_{aq}=2.959\times10^{-5}\times0.0250=7.40\times10^{-7}\ \text{mol}; nhex=2.426×10−3×0.0100=2.426×10−5 moln_{hex}=2.426\times10^{-3}\times0.0100=2.426\times10^{-5}\ \text{mol}. Sum =7.40×10−7+2.426×10−5=2.500×10−5 mol=7.40\times10^{-7}+2.426\times10^{-5}=2.500\times10^{-5}\ \text{mol}, matching the total from above, consistent. As expected, caqc_{aq} is far smaller than the original 1.00×10−3 mol dm−31.00\times10^{-3}\ \text{mol dm}^{-3}, since Kpc=82.0K_{pc}=82.0 strongly favours the hexane layer.)

Final answers

  • (a) Kpc=[I2(hexane)][I2(aq)]K_{pc}=\dfrac{[\text{I}_2\text{(hexane)}]}{[\text{I}_2\text{(aq)}]}
  • (b) Kpc=82.0K_{pc}=82.0
  • (c) [I2(aq)]=2.96×10−5 mol dm−3[\text{I}_2\text{(aq)}]=2.96\times10^{-5}\ \text{mol dm}^{-3}