Chemistry A-Level Equilibria

Equilibria: Distribution of a Solute Between Immiscible Systems

• When a solute S is shaken with two immiscible solvents, it will distribute itself between the solvents until equilibrium is reached.
• At equilibrium, the concentration of the solute in each solvent remains constant.
• The distribution depends on the relative solubility of the solute in the two solvents.

Distribution/Partition Coefficient (Kd)

• Kd is the ratio of the concentrations of the solute in the two solvents at equilibrium.
• Kd = \frac{[S]e}{[S]w}, where:
  • [S]_e is the concentration of S in ether (mass or molar).
  • [S]_w is the concentration of S in water (mass or molar).
• Alternative formula using mass and volume:
  • Kd = \frac{me \times Vw}{mw \times Ve}, where:
    • m_e is the mass of solute in ether.
    • m_w is the mass of solute in water.
    • V_e is the volume of ether.
    • V_w is the volume of water.
• Kd is an equilibrium constant dependent on temperature.
• The value of Kd depends on the solubility of the solute in the two solvents.
• Kd is dimensionless (no units).

Example Calculations:

Question 14

• Given: Kd = \frac{[X]e}{[X]w} = 10, Ve = 200 \text{ cm}^3, Vw = 400 \text{ cm}^3
• Calculation:
  • 10 = \frac{x}{6-x} \times \frac{400}{200}
  • 10 = \frac{2x}{6-x}
  • 60 - 10x = 2x
  • 60 = 12x
  • x = 5 \text{ g}

Question 15

• Two extractions are performed.
• 1st Extraction:
  • 10 = \frac{x}{6-x} \times \frac{400}{200}
  • 10 = \frac{2x}{6-x}
  • 60 - 10x = 2x
  • 60 = 14x
  • x = 4.29 \text{ g}
• 2nd Extraction:
  • 10 = \frac{x}{1.71-x} \times \frac{400}{200}
  • 10 = \frac{2x}{1.71-x}
  • 17.1 - 10x = 2x
  • 17.1 = 12x
  • x = 1.425 \text{ g}
• Total mass of X extracted = 4.29 + 1.425 = 5.715 \text{ g}

Multiple Extractions

• Multiple extractions are more efficient than a single extraction, even if the total volume of solvent used is the same.
• This principle is used in solvent extraction, a purification technique in which a substance is extracted from one solvent to another immiscible solvent, leaving impurities behind.

More Kd Questions

Question 16

• Given: Kd = \frac{[Y]e}{[Y]w} = 4, Ve = 10 \text{ cm}^3, Vw = 100 \text{ cm}^3
• Calculation:
  • 4 = \frac{x}{5-x} \times \frac{100}{10}
  • 4 = \frac{10x}{5-x}
  • 20 - 4x = 10x
  • 20 = 14x
  • x = 1.43 \text{ g}

Question 17

• Given: Kd = \frac{[S]e}{[S]w} = 12, Ve = 10 \text{ cm}^3, Vw = 100 \text{ cm}^3
• Calculation:
  • 12 = \frac{x}{25-x} \times \frac{100}{10}
  • 12 = \frac{10x}{25-x}
  • 300 - 12x = 10x
  • 300 = 22x
  • x = 13.64 \text{ g}

Question 18

• Given: Kd = \frac{[Z]e}{[Z]w} = 400, Ve = 100 \text{ cm}^3, Vw = 250 \text{ cm}^3
• No further calculation is shown.

Question 19

• Initial moles of A in 500 cm³ = 0.060 moles
• Given: Kd = \frac{[A]e}{[A]w} = 60, Ve = 50 \text{ cm}^3, Vw = 500 \text{ cm}^3
• Calculation:
  • 60 = \frac{x}{0.06-x} \times \frac{500}{50}
  • 60 = \frac{10x}{0.06-x}
  • 3.6 - 60x = 10x
  • 3.6 = 70x
  • x = 0.051 \text{ moles}

Question 20

• HX + NaOH titration:
  • 25 cm³ of HX requires 22.5 cm³ of NaOH.
  • Moles of NaOH in 22.5 cm³ = 0.0225 moles.
  • [HX]_w = \frac{0.0225 \text{ moles}}{25 \text{ cm}^3} \times 1000 = 0.9 \text{ mol/dm}^3
  • [HX]_e = \frac{9 \times 10^{-4} \text{ moles}}{25 \text{ cm}^3} \times 1000 = 0.036 \text{ mol/dm}^3
• Kd = \frac{[HX]e}{[HX]w} = \frac{0.9}{0.036} = 25

Question 21

• I2 + 2S2O3^{2-} \rightarrow 2I^- + S4O_6^{2-}
  • 25 cm³ of I2 requires 27.2 cm³ of S2O_3^{2-}.
  • Moles of S2O3^{2-} in 27.2 cm³ = 0.0550 mol/dm³.
• Iodine distribution between CCl_4 and water:
  • [I2](\text{CCl}_4) = \frac{0.01564 \text{ moles}}{50 \text{ cm}^3} \times 1000 = 0.3128 \text{ mol/dm}^3
  • [I2](\text{H}_2\text{O}) = \frac{1.8425 \times 10^{-4} \text{ moles}}{500 \text{ cm}^3} \times 1000 = 3.685 \times 10^{-4} \text{ mol/dm}^3
• Kd = \frac{[I2](\text{CCl}4)}{[I2](\text{H}2\text{O})} = 0.012
• Iodine is much more soluble in tetrachloromethane than in water, since the ratio is 0.012:1