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.
• $Kd = \frac{[S]<em>e}{[S]</em>w}$, 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:
  • $Kd = \frac{m<em>e \times V</em>w}{m<em>w \times V</em>e}$, 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: $Kd = \frac{[X]<em>e}{[X]</em>w} = 10$, $V<em>e = 200 \text{ cm}^3$, $V</em>w = 400 \text{ cm}^3$
• 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: $Kd = \frac{[Y]<em>e}{[Y]</em>w} = 4$, $V<em>e = 10 \text{ cm}^3$, $V</em>w = 100 \text{ cm}^3$
• 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: $Kd = \frac{[S]<em>e}{[S]</em>w} = 12$, $V<em>e = 10 \text{ cm}^3$, $V</em>w = 100 \text{ cm}^3$
• 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: $Kd = \frac{[Z]<em>e}{[Z]</em>w} = 400$, $V<em>e = 100 \text{ cm}^3$, $V</em>w = 250 \text{ cm}^3$
• No further calculation is shown.

Question 19

• Initial moles of A in 500 cm³ = 0.060 moles
• Given: $Kd = \frac{[A]<em>e}{[A]</em>w} = 60$, $V<em>e = 50 \text{ cm}^3$, $V</em>w = 500 \text{ cm}^3$
• 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.
  • $[HX]_w = \frac{0.0225 \text{ moles}}{25 \text{ cm}^3} \times 1000 = 0.9 \text{ mol/dm}^3$
  • $[HX]_e = \frac{9 \times 10^{-4} \text{ moles}}{25 \text{ cm}^3} \times 1000 = 0.036 \text{ mol/dm}^3$
• $Kd = \frac{[HX]<em>e}{[HX]</em>w} = \frac{0.9}{0.036} = 25$

Question 21

• $I<em>2 + 2S</em>2O<em>3^{2-} \rightarrow 2I^- + S</em>4O_6^{2-}$
  • 25 cm³ of $I<em>2$ requires 27.2 cm³ of $S</em>2O_3^{2-}$.
  • Moles of $S<em>2O</em>3^{2-}$ in 27.2 cm³ = 0.0550 mol/dm³.
• Iodine distribution between $CCl_4$ and water:
  • $[I<em>2]</em>(\text{CCl}_4) = \frac{0.01564 \text{ moles}}{50 \text{ cm}^3} \times 1000 = 0.3128 \text{ mol/dm}^3$
  • $[I<em>2]</em>(\text{H}_2\text{O}) = \frac{1.8425 \times 10^{-4} \text{ moles}}{500 \text{ cm}^3} \times 1000 = 3.685 \times 10^{-4} \text{ mol/dm}^3$
• $Kd = \frac{[I<em>2]</em>(\text{CCl}<em>4)}{[I</em>2]<em>(\text{H}</em>2\text{O})} = 0.012$
• Iodine is much more soluble in tetrachloromethane than in water, since the ratio is 0.012:1