60. A student decided to perform liquid-liquid extraction of solute A and solute B separately. Based
on the table below, indicate which of the following options is CORRECT about their partition coefficient
(K)?
Solute A Solute B
Concentration of solute in extract (g/L) 50 1
Concentration of solute in raffinate (g/L) 1 50
1.𝐾𝐴 = 𝐾𝐵
2. 𝐾𝐴 > 𝐾𝐵
3. 𝐾𝐴 < 𝐾𝐵
4. 𝐾 cannot be determined with the given information

Answer: 3. 𝐾𝐴 < 𝐾𝐵

Explanation:

The partition coefficient $K$ is defined as the ratio of the concentration of a solute in the extract phase to the concentration of the solute in the raffinate phase:

K=Concentration of solute in extractConcentration of solute in raffinateK = \frac{\text{Concentration of solute in extract}}{\text{Concentration of solute in raffinate}}K=Concentration of solute in raffinateConcentration of solute in extract​

For each solute:

• For Solute A:

  KA=50 g/L1 g/L=50K_A = \frac{50 \, \text{g/L}}{1 \, \text{g/L}} = 50KA​=1g/L50g/L​=50

• For Solute B:

  KB=1 g/L50 g/L=0.02K_B = \frac{1 \, \text{g/L}}{50 \, \text{g/L}} = 0.02KB​=50g/L1g/L​=0.02

Conclusion:

• The partition coefficient for Solute A (KA=50K_A = 50KA​=50) is much larger than that for Solute B (KB=0.02K_B = 0.02KB​=0.02).

• Therefore, KA>KBK_A > K_BKA​>KB​, which means Option 2 is incorrect, and Option 3 is correct.