GATE 2020 BT Q.20: solution and answer

GATE 2020 · BTQ.20 · 1 mark · NAT

Solvents A and B are completely immiscible. Solute S is soluble in both these solvents. 100 g of S was added to a container which has 2 kg each of A and B. The solute is 1.5 times more soluble in solvent A than in solvent B. The mixture was agitated thoroughly and allowed to reach equilibrium. Assuming that the solute has completely dissolved, the amount of solute in solvent A phase is ________ g.

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Answer and solution

ANSWER

At equilibrium, solvent phase A holds 60 g of the solute (and phase B holds 40 g).

Official answer: 60Source: Official answer key, GATE 2020 BT (IIT Delhi)

Define the partition (distribution) coefficient $K = \frac{C _{A}}{C _{B}}$ , where $C_{A}$ and $C_{B}$ are the equilibrium solute concentrations in solvents A and B, in $g solute per kg solvent$ . "1.5 times more soluble in A than in B" gives $K = 1.5$ .
Let $x$ g of S end up in phase A. All 100 g dissolves and the solvents are completely immiscible, so phase B holds $(100 - x)$ g.
Concentrations: $C_{A} = \frac{x}{2} g kg^{- 1}$ and $C_{B} = \frac{100 - x}{2} g kg^{- 1}$ . Both phases contain 2 kg of solvent, so the solvent masses cancel.
Equilibrium condition: $\frac{x /2}{( 100 - x ) /2} = 1.5 \Rightarrow x = 1.5 (100 - x)$
Solve: $x = 150 - 1.5 x \Rightarrow 2.5 x = 150 \Rightarrow x = 60 g$ .
Check: phase B holds $100 - 60 = 40 g$ . Then $C_{A} = 30 g kg^{- 1}$ and $C_{B} = 20 g kg^{- 1}$ , and $C_{A} / C_{B} = 30/20 = 1.5$ , as required.

KEY IDEA

In single-stage liquid-liquid extraction between two immiscible solvents, the solute divides so that the ratio of its concentrations follows the partition coefficient: $K = C_{A} / C_{B}$ . Combine that with a mass balance on the solute. When the two solvent masses are equal, the solute masses themselves come out in the ratio $K : 1$ , here $1.5 : 1$ , so A gets $\frac{1.5}{2.5} \times 100 = 60 g$ .

WATCH OUT FOR

Reading "1.5 times more soluble" as $K = 2.5$ (that is, $1 + 1.5$ ). That gives $100 \times 2.5/3.5 \approx 71.4 g$ , which is wrong: GATE means $C_{A} / C_{B} = 1.5$ . A second mistake is to give the whole 100 g to the better solvent, or to split it 50:50. A third is to apply $K$ to the solute masses when the solvent masses are unequal. That shortcut works here only because both phases hold 2 kg of solvent.

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GATE 2020 BT Q.20: solution and answer

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