If a solute such as iodine is added to a mixture of two immiscible liquids $C C l_{4}$ and $H_{2} O$ and the vessel is shaken to distribute the iodine through the container, the iodine is partitioned between the two phases at equilibrium with a characteristic concentration ratio, $\frac{{[I_{2}]}_{C C l_{4}}}{{[I_{2}]}_{H_{2} O}}$ , the partition coefficient K. The partition coefficient for the distribution of iodine between the two liquids is 85 at 298 K. An aqueous solution has an iodine concentration of $2.00 \times 10^{- 3} M$ . Calculate the percentage of iodine remaining in the aqueous phase after extraction of 0.100 L of this aqueous solution with 0.050 L of $C C l_{4}$ at $25^{\circ} C$ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$48.85 %$

$2.3 %$

$97.7 %$

$23 %$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The number of moles of $I_{2}$ present is
$(2.00 \times 10^{- 3} m o l L^{- 1}) (0.100 L) = 2.00 \times 10^{- 4} m o l$
Suppose that y mol remains in the aqueous phase and $(2.00 \times 10^{- 4} - y)$ mol passes into the $C C l_{4}$ phase. Then.
$\frac{{[I_{2}]}_{C C l_{_{4}}}}{{[I_{2}]}_{a q}} = K = 85$
$= \frac{(2.00 \times 10^{- 4} - y) / 0.050}{y / 0.100}$ $y = 4.6 \times 10^{- 6} m o l$
$= \frac{2 (2.00 \times 10^{- 4} - y)}{y}$
Note that the volumes of the two solvents used are unchanged because they are immiscible. Solving for y gives
The fraction remaining in the aqueous phase is $(4.6 \times 10^{- 6}) / (2.0 \times 10^{- 4}) = 0.023$ , or $2.3 %$ .