Question

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of the compound were dissolved in 100mL of water, what weight of compound could be extracted by THREE sequential 10-mL portions of benzene?

Answer 1

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 $K_{p}$ = 2.7,         $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  $f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

                                  = $[1 + 2.7(\frac{100}{10})]^{-3}$

                                  = $(28)^{-3}$

                                  = $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - $4.55 \times 10^{-5}$

                                  = 0.00001

or,                               = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$.

Answer 2

Answer:

It could be extracted 0.512 g of solute

Explanation:

The equation that relates the $K_{D}$ and the volumes of organic and aqueous phases is:

$q_{solute-aq} =\frac{V_{aq} }{K_{D}xV_{org} + V_{aq}  }$

Where q_{solute-aq} refers to the fraction of solute remaining in the aqueous phase, V_{aq} is the aqueos phase volume, V_{org} is the organic phase volume and K_{D} is the partition coefficient of the solute in the solvents.

Moreover,for the three consecutive extractions of the same volume of organic phase we can write:

$q_{solute-aq} =(\frac{V_{aq} }{K_{D}xV_{org} + V_{aq}  })^{ 3}$

So, plugging the values given into the equation we get:

$q_{solute-aq} =(\frac{100 mL }{2.7x10 mL + 100 mL  })^{ 3}$

$q_{solute-aq} =0.488$

The result obtained indicates that a fraction of 0.488 of solute remains in the aqueous phase.

Taking in account that the fraction formula is:

$q_{solute-aq} = \frac{mass- of- solute- aq}{initial-mass- of -solute}$

$0.488= \frac{mass- of- solute- aq}{1.0 g}\\\\0.488 x 1.0 g= {mass- of- solute- aq}\\0.488 g= {mass- of- solute- aq}$

Finally we substract the solute in the aqueous phase form the initial to get the amount in the organic phase:

$1.0g - 0.488g = 0.512 g$