The question is in the 3rd picture. Thank you solute solute where K, is the dist
ID: 229421 • Letter: T
Question
The question is in the 3rd picture. Thank you
solute solute where K, is the distribution coefficient, [solutel is the concentration of solute (in moles/L) in solvent A (usually the organic solvent) and [solutel is the concentration of solute (in moles/L) in solvent B (usually water) For example, 6.67 g of phenol will dissolve in 100 mL of water, while 8.33 g will dissolve in 100 mL of benzene. From this, the distribution coe ficient is calculated to be (8.33 g/100 mL) (6.67 g/100 mL) ene = Note that in this example, g/100 mL has replaced the units of moles/L that appeared in the original definition. This is because units cancel out in such a ratio Now assume that we have a solution composed of 1 g of phenol in 50 mL of water, and we extract that solution with 50 mL of benzene. The phe- nol will distribute itself between the water and the benzene in the follow ing way (x g/50milienzene, = 1.25 (1-x)/50 mL| where mass (in g) of phenol dissolved in the benzene layerExplanation / Answer
We have deduced the distribution coefficient of phenol in water/benzene system as Kp = 1.25.
Let x g phenol dissolve in benzene; hence, we can write
Kp = (x g/50 mL)benzene/[(1 – x)g/50 mL)]water
=====> 1.25 = x/(1 – x)
=====> 1.25 – 1.25x = x
=====> 1.25 = 1.25x + x = 2.25x
=====> x = 1.25/2.25 = 0.555 0.56
Therefore, 0.56 g phenol will dissolve in benzene and the amount of phenol dissolved in water is (1 – 0.56) g = 0.44 g.
Next, let y be the amount of phenol dissolved in 25 mL benzene; therefore, we have,
1.25 = (y g/25 mL)benzene/[(1 – y)g/50 mL]water
=====> 1.25 = (y/25)/[(1 – y)/50]
=====> 1.25 = 2y/(1 – y)
=====> 1.25 – 1.25y = 2y
=====> 1.25 = 1.25y + 2y = 3.25y
=====> y = 1.25/3.25 = 0.3846 0.38
Therefore, 0.38 g phenol will be extracted in benzene with the first 25 mL portion of benzene; the amount of phenol remaining in water is (1 – 0.38) g = 0.62 g. The above mass of phenol is again extracted with 25 mL portion of benzene. Let z g of phenol be extracted in the benzene layer. Therefore,
1.25 = (z g/25 mL)benzene/[(0.62 – z)g/50 mL]water
=====> 1.25 = 2z/(0.62 – z)
=====> 1.25*(0.62 – z) = 2z
=====> 0.775 – 1.25z = 2z
=====> 0.775 = 1.25z + 2z
=====> 3.25z = 0.775
=====> z = 0.775/3.25 = 0.2384 0.24
Therefore, 0.24 g phenol will dissolve in the benzene layer on the second extraction with the mass of phenol retained in water is (0.62 – 0.24) g = 0.38 g.
The total mass of phenol extracted in the benzene layer with two successive 25 mL portions of benzene is (0.38 + 0.24) g = 0.62 g (ans); this is clearly higher than the mass of phenol extracted with one 50 mL portion of benzene (as determined at the beginning and is equal to 0.56 g).