MACROSCALE+MICRO.ORG...-OWLV2 ACCESS
MACROSCALE+MICRO.ORG...-OWLV2 ACCESS
7th Edition
ISBN: 9781305884175
Author: Williamson
Publisher: CENGAGE L
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Chapter 7, Problem 1Q

(a)

Interpretation Introduction

Interpretation:Weight of malononitrile recovered upon extraction with three 100 mL portions of ether should be calculated.

Concept introduction:The distribution coefficient is a parameter that indicates the extent of solubility of two immiscible components in two different phases. One of the phases is aqueous while the other is the organic phase.

Mathematically,

  KD=[solute]org[solute]aq

(a)

Expert Solution
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Answer to Problem 1Q

Weight of malononitrile recovered upon extraction with three 100 mL portions of ether is 29.2365 g .

Explanation of Solution

Since ether is an organic solvent, the amount of solute extracted from hexane indicates [solute]org .

Since the total amount of solute A in 100 mL of water is 10 g , the amount in aqueous layer is (10x) g .

Total volume of solution is 100 mL.

The expression fordistribution coefficient is given as:

  KD=[solute]org[solute]aq

Where,

  • [solute]orgindicates solubility of solute in organic phase.
  • [solute]aq indicates solubility of solute in aqueous phase.
  • KD denotes equilibrium constant for the partition equilibrium.

The solubility of malononitrile in ether is 20 g/100 mL .

The solubility of malononitrile in water is 13.3 g/100 mL .

Substitute these values in above expression of KD .

  KD= [ solute] org [ solute] aq= 20 g 100 mL 13.3 g 100 mL=1.5037

Solubility ofsolute A in organic phase is given as follows:

  [solute]org=x300

Solubility ofsolute A in aqueous phase is given as follows:

  [solute]aq=(30x)300

Substitute the values in the above expression of distribution coefficient as follows:

  KD= [ solute] org [ solute] aq1.5037=x 300 ( 30x ) 300

Simplify the above expression as follows:

  1.5037=x30x45.1111.5037x=x

Rearrange and obtain the value of x .

  45.111= 2.5037xx=45.1112.5037=18.0177 g

For second extraction repeat the same procedure with volume being 100 mL .

The first cycle of extraction eliminates 18.0177 g out of organic layer and so 11.98 g is left in the aqueous layer.

Substitute the values in the above expression of distribution coefficient as follows:

  KD= [ solute] org [ solute] aq1.5037=x 100 ( 11.98x ) 100

Simplify the above expression as follows:

  1.5037=(x11.98x)

Rearrange and obtain the value of x .

  x=18.0141.5037x2.5037x=18.014x=18.0142.5037=7.195 g

The second cycle of extraction eliminates 7.195 g and so 4.785 g is left in the aqueous layer.

For the next cycle of extraction the amount of solute gets changed from (11.98x) g to (4.785x) g . Thus the expression modifies as follows:

  KD= [ solute] org [ solute] aq1.5037=x 300 ( 4.785x ) 300

Simplify the above expression as follows:

  KD= [ solute] org [ solute] aq1.5037=x 300 ( 4.785x ) 3001.5037=(x 4.785x)

Rearrange and obtain the value of x .

  x=7.19521.5037x2.5037x=7.1952x=7.19522.5037=2.8738 g

The second cycle of extraction eliminates 2.8738 g and so 1.9112 g is left in the aqueous layer.

For the next cycle of extraction the amount of solute gets changed from (11.98x) g to (1.9112x) g . Thus the expression modifies as follows:

  KD= [ solute] org [ solute] aq1.5037=x 300 ( 1.9112x ) 300

Simplify the above expression as follows:

  1.5037=(x1.9112x)

Rearrange and obtain the value of x .

  x=2.87381.5037x2.5037x=2.8738x=2.87382.5037=1.1478 g

Thus, the total weight of malononitrile that has been eliminated in four successive extractions with 25 mL portions of hexane is obtained by the sum of the weights obtained for each cycle.

  Total weight extracted =18.02 g+ 7.1949 g+2.8738 g+1.1478 g=29.2365 g

(b)

Interpretation Introduction

Interpretation:Weight of malononitrile recovered upon extraction with one 300 mL portions of ether should be calculated.

Concept Introduction:The distribution coefficient is a parameter that indicates the extent of solubility of two immiscible components in two different phases. One of the phases is aqueous, while the other is the organic phase.

Mathematically,

  KD=[solute]org[solute]aq

The expression to calculate fraction of solute left after n extractions is given as:

  qn=( V 1 V 1 + K d V 2 )n

(b)

Expert Solution
Check Mark

Answer to Problem 1Q

Weight of malononitrile recovered upon extraction with one 300 mL portion of ether is

found to be 29.6 g .

Explanation of Solution

The expression to calculate fraction of solute left after n extractions is given as:

  qn=( V 1 V 1 + K d V 2 )n

Where,

  • V1denotes volume used for the extraction cycle;
  • Kd denotes distribution coefficient ;
  • n represents the number of cycles of extraction carried out;
  • q denotes the fraction of weight that remains after n extractions.

The volume used for the extraction cycle is 300 mL .

  Kd is found to be 1.5037.

Since only single extraction is carried out with 300 mL portion of ether,the value of n is 1.

Substitute these values in above expression to calculate the fraction of malononitrile recovered as:

  q1=( 300 ( 300 )+( 1.5037 )( 300 ))1=0.399 g

Thus, weight of malononitrile recovered upon extraction with one 300 mL portion of ether is found as follows:

  weight of malononitrile recovered=(30 g)(0.40 g)=29.6 g

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