Connect 1-semester Access Card For Chemistry
Connect 1-semester Access Card For Chemistry
4th Edition
ISBN: 9781259636936
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 15, Problem 124AP

Iodine is sparingly soluble in water but much more so in carbon tetrachloride ( CCl 4 ) . The equilibrium constant, also called the partition coefficient, for the distribution of I 2 between these two phases:

I 2 ( a q ) I 2 ( CCI 4 )

is 83  at  20 ° C . (a) A student adds 0.030 L of CCl 4 to 0.200 L of an aqueous solution containing 0.032 g of I 2 The mixture at 20 ° C is shaken, and the two phases are then allowed to separate. Calculate the fraction of I 2 remaining in the aqueous phase, (b) The student now repeats the extraction of I 2 with another 0.030 L of CCl 4 . Calculate the fraction of the I 2 from the original solution that remains in the aqueous phase, (c) Compare the result in part (b) with a single extraction using 0.060 L of CCl 4 . Comment on the difference.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The fraction of iodine remaining in the aqueous solution under two conditions is to be calculated and their results are to be compared.

Concept introduction:

Equilibrium constant is the ratio of the concentration of reactants and products present in the chemical reaction.

For a general reaction: xA+yBzC

The general formula for writing equilibrium expression for the reaction is given as:

KC=[ C ]Zeqm[ A ]Xeqm[ B ]Yeqm

Here, KC is the equilibrium constant and C in KC stands for the concentration. [ A ], [ B ], and [ C ] are the equilibrium concentration of reactants A and B and product C.

A and B are reactants, C is products, and x,y, and z are their respective stoichiometric coefficients.

The ratio between the number of moles of a substance and the total number of moles of a solution is called mole fraction.

fraction(f)=molesofsolutetotalnumbersofmolesofsolution

Moles of a substance can be calculated by: moles =given massmolar mass

Concentration of a substance can be calculated as: Concentration =molesof substanceVolumeofsolvent

Answer to Problem 124AP

Solution:

0.07

0.5%

4%, which is more than the two individual extractions with 0.030L solutions each.

Explanation of Solution

Given Information: The mass of iodine in the solution is 0.032g.

The volume of the solution is 0.200L.

The volume of the CCl4 solution is 0.030L.

The equilibrium constant of the reaction is 83.

a) The fraction of I2 remaining in the aqueous phase.

The fraction of iodine in the aqueous phase is given below.

The number of moles of iodine is calculated as follows:

moles =given massmolar mass

MolesofI2=0.032g(1mol253.8g)=0.032(3.940×103)mol=1.26×104mol

The number of moles of iodine that dissolves in CCl4 is considered to be x. So, (1.26×104x) moles remains dissolved in water.

The concentration of iodine in carbon tetrachloride and water is given as follows:

Concentration =molesof substanceVolumeofsolvent

I2(aq)=(1.26×104x)mol0.200L[ I2(CCl4) ]=xmol0.030L

The concentration of iodine dissolved in tetrachloride and water is calculated by the expression as follows:

KC=[ C ]Zeqm[ A ]Xeqm[ B ]Yeqm

K=[ I2(CCl4) ][ I2(aq) ]

Substitute the values of K,[ I2(CCl4) ] and [ I2(aq) ] in the above equation,

83=x0.0301.26×104x0.20083(1.26×104x)=6.67xx=1.166×104

The fraction of iodine in the aqueous phase is,

fraction(f)=molesofsolutetotalnumbersofmolesofsolution

fraction(f)=(1.26×104)(1.166×104)1.26×104=9.4×1061.26×104=0.07

Hence, the fraction of iodine that remains in the aqueous phase is 0.07.

b) The fraction of I2

from the original solution that remains in the aqueous phase

The solution contains only 7.0% iodine after the first extraction.

The fraction of iodine left in the next extraction with 0.030L of CCl4 is calculated as follows:

Fractionofiodineleft=(0.07)(0.07)=5.0×103=0.5%

Hence, the fraction remaining after the second extraction is 0.5%.

c) Compare the result in part (b) with a single extraction using 0.060L of CCl4

and comment on difference.

The single extraction of iodine with 0.060L of CCl4.

The number of moles of iodine in CCl4

(y) is calculated by the expression as follows:

KC=[ C ]Zeqm[ A ]Xeqm[ B ]Yeqm

K=[ I2(CCl4) ][ I2(aq) ]

Substitute the values of K,[ I2(CCl4) ] and [ I2(aq) ] in the above equation.

83=y0.0601.26×104y0.20083(1.26×104y)=3.33yy=1.211×104

The fraction of iodine remaining in the aqueous phase is,

fraction(f)=molesofsolutetotalnumbersofmolesofsolution

fraction(f)=(1.26×104)(1.211×104)1.26×104=4.9×1061.26×104=0.04

Hence, the fraction of iodine that remains in the aqueous phase is 0.04, which is equal to 4%.

The fraction of iodine that is dissolved in water is 0.04. The extraction with 0.060L of CCl4 is not effective as compared to the two separate extractions with 0.030L solution each.

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Chapter 15 Solutions

Connect 1-semester Access Card For Chemistry

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