Chapter 15, Problem 124AP

Iodine is sparingly soluble in water but much more so in carbon tetrachloride $({\text{CCl}}_4)$. The equilibrium constant, also called the partition coefficient, for the distribution of ${\text{I}}_2$ between these two phases:

${\text{I}}_2(aq)\stackrel{}{\rightleftarrows }{\text{I}}_2({\text{CCI}}_4)$

is $83\text{ at }20°\text{C}$. (a) A student adds 0.030 L of ${\text{CCl}}_4$ to 0.200 L of an aqueous solution containing 0.032 g of ${\text{I}}_2$ The mixture at $20°\text{C}$ is shaken, and the two phases are then allowed to separate. Calculate the fraction of ${\text{I}}_2$ remaining in the aqueous phase, (b) The student now repeats the extraction of ${\text{I}}_2$ with another 0.030 L of ${\text{CCl}}_4$. Calculate the fraction of the ${\text{I}}_2$ 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 ${\text{CCl}}_4$. Comment on the difference.

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: $\text{xA}\text{+}\text{yB}\rightleftarrows \text{zC}$

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

$K_C=\frac{{\left[\text{C}\right]}^Z{}_{eqm}}{{\left[\text{A}\right]}^X{}_{eqm}{\left[\text{B}\right]}^Y{}_{eqm}}$

Here, $K_C$ is the equilibrium constant and $C$ in $K_C$ stands for the concentration. $\left[\text{A}\right]$, $\left[\text{B}\right]$, and $\left[\text{C}\right]$ 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.

$\text{fraction(f)}\text{}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{numbers}\text{of}\text{moles}\text{of}\text{solution}}$

Moles of a substance can be calculated by: $\text{moles =}\frac{\text{given mass}}{\text{molar mass}}$

Concentration of a substance can be calculated as: $\text{Concentration =}\frac{\text{moles}\text{of substance}}{\text{Volume}\text{}\text{of}\text{solvent}}$

Answer to Problem 124AP

Solution:

$0.07$

$0.5\%$

$4\%$, which is more than the two individual extractions with $0.030\text{L}$ solutions each.

Explanation of Solution

Given Information: The mass of iodine in the solution is $0.032\text{g}$.

The volume of the solution is $0.200\text{L}$.

The volume of the ${\text{CCl}}_4$ solution is $0.030\text{L}$.

The equilibrium constant of the reaction is $83$.

a) The fraction of ${\text{I}}_{\text{2}}$ 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:

$\text{moles =}\frac{\text{given mass}}{\text{molar mass}}$

$\begin{array}{c}Molesof{I}_2=0.032\cancel{g}\left(\frac{1mol}{253.8\cancel{g}}\right)\\ =0.032\left(3.940\times {10}^{-3}\right)mol\\ =1.26\times {10}^{-4}mol\end{array}$

The number of moles of iodine that dissolves in $CC{l}_4$ is considered to be $x$. So, $\left(1.26\times {10}^{-4}-x\right)$ moles remains dissolved in water.

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

$\text{Concentration =}\frac{\text{moles}\text{of substance}}{\text{Volume}\text{}\text{of}\text{solvent}}$

$\begin{array}{c}{I}_2\left(aq\right)=\frac{\left(1.26\times {10}^{-4}-x\right)mol}{0.200L}\\ \left[I_2\left(CC{l}_4\right)\right]=\frac{xmol}{0.030L}\end{array}$

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

$K_C=\frac{{\left[\text{C}\right]}^Z{}_{eqm}}{{\left[\text{A}\right]}^X{}_{eqm}{\left[\text{B}\right]}^Y{}_{eqm}}$

$K=\frac{\left[I_2\left(CC{l}_4\right)\right]}{\left[I_2\left(aq\right)\right]}$

Substitute the values of $K,\left[I_2\left(CC{l}_4\right)\right]$ and $\left[I_2\left(aq\right)\right]$ in the above equation,

$\begin{array}{c}83=\frac{\frac{x}{0.030}}{\frac{1.26\times {10}^{-4}-x}{0.200}}\\ 83\left(1.26\times {10}^{-4}-x\right)=6.67x\\ x=1.166\times {10}^{-4}\end{array}$

The fraction of iodine in the aqueous phase is,

$\text{fraction(f)}\text{}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{numbers}\text{of}\text{moles}\text{of}\text{solution}}$

$\begin{array}{c}fraction\left(f\right)=\frac{\left(1.26\times {10}^{-4}\right)-\left(1.166\times {10}^{-4}\right)}{1.26\times {10}^{-4}}\\ =\frac{9.4\times {10}^{-6}}{1.26\times {10}^{-4}}\\ =0.07\end{array}$

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

b) The fraction of ${\text{I}}_{\text{2}}$

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 $CC{l}_4$ is calculated as follows:

$\begin{array}{c}Fractionofiodineleft=\left(0.07\right)\left(0.07\right)\\ =5.0\times {10}^{-3}\\ =0.5\%\end{array}$

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 $CC{l}_4$

and comment on difference.

The single extraction of iodine with $0.060L$ of $CC{l}_4$.

The number of moles of iodine in $CC{l}_4$

$\left(y\right)$ is calculated by the expression as follows:

$K_C=\frac{{\left[\text{C}\right]}^Z{}_{eqm}}{{\left[\text{A}\right]}^X{}_{eqm}{\left[\text{B}\right]}^Y{}_{eqm}}$

$K=\frac{\left[I_2\left(CC{l}_4\right)\right]}{\left[I_2\left(aq\right)\right]}$

Substitute the values of $K,\left[I_2\left(CC{l}_4\right)\right]$ and $\left[I_2\left(aq\right)\right]$ in the above equation.

$\begin{array}{c}83=\frac{\frac{y}{0.060}}{\frac{1.26\times {10}^{-4}-y}{0.200}}\\ 83\left(1.26\times {10}^{-4}-y\right)=3.33y\\ y=1.211\times {10}^{-4}\end{array}$

The fraction of iodine remaining in the aqueous phase is,

$\text{fraction(f)}\text{}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{numbers}\text{of}\text{moles}\text{of}\text{solution}}$

$\begin{array}{c}fraction\left(f\right)=\frac{\left(1.26\times {10}^{-4}\right)-\left(1.211\times {10}^{-4}\right)}{1.26\times {10}^{-4}}\\ =\frac{4.9\times {10}^{-6}}{1.26\times {10}^{-4}}\\ =0.04\end{array}$

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 $CC{l}_4$ is not effective as compared to the two separate extractions with $0.030L$ solution each.