Question

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases: $$\mathrm{I}_{2}(a q) \rightleftarrows \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right)$$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CC} 1_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) of \(\mathrm{I}_{2}\). The mixture at \(20^{\circ} \mathrm{C}\) is shaken, and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(\mathrm{I}_{2}\) with another \(0.030 \mathrm{~L}\) of \(\mathrm{CC} 1_{4} .\) Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in part (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CC} 1_{4}\). Comment on the difference.

Short answer

Two extractions decrease iodine more than one single extraction. About 0.5% remains after two extractions, while 0.95% remains after one.

Step-by-step solution

Understanding the Partition Coefficient

The partition coefficient, denoted as \( K \), is the ratio of concentrations of a solute present in the two immiscible liquids. For this case: \[ K = \frac{[\mathrm{I_2}]_{CCl_4}}{[\mathrm{I_2}]_{aq}} = 83 \] This means iodine is 83 times more concentrated in carbon tetrachloride than in water.

Calculate Initial Concentrations

Calculate the initial concentration of \( \mathrm{I_2} \) in the aqueous solution:\[ \text{mass of } \mathrm{I_2} = 0.032 \, \text{g, Molar mass of } \mathrm{I_2} = 253.8 \, \text{g/mol} \] \[ \text{moles of } \mathrm{I_2} = \frac{0.032}{253.8} = 0.000126 \text{ mol} \] \[ [\mathrm{I_2}]_{aq}^{initial} = \frac{0.000126}{0.200} = 0.00063 \text{ mol/L} \] The initial concentration of \( \mathrm{I_2} \) in \( \mathrm{CCl_4} \) is zero since it starts from pure \( \mathrm{CCl_4} \).

Set Up the Distribution Equation for First Extraction (Part a)

Assuming \( x \) moles of \( \mathrm{I_2} \) go into \( \mathrm{CCl_4} \), then:- \([\mathrm{I_2}]_{CCl_4} = \frac{x}{0.030} \text{ mol/L} \)- \([\mathrm{I_2}]_{aq} = \frac{0.000126 - x}{0.200} \text{ mol/L} \) Substitute into the partition coefficient equation:\[ 83 = \frac{\frac{x}{0.030}}{\frac{0.000126 - x}{0.200}} \] Solve for \( x \).

Solve the Equation for First Extraction

Solving for \( x \) from step 3 gives:\[ 83 = \frac{200x}{0.03(0.000126 - x)} \Rightarrow x = \frac{83 \times 0.000126 \times 0.03}{83 \times 0.03 + 200} \approx 0.000116 \] This means 0.000116 mol goes into \( \mathrm{CCl}_4 \), leaving \( 0.000126 - 0.000116 = 0.00001 \) mol in water.

Calculate Fraction Remaining After First Extraction

The fraction of \( \mathrm{I_2} \) remaining in the aqueous solution is:\[ \text{Fraction} = \frac{0.00001}{0.000126} \approx 0.079 \] This is the fraction left in the aqueous phase after the first extraction.

Perform Second Extraction (Part b)

For the second extraction, we treat \( 0.00001 \) mol as the new initial amount in the aqueous phase. Repeat Step 3 with this new amount:- Initial \([\mathrm{I_2}]_{aq} = \frac{0.00001}{0.200} \text{ mol/L} \)Substitute:\[ 83 = \frac{x}{0.030}/\frac{0.00001 - x}{0.200} \]

Solve Equation for Second Extraction

Solving for \( x \) in the new equation:\[ 83 = \frac{200x}{0.03(0.00001 - x)} \Rightarrow x \approx 0.0000093 \] Then \( 0.00001 - 0.0000093 = 0.0000007 \) mol remains in water. Calculate the new fraction:\[ \text{Fraction left} = \frac{0.0000007}{0.000126} \approx 0.0056 \]

Comparisons with Single Extraction (Part c)

For a single extraction using 0.060 L \( \mathrm{CCL_4} \), redo Step 3 using this volume:Substitute into the equation:\[ 83 = \frac{x}{0.060}/\frac{0.000126 - x}{0.200} \]Solving gives \( 0.0000012 \) mol remaining, fraction:\[ \text{Fraction} = \frac{0.0000012}{0.000126} \approx 0.0095 \] Two extractions with 0.030 L each results in less \( \mathrm{I_2} \) remaining than one with 0.060 L.

Key concepts

Iodine Solubility

Iodine is a halogen element that is minimally soluble in water. This means only a small amount of iodine can dissolve in water, making it "sparingly soluble." However, when iodine encounters organic solvents, such as carbon tetrachloride (CCl₄), its solubility significantly increases. This change happens because iodine is non-polar, just like CCl₄, allowing them to mix better compared to water, which is polar.

This difference in solubility becomes important when dealing with separations and extractions in chemistry. If you want to efficiently extract iodine from an aqueous solution, you'll prefer an organic solvent like carbon tetrachloride, where it dissolves much more readily. Understanding this concept is crucial for students learning about the behavior of different substances, and how they interact with various solvents.

Carbon Tetrachloride

Carbon tetrachloride, with the chemical formula CCl₄, is a colorless liquid commonly used in laboratories for extraction processes. Known for its non-polar properties, CCl₄ serves as an ideal solvent for substances like iodine, which do not dissolve well in water.

It's important to understand that CCl₄ has a higher affinity for non-polar solutes. This makes it very effective in partitioning experiments where you measure how a solute distributes itself between different phases—aqueous and organic.

• CCl₄’s non-polar nature ensures better solute solubility than water for non-polar substances.
• Caution: CCl₄ is also toxic and should be handled with care in controlled environments.

Understanding these properties makes it easier to predict and control the outcomes when using CCl₄ as part of a chemical extraction.

Two-Phase Equilibrium

Two-phase equilibrium refers to a state where two immiscible liquids, like water and carbon tetrachloride, come into contact and reach a balance in terms of solute distribution. In our context, iodine distributes itself between these two phases based on its solubility in each.

The equilibrium is governed by the partition coefficient, denoted as K. This coefficient is a fixed value for a specific solute-solvent pair at a given temperature. When iodine solutes mix with water and CCl₄, they attain a state of equilibrium represented by the ratio of their concentrations in each phase:

• High K value (e.g., 83): Indicates a higher concentration in CCl₄ compared to water.
• Reaching equilibrium demands careful calculation of concentrations and the use of principles like Le Chatelier's.

Being able to predict and measure how solutes act across different phases aids in performing efficient chemical extractions and is a core aspect of analytical chemistry.

Chemical Concentration

Chemical concentration is a measure of the amount of solute present in a given volume of solution. When dealing with partitioning problems, like the one involving iodine and carbon tetrachloride, understanding concentration is key to solving the problem effectively.

To calculate concentrations precisely:

• Determine the initial amount of solute (e.g., in moles) using its mass and molar mass.
• Consider the volume of each phase to find corresponding molar concentrations.

In the given problem, students calculate the iodine concentration in both water and CCl₄ to find how the iodine is distributed. These concentrations are then compared using the partition coefficient to analyze the equilibrium state of the system.

Mastery of these calculations allows students to gauge how changes in solvent volume or repeated extractions affect the concentration of solutes, ultimately leading them to a better understanding of chemical separation techniques.