Question

Figure \(15-1\) (page 656 ) shows that \(I_{2}\) is considerably more soluble in \(\mathrm{CCl}_{4}(1)\) than it is in \(\mathrm{H}_{2} \mathrm{O}(1) .\) The concentration of \(I_{2}\) in its saturated aqueous solution is \(1.33 \times 10^{-3} \mathrm{M},\) and the equilibrium achieved when \(\bar{I}_{2}\) distributes itself between \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CCl}_{4}\) is $$\mathrm{I}_{2}(\mathrm{aq}) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) \quad K_{\mathrm{c}}=85.5$$ (a) \(\mathrm{A} 10.0 \mathrm{mL}\) sample of saturated \(\mathrm{I}_{2}(\mathrm{aq})\) is shaken with \(10.0 \mathrm{mL} \mathrm{CCl}_{4} .\) After equilibrium is established, the two liquid layers are separated. How many milligrams of \(I_{2}\) will be in the aqueous layer? (b) If the \(10.0 \mathrm{mL}\) of aqueous layer from part (a) is extracted with a second \(10.0 \mathrm{mL}\) portion of \(\mathrm{CCl}_{4}\) how many milligrams of \(\mathrm{I}_{2}\) will remain in the aqueous layer when equilibrium is reestablished? (c) If the 10.0 mL sample of saturated \(I_{2}(\) aq) in part (a) had originally been extracted with \(20.0 \mathrm{mL} \mathrm{CCl}_{4}\) would the mass of \(I_{2}\) remaining in the aqueous layer have been less than, equal to, or greater than that in part (b)? Explain.

Short answer

The mass of \(I_{2}\) remaining in the aqueous layer after the first extraction is approximately \(0.18 \, mg\). After the second extraction, \(0.17 \, mg\) of \(I_{2}\) remains in the aqueous layer. If a single extraction with \(20.0 \, mL \, CCl_{4}\) was done instead of two separate \(10.0 \, mL\) extractions, the mass of \(I_{2}\) left in the aqueous layer would have been less than \(0.17 \, mg\).

Step-by-step solution

Calculate initial mass of \(I_{2}\) in \(H_{2}O\)

The initial concentration of \(I_{2}\) given is \(1.33 \times 10^{-3} \, M\). This is the amount of \(I_{2}\) in a \(1.0 \, L\) or \(1000.0 \, mL\) of water. If we have \(10.0 \, mL\) of the solution, the initial moles of \(I_{2}\) will be \(\frac{10.0}{1000.0} \times 1.33 \times 10^{-3} = 1.33 \times 10^{-5} \, mol\). Since the molar mass of \(I_{2}\) is approximately \(253.8 \, g/mol\), the initial mass of \(I_{2}\) in the solution is \(1.33 \times 10^{-5} \, mol \times 253.8 \, g/mol \approx 3.38 \, mg\).

Calculate quantity of \(I_{2}\) transferred to \(CCl_{4}\) layer

The equilibrium constant \(K_{c}\) is 85.5, which means the reaction favors the right side; implying that more \(I_{2}\) will dissolve in \(CCl_{4}\) than in \(H_{2}O\). The equilibrium equation is \([I_{2}(CCl_{4})] = K_{c} \times [I_{2}(H_{2}O)]\). As the volumes of \(H_{2}O\) and \(CCl_{4}\) are the same, the equilibrium moles of \(I_{2}\) in \(CCl_{4}\) is \(85.5 \times [I_{2}(H_{2}O)]\). From the equilibrium equation, we can calculate the mass of \(I_{2}\) transferred to \(CCl_{4}\), which is \(3.38 \, mg - 3.38 \, mg / (85.5 + 1) \approx 3.20 \, mg\). Thus, \(0.18 \, mg\) of \(I_{2}\) remains in \(H_{2}O\). This is the answer to part (a).

Adjust for a second extraction

After the second \(10.0 \, mL\) portion of \(CCl_{4}\) is introduced, the process repeats itself. We apply the same equilibrium calculation to the remaining mass of \(I_{2}\) in the \(H_{2}O\). So, the mass of \(I_{2}\) remaining in \(H_{2}O\) after the second extraction is \(0.18 \, mg - 0.18 \, mg / (85.5 + 1) \approx 0.17 \, mg\). This is the answer to part (b).

Compare to a single extraction with \(20.0 \, mL\) \(CCl_{4}\)

In part (c), instead of two separate \(10.0 \, mL\) extractions of \(CCl_{4}\), if \(20.0 \, mL\) was used all at once, the mass of \(I_{2}\) left in \(H_{2}O\) would have been less. This is due to the increased quantity of \(CCl_{4}\) which will absorb more iodine based on the equilibrium constant. That's because the volume of \(CCl_{4}\) is a factor in the equilibrium calculation, so using a larger volume once would result in more \(I_{2}\) being extracted.

Key concepts

Equilibrium Constant

The equilibrium constant represents the ratio of concentrations of products to reactants at equilibrium in a reversible chemical reaction. In this exercise, the equilibrium constant \( K_c \) for the distribution of iodine between water and carbon tetrachloride (\( \text{CCl}_4 \)) is given as 85.5. This number indicates how much more the iodine prefers to dissolve in \( \text{CCl}_4 \) than in water. The mathematical expression for this equilibrium is:\[ K_c = \frac{[I_2(\text{CCl}_4)]}{[I_2(\text{H}_2O)]} \]A large \( K_c \) value, like 85.5, suggests that iodine is significantly more soluble in \( \text{CCl}_4 \) than in water. Understanding \( K_c \) helps in predicting how much of a solute will remain in each layer after reaching equilibrium. A larger \( K_c \) means a greater concentration of iodine in the \( \text{CCl}_4 \).

Aqueous Solution

An aqueous solution is simply a solution where water is the solvent. In this particular context, it pertains to the water layer in which iodine is dissolved. At the beginning, the concentration of iodine in the aqueous solution is quantified as \( 1.33 \times 10^{-3} \, M \). A few key points about aqueous solutions:

• Water as a solvent has polar characteristics, which can influence the solubility of a solute.
• The solubility equilibrium reached in aqueous solutions can be shifted by adding another solvent, like \( \text{CCl}_4 \) in our case.
• In this experiment, the concern is how much iodine remains in this aqueous layer after the distribution equilibrium is reached with \( \text{CCl}_4 \).

Remember, the concentration of iodine in this practice helps us determine the starting point for calculating how much iodine eventually migrates to \( \text{CCl}_4 \) once equilibrium is achieved.

CCl4 Solvent

Carbon tetrachloride (\( \text{CCl}_4 \)) serves as the second solvent in this problem, differing from water by being non-polar. This quality makes it an effective medium for dissolving non-polar solutes such as iodine. Here’s why \( \text{CCl}_4 \) is a crucial part of the process:

• \( \text{CCl}_4 \) increases the solubility of iodine compared to water.
• The equilibrium constant indicates a preference for \( I_2 \) to dissolve in \( \text{CCl}_4 \).
• \( \text{CCl}_4 \) allows for extraction processes to occur efficiently, as discussed in parts (a) and (b) of the problem.

Using \( \text{CCl}_4 \) emphasizes understanding the importance of choosing the right solvent based on the solute's properties when working with equilibrium systems.

Extraction Process

The extraction process is a technique in chemistry where a particular solute is removed from one liquid layer into another. In this case, it involves transferring iodine from an aqueous layer to a \( \text{CCl}_4 \) layer. Here’s how the extraction plays out and what to consider:- **First Extraction**: Shaking a sample of iodine-saturated water with \( \text{CCl}_4 \) helps iodine migrate to \( \text{CCl}_4 \), guided by the equilibrium constant value of 85.5.- **Second Extraction**: A subsequent addition of \( \text{CCl}_4 \) further pulls iodine from the aqueous layer, reducing the iodine concentration in water even more.Deciding between a single extraction with a larger amount of \( \text{CCl}_4 \) or multiple extractions with smaller volumes can influence how effectively the iodine is removed. As shown in part (c), a single extraction with a larger volume can decrease the amount of iodine remaining in the aqueous layer even more due to increased solvent capacity. The planning of these steps underlines the strategy needed to optimize the extraction based on equilibrium principles.