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Magazine Chapter 31: Problem 11
The distribution constant for \(\mathrm{X}\) berween \(n\)-hexane and water is 8.9. Calculate the concentration of \(\mathrm{X}\) remaining in the aqueous phase after \(50.0 \mathrm{~mL}\). of \(0.200 \mathrm{MX}\) is treated by extraction with the following quantities of \(n\)-hexane: (a) one 40.0-mL portion. (b) two 20.0-mL portions. (c) four 10.0-mL portions. (d) eight 5.00-mL portions.
Short Answer
Expert verified
The concentrations are (a) 0.0294 M, (b) 0.0220 M, (c) 0.0167 M, (d) 0.0127 M.
Step by step solution
01
Determine the Initial Moles of X in Aqueous Phase
The initial concentration of X is \(0.200 \text{ M}\) in \(50.0 \text{ mL}\) of water. First, convert the volume to liters: \(50.0 \text{ mL} = 0.050 \text{ L}\). Calculate the initial moles of X: \[ n = C \cdot V = 0.200 \text{ M} \times 0.050 \text{ L} = 0.010 \text{ moles} \]
02
Understand Distribution Constant and Formula
The distribution constant \(K_d\) is given as 8.9. For the extraction, \[ K_d = \frac{C_{hex}}{C_{water}} = \frac{x}{0.200-x} \times \frac{0.050}{V_{hex}} \]where \(x\) is the moles of X in \(n\)-hexane, and \(V_{hex}\) is the volume of \(n\)-hexane used in liters.
03
Calculate (a) One 40.0-mL Portion
Convert the volume of \(n\)-hexane to liters: \(40.0 \text{ mL} = 0.040 \text{ L}\). Using the formula:\[ 8.9 = \frac{x}{0.010-x} \times \frac{0.050}{0.040} \] Rearrange to find \(x\), then calculate \(m_{after} = 0.010 - x\). Solving gives \(x = 0.008529\), and thus, \[ m_{after} = 0.010 - 0.008529 = 0.001471 \text{ moles}\] Concentration after extraction: \[ C_{water} = \frac{0.001471}{0.050} = 0.0294 \text{ M}\]
04
Calculate (b) Two 20.0-mL Portions
In the first extraction, using \(20.0 \text{ mL} = 0.020 \text{ L}\): \[ 8.9 = \frac{x_1}{0.010 - x_1} \times \frac{0.050}{0.020} \] Solve for \(x_1\), then remaining moles \(m_1 = 0.010 - x_1 = 0.002976\).For the second extraction, using remaining concentration and same formula, solve for \(x_2\) with \(m_1\). Resulting \(m_2 = m_1 - x_2\), and final concentration: \[ C_{water} = \frac{m_2}{0.050}\] Final \(x_2\) and concentration after two extraction steps gives \[C_{water} = 0.0220 \text{ M}\].
05
Calculate (c) Four 10.0-mL Portions
For each 10.0-mL extraction step (\(0.010 \text{ L}\) hexane):1. Solve \[ 8.9 = \frac{x_1}{0.010-x_1} \times \frac{0.050}{0.010} \] First \(m_1\) from \(x_1\).2. Use \(m_1\) in second extraction to find \(m_2\).3. Repeat for next two steps using the recursive process.Resulting concentration after four extractions: \[C_{water} = 0.0167 \text{ M}\].
06
Calculate (d) Eight 5.00-mL Portions
For each 5.00-mL extraction step (\(0.005 \text{ L}\) hexane):1. Use initial conditions and solve for first extraction moles.2. Repeat for each subsequent extraction, recalculating from the latest moles in water.After eight steps, the resulting concentration:\[C_{water} = 0.0127 \text{ M}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
extraction
Extraction is a key concept in chemistry used for separating compounds based on their different solubilities in two immiscible liquids. One liquid is usually water (the aqueous phase), and the other is an organic solvent, like n-hexane. Here's how this process works:
- The mixture is treated with a solvent that dissolves one or more of its components. This process helps in isolating desired substances by taking advantage of their solubility differences.
- The extraction process can be repeated several times to increase the purity or concentration of the extracted compound. This is known as multiple extractions, which can be more efficient than a single extraction.
- The distribution constant, or partition coefficient (\(K_d\)), expresses the method's foundation, as it determines how much of the compound moves into each phase.
aqueous phase
The aqueous phase in a chemical extraction refers to the water layer in which dissolved substances are extracted from using an organic solvent. It is part of a two-phase system where both phases do not mix. Understanding this concept is crucial:
- This phase serves as the starting point for many separation techniques, holding the compounds before the introduction of an extracting solvent (like n-hexane in this exercise).
- Compounds distributed between phases are in equilibrium, determined by their solubility and affinity for each phase.
- The movement of a compound from the aqueous phase to the organic phase relies heavily on the concentration difference and the distribution constant (\(K_d\)). The concept of equilibrium plays a significant role here.
concentration calculation
Concentration calculation is essential for determining how successful an extraction has been and is widely used in chemistry to describe how much of a solute is present in a solution. The exercise outlines multiple steps to achieve this:
- First, calculate the initial amount of the solute (X) in the system using the formula: \( n = C \cdot V \), where \(n\) is the number of moles, \(C\) is the concentration, and \(V\) is the volume.
- With each extraction step, the remaining concentration in the aqueous phase is determined by applying the distribution constant, which represents the equilibrium between the phases.
- Understanding equilibrium equations is crucial as they enable prediction of the concentrations after each stage of extraction, aiding in planning further steps or extractions if needed.
- Using recursive calculations, we refine each concentration until optimal extraction efficiency is achieved.
analytical chemistry problem
Analytical chemistry problems often involve exercises like the one discussed, where understanding various practical and theoretical concepts is tested.
- The goal is to grasp the underlying principles of separation, solubility, and reaction mechanisms involved in chemical processes.
- Analyzing complex problems also aids in developing skills for conducting quantitative and qualitative assessments of chemical substances.
- These problems usually require stepwise calculations and methodical approaches to break down complex equations into manageable parts.
- In this task, understanding the purpose and application of the distribution constant and the importance of precision in concentration calculation is pivotal.
