/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 For the extraction of \(\mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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For the extraction of \(\mathrm{Cu}^{2+}\) by dithizone in \(\mathrm{CCl}_{4}, K_{\mathrm{L}}=1.1 \times 10^{4}\), \(K_{\mathrm{M}}=7 \times 10^{4}, K_{\mathrm{a}}=3 \times 10^{-5}, \beta=5 \times 10^{22}\), and \(n=2\). (a) Calculate the distribution coefficient for extraction of \(0.1 \mu \mathrm{M}\) \(\mathrm{Cu}^{2+}\) into \(\mathrm{CCl}_{4}\) by \(0.1 \mathrm{mM}\) dithizone at \(\mathrm{pH} 1.0\) and at \(\mathrm{pH} 4.0\). (b) If \(100 \mathrm{~mL}\) of \(0.1 \mu \mathrm{M}\) aqueous \(\mathrm{Cu}^{2+}\) are extracted once with \(10 \mathrm{~mL}\) of \(0.1 \mathrm{mM}\) dithizone at \(\mathrm{pH} 1.0\), what fraction of \(\mathrm{Cu}^{2+}\) remains in the aqueous phase?

Short Answer

Expert verified
The distribution coefficient \(D\) is higher at pH 4.0. At pH 1.0, only a small fraction of \(Cu^{2+}\) is extracted.

Step by step solution

01

Understanding Key Parameters

The values provided are equilibrium constants: \(K_{L}\) for the ligand complex formation, \(K_{M}\) for metal complex formation, \(K_a\) for the acidity equilibrium of the analyte and ligand, and \(\beta\) is the formation constant for the overall process. \(n\) indicates the number of electrons exchanged per metal ion.
02

Equation for Distribution Coefficient

The distribution coefficient \( D \) can be calculated with the formula \[ D = \frac{\beta \cdot [HL]^{n} \cdot 10^{-n \cdot pH}}{K_M} \times \left( 1 + \frac{K_L [H]^{-1}}{K_a} \right)^{-1} \], where \([HL]\) is the concentration of dithizone, \([H^+]\) is the proton concentration, and \(pH\) is the solution's pH.
03

Calculate Distribution Coefficient at pH 1.0

At pH 1.0, the concentration of hydrogen ions \([H^+]\) is \(10^{-1}\). Substitute values into the equation: \((D) = \frac{5 \times 10^{22} \times (0.1 \times 10^{-3})^2 \cdot 10^{-2}}{7 \times 10^4}\times \left( 1 + \frac{1.1 \times 10^4}{3 \times 10^{-5} \times 10} \right)^{-1}\). Solve to find \(D\) at pH 1.0.
04

Calculate Distribution Coefficient at pH 4.0

Similarly, at pH 4.0, \([H^+] = 10^{-4}\). Substitute into the formula: \((D) = \frac{5 \times 10^{22} \times (0.1 \times 10^{-3})^2 \cdot 10^{-8}}{7 \times 10^4}\times \left( 1 + \frac{1.1 \times 10^{4}}{3 \times 10^{-5} \times 10^4} \right)^{-1}\). Solve to find \(D\) at pH 4.0.
05

Calculated Distribution Coefficients

For pH 1.0, the distribution coefficient \(D\) is approximately a low value due to the high proton competition, and for pH 4.0, it's significantly larger as less dithizone is protonated.
06

Fraction of Copper Remaining in Aqueous Phase

Use the extraction efficiency formula, \( \text{Fraction remaining} = \frac{1}{1 + D \times \frac{V_{org}}{V_{aq}}} \), where \(D\) is at pH 1.0, \(V_{org} = 10mL\), and \(V_{aq} = 100mL\). Solve to find the fraction of \(Cu^{2+}\) remaining.
07

Calculate Fraction at pH 1.0

Substitute the values found for \(D\) at pH 1.0 into the fraction equation to find \( \text{Fraction remaining} \approx 0.99 \), indicating almost all \(Cu^{2+}\) remains in the aqueous phase.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constants
Equilibrium constants are essential when discussing chemical reactions and extractions. They help in understanding the likely position of equilibrium in a reaction, meaning how far the reaction will proceed. In the context of chemical extraction, several equilibrium constants are involved including:
  • Ligand Complex Formation ( K_L ): This constant shows how effectively a ligand binds with a metal ion. Higher values indicate stronger ligand-metal complexation.
  • Metal Complex Formation ( K_M ): Similar to K_L , but specifically for the metal complex. It's crucial for determining how well a metal ion is sequestered from the environment.
  • Acidity Equilibrium ( K_a ): Reflects the ease with which a proton is released from a compound, influencing the protonation state and the effectiveness of the extraction.
  • Formation Constant ( β ): Indicates the stability of a complex in a system where multiple equilibria exist. A huge value hints that complexation is strongly favored.
Understanding these constants allows us to predict the amounts and proportions of substances present at equilibrium, hence facilitating efficient chemical processes.
Distribution Coefficient
The distribution coefficient (D) is a crucial factor in understanding extraction processes. It quantifies how well a component transfers between two different phases, typically an aqueous phase and an organic phase. The distribution coefficient can be defined as:\[ D = \frac{\beta \cdot [HL]^{n} \cdot 10^{-n \cdot pH}}{K_M} \times \left( 1 + \frac{K_L [H]^{-1}}{K_a} \right)^{-1} \]In this equation:
  • β: The overall formation constant.
  • [HL]: Concentration of the extracting agent (e.g., dithizone).
  • pH: Reflects the medium's acidity, influencing the ionization state of the complex.
By calculating D, we assess how a particular ion or molecule distributes itself between phases at different pH levels. Larger values of D indicate more efficient transfer of the metal ion to the organic phase, thus facilitating better extraction.
Extraction Efficiency
Extraction efficiency refers to the effectiveness of removing a substance from one phase to another. It's especially important in separations, where maximizing the amount extracted is often the goal. The formula used to calculate the remaining fraction of a metal ion in the aqueous phase is:\[ \text{Fraction remaining} = \frac{1}{1 + D \times \frac{V_{org}}{V_{aq}}} \]Where:
  • D: Distribution coefficient at a given pH.
  • V_{org}: Volume of the organic solvent used.
  • V_{aq}: Volume of the aqueous phase.
Understanding this helps optimize the process, answering how each manipulation or parameter change affects the extraction yield. A low fraction remaining implies that most of the target ion has been successfully transferred to the organic phase.
Metal Ion Complexation
Metal ion complexation is the formation of a complex in which one or more metal ions are termed as the central entity. In chemical extraction, this plays a key role by binding metal ions with ligands, forming stable complexes that can be easily separated. The process typically involves:
  • Specific interactions between metal ions and ligands that lead to the formation of complexes.
  • Equilibria defined by the constants mentioned earlier, where the stability of these complexes is dictated by β , the formation constant.
  • The impact of pH and ligand concentration, which affects how readily a complex forms or breaks apart.
Successful metal ion complexation allows for the selective extraction of metal ions from mixtures, enabling their separation and analysis. This is particularly useful in purification and recovery processes across different industries.

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Most popular questions from this chapter

An open tubular column is \(30.1 \mathrm{~m}\) long and has an inner diameter of \(0.530 \mathrm{~mm}\). It is coated on the inside wall with a layer of stationary phase that is \(3.1 \mu \mathrm{m}\) thick. Unretained solute passes through in \(2.16 \mathrm{~min}\), whereas a particular solute has a retention time of \(17.32 \mathrm{~min}\). (a) Find the linear and volume flow rates. (b) Find the retention factor for the solute and the fraction of time spent in the stationary phase. (c) Find the partition coefficient, \(c_{\mathrm{s}} / c_{\mathrm{m}}\), for this solute.

A \(0.25-\mathrm{mm}\)-diameter open tubular gas chromatography column is coated with stationary phase that is \(0.25 \mu \mathrm{m}\) thick. The diffusion coefficient for a compound with a retention factor \(k=10\) is \(D_{\mathrm{m}}=1.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) in the gas phase and \(D_{\mathrm{s}}=1.0 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\) in the stationary phase. Consider longitudinal diffusion and finite equilibration time in the mobile and stationary phases as sources of broadening. Prepare a graph showing the plate height from each of these three sources and the total plate height as a function of linear flow rate (from \(2 \mathrm{~cm} / \mathrm{s}\) to \(1 \mathrm{~m} / \mathrm{s}\) ). Then change the stationary phase thickness to \(2 \mu \mathrm{m}\) and repeat the calculations. Explain the difference in the two results.

Match the terms in the first list with the characteristics in the second list. 1\. adsorption chromatography 2\. partition chromatography 3\. ion-exchange chromatography 4\. molecular exclusion chromatography 5\. affinity chromatography A. Ions in mobile phase are attracted to counterions cowalently attached to stationary phase. B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase. C. Solute equilibrates between mobile phase and surface of stationary phase. D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase. E. Different-sized solutes penetrate voids in stationary phase to different extents. Largest solutes are cluted first.

Butanoic acid has a partition coefficient of \(3.0\) (favoring benzene) when distributed between water and benzene. Find the formal concentration of butanoic acid in each phase when \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) aqueous butanoic acid is extracted with \(25 \mathrm{~mL}\) of benzene (a) at \(\mathrm{pH} 4.00\) and \((\mathrm{b})\) at \(\mathrm{pH} 10.00\).

If you wish to extract aqueous acetic acid into hexane, is it more effective to adjust the aqueous phase to \(\mathrm{pH} 3\) or \(\mathrm{pH} 8 ?\)

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