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Distribution Coeffecient without concentrations

Standard solution and extracted solution : same concentration?Equilibrium Concentrations of Products/Reactantssolution concentrationsSolubility of Gold in low concentrationsHow to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture?Chemical Reagent Test sensitive at low concentrationsHow to find pH of an aqueous ammonia solution at high pressures?Increasing partition coefficient in an extractionWhy do my equilibrium calculations on this HF/NH4OH buffer system not match those in literature?How to without calculations understand the liquid-liquid extraction efficiency?

2

$begingroup$

From what I understand about distribution coefficient is straight from my book-- which does NOT give any practice examples-- is that:

D= CA(ext) ÷ CA(orig)

where CA(ext) and CA(orig) represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

In an extraction experiment, it is found that 0.0376 g of an analyte are extracted into 50 mL of solvent from 150 mL of a water sample. If there was originally 0.192 g of analyte in this volume of the water sample, what is the distribution coefficient?

The answer for this problem is 0.731%, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

equilibrium solutions analytical-chemistry extraction

share|improve this question

edited 1 hour ago

andselisk

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asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2

$begingroup$

From what I understand about distribution coefficient is straight from my book-- which does NOT give any practice examples-- is that:

D= CA(ext) ÷ CA(orig)

where CA(ext) and CA(orig) represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

In an extraction experiment, it is found that 0.0376 g of an analyte are extracted into 50 mL of solvent from 150 mL of a water sample. If there was originally 0.192 g of analyte in this volume of the water sample, what is the distribution coefficient?

The answer for this problem is 0.731%, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

equilibrium solutions analytical-chemistry extraction

share|improve this question

edited 1 hour ago

andselisk

16.6k654115

asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

From what I understand about distribution coefficient is straight from my book-- which does NOT give any practice examples-- is that:

D= CA(ext) ÷ CA(orig)

where CA(ext) and CA(orig) represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

In an extraction experiment, it is found that 0.0376 g of an analyte are extracted into 50 mL of solvent from 150 mL of a water sample. If there was originally 0.192 g of analyte in this volume of the water sample, what is the distribution coefficient?

The answer for this problem is 0.731%, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

equilibrium solutions analytical-chemistry extraction

share|improve this question

edited 1 hour ago

andselisk

16.6k654115

asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 1 hour ago

andselisk

16.6k654115

asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

andselisk

16.6k654115

asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

andselisk

16.6k654115

edited 1 hour ago

andselisk

16.6k654115

asked 2 hours ago

Molly HahnMolly Hahn

113

New contributor

Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

Molly HahnMolly Hahn

113

1 Answer
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2

$begingroup$

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = frac{n_i}{V_i} = frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = frac{C_mathrm{s}}{C_mathrm{w}} = frac{m_mathrm{s}V_mathrm{w}}{m_mathrm{w}V_mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_mathrm{w} = m_0 - m_mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$

share|improve this answer

answered 1 hour ago

andseliskandselisk

16.6k654115

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2

$begingroup$

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = frac{n_i}{V_i} = frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = frac{C_mathrm{s}}{C_mathrm{w}} = frac{m_mathrm{s}V_mathrm{w}}{m_mathrm{w}V_mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_mathrm{w} = m_0 - m_mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$

share|improve this answer

answered 1 hour ago

andseliskandselisk

16.6k654115

$endgroup$

add a comment | 

2

$begingroup$

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = frac{n_i}{V_i} = frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = frac{C_mathrm{s}}{C_mathrm{w}} = frac{m_mathrm{s}V_mathrm{w}}{m_mathrm{w}V_mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_mathrm{w} = m_0 - m_mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$

share|improve this answer

answered 1 hour ago

andseliskandselisk

16.6k654115

$endgroup$

add a comment | 

$begingroup$

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = frac{n_i}{V_i} = frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = frac{C_mathrm{s}}{C_mathrm{w}} = frac{m_mathrm{s}V_mathrm{w}}{m_mathrm{w}V_mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_mathrm{w} = m_0 - m_mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$

share|improve this answer

answered 1 hour ago

andseliskandselisk

16.6k654115

$endgroup$

share|improve this answer

answered 1 hour ago

andseliskandselisk

16.6k654115

answered 1 hour ago

andseliskandselisk

16.6k654115

answered 1 hour ago

andseliskandselisk

16.6k654115