Distribution Coeffecient without concentrationsStandard solution and extracted solution : same...
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Distribution Coeffecient without concentrations
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$begingroup$
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
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
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
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.
$endgroup$
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
equilibrium solutions analytical-chemistry extraction
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
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
edited 1 hour ago
andselisk
16.6k654115
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
asked 2 hours ago
Molly HahnMolly Hahn
113
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.
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.
Molly Hahn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
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
$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$
answered 1 hour ago
andseliskandselisk
16.6k654115
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
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
$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$
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
$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$
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
$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$
answered 1 hour ago
andseliskandselisk
16.6k654115
$endgroup$
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
$$D = frac{m_mathrm{s}V_mathrm{w}}{(m_0 - m_mathrm{s})V_mathrm{s}} = frac{pu{0.0376 g}cdotpu{150 mL}}{(pu{0.192 g} - pu{0.0376 g})cdotpu{50 mL}} = 0.731$$
answered 1 hour ago
andseliskandselisk
16.6k654115
answered 1 hour ago
andseliskandselisk
16.6k654115
answered 1 hour ago
andseliskandselisk
16.6k654115
answered 1 hour ago
andseliskandselisk
16.6k654115
16.6k654115
add a comment |
add a comment |
Molly Hahn is a new contributor. Be nice, and check out our Code of Conduct.
Molly Hahn is a new contributor. Be nice, and check out our Code of Conduct.
Molly Hahn is a new contributor. Be nice, and check out our Code of Conduct.
Molly Hahn is a new contributor. Be nice, and check out our Code of Conduct.
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