Distribution Coeffecient without concentrations












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







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$endgroup$

















    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









    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.







    $endgroup$















      2












      2








      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









      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






      share|improve this question









      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









      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




      share|improve this question








      edited 1 hour ago









      andselisk

      16.6k654115




      16.6k654115






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      asked 2 hours ago









      Molly HahnMolly Hahn

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          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



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






          share|improve this answer









          $endgroup$













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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



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






            share|improve this answer









            $endgroup$


















              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









              $endgroup$
















                2












                2








                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









                $endgroup$



                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



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







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                andseliskandselisk

                16.6k654115




                16.6k654115






















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