Question

0.070 g of caffeine is dissolved in 4.0 mL of water. The caffeine is extracted from the aqueous solution three times in 2.0 mL portions of methylene chloride. Calculate the total amount of caffeine that can be extracted into the three portions of methylene chloride. Caffeine has a distribution coefficient of 4.6, between methylene chloride and water.

Answer 1

Concepts and reason

The concept used to solve this problem is based on distribution coefficient.

The equilibrium constant for distribution of analyte in two immiscible solvents is called distribution constant.

Fundamentals

The distribution coefficient ${K_{\rm{D}}}$ can be written as follow.

${K_{\rm{D}}} = \frac{{{{\left[ X \right]}_{{\rm{org}}}}}}{{{{\left[ X \right]}_{{\rm{aq}}}}}}$

Here ${\left[ X \right]_{{\rm{org}}}}$is concentration of X in organic solvent and ${\left[ X \right]_{{\rm{aq}}}}$is concentration of X in aqueous solution.

Mass of caffeine is as follow.

${m_{\rm{C}}} = {m_{{{\rm{H}}_2}{\rm{O}}}} + {m_{{\rm{meth}}}}$

Let the mass of water$\left( {{m_{{{\rm{H}}_2}{\rm{O}}}}} \right)$ is x grams. Substitute 0.070 grams for ${m_{\rm{C}}}$ and x grams for ${m_{{{\rm{H}}_2}{\rm{O}}}}$.Thus,

${m_{{\rm{meth}}}} = \left( {0.070 - x} \right){\rm{ g}}$

Concentration of water ${\left[ C \right]_1}$ can be written as ratio of mass of water $\left( {{m_{{{\rm{H}}_2}{\rm{O}}}}} \right)$and volume of water ${V_{{{\rm{H}}_2}{\rm{O}}}}$.

${\left[ C \right]_1} = \frac{{{m_{{{\rm{H}}_2}{\rm{O}}}}}}{{{V_{{{\rm{H}}_2}{\rm{O}}}}}}$

Substitute x grams for ${m_{{{\rm{H}}_2}{\rm{O}}}}$and 4.0 mL for ${V_{{{\rm{H}}_2}{\rm{O}}}}$.

${\left[ C \right]_1} = \frac{{x{\rm{ g}}}}{{4.0{\rm{ mL}}}}$

Concentration of methylene chloride ${\left[ C \right]_2}$ can be written as ratio of mass of methylene chloride $\left( {{m_{{\rm{meth}}}}} \right)$and volume of methylene chloride ${V_{{\rm{meth}}}}$.

${\left[ C \right]_2} = \frac{{{m_{{\rm{meth}}}}}}{{{V_{{\rm{meth}}}}}}$

Substitute $\left( {0.070 - x} \right){\rm{ g}}$ for ${m_{{\rm{meth}}}}$and 2.0 mL for ${V_{{\rm{meth}}}}$.

${\left[ C \right]_2} = \frac{{\left( {0.070 - x} \right){\rm{ g}}}}{{2.0{\rm{ mL}}}}$

The distribution coefficient of caffeine can be written as follow.

${K_D} = \frac{{{{\left[ C \right]}_2}}}{{{{\left[ C \right]}_1}}}$

Here ${\left[ C \right]_2}$is concentration of caffeine in methylene chloride and ${\left[ C \right]_1}$is concentration of caffeine in water.

Substitute $\frac{{x{\rm{ g}}}}{{4.0{\rm{ mL}}}}$for ${\left[ C \right]_1}$, $\frac{{\left( {0.070 - x} \right){\rm{ g}}}}{{2.0{\rm{ mL}}}}$ for ${\left[ C \right]_2}$and 4.6 for ${K_D}$. Thus,

$\begin{array}{c}\\4.6 = \frac{{\frac{{\left( {0.070 - x} \right){\rm{ g}}}}{{2.0{\rm{ mL}}}}}}{{\frac{{x{\rm{ g}}}}{{4.0{\rm{ mL}}}}}}\\\\4.6 = \left( {\frac{{\left( {0.070 - x} \right){\rm{ g}}}}{{2.0{\rm{ mL}}}}} \right)\left( {\frac{{4.0{\rm{ mL}}}}{{x{\rm{ g}}}}} \right)\\\\2.3x = 0.070 - x\\\\x = 0.021\\\end{array}$

Mass of water $\left( {{m_{{{\rm{H}}_2}{\rm{O}}}}} \right)$ is 0.021 grams and mass of methylene chloride is as follow.

${m_{{\rm{meth}}}} = \left( {0.070 - x} \right){\rm{ g}}$

Substitute 0.0212 g for x.

$\begin{array}{c}\\{m_{{\rm{meth}}}} = \left( {0.070 - 0.021} \right){\rm{ g}}\\\\{\rm{ = 0}}{\rm{.049 g}}\\\end{array}$

The total amount of caffeine that can be extracted into three portion of methylene chloride is 0.049 g.

Ans:

The amount of caffeine for the extraction of three portion of methylene chloride is 0.049 g.