Chapter 7, Problem 1Q

(a)

Interpretation Introduction

Interpretation:Weight of malononitrile recovered upon extraction with three $100\text{ mL}$ portions of ether should be calculated.

Concept introduction:The distribution coefficient is a parameter that indicates the extent of solubility of two immiscible components in two different phases. One of the phases is aqueous while the other is the organic phase.

Mathematically,

  $K_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}$

Answer to Problem 1Q

Weight of malononitrile recovered upon extraction with three $100\text{ mL}$ portions of ether is $29.2365\text{ g}$ .

Explanation of Solution

Since ether is an organic solvent, the amount of solute extracted from hexane indicates ${\left[\text{solute}\right]}_{\text{org}}$ .

Since the total amount of solute A in $100\text{ mL}$ of water is $10\text{ g}$ , the amount in aqueous layer is $\left(10-x\right)\text{ g}$ .

Total volume of solution is $100\text{ mL}$.

The expression fordistribution coefficient is given as:

  $K_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}$

Where,

• ${\left[\text{solute}\right]}_{\text{org}}$indicates solubility of solute in organic phase.
• ${\left[\text{solute}\right]}_{\text{aq}}$ indicates solubility of solute in aqueous phase.
• $K_{\text{D}}$ denotes equilibrium constant for the partition equilibrium.

The solubility of malononitrile in ether is $20\text{ g/}100\text{ mL}$ .

The solubility of malononitrile in water is $\text{13}\text{.3 g/}100\text{ mL}$ .

Substitute these values in above expression of $K_{\text{D}}$ .

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ =\frac{\frac{20\text{ g}}{100\text{ mL}}}{\frac{\text{13}\text{.3 g}}{100\text{ mL}}}\\ =1.5037\end{array}$

Solubility ofsolute A in organic phase is given as follows:

  ${\left[\text{solute}\right]}_{\text{org}}=\frac{x}{300}$

Solubility ofsolute A in aqueous phase is given as follows:

  ${\left[\text{solute}\right]}_{aq}=\frac{\left(30-x\right)}{300}$

Substitute the values in the above expression of distribution coefficient as follows:

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ 1.5037=\frac{\frac{x}{300}}{\frac{\left(30-x\right)}{300}}\end{array}$

Simplify the above expression as follows:

  $\begin{array}{c}1.5037=\frac{x}{30-x}\\ 45.111-1.5037x=x\end{array}$

Rearrange and obtain the value of $x$ .

  $\begin{array}{c}45.111=\text{ 2}\text{.5037}x\\ x=\frac{45.111}{\text{2}\text{.5037}}\\ =18.0177\text{ g}\end{array}$

For second extraction repeat the same procedure with volume being $\text{100 mL}$ .

The first cycle of extraction eliminates $\text{18}\text{.0177 g}$ out of organic layer and so $\text{11}\text{.98 g}$ is left in the aqueous layer.

Substitute the values in the above expression of distribution coefficient as follows:

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ 1.5037=\frac{\frac{x}{100}}{\frac{\left(11.98-x\right)}{100}}\end{array}$

Simplify the above expression as follows:

  $1.5037=\left(\frac{x}{11.98-x}\right)$

Rearrange and obtain the value of $x$ .

  $\begin{array}{c}x=18.014-1.5037x\\ 2.5037x=18.014\\ x=\frac{18.014}{2.5037}\\ =7.195\text{ g}\end{array}$

The second cycle of extraction eliminates $\text{7}\text{.195 g}$ and so $\text{4}\text{.785 g}$ is left in the aqueous layer.

For the next cycle of extraction the amount of solute gets changed from $\left(11.98-x\right)\text{ g}$ to $\left(4.785-x\right)\text{ g}$ . Thus the expression modifies as follows:

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ 1.5037=\frac{\frac{x}{300}}{\frac{\left(4.785-x\right)}{300}}\end{array}$

Simplify the above expression as follows:

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ 1.5037=\frac{\frac{x}{300}}{\frac{\left(4.785-x\right)}{300}}\\ 1.5037=\left(\frac{x}{4.785-x}\right)\end{array}$

Rearrange and obtain the value of $x$ .

  $\begin{array}{c}x=7.1952-1.5037x\\ 2.5037x=7.1952\\ x=\frac{7.1952}{2.5037}\\ =2.8738\text{ g}\end{array}$

The second cycle of extraction eliminates $2.8738\text{ g}$ and so $\text{1}\text{.9112 g}$ is left in the aqueous layer.

For the next cycle of extraction the amount of solute gets changed from $\left(11.98-x\right)\text{ g}$ to $\left(1.9112-x\right)\text{ g}$ . Thus the expression modifies as follows:

  $\begin{array}{c}{K}_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}\\ 1.5037=\frac{\frac{x}{300}}{\frac{\left(1.9112-x\right)}{300}}\end{array}$

Simplify the above expression as follows:

  $1.5037=\left(\frac{x}{1.9112-x}\right)$

Rearrange and obtain the value of $x$ .

  $\begin{array}{c}x=2.8738-1.5037x\\ 2.5037x=2.8738\\ x=\frac{2.8738}{2.5037}\\ =1.1478\text{ g}\end{array}$

Thus, the total weight of malononitrile that has been eliminated in four successive extractions with $25\text{ mL}$ portions of hexane is obtained by the sum of the weights obtained for each cycle.

  $\begin{array}{c}\text{Total weight extracted }=18.02\text{ g}+\text{ 7}\text{.1949 g}+2.8738\text{ g}+1.1478\text{ g}\\ =29.2365\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation:Weight of malononitrile recovered upon extraction with one $300\text{ mL}$ portions of ether should be calculated.

Concept Introduction:The distribution coefficient is a parameter that indicates the extent of solubility of two immiscible components in two different phases. One of the phases is aqueous, while the other is the organic phase.

Mathematically,

  $K_{\text{D}}=\frac{{\left[\text{solute}\right]}_{\text{org}}}{{\left[\text{solute}\right]}_{\text{aq}}}$

The expression to calculate fraction of solute left after $n$ extractions is given as:

  $q^n={\left(\frac{{\text{V}}_{\text{1}}}{{\text{V}}_1+K_{\text{d}}{\text{V}}_2}\right)}^n$

Answer to Problem 1Q

Weight of malononitrile recovered upon extraction with one $300\text{ mL}$ portion of ether is

found to be $29.6\text{ g}$ .

Explanation of Solution

The expression to calculate fraction of solute left after $n$ extractions is given as:

  $q^n={\left(\frac{{\text{V}}_{\text{1}}}{{\text{V}}_1+K_{\text{d}}{\text{V}}_2}\right)}^n$

Where,

• ${\text{V}}_1$denotes volume used for the extraction cycle;
• $K_{\text{d}}$ denotes distribution coefficient ;
• $n$ represents the number of cycles of extraction carried out;
• $q$ denotes the fraction of weight that remains after $n$ extractions.

The volume used for the extraction cycle is $300\text{ mL}$ .

  $K_{\text{d}}$ is found to be 1.5037.

Since only single extraction is carried out with $300\text{ mL}$ portion of ether,the value of $n$ is 1.

Substitute these values in above expression to calculate the fraction of malononitrile recovered as:

  $\begin{array}{c}{q}^1={\left(\frac{300}{\left(300\right)+\left(1.5037\right)\left(300\right)}\right)}^1\\ =0.399\text{ g}\end{array}$

Thus, weight of malononitrile recovered upon extraction with one $300\text{ mL}$ portion of ether is found as follows:

  $\begin{array}{c}\text{weight of malononitrile recovered}=\left(30\text{ g}\right)-\left(0.40\text{ g}\right)\\ =29.6\text{ g}\end{array}$