Question

The weak base $\mathbf{B}\left(K_b=1.0\times {10}^{-5}\right)$equilibrates between water (phase 1) and benzene (phase 2)

(a) Define the distribution coefficient, D for this system.

(b) Explain the difference between D and K the partition coefficient.

(c) Calculate D at $\mathbf{pH}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{if}\mathbf{}\mathbf{k}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}$

(d) Will D be greater or less at $\mathbf{pH}\mathbf{10}\mathbf{}\mathbf{than}\mathbf{}\mathbf{at}\mathbf{}\mathbf{pH}\mathbf{8}$? Explain why.

Short answer

• In this task, we have a weak base $\mathrm{B}({\mathrm{K}}_{\mathrm{b}}=1.0\times {10}^{-5})$that equilibrates between water (phase 1) and benzene (phase 2) .

Step-by-step solution

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(a)

• Let us write distribution coefficient for this system.

• The distribution coefficient, D for this system is$\mathrm{D}=\frac{\left[\mathrm{B}\right]{\mathrm{C}}_6{\mathrm{H}}_6}{{\left[\mathrm{B}\right]}_{{\mathrm{H}}_2}\mathrm{O}+{\left[{\mathrm{BH}}^{+}\right]}_{{\mathrm{H}}_2\mathrm{O}}}$

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(b)

• From the distribution coefficient$\mathrm{D}=\frac{\left[\mathrm{B}\right]{\mathrm{C}}_6{\mathrm{H}}_6}{{\left[\mathrm{B}\right]}_{{\mathrm{H}}_2}\mathrm{O}+{\left[{\mathrm{BH}}^{+}\right]}_{{\mathrm{H}}_2\mathrm{O}}}$, let us find the difference between D and K the partition coefficient.
• Here the difference is distribution coefficient D- quotient of total concentrations of species in phases
• Partition coefficient K - quotient of neutral species concentrations {B} in phases.

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(c)

• Let us recall the following equation before moving on to solve the (c) part. $$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{K}}_{\mathrm{b}}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{10}^{-14}}{{10}^{-5}} \\ {\mathrm{K}}_{\mathrm{a}}={10}^{-9} \end{gathered}$$

• Using the and calculate the concentration of,

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}} \\ \left[{\mathrm{H}}^{+}\right]={10}^{-8} \end{gathered}$$

• Now, we shall calculate at if while using the equation below.

$$\begin{gathered} \mathrm{D}=\frac{\mathrm{K}\times \mathrm{K}}{{\mathrm{K}}_{\mathrm{a}}+\left[{\mathrm{H}}^{+}\right]} \\ \mathrm{D}=\frac{50\times \left({10}^{-9}\right)}{{10}^{-9}+{10}^{-8}} \\ \mathrm{D}=4.545 \end{gathered}$$

• Calculate D at pH 8.00 if K = 50.0 , we get D = 4.54

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(d)

• Here, let us explain will be greater or less at pH = 10 than at pH = 8 .
• D would be greater as it is inversely proportional to the concentration of $\left[{\mathrm{H}}^{+}\right]$
• This can be better explained by calculating the distribution coefficient:

$$\begin{gathered} \mathrm{D}=\frac{50\times \left[{10}^{-9}\right]}{{10}^{-9}+{10}^{-8}} \\ \mathrm{D}=\frac{\mathrm{K}\times \mathrm{K}}{{\mathrm{K}}_{\mathrm{a}}+\left[{\mathrm{H}}^{+}\right]} \\ \mathrm{D}=\frac{50\times \left({10}^{-9}\right)}{{10}^{-9}+{10}^{-10}} \\ \mathrm{D}=45.45 \end{gathered}$$

Result:

• (a) The distribution coefficient, D for this system is $\mathrm{D}=\frac{\left[\mathrm{B}\right]{\mathrm{C}}_6{\mathrm{H}}_6}{{\left[\mathrm{B}\right]}_{{\mathrm{H}}_2}\mathrm{O}+{\left[{\mathrm{BH}}^{+}\right]}_{{\mathrm{H}}_2\mathrm{O}}}$.
• (b) distribution coefficient D - quotient of total concentrations of species in phases
• Partition coefficient - quotient of neutral species concentrations in phases.
• (c) D at pH 8.00 if K = 50.0 is 4.545 .
• (d) D would be greater as it is inversely proportional to the concentration of $\left[{\mathrm{H}}^{+}\right]$