91Ó°ÊÓ

(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)

• 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.

(c)

(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:

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

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 $\left[{\mathrm{H}}^{+}\right]$