91Ó°ÊÓ

$\mathrm{K}=3,{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}}=100\mathrm{mL},\left[\mathrm{butanoic}\mathrm{acid}\right]=0.1\mathrm{M},{\mathrm{V}}_{\mathrm{benzene}}=25\mathrm{mL},\mathrm{pH}=4,\mathrm{pH}=10.$

Distribution coefficient is given by

$\mathrm{D}=\frac{\mathrm{K}×\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}^{+}\right]+{\mathrm{K}}_{\mathrm{a}}}$

Substitute the given values are

localid="1654765945821" $$\begin{gathered} \mathrm{D}=\frac{3×{10}^{-4}}{{10}^{-4}+\left(1.52×{10}^{-5}\right)} \\ \mathrm{D}=2.6 \end{gathered}$$

The concentration of $\left[{\mathrm{H}}^{+}\right]$is calculated as ${10}^{-\mathrm{pH}}$

To calculate the fraction remaining in the water by using the formula as

$$\begin{gathered} \mathrm{q}=\frac{{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}}}{{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}}+\left(\mathrm{D}×{\mathrm{V}}_{\mathrm{benzene}}\right)} \\ \mathrm{q}=\frac{100\mathrm{mL}}{100\mathrm{mL}+\left(2.6×25\mathrm{mL}\right)} \\ \mathrm{q}=0.606 \end{gathered}$$

Molarity in water is

localid="1654769344387" $$\begin{gathered} \mathrm{q}×\left[\mathrm{butanoic}\mathrm{acid}\right]=0.606×0.1\mathrm{M} \\ \mathrm{q}×\left[\mathrm{butanoic}\mathrm{acid}\right]=0.0606\mathrm{M} \end{gathered}$$

The total number of moles of butanoic acid

localid="1654769351603" $$\begin{gathered} {\mathrm{n}}_{\mathrm{s}}=\left[\mathrm{butanoic}\mathrm{acid}\right]×{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}} \\ {\mathrm{n}}_{\mathrm{s}}=0.1\mathrm{M}×0.1\mathrm{L} \\ {\mathrm{n}}_{\mathrm{s}}=0.01\mathrm{mol} \end{gathered}$$

Molarity in benzene is; to determine the fraction in benzene as

localid="1654769103015" $$\begin{gathered} {\mathrm{q}}_2=1-0.606=0.394 \\ {\mathrm{q}}_2×\frac{{\mathrm{n}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{benzene}}}=0.394×\frac{0.01\mathrm{mol}}{0.025\mathrm{L}} \\ {\mathrm{q}}_2×\frac{{\mathrm{n}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{benzene}}}=0.16\mathrm{M} \end{gathered}$$

Distribution coefficient is given by

$\mathrm{D}=\frac{\mathrm{K}×\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}^{+}\right]+{\mathrm{K}}_{\mathrm{a}}}$

Substitute the given values are

localid="1654769394777" $\mathrm{D}=\frac{3×{10}^{-10}}{{10}^{-10}+\left(1.52×{10}^{-5}\right)}$

localid="1654769398481" $\mathrm{D}=1.97×{10}^{-5}$

The concentration of $\left[{\mathrm{H}}^{+}\right]$is calculated as ${10}^{-\mathrm{pH}}$

To calculate the fraction remaining in the water by using the formula as

localid="1654768500788" $$\begin{gathered} \mathrm{q}=\frac{{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}}}{{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}}+\left(\mathrm{D}×{\mathrm{V}}_{\mathrm{benzene}}\right)} \\ \mathrm{q}=\frac{100\mathrm{mL}}{100\mathrm{mL}+\left(\left(1.97×{10}^{-5}\right)×25\mathrm{mL}\right)} \\ \mathrm{q}=0.9999951 \end{gathered}$$

Molarity in water is

$$\begin{gathered} \mathrm{q}×\left[\mathrm{butanoic}\mathrm{acid}\right]=0.999×0.1\mathrm{M} \\ \mathrm{q}×\left[\mathrm{butanoic}\mathrm{acid}\right]=0.9999951\mathrm{M} \end{gathered}$$

The total number of moles of butanoic acid is

$$\begin{gathered} {\mathrm{n}}_{\mathrm{s}}=\left[\mathrm{butanoic}\mathrm{acid}\right]×{\mathrm{V}}_{\mathrm{butanoic}\mathrm{acid}} \\ {\mathrm{n}}_{\mathrm{s}}=0.1\mathrm{M}×0.1\mathrm{L} \\ {\mathrm{n}}_{\mathrm{s}}=0.01\mathrm{mol} \end{gathered}$$

Molarity in benzene is

$$\begin{gathered} {\mathrm{q}}_2=1-0.9999951 \\ {\mathrm{q}}_2=4.9×{10}^{-6} \end{gathered}$$

To calculate as

$$\begin{gathered} {\mathrm{q}}_2×\frac{{\mathrm{n}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{benzene}}}=\left(4.9×{10}^{-6}\right)×\frac{0.01\mathrm{mol}}{0.025\mathrm{L}} \\ {\mathrm{q}}_2×\frac{{\mathrm{n}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{benzene}}}=1.96×{10}^{-6}\mathrm{M} \end{gathered}$$