Question

Butanoic acid has a partition coefficient of 3.0 (favouring benzene) when distributed between water and benzene. Find the formal concentration of butanoic acid in each phase when 100mL of 0.10M aqueous butanoic acid is extracted with 25mL of benzene (a) at pH 4.00 and (b) at pH 10.00.

Short answer

$$\begin{gathered} \mathrm{a}){\mathrm{q}}_2\times \frac{{\mathrm{n}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{benzene}}}=0.16\mathrm{M} \\ \mathrm{b}){\mathrm{q}}_2\times \frac{{\mathrm{n}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{benzene}}}=1.96\times {10}^{-6}\mathrm{M} \end{gathered}$$

Step-by-step solution

Given Information:

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

To Find the formal concentration at  :

Distribution coefficient is given by

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

Substitute the given values are

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

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

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

Molarity in water is

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

The total number of moles of butanoic acid

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

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

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

To Find the formal concentration at pH=10  :

Distribution coefficient is given by

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

Substitute the given values are

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

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

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

Molarity in water is

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

The total number of moles of butanoic acid is

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

Molarity in benzene is

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

To calculate as

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