At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and the pH of the resulting solution is 3.0. Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm,

$$\dfrac{x}{m}=kC^{1/n}$$

If the plot of $$\log_{10}(x/m)$$ against $$\log_{10}C$$ gives a straight line with slope 1, the value of $$k$$ in L mol$$^{-1}$$ is ___.

Given: The molar mass of acetic acid is 60 g mol$$^{-1}$$.
The acid dissociation constant of acetic acid is $$1.0\times 10^{-5}$$ at the given temperature.
$$x$$ is the mass (in grams) of acetic acid adsorbed. $$m$$ is the mass (in grams) of charcoal.
$$C$$ is the equilibrium concentration of acetic acid in the solution after the adsorption is complete.
$$k$$ and $$n$$ are constants for acetic acid$$-$$charcoal system at the given temperature.

The initial mass of acetic acid is $$0.45\ \text{g}$$. With molar mass $$60\ \text{g mol}^{-1}$$, the initial number of moles is

$$n_0=\frac{0.45}{60}=0.0075\ \text{mol}$$

The solution volume is $$50\ \text{mL}=0.05\ \text{L}$$, so the initial concentration is

$$C_0=\frac{n_0}{0.05}=0.15\ \text{mol L}^{-1}$$

After the mixture is shaken with $$1.0\ \text{g}$$ of charcoal, the pH of the equilibrium solution is 3.0, i.e.

$$[H^+]=10^{-3}\ \text{mol L}^{-1}$$

For a weak monoprotic acid $$\mathrm{HA}$$ of equilibrium concentration $$C$$,

$$K_a=\frac{[H^+]^2}{C-[H^+]} \quad -(1)$$

Using $$K_a = 1.0\times10^{-5}$$ and $$[H^+]=10^{-3}$$, equation $$-(1)$$ gives

$$1.0\times10^{-5}=\frac{(10^{-3})^2}{C-10^{-3}}$$

$$\Rightarrow\;1.0\times10^{-5}(C-10^{-3})=10^{-6}$$

$$\Rightarrow\;C-10^{-3}=0.10$$

$$\Rightarrow\;C\approx0.101\ \text{mol L}^{-1}\;(\text{very close to }0.10\ \text{mol L}^{-1})$$

Moles of acetic acid that remain in solution:

$$n_{\text{eq}} = C \times 0.05 \approx 0.005\ \text{mol}$$

Moles adsorbed on charcoal:

$$n_{\text{ads}} = n_0 - n_{\text{eq}} = 0.0075-0.005 = 0.0025\ \text{mol}$$

Mass adsorbed:

$$x = n_{\text{ads}}\times60 = 0.0025\times60 = 0.15\ \text{g}$$

Given $$m = 1.0\ \text{g}$$ of charcoal,

$$\frac{x}{m} = \frac{0.15}{1.0}=0.15$$

The Freundlich isotherm is $$\dfrac{x}{m}=kC^{1/n}$$. The slope of the plot of $$\log_{10}(x/m)$$ versus $$\log_{10}C$$ is the exponent $$1/n$$. Because the slope is given to be 1,

$$\frac{1}{n}=1\;\Rightarrow\;n=1$$

Hence the equation reduces to

$$\frac{x}{m}=kC$$

Therefore,

$$k=\frac{x/m}{C}=\frac{0.15}{0.10}=1.50\ \text{L mol}^{-1}$$

Final answer: $$1.50$$