Calculation of Freundlich Constant k for Acetic Acid Adsorption

To find the value of the Freundlich constant $k$, we follow a step-by-step approach based on chemical equilibrium and adsorption principles:

1. Initial Moles of Acetic Acid

The initial mass of acetic acid ($\text{CH}_3\text{COOH}$) is $m_{\text{initial}} = 0.45\text{ g}$, and its molar mass is $M = 60\text{ g mol}^{-1}$.

$\text{Initial moles of acetic acid, } n_{\text{initial}} = \frac{0.45\text{ g}}{60\text{ g mol}^{-1}} = 0.0075\text{ mol}$

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2. Equilibrium Concentration of Acetic Acid in Solution

The $\text{pH}$ of the solution after equilibrium is reached is $3.0$. Therefore: $[\text{H}^+] = 10^{-\text{pH}} = 10^{-3}\text{ M}$

Acetic acid dissociates according to the reaction: $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$

The acid dissociation constant expression is: $K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$

Since $[\text{CH}_3\text{COO}^-] = [\text{H}^+] = 10^{-3}\text{ M}$ and $K_a = 1.0 \times 10^{-5}$: $1.0 \times 10^{-5} = \frac{(10^{-3})^2}{[\text{CH}_3\text{COOH}]}$

$[\text{CH}_3\text{COOH}] = \frac{10^{-6}}{10^{-5}} = 0.1\text{ M}$

Thus, the equilibrium concentration of acetic acid in the solution is $C = 0.1\text{ M}$ (or $C_{\text{total}} = [\text{CH}_3\text{COOH}] + [\text{CH}_3\text{COO}^-] = 0.1 + 0.001 = 0.101\text{ M}$).

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3. Amount of Acetic Acid Adsorbed ($x$)

The volume of the solution is $V = 50\text{ mL} = 0.050\text{ L}$.

• Moles of acetic acid remaining in solution: $n_{\text{remaining}} = C \times V = 0.1\text{ mol L}^{-1} \times 0.050\text{ L} = 0.0050\text{ mol}$

• Moles of acetic acid adsorbed: $n_{\text{adsorbed}} = n_{\text{initial}} - n_{\text{remaining}} = 0.0075\text{ mol} - 0.0050\text{ mol} = 0.0025\text{ mol}$

• Mass of acetic acid adsorbed ($x$): $x = n_{\text{adsorbed}} \times M = 0.0025\text{ mol} \times 60\text{ g mol}^{-1} = 0.15\text{ g}$

(Note: Considering complete dissociation $[\text{CH}_3\text{COO}^-]$, $n_{\text{remaining}} = 0.00505\text{ mol} \implies x = 0.147\text{ g}$)

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4. Calculation of Freundlich Constant $k$

The Freundlich adsorption isotherm is given by: $\frac{x}{m} = k C^{1/n}$

Taking $\log_{10}$ on both sides: $\log_{10}\left(\frac{x}{m}\right) = \log_{10} k + \frac{1}{n}\log_{10} C$

Given that the plot of $\log_{10}(x/m)$ against $\log_{10} C$ gives a straight line with a slope equal to $1$: $\frac{1}{n} = 1 \implies n = 1$

Therefore, the isotherm simplifies to: $\frac{x}{m} = k C \implies k = \frac{x}{m \cdot C}$

Given $m = 1.0\text{ g}$ of charcoal: $k = \frac{0.15\text{ g}}{1.0\text{ g} \times 0.1\text{ mol L}^{-1}} = 1.5\text{ L mol}^{-1}$

(Including the minor contribution of dissociated acid yields $k \approx 1.47\text{ L mol}^{-1}$, which is also within the accepted range).

Final Answer: The value of $k$ is 1.5 (acceptable range: 1.45 to 1.52).