Calculate Adsorbed Phenol Concentration Using Freundlich Isotherm

According to the Freundlich adsorption isotherm for adsorption from solution: $\frac{x}{m} = k C^{1/n}$

where:

• $\frac{x}{m}$ is the concentration of adsorbed solute per unit mass of adsorbent (in $\text{mg g}^{-1}$),
• $C$ is the equilibrium concentration of the solution (in $\text{mg g}^{-1}$),
• $k$ and $n$ are constants.

Let $a = \frac{x}{m}$. Using the given data:

1. For $C_1 = 10\text{ mg g}^{-1}$, $a_1 = 4\text{ mg g}^{-1}$: $4 = k (10)^{1/n} \quad \text{--- (1)}$

2. For $C_2 = 16\text{ mg g}^{-1}$, $a_2 = 10\text{ mg g}^{-1}$: $10 = k (16)^{1/n} \quad \text{--- (2)}$

Dividing Equation (2) by Equation (1): $\frac{10}{4} = \left(\frac{16}{10}\right)^{1/n}$ $2.5 = (1.6)^{1/n}$

Taking $\log_{10}$ on both sides: $\log_{10}(2.5) = \frac{1}{n} \log_{10}(1.6)$

Using the given value $\log_{10} 2 = 0.3$: $\log_{10}(2.5) = \log_{10}\left(\frac{10}{4}\right) = \log_{10} 10 - \log_{10} 4 = 1 - 2\log_{10} 2 = 1 - 2(0.3) = 0.4$ $\log_{10}(1.6) = \log_{10}\left(\frac{16}{10}\right) = \log_{10} 16 - \log_{10} 10 = 4\log_{10} 2 - 1 = 4(0.3) - 1 = 0.2$

Substituting these values back: $0.4 = \frac{1}{n} (0.2) \implies \frac{1}{n} = \frac{0.4}{0.2} = 2$

Now, for $C_3 = 20\text{ mg g}^{-1}$, let $a_3$ be the concentration of adsorbed phenol: $a_3 = k (20)^{1/n} \quad \text{--- (3)}$

Dividing Equation (3) by Equation (1): $\frac{a_3}{a_1} = \left(\frac{C_3}{C_1}\right)^{1/n}$ $\frac{a_3}{4} = \left(\frac{20}{10}\right)^2 = 2^2 = 4$ $a_3 = 4 \times 4 = 16\text{ mg g}^{-1}$

The concentration of adsorbed phenol from $20\text{ mg g}^{-1}$ aqueous solution is 16.