Question

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$

Hint:

Freundlich isotherm: \(x/m = k C^{1/n}\). Use logarithms to find \(n\) and \(k\) from two data points, then predict adsorption at any concentration.

Answer 1

Adsorption obeys Freundlich isotherm, which can be written as: \[ x/m = k C^{1/n} \] where, \( x/m \) = amount adsorbed per unit mass of adsorbent (mg/g), \( C \) = equilibrium concentration of adsorbate in solution (mg/L or mg/g), \( k \), \( n \) = constants depending on temperature and adsorbate-adsorbent system. Given data: \[ \begin{cases} C_1 = 10 \text{ mg/g}, & (x/m)_1 = 4 \text{ mg/g} \\ C_2 = 16 \text{ mg/g}, & (x/m)_2 = 10 \text{ mg/g} \end{cases} \] From Freundlich isotherm, for the two points: \[ (x/m)_1 = k C_1^{1/n} \quad \Rightarrow \quad 4 = k \times 10^{1/n} \] \[ (x/m)_2 = k C_2^{1/n} \quad \Rightarrow \quad 10 = k \times 16^{1/n} \] Dividing the two equations to eliminate \(k\): \[ \frac{10}{4} = \frac{16^{1/n}}{10^{1/n}} = \left(\frac{16}{10}\right)^{1/n} = \left(\frac{8}{5}\right)^{1/n} \] Taking logarithm base 10 on both sides: \[ \log_{10}\left(\frac{10}{4}\right) = \frac{1}{n} \log_{10}\left(\frac{8}{5}\right) \] Calculate the logs: \[ \log_{10} \frac{10}{4} = \log_{10} 2.5 = \log_{10}(5/2) = \log_{10} 5 - \log_{10} 2 \] Using \(\log_{10} 2 = 0.3\) and \(\log_{10} 5 = \log_{10}(10/2) = 1 - 0.3 = 0.7\), so: \[ \log_{10} 2.5 = 0.7 - 0.3 = 0.4 \] Similarly, \[ \log_{10} \frac{8}{5} = \log_{10} 8 - \log_{10} 5 \] \[ \log_{10} 8 = \log_{10}(2^3) = 3 \times 0.3 = 0.9 \] \[ \log_{10} 5 = 0.7 \] \[ \log_{10} \frac{8}{5} = 0.9 - 0.7 = 0.2 \] So, \[ 0.4 = \frac{1}{n} \times 0.2 \quad \Rightarrow \quad \frac{1}{n} = \frac{0.4}{0.2} = 2 \] \[ \therefore n = \frac{1}{2} = 0.5 \] Now, find \(k\) using one of the points, say \(C_1=10\), \((x/m)_1=4\): \[ 4 = k \times 10^{1/n} = k \times 10^{2} = k \times 100 \] \[ k = \frac{4}{100} = 0.04 \] We are asked to find adsorbed amount \(x/m\) when \(C = 20\) mg/g: \[ x/m = k C^{1/n} = 0.04 \times 20^{2} = 0.04 \times 400 = 16 \text{ mg/g} \] % Final answer Answer: \(\boxed{16 \text{ mg/g}}\)