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: $$\log_{10} 2 = 0.3$$

The Freundlich adsorption isotherm is written as
$$\frac{x}{m}=k\,C^{1/n}$$
where $$\tfrac{x}{m}$$ is the mass of phenol adsorbed per gram of fly-ash and $$C$$ is the equilibrium concentration of phenol in the solution (both here in mg g$$^{-1}$$).

Taking common logarithm (base 10):
$$\log\!\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log C \quad -(1)$$

The two experimental data pairs are

$$C_1=10\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_1 = 4\;\text{mg g}^{-1}$$
$$C_2=16\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_2 = 10\;\text{mg g}^{-1}$$

Substituting in equation $$-(1)$$:

For the first pair:
$$\log 4 = \log k + \frac{1}{n}\,\log 10 \quad -(2)$$

For the second pair:
$$\log 10 = \log k + \frac{1}{n}\,\log 16 \quad -(3)$$

Using the given value $$\log 2 = 0.3$$:

$$\log 4 = \log(2^2)=2\times0.3 = 0.6$$
$$\log 10 = 1$$
$$\log 16 = \log(2^4)=4\times0.3 = 1.2$$

Equations $$-(2)$$ and $$-(3)$$ become

$$0.6 = \log k + \frac{1}{n}(1) \quad -(4)$$
$$1.0 = \log k + \frac{1}{n}(1.2) \quad -(5)$$

Subtract $$-(4)$$ from $$-(5)$$:

$$1.0 - 0.6 = \frac{1}{n}(1.2 - 1.0)$$
$$0.4 = \frac{1}{n}\times0.2$$
$$\frac{1}{n} = 2 \;\;\Longrightarrow\;\; n = 0.5$$

From equation $$-(4)$$:
$$\log k = 0.6 - \frac{1}{n}(1) = 0.6 - 2 = -1.4$$
Hence $$k = 10^{-1.4}$$

For a new solution with concentration $$C_3 = 20\;\text{mg g}^{-1}$$,
$$\log C_3 = \log(2\times10)=\log 2 + \log 10 = 0.3 + 1 = 1.3$$

Using equation $$-(1)$$:
$$\log\!\left(\frac{x}{m}\right)_3 = -1.4 + 2 \times 1.3 = -1.4 + 2.6 = 1.2$$

Therefore,
$$\left(\frac{x}{m}\right)_3 = 10^{1.2}$$

An antilog of $$1.2$$ is
$$10^{1.2} = 10^{1}\times10^{0.2} \approx 10 \times 1.58 \approx 15.8\;\text{mg g}^{-1}$$

Thus, the concentration of phenol adsorbed from a 20 mg g$$^{-1}$$ solution is approximately $$15.8\;\text{mg g}^{-1}$$, which lies in the required range 15.5 - 16.5.