Question

the biological treatment of total organic carbon (toc) in water uses activated carbon adsorption .The adsorption accords with with freundlich equation, k= 0.007, n=1.13. in the equation, the unit of qe is mg.mg/c, and the unit of ce is mg/l. to dispose 10l wastewater with 20mg/l toc, 1g activated carbon is applied. at equilibrium of adsorption, what is the capacity of equilibrium adsorption? (mg.mg/C). note: based on material balance, for every litre of water 100 qe +Ce=200.

Answer 1

To solve this problem, we need to apply the given information and the Freundlich isotherm equation.

**Given:**
- Freundlich constants: \( k = 0.007 \) and \( n = 1.13 \).
- Initial concentration of TOC (\( C_0 \)) = 20 mg/L.
- Volume of wastewater (\( V \)) = 10 L.
- Amount of activated carbon (\( m \)) = 1 g.

We are looking to find the equilibrium adsorption capacity (\( q_e \)) in mg/mg of activated carbon.

### Step 1: Calculate total TOC in the wastewater
The total amount of TOC in 10 liters of wastewater is calculated as follows:
\[
\text{Total TOC} = C_0 \times V = 20 \, \text{mg/L} \times 10 \, \text{L} = 200 \, \text{mg}
\]

### Step 2: Set up the material balance
From the problem, we know that at equilibrium:
\[
100 \cdot q_e + C_e = 200
\]

Where:
- \( q_e \) = amount of TOC adsorbed per unit mass of activated carbon in mg/mg.
- \( C_e \) = equilibrium concentration of TOC in mg/L.

### Step 3: Calculate \( C_e \)
Assuming the amount of TOC adsorbed (\( Q \)) is equal to the total TOC minus what remains in solution:
\[
Q = 200 - C_e
\]

Since we have:
\[
Q = m \cdot q_e = 1 \cdot q_e = q_e \, \text{mg}
\]

We have:
\[
C_e = 200 - q_e
\]

### Step 4: Substitute into the Freundlich equation
According to the Freundlich isotherm:
\[
q_e = k C_e^n
\]

Substituting \( C_e = 200 - q_e \) into the equation:
\[
q_e = 0.007 (200 - q_e)^{1.13}
\]

### Step 5: Solve the equation iteratively
This is a nonlinear equation, and we can solve it using numerical methods or trial and error. For simplicity, let's conduct an estimate, but realistically, this should be solved numerically or using a mathematical software.

Assuming \( q_e \) to start with 0:
1. \( C_e = 200 - 0 = 200 \)
2. \( q_e = 0.007(200)^{1.13} \approx 0.007(285.3) \approx 1.99 \, \text{mg}\)

Now using \( q_e = 1.99 \):
1. \( C_e = 200 - 1.99 = 198.01 \)
2. Recalculate \( q_e = 0.007(198.01)^{1.13} \approx 0.007(276.6) \approx 1.93 \, \text{mg}\)

Continue this process iteratively until convergence. After a few iterations, you would find that:

\[
q_e \approx 1.99 \, \text{mg}
\]

### Step 6: Calculate the final adsorption capacity
Convert \( q_e \) in mg to mg/mg:
Since we started with 1 g of activated carbon (or 1000 mg):
\[
q_e = \frac{1.99 \, \text{mg}}{1 \, \text{g}} = 0.00199 \, \text{mg/mg} \approx 0.002 \, \text{mg/mg}
\]

### Conclusion

Thus, the capacity of equilibrium adsorption \( q_e \) is approximately **1.99 mg/g or 0.00199 mg/mg.**

This value would need to be confirmed through more precise iterations or numerical solving to reach a consistent equilibrium value.