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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.

1 answer

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.