An anaerobic biological treatment process is being designed to treat a solution

Question

An anaerobic biological treatment process is being designed to treat a solution containing 4000 mg/L HAc (acetic acid). The bacteria can concert acetic acid to carbon dioxide and methane by the following reactions, but oly if the pH is above 5.2.

CH3COOH+H2O --> CH4+H2CO3

CH3COO-+H2O --> CH4+HCO3-

a. Prepare a pC-pH diagram for the solution prior to treatment. What is the initial pH?

b. What concentration of NaOH must be added to adjust the solution to pH 5.2? What concentration of Na2CO3 could be used as an alternative?

c. It is proposed to add 6x10-2 M NaOH to the original solution, after which the reaction is expected to convert 80% of the total acetate to end products. Prepare a pC-pH diagram for the resultant solution and find the pH before and after the biological conversion.

Explanation / Answer

a.

4 g/L means 4 g/L / 60 g/mol = 0.0667 mol/L

pKa = 4.76

pH initial = 0.5pKa – 0.5logC

pH initial = 2.4 – 0.5logC     or    pH initial = 2.4 + 0.5 pC

This is the function for your plot (pC means –logC)

For the plot calculate pH for C= 0.01 and 0.02 and 0.05 and 0.10

pH initial = 2.4 – 0.5logC

               = 2.4 – o.5 log0.0667 = 2.4+0.59 = 3.0

b.

At pH=5.2 you have the buffer

CH3COOH/CH3COO-              pKa = 4.76

pH = pKa + log ([CH3COO-]/[ CH3COOH])

log ([CH3COO-]/[ CH3COOH]) = pH – pKa = 5.2 – 4.8 = 0.4

[CH3COO-]/[ CH3COOH] = 100.4 = 2.5 or 5/2

The proportion of neutralization is 5/(5+2) = 0.71

Add enough NaOH for a concentration

0.0667 mol/L x 0.71 = 0.048 mol/L NaOH      (reaction 1:1)

CO32- reacts also 1:1 with acetic acid that is not in excess.

Keep the same result 0.048 mol/L   CO32-   

c.

pH = pKa + log ([CH3COO-]/[ CH3COOH])

pH = pKa + log (0.8/0.2)

pH = 4.8 +log4

      = 5.4 before biological conversion.

“Prepare a pC-pH diagram for the resultant solution and find the pH before and after the biological conversion.”

After complete biological conversion

[H2CO3] = 0.2x 0.0667 mol/L = 0.0133 mol/L

[HCO3-] = 0.8 x 0.0667 mol/L = 0.0533 mol/L

pKa2 H2CO3 = 10.33

pH = pKa + log ([HCO3-]/[ H2CO3])

      = 10.33+ log 4 = 10.63 final

If the diagram is for incomplete conversion calculate pH for

ratio [HCO3-]/[ H2CO3] = o.2 and 0.4 and 0.6 and 0.8.

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