It has been shown that the undissociated form (i.e., the acid form) of acetic ac

Question

It has been shown that the undissociated form (i.e., the acid form) of acetic acid (HAc = CH3COOH) is toxic to the bacteria that make methane. The toxic level is reported to be 10 mg/L (= 1.67 x 10-4 M). The total acetic acid concentration (e.g., TOTAc) in an anaerobic digester is 6,000 mg/L as CH3COOH (= 0.10 M) and the pH is 6.5. Is the resulting level of HAc toxic to the bacteria? Show all calculations necessary to justify your answer. A yes or no without supporting calculations will receive no credit. The MW of CH3COOH is 60 g/mol.

Explanation / Answer

CH3COOH (aq) <---> CH3COO- (aq) + H+ (aq)

given pH = 6.5 , [H+] = 10^ -6.5 = 3.16 x 10^ -7 M = [CH3COO-]

Initially [CH3COOH] = 0.1 M , [H+] = [CH3COO-] = 0

at equilibrium [CH3COOH] = 0.1-X , [H+] =[CH3COO-] = X , where X is concentration fo CH3COOH dissociated

hence X = 3.16x10^ -7 M

hence [CH3COOH] = 0.1-3.16x10^-7 = 0.1   nearly

hence most CH3COOH is in undissociated form

Hence CH3COOH levels is above toxicity level ie 1.67x10^ -4 M

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