Question

The enzyme acetyl-CoA synthetase does the two step reaction below to attach acetate to coenzyme A. The first step requires nucleophilic attack by the deprotonated carboxylic acid of acetate on ATP. Assume the cell cytosol is at pH 6.76 and the pKa of acetic acid/acetate is 4.76. Question 1A: what percentage of the acetic acid/acetate conjugate pair is in the deprotonated acetate state? Question 1B: What would the pH of the cell be, if 50% of the acetate is protonated?

Answer 1

1A) given the pH = -log[H+(aq)] = 6.76

=> [H+(aq)] = 10^-6.76 = 1.74x10^-7

The dissociation reaction of CH3COOH is

CH3COOH <-------> CH3COO^-(aq) + H+(aq), Ka = 10^-4.76 = 1.74x10^-5

Ka = 1.74x10^-5 = [CH3COO^-(aq)]x[H+(aq)] / [CH3COOH]

=> [CH3COO^-(aq)] / [CH3COOH] = 1.74x10^-5 / [H+(aq)] = 1.74x10^-5 /1.74x10^-7 = 100

=> [CH3COO^-(aq)] = 100x [CH3COOH] -------(1)

Hence  percentage of the acetic acid/acetate conjugate pair in the deprotonated acetate state

= [CH3COO^-(aq)] / ([CH3COO^-(aq)] + [CH3COOH(aq)])

= 100x [CH3COOH] / (100x [CH3COOH] + [CH3COOH(aq)])

= 100 / 101 = 0.99 = 99 % (answer)

1B): When 50% acetate is deprotonated, [CH3COO^-(aq)] = [CH3COOH]

=>Ka = 1.74x10^-5 = [CH3COO^-(aq)]x[H+(aq)] / [CH3COOH] = [H+(aq)]

=> [H+(aq)] = 1.74x10^-5

=> pH = - log[H+(aq)] = - log(1.74x10^-5 ) = 4.76 (answer)