Question

To the nearest tenths, what is the equilibrium constant for the enzymatic hydrolysis of 0.1 M A-O-PO32- to A-OH and inorganic phosphate (where A is some generalized molecule) given that 0.06% of the original A-O-PO32- remained after reaching equilibrium and the activity of water is unity.

Answer 1

given the reaction is

A-0-P032- + H20 --> A-OH + inorganic phosphate

given

0.06 % of original A-0-P032- remained

so

A-0-P032- reacted = 0.1 - ( 6 x 10-5) = 0.09994

now

from the reaction we can see that

A-OH formed = A-0-P032- reacted = 0.09994

inorganic phosphate formed = A-O-P032- reacted = 0.09994

now

A-O-P032- + H20 --> A-OH + inorganic phosphate

the equilibrium constant is given by

Keq = [A-OH] [inorganic phosphate ] / [A-0-P032-]

so

Keq = 0.09994 x 0.09994 / 6 x 10-5

Keq = 166.47

so

the value of equilibrium constant is 166.47