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 0.04% of the original A-O-PO32- remained after reaching equilibrium and the activity of water is unity. Assume a pH of 7.0.

Answer 1

In firt place, it should be defined the equilibrium constant in molar concentration because the only clue in this exercise is the initial concentration [A-O-PO₃^2-]​= 0.10M.

K_M​ =​ {[AOH][PO_4​^3-]​ ​}/[A-O-PO_3​​ ^2-]

Next, it should be written this reaction to make its equilibrium scheme:

A-O-PO3​2-(ac)​ + H2O(l)           => AOH(ac)​+ PO4​3-​​​(ac)​ ​​

Water concentration remains constant and the sheme is:

	[A-OPO3​2- ​]​ (M)	[AOH] (M)	[PO4​3-]​ (M)
Initial	0.10	0.00	0.00
Change	-x   (because it decreases)	+x (because it increases)	+x (because it increases)
At equilibrium concentration	0.10-x	+x	+x

If the initial concentration of A-O-PO_3​^2-​ is 0.10M and remains 0.04% at equilibrium, it must be calculated its final concentration.

0.10M _________100%

y        __________0.04%

y= 0.04x0.10M/100= 4.0.10^-5​M

​0.10M - x = 4.0.10^-5​M

​-x = (4.0.10^-5​- 0.10)M

-x= -0.09996M

x= 0.09996M

For our convenience; it should be defined the K_{M ​}again but using x at the equilibrium.

K_M​= x.x/(0.10-x)

​K_M =​ ​x^2​/(0.10-x)     x= 0.09996M​​

K_M​= 0.09996^2​M^2​/(0.10-0.09996)M= 249.8M​​​

Finally:   K_M​= 249.8 M​_​

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