Question

In: Chemistry

To the nearest tenths, what is the equilibrium constant for the enzymatic hydrolysis of 0.1 M...

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.

Expert Solution

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][PO₄^3-] }/[A-O-PO₃ ^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-PO₃^2- 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^-5M

0.10M - x = 4.0.10^-5M

-x = (4.0.10^-5- 0.10)M

-x= -0.09996M

x= 0.09996M

For our convenience; it should be defined the K_Magain but using x at the equilibrium.

K_M= x.x/(0.10-x)

K_M = x²/(0.10-x) x= 0.09996M

K_M= 0.09996²M²/(0.10-0.09996)M= 249.8M

Finally: K_M= 249.8 M

queen_honey_blossom answered 2 years ago

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