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equilibrium constant

Submitted by Anonymous (not verified) on Thu, 05/20/2010 - 06:08

Given the following table of equilibrium constants at 25oC:
H3PO4  H+ + H2PO4-        6.9 x 10-3
H+ + PO43-  HPO42-        2.1 x 1012
H2PO4-  HPO42- + H+        6.2 x 10-8
calculate the equilibrium constant for the following reaction.
H3PO4 + 3 H2O  3H3O+ + PO43-

Note the water in this equation is a solvent in a dilute solution.

a. 2.1 x 10-22

b. 3.3 x 10-15

c. 5.2 x 10-9

d. 4.3 x 10-10

e. 9.0 x 102


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do i need to manipulate the given equilibrum equations to match our unknown equation???

cheers


You need to manipulate the equations to get the desired equation. If you reverse an equation, Kc becomes 1/Kc
Also H+ is also the same as H3O+

H3PO4 + H2O ---> H3O+ + H2PO4-      6.9 x 10-3
H3O+ + PO43-  --> HPO42- + H2O      2.1 x 1012  = reverse this one to get HPO42- on the LHS
H2PO4- + H2O ---> HPO42- + H3O+        6.2 x 10-8
calculate the equilibrium constant for the following reaction.
H3PO4 + 3 H2O  3H3O+ + PO43-

H3PO4  + H2O---> H3O+ + H2PO4-        6.9 x 10-3
HPO42- + H2O ---> H3O+ + PO43-          1/2.1 x 1012
H2PO4- + H2O ---> HPO42- + H3O+        6.2 x 10-8

Adding the equation gives
H3PO4  + H2O + HPO42- + H2O + H2PO4- + H2O  ---> H3O+ + H2PO4- + H3O+ + PO43- + HPO42- + H3O+

H3PO4  + 3H2O  ---> 3H3O+ + PO43-

Kc = 6.9 x 10-3 x 6.2 x 10-8/2.1 x 1012