Question

What is Kc for the following equilibrium? For phosphoric acid, Ka1=6.9x10^-3, Ka2=6.2x10^-8 and Ka3=4.8x10^-13
HPO4^-2 + OH^- <==> PO4^-3 + H2O

Answer 1

Sara, this is a reverse hydrolysis problem. See your problem above for how that works.
So Kb for PO4^3- = (Kw/k3 for H3PO4) = ?
Then for the reverse hydrolysis it will be Kc = 1/Kb.

Answer 2

I plug in the numbers put it doesn't match any of the choices.... I get 4.81x10^-29
The choices are
A)2.3x10^-35
B)6.4x10^-23
C)7.3x10^-16
D)48
E)491
Please help!!

Answer 3

I don't know which buttons you punched but when I crunched the numbers I got 48 and that is one of the answers listed. I'd be interested in knowing what you did? Did I say something that was difficult to understand?
Kb phosphate = (kw/k3) = 1E-14/4.8E-13 = 0.02083 = Kb for phosphate.
1/Kb = 1/0.0203 = 48.0

Answer 4

I don't press the E button i press the 10^x
Button but when I divided the kw/Kb it gives me 2.083x10^-28

Answer 5

Take a look at your calculator instruction manual for dividing one number by the other.
(Kw/k3) = 1 x 10^-14/4.8 x 10^-13 = 0.02038
1/0.02038 = 48.0