The equilibrium constant for the reaction H2O Arrows H OH i

The equilibrium constant for the reaction H2O (Arrows) (H+) + (OH-) is 0.1*10^-14 at 25 degrees C. What is the value of K for the reaction 4H2O (Arrows) 4(H+) + 4(OH-)?

H2O (l) <---> H+ + OH-

Kw = [H+] [OH-] = 0.1 x 10-14

now

4 H2O (l) <--> 4 H+ + 4OH-

K = [H+]^4 [OH-]^4

K = ( [H+] [OH-])^4

K = ( Kw)^4

K = ( 0.1 x 10-14)^4

K = 1 x 10^-60

The equilibrium constant for the reaction H2O (Arrows) (H+) + (OH-) is 0.1*10^-14 at 25 degrees C. What is the value of K for the reaction 4H2O (Arrows) 4(H+) +

The equilibrium constant for the reaction H2O Arrows H OH i

Solution

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