Question

The equilibrium constant for the reaction H2O⇄H++OH-is called KW and has the value 1.0×10-14 at 25°C...

The equilibrium constant for the reaction H2O⇄H++OH-is called KW and has the value 1.0×10-14 at 25°C Given that K NH3= 1.8×10-5 for the reaction NH3(aq) +H2O⇄NH4++ OH-, find K for the reaction NH4+⇄NH3(aq)+ H+.

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Answer #1

from the given data

             H2O <=====> H+ + OH-   Kw = 1*10^-14

       NH4++ OH- <=====> NH3(aq) + H2O (1/KNH3 = (1/(1.8*10^-5))
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             NH4+ <====> NH3(aq) + H+ K = Kw*(1/KNH3)
--------------------------------------------
     K = Kw*(1/KNH3)

       = ( 1*10^-14)*(1/(1.8*10^-5))

       = 5.55 *10^-10

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