Ka (HNO3) = [H3O][NO3] / [HNO3] = 20 Ka (NH4) = [H3O][NH3] / [NH4] = 5.6E-16 Ok
ID: 686445 • Letter: K
Question
Ka (HNO3) = [H3O][NO3] / [HNO3] = 20Ka (NH4) = [H3O][NH3] / [NH4] = 5.6E-16
Ok my question is where do they get the values of the productsand reactants to get these answers. I looked in my book where theionization constants are, but didn't find anything. Am I notlooking in the right place? Thank you. Need this so I can go to sleep.
Ka (NH4) = [H3O][NH3] / [NH4] = 5.6E-16
Ok my question is where do they get the values of the productsand reactants to get these answers. I looked in my book where theionization constants are, but didn't find anything. Am I notlooking in the right place? Thank you. Need this so I can go to sleep.
Explanation / Answer
We Know that : HNO3(aq) + H2O ---------> H3O+ (aq) + NO3-(aq) Ka = [H3O+] [NO3-]/ [HNO3] at standard temperature Ka = 2.4 x 101 NH4+ (aq) + H2O ---------> NH3 (aq)+ H3O+ (aq) Ka = [H3O+] [NH3] / [NH4+] at standard temperature Ka = 5.52 x 10-10 [ Kb of NH3 1.81 x 10^-5 Ka = Kw / Kb = 5.52 x10-10 ] NH4+ (aq) + H2O ---------> NH3 (aq)+ H3O+ (aq) Ka = [H3O+] [NH3] / [NH4+] at standard temperature Ka = 5.52 x 10-10 [ Kb of NH3 1.81 x 10^-5 Ka = Kw / Kb = 5.52 x10-10 ]Related Questions
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