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Given that Ka for HBrO is 2.8 x 109 at 25 oC, what is the value of Kb for Bro at 25 oC? Number Given that Kb for C6H5NH2 is 1.7 x 109 at 25 oC, what is the value of Ka for C6H5NH3 at 25 oC? Number

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Answer #1

Kw = Ka x Kb

Kb = Kw /Ka

Kb = 1.0 x 10-14  / 2.8 x 10-9

Kb = 3.57 x 10-6

2) Kw = Ka x Kb

Ka = Kw / Kb

Ka = 1.0 x 10-14 / 1.7 x 10-9

Ka = 5.9 x 10-6

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