If Kb for NX3 is 8.5×10−6 , what is the the pKa for the following reaction ?

Question

HNX3² (aq) +H2O (l) ⇌ NX3 (aq) + H3O⁺(aq)

Answer 1

Not sure what you mean by HNX₃^2, so I'll assume you meant HNX₃⁺. That being the case, we see that

HNX₃⁺ is acting as an acid by donating a proton (H⁺) to H₂O. Thus, to find the pKa of HNX₃⁺, we first find the Ka and then take the negative log:

Kw = 1x10^-14 = KaKb

Ka = 1x10^-14 / Kb = 1x10^-14 / 8.5x10^-6

Ka = 1.2x10^-9

pKa = -log 1.2x10^-9

pKa = 8.9

Another way to find the same answer, is to use the relationship of pKa + pKb = 14

pKb = -log 8.5x10^-6 = 5.1

pKa = 14 - 5.1 = 8.9