If K b for NX3 is 5.5×10 ?6 , what is the the p K a for the following reaction?

Question

If Kb for NX3 is 5.5×10?6 , what is the the pKa for the following reaction?

HNX3+(aq)+H2O(l)?NX3(aq)+H3O+(aq)

± Weak Base Calculations Correct Many common weak bases are derivatives of NH3 where one or more of the hydrogen atoms have been replaced by another substituent. Such reactions carn be Part B generically symbolized as If Kb, for NX3 is 5.5x10-6, what is the percent ionization of a 0.325 Maqueous solution of NX3? Express your answer numerically to three significant figures. NX3 (aq) + H2O(l) HNXs' (aq) + OH (aq) where NX3 is the base and HNXis the conjugate acid. The equilibrium-constant expression for this reaction is percent ionization = 0.411 % Submit Hints My Answers Give Up Review Part NX3] Correct where Kb is the base ionization constant. The extent of ionization, and thus the strength of the base increases as the value of Kb incre Since the Kb value has two signific ant figures, the percent ionization should actually have only two significant figures as well. It would be rounded to 0.41 % ases Ka and Kb are related through the equation Part C As the strength of an acid increases, its Ka value increase and the strength of the conjugate base decreases (smaller Kb value). If Kb for NX3 is 5.5x10-6, what is the the pKa for the following reaction? HNX3 + (aq) + H2O(l) NX3 (aq) + H30' (aq) Express your answer numerically to two decimal places. Submit Hints My Answers Give Up Review Part Incorrect; Iry Again, 4 attempts remaining

Explanation / Answer

Kb for NX3 is for the followin reaction ....

NX3 + H2O <---------> OH- + HNX3+ .....

Kb = [OH-] [HNX3+] / [NX3] = 5.5 X 10^-6

pure liquids like H2O are not written in Kb or Ka expression

and you have to find Ka for this :-

HNX3+ + H2O <-----------> NX3 + H3O+ ........

Ka = [NX3] [H3O+] / [HNX3+] = ?

now lets see what happen when you multiply Ka and Kb

Ka X Kb = [NX3] [H3O+] / [HNX3+] X [OH-] [HNX3+] / [NX3] = [H3O+] [OH-]

now if you recall [H3O+] [OH-] = Kw = 10^-14

so Ka X Kb = 10^-14

Ka = 10^-14/Kb = 10^-14 / (5.5 X 10^-6) =5.5 X 10^-9

and pKa = -log Ka = -log 5.5 X 10^-9 =

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