part A) if Kb for NX3 is 2.0x10^-6 what is the pOH of a 0.175 M aqueous solution

Question

part A) if Kb for NX3 is 2.0x10^-6 what is the pOH of a 0.175 M aqueous solution of NX3?

part B) if Kb for NX3 is 2.0x10^-6 what percent ionization of a 0.325M aqueous solution of NX3?

part C) if KB for NX3 is 2.0x10^-6 what is the pKa for the following reaction: HNX3^+(aq)+H2O->NX3 (aq)+H3O^+(aq)

Queslion 16 Many common weak bases are derivatives of NH, where one or more of the hydrogen atoms have been replaced by another substituent Such reaclions can be generically symbolized as where NX3 is the base and HNX3t is the conjugate acid. The equilibrium constant expression for this reaction is where Kb is the base ionization constant The extent of ionization, and thus the strength of the base. increases as the value of K b increases Ka and Kb are related through the equalion As the strength of an acid increases, its Ka value increase and the strength of the conjugate base decreases (smaller Kb value)

Explanation / Answer

a)

NX3 --> HNX3 + OH- / NX3

Kb = [HNX3 ][OH-]/[NX3]

2.0*10^-6= (x*x)/(0.175-x)

x = OH = 5.91*10^-4

pOH = -log(OH) = -log( 5.91*10^-4) = 3.22841

b)

% ionizaiotn = [OH-] / M*100%

Kb = [HNX3 ][OH-]/[NX3]

Kb = x*x/(0.325-x)

2*10^-6= x*x/(0.325-x)

x = [OH-] = 8.05*10^-4

% ion = ( 8.05*10^-4 )/ 0.325 * 100 = 0.24769%

C)

Kw = KA*KB

Ka = Kw/Kb

Ka = (10^-14)/(2*10^-6) = 5*10^-9

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