NX3(aq)+H2O(l)HNX3+(aq)+OH(aq) K b=[HNX3+][OH][NX3] where K b is the base ioniza

Question

NX3(aq)+H2O(l)HNX3+(aq)+OH(aq)

Kb=[HNX3+][OH][NX3]

where Kb is the base ionization constant. The extent of ionization, and thus the strength of the base, increases as the value of Kb increases.

Ka and Kb are related through the equation

Ka×Kb=Kw

As the strength of an acid increases, its Ka value increase and the strength of the conjugate base decreases (smaller Kb value).

Part A

If Kb for NX3 is 8.5×106, what is the pOH of a 0.175 M aqueous solution of NX3?

Express your answer numerically.

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Part B

If Kb for NX3 is 8.5×106, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Express your answer numerically to three significant figures.

Part C

If Kb for NX3 is 8.5×106 , what is the the pKa for the following reaction? HNX3+(aq)+H2O(l)NX3(aq)+H3O+(aq)

Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent. Such reactions can be generically symbolized as

NX3(aq)+H2O(l)HNX3+(aq)+OH(aq)

where NX3 is the base and HNX3+ is the conjugate acid. The equilibrium-constant expression for this reaction is

Kb=[HNX3+][OH][NX3]

where Kb is the base ionization constant. The extent of ionization, and thus the strength of the base, increases as the value of Kb increases.

Ka and Kb are related through the equation

Ka×Kb=Kw

As the strength of an acid increases, its Ka value increase and the strength of the conjugate base decreases (smaller Kb value).

Part A

If Kb for NX3 is 8.5×106, what is the pOH of a 0.175 M aqueous solution of NX3?

Express your answer numerically.

Explanation / Answer

The reaction can be represented as NX3+H2O-->NX3H+ +OH-

Kb= [NX3H+] [OH-]/ [NX3]

at Equilibrium let x= [OH-] =[NX3H+]

Kb= x2/(0.175-x)= 8.5*10-6 This when solved using solver gives x= 0.001216

pOH= -log(0.001216)= 2.92

2, again, x2/(0.325-x) =8.5*10-6 gives x= 0.00166 ( x = drop in concentration of NX3 to establish equilibrium)

Percen tionizaation =100*0.00166/0.325 =0.52%

3. Kb for NX3+H2O---> NX3H+ +OH- Kb= [NX3H+] [OH-]/ [NX3]

Ka for HNX2+ + H2O ----> NX3(aq)+ H3O+ Ka= [NX3] [H3O+]/ [HNX3+]

Ka*Kb= [H3O+] [OH-] =10-14

Ka*Kb= 10-14 and given Kb= 8.5*10-6

ka= 10-14/(8.5*10-6)=1.76*10-9

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