Question

Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent.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 5.0×10−6, what is the pOH of a 0.175 M aqueous solution ofNX3? Express your answer numerically. View Available Hint(s) pOH = nothing Submit Part B If Kb for NX3 is 5.0×10−6, what is the percent ionization of a 0.325 M aqueous solution of NX3? Express your answer numerically to three significant figures. View Available Hint(s) percent ionization = nothing   % Submit Part C If Kb for NX3 is 5.0×10−6 , what is the the pKa for the following reaction? HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq) Express your answer numerically to two decimal places.

Answer 1

Part-A

Kb= 5.0x10^-6

          NX3(aq)    + H2O(l) -----------------H NX3+(aq) +    OH-(aq)

          0.175                                          0                     0

            -x                                               +x                  +x

         0.175-x                                        +x                +x

         Kb= [HNX3+][OH-]/[NX3]

      5.0x10^-6 = x*x/(0.175-x)

( 5.0x10^-6)x( 0.175-x) = x^2

0.875x10^-6 - 5.0x10^-6 x= x^2

x^2 + 5.0x10^-6 - 0.875x10^-6 = 0

for solving the quardratic equation

x= 0.000933

[OH-]= 0.000933 M

-log[OH-]= -log( 0.000933)

POH= 3.03

Part-B

Concentration of NX3= 0.325M

             NX3(aq)    + H2O(l) -----------------H NX3+(aq) +    OH-(aq)

          0.325                                         0                     0

            -x                                               +x                  +x

         0.325-x                                        +x                +x

         Kb= [HNX3+][OH-]/[NX3]

      5.0x10^-6 = x*x/(0.325-x)

( 5.0x10^-6)x( 0.325-x) = x^2

1.625x10^-6 - 5.0x10^-6 x= x^2

x^2 + 5.0x10^-6 - 1.625x10^-6=0

for solving the equation

x=0.00127

[OH-] = 0.00127M

C= 0.325M

for weak bases

[OH-]= c

= [OH-]/C = 0.00127/0.325

= 0.0039

Percent ionisation

= 0.0039x100 = 0.390%

= 0.390%

Part-C

Kb= 5.0x10^-6

KaxKb= Kw

Ka = Kw/Kb                  where Kw= 1.0x10^-14

Ka= 1.0x10^-14/5.0x10^-6

Ka= 0.2x10^-8

Ka= 2.0x10^-9

-log(Ka)= -log(2.0x10^-9)

PKa= 8.69