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Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent.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 9.0×10−6, what is the pOH of a 0.175 M aqueous solution of NX3? Express your answer numerically.

Answer 1

2.90 is the pOH of a 0.175 M aqueous solution of
`NX_3`.

Step-by-step explanation:

The ionization constant of the base =
`K_b=9.0* 10^(-6)`

Concentration of the base present initially,c = 0.175 M

`NX_3(aq)\rightleftharpoons HNX_3+(aq)+OH^-(aq)`

c 0 0

c-x x x

The ionization constant of the base is given by an expression:

`K_b=([HNX_3][OH^-])/([NX_3])`

`9.0* 10^(-6)=(x^2)/(c-x)`

`9.0* 10^(-6)=(x^2)/(0.175-x)`

`1.575* 10^(-6)-9.0x* 10^(-6)=x^2`

On solving:

x = 0.0012505 M

`[HNX_3]=[OH^-]=0.0012505 M`

`pOH=-\log[OH^-]`

`pOH=-\log[0.0012505 ]=2.90`

2.90 is the pOH of a 0.175 M aqueous solution of
`NX_3`.