Question

QUESTION 25

± Weak Base Calculations

Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent.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 8.0×10⁻⁶, what is the pOH of a 0.175 M aqueous solution of NX3?

Express your answer numerically.

Part B

If Kb for NX3 is 8.0×10⁻⁶, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Express your answer numerically to three significant figures.

Part C

If Kb for NX3 is 8.0×10⁻⁶ , 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

A)

Lets write the dissociation equation of NX3

NX3 +H2O -----> HNX3+ + OH-

0.175 0 0

0.175-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((8*10^-6)*0.175) = 1.183*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.183*10^-3 M

So, [OH-] = x = 1.183*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.183*10^-3)

= 2.93

Answer: 2.93

B)

Lets write the dissociation equation of NX3

NX3 +H2O -----> HNX3+ + OH-

0.325 0 0

0.325-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((8*10^-6)*0.325) = 1.612*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.612*10^-3 M

% dissociation = (x*100)/c

= 1.612*10^-3*100/0.325

= 0.4961 %

Answer: 0.496 %

C)

we have below equation to be used:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/8*10^-6

Ka = 1.25*10^-9

we have below equation to be used:

pKa = -log Ka

= -log (1.25*10^-9)

= 8.9031

Answer: 8.90