± Weak Base Calculations NX3(aq)+H2O(l)HNX3+(aq)+OH(aq) K b=[HNX3+][OH][NX3] whe
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Question
± Weak Base Calculations
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 2.0×106, what is the pOH of a 0.175 M aqueous solution of NX3?
Express your answer numerically.
3.23
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Part B
If Kb for NX3 is 2.0×106, what is the percent ionization of a 0.325 M aqueous solution of NX3?
Express your answer numerically to three significant figures.
0.33
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Part C
HNX3+(aq)+H2O(l)NX3(aq)+H3O+(aq)
Express your answer numerically to two decimal places.
8.54
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± 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Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent.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 asNX3(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 isKb=[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 2.0×106, what is the pOH of a 0.175 M aqueous solution of NX3?
Express your answer numerically.
pOH =3.23
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Correct
Part B
If Kb for NX3 is 2.0×106, what is the percent ionization of a 0.325 M aqueous solution of NX3?
Express your answer numerically to three significant figures.
percent ionization =0.33
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Part C
If Kb for NX3 is 2.0×106 , 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.
pKa =8.54
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Explanation / Answer
Kb = [NHX3+] [OH-] / [NH3]; or Kb = x^2 / [NX3]
a) Substitute:
2×10^6 = x^2 / 0.175;
Solving, x = [OH-] = 0.0006 M, so pOH = 3.22
b) [OH-] = 0.0006 M, so the percent ionization is 100% * 0.0006 M / 0.325 M = 0.18%
c) For a weak acid / weak base conjugate pair,
KaKb = Kw. Therefore,
Ka = (1x10^-14) / (2x10^-6) = 0.5*10^-8
Pka= -log[ka]= -log[0.5*10^-8]= 8.54
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