Many common weak bases are derivatives of NH3, where one or more of the hydrogen

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Many common weak bases are derivatives of NH3, where one or more of the hydrogen atoms have been replaced by another substituent.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. If Kb for NX3 is 3.0 times 10-6, what is the pOH of a 0.175 M aqueous solution of NX3? Express your answer numerically. pOH = My Answers Give Up Review Part If Kb for NX3 is 3.0 times 10-6, what is the percent ionization of a 0.325 M aqueous solution of NX3? Express your answer numerically to three significant figures. percent ionization = My Answers Give Up Review Part If Kb for NX3 is 3.0 times 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.

Explanation / Answer

A)

ka x kb = kw
ka = kw / kb
ka = 1 x 10^-14 / 3.0x10-6
ka = 3.33 x 10^-9


ka = ( equilibrium concentration ) ^2 / initial concentration
5.56 x 10^-10 = ( equilibrium concentration ) ^2 / 0.175
equilibrium concentration = 1.38 x 10^-4 M

pH = -log ( equilibrium concentration )
pH = 3.86


Actually
pH + pOH = 14
pOH = 14 - 3.86
pOH = 8.14

B)

NX3 + H2O <----> HNX3+ + OH-
Kb = 3.0 10^-6
Kb = ([HNX3+]*[OH-]) / [NX3]
Kb = [x^2]/[.325-x]
7E-6 = [x^2]/[.325-x]
Solve for x using quadratic formula, the plug x into .325-x to get the concentration of NX3, and plug into the equation below
----------------------[x] / [.325-x]
% ionization = [HNX3+] / [NX3] *100

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