Part A If K b for NX3 is 4.5×10 6 , what is the pOH of a 0.175 M aqueous solutio
ID: 504013 • Letter: P
Question
Part A
If Kb for NX3 is 4.5×106, what is the pOH of a 0.175 M aqueous solution of NX3?
Express your answer numerically.
Part B
If Kb for NX3 is 4.5×106, what is the percent ionization of a 0.325 M aqueous solution of NX3?
Express your answer numerically to three significant figures.
Part B
If Kb for NX3 is 4.5×106, 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 4.5×106 , 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.
Part B
If Kb for NX3 is 4.5×106, 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 4.5×106 , 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.
Explanation / Answer
part-A
NX3 + H2O ---------> HNX3+ + OH-
I 0.175 0 0
C -x +x +x
E 0.175-x +x +x
Kb = [HNX3+][OH-]/[NX3]
4.5*10-6 = x*x/0.175-x
4.5*10-6 *(0.175-x) = x^2
x = 0.000885
[OH-] = x= 0.000885M
POH =-log[OH-]
= -log0.000885
= 3.053
part-B
NX3 + H2O ---------> HNX3+ + OH-
I 0.325 0 0
C -x +x +x
E 0.325-x +x +x
Kb = [HNX3+][OH-]/[NX3]
4.5*10-6 = x*x/0.325-x
4.5*10-6 *(0.325-x) = x^2
x = 0.0012
[OH-] = x= 0.0012M
percent ionisation = Final [OH-] *100/conc of base
= 0.0012*100/0.325 =0.37%
part-C
Ka = Kw/Kb
= 1*10^-14/4.5*10^-6 = 2.23*10^-9