Part A If Kb for NX3 is 5.0 × 10-6, what is the pOH of a 0.175 mol L-1 aqueous s
ID: 694640 • Letter: P
Question
Part A If Kb for NX3 is 5.0 × 10-6, what is the pOH of a 0.175 mol L-1 aqueous solution of NXs? Express your answer numerically. Hints Submit My Answers Give Up Part B If Kb for NXs is 5.0 x 10-6, what is the percent ionization of a 0.325 molL-1 aqueous solution of NX3? Express your answer numerically to three significant figures. Hints 0 percent ionization- Submit My Answers Give Up Part C If K, for NX is 5.0 x 10-6, what is the pKa for the following reaction? HNX3 + (aq)-+ H2O(l) NX3 (aq) +-H,0+ (aq) Express your answer numerically to two decimal places. Hints Submit My Answers Give Up ContinueExplanation / Answer
A)
B + H2O <-> HB+ + OH-
The equilibrium Kb:
Kb = [HB+][OH-]/[B]
initially:
[HB+] = 0
[OH-] = 0
[B] = M
the change
[HB+] = x
[OH-] = x
[B] = - x
in equilibrium
[HB+] = 0 + x
[OH-] = 0 + x
[B] = M - x
Now substitute in Kb
Kb = [HB+][OH-]/[B]
Kb = x*x/(M-x)
x^2 + Kbx - M*Kb = 0
x^2 + (5*10^-6)x - (0.175)(5*10^-6) = 0
solve for x
x = 9.33*10^-4
substitute:
[HB+] = 0 + x = 9.33*10^-4M
[OH-] = 0 + x = 9.33*10^-4M
pOH = log(9.33*10^-4) = 3.03
B.
apply the same:
This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"
there are free OH- ions so, expect a basic pH
B + H2O <-> HB+ + OH-
The equilibrium Kb:
Kb = [HB+][OH-]/[B]
initially:
[HB+] = 0
[OH-] = 0
[B] = M
the change
[HB+] = x
[OH-] = x
[B] = - x
in equilibrium
[HB+] = 0 + x
[OH-] = 0 + x
[B] = M - x
Now substitute in Kb
Kb = [HB+][OH-]/[B]
Kb = x*x/(M-x)
x^2 + Kbx - M*Kb = 0
x^2 + (5*10^-6)x - (0.325)(5*10^-6) = 0
solve for x
x = 0.003315
substitute:
[OH-] = 0 + x = 0.00127 M
%ionization = [OH-]/[B]*!00% = 0.00127 /0.325*100 = 0.39%
C)
pKb = -log(Kb)= -log(5*10^-6 ) = 5.30