Question

Background: A 102.4 mL sample of 0.110 M methylamine (CH3NH2, Kb=3.7×10−4) is titrated with 0.230 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.

A.) 0.0 mL

B.) 24.5 mL

C.) 49.0 mL

D.) 73.5 mL

Answer 1

HNO3 + CH3NH2 <--> CH3NH3+ + NO3-
CH3NH2 accepts a H+ forming its conj. acid
A) Kb = x^2/0.105
3.7*10^-4 = x^2/0.105
x^2 = 3.885 x 10^-5
x = [OH-] = 0.00623 M
-log[OH-] = pOH = 2.2055
pH = 14 - pOH
pH = 14 - 2.2055 = 11.79
B) Moles HNO3: 0.0223 L x 0.270 mol/L = 0.006021 mol
Moles initial CH3NH2: 0.1148 L x 0.105 M = 0.012054 mol
Molar ratios 1:1, HNO3 limiting
It reacts w/ base to form conj. acid, moles equal
HA after titration: 0.006021 mol
A- (CH3NH2) left: 0.012054 - 0.006021 = 0.006033 moles
Change Kb to Ka
-log(Kb) = pKb
= 3.4318
14 - pKb = pKa
pKa = 10.57
Since there is a weak base and its conj. acid,
it forms a buffer solution. Thus, use
the Henderson-Hasselbalch equation for buffers:
pH = pKa + log(A-/HA)
pH = 10.57 + log(0.006033/0.006021)
pH = 10.57
C) Moles HNO3: 0.0446 L x 0.270 M = 0.012042 mol
Moles CH3NH2: 0.012054
Moles CH3NH2 left: 1.2*10^-5
Moles CH3NH3+: 0.012042
pH = 10.57 + log(1.2*10^-5/0.012042)
pH = 7.57
Most of CH3NH2 was converted
into its conj. acid, CH3NH3+, so the pH Is lower
But it is still a buffer solution.
D) There will be excess HNO3.
for titration will have passed
equivalence point.
0.067 L x 0.270 M = 0.01809 mol
Moles CH3NH2 = CH3NH3+
0.012054 moles CH3NH2 converted
0.01809 mol - 0.012054 mol =
0.006036 moles strong acid excess
Since HNO3 is a very strong acid,
it will dissociate completely and hence
determine the overall pH of the resulting solution.
[H+] = moles H+/ total volume
[H+] = 0.006036 mol/(0.067 L + 0.1148 L)
= 0.0332 M
-log[H+] = pH = 1.48