Question

1.Consider the titration of a 23.0-mL sample of 0.180 M CH3NH2 with 0.150 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

A.Determine the initial pH.
B.Determine the volume of added acid required to reach the equivalence point.

C.Determine the pH at 4.0 mL of added acid.

D. Determine the pH at one-half of the equivalence point.

E.Determine the pH at the equivalence point.

F.Determine the pH after adding 6.0 mL of acid beyond the equivalence point.

2.For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

A. 250.0 mL of pure water

B.250.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.285 M in KCHO2

C.250.0 mL of a buffer solution that is 0.290 M in CH3CH2NH2 and 0.260 M in CH3CH2NH3Cl

Answer 1

a) Realize that currently in the beaker, there is only base in the solution at the initIal value.

Make an I.C.E Table

.........CH3NH2(aq) <--> CH3NH3+(aq) + OH-(aq)

I...........0.180.............0.....0
C...........-x...............+x......+x
E........0.180-x...........x.........x

Kb = x^2 / (0.180-x)

4.4×10^-4 = x^2 / (0.180-x)
x = 8.90×10^(-3)

[OH-] = ~8.90×10^(-3)
pOH = -log(~8.90×10^(-3))

pH = 14 - pOH
pH = 11.949

b) The equivalence point is reached when moles of acid equals moles of base.

M1V1 = M2V2
(0.180)(23) = (0.150)(V2)
27.6 mL = V2

c) When you add 4mL of acid, the volume changes.
Total volume = 23mL + 4mL

Initial mol CH3NH2
0.180 mol CH3NH2 / L × 0.023 L = 0.00414 mol CH3NH2

Added mol HBr
0.150 mol HBr / L × 0.004L = 6.21×10^-4 mol HBr

Determine the Ka value
Kw = Ka ×Kb
Ka = (1.0×10^-14) / (4.4 ×10^-4) = 2.27×10^-11

pH = pKa + log ([base]/[acid])
pH=-log(2.27×10^-11)+log[(1.466×10^4)/(2.0690×10^-5)]
pH = 10.64 + 0.8503
pH = ~11.49

d) Half of the equivalence point.
The mol ratio of base : acid will be equal.
Therefore it's 1:1. pH = pKa

pH = pKa + log ([base]/[acid])
pH = 10.64 + log (1/1)
pH = 10.64

e) Equivalence point
At the equivalence point, you have reacted all of the 0.00414 moles that you had initially to form 0.00414 moles of CH3NH3+. The final volume = 23.0 mL + 24.3 mL = 54.3 mL = 0.0543 L.
[CH3NH3+] = moles CH3NH3+ / L = 0.00414 / 0.0543 = 0.0762 M

The CH3NH3+ solution will be slightly acidic (since it was made from a weak base and a STRONG acid). Hence, it hydrolyzes in water:

Molarity . . . . . .CH3NH3+ + H2O <==> H3O+ + CH3NH2
initial . . . . . . . . . .0.0762. . . . . . . . . . . . .0 . . . . . .0
change . . . . . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
at equilibrium . .0.0762 - x . . . . . . . . . . . . x . . . . . .x

Ka CH3NH3+ = [H3O+][CH3NH2] = x^2 / (0.0762- x) = 2.3 x 10^-11

Because Ka is so small, the -x term after 0.0762 can be ignored, leaving

x^2 / 0.07624 = 2.3 x 10^-11
x^2 = 1.75 x 10^-12
x = 1.324x 10^-6 = [H3O+]

pH = -log [H3O+] = -log (1.3 x 10^-6) = 5.87 . . .which is slightly acidic.

f) pH after 6mL of acid is added beyond the equivalence point
After the endpoint, the pH is controlled by the amount of excess strong acid (HBr) added.

mL HBr after endpoint = 6.0
moles excess HBr = M HBr x L HBr = (0.150)(0.0060) = 9.0×10^-4 moles HBr

Since HBr is a strong acid,
[HBr] = [H3O+] = 9.0×10^-4 moles

The volume is now 54.3 mL + 6.0 mL = 63.3 mL = 0.0633 L

[H3O+] = moles H3O+ / L
= 9.0×10^-4 moles / 0.0633 L
= 1.422×10^-2 M

pH = -log [H3O+]
= -log (1.422×10^-2 M)
= 1.85