Question

Consider the titration of a 28.0-mL sample of 0.170 M CH3NH2 with 0.150 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

Determine the pH at 5.0 mL of added acid.

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

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

Answer 1

a) When you add 5mL of acid, the volume changes.
Total volume = 28mL + 5mL = 32 mL

Initial mol CH3NH2
0.170 mol CH3NH2 / L * 0.028 L = 0.00476 mol CH3NH2

Added mol HBr
0.150 mol HBr / L * 0.005L = 7.5*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 [(0.00476)/(7.5*10^-4)]
pH = 10.64 + 0.8025
pH = 11.44

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

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

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

M1V1 = M2V2
(0.170)(28) = (0.150)(V2)
31.73 mL = V2

Equivalence point
At the equivalence point, you have reacted all of the 0.00476 moles that you had initially to form 0.00476 moles of CH3NH3+. The final volume = 28.0 mL + 31.73 mL = 59.73 mL = 0.05973 L.
[CH3NH3+] = moles CH3NH3+ / L = 0.00476 / 0.05973 = 0.0797 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 . . . . . . . . . .0.0797 . . . . . . . . . . . . .0 . . . . . .0
change . . . . . . . . . .-x . . . . . . . . . . . . . . .x . . . . . .x
at equilibrium . .0.0797 - x . . . . . . . . . . . . x . . . . . .x

Ka CH3NH3+ = [H3O+][CH3NH2] = x^2 / (0.0797 - x) = 2.27 x 10^-11

Because Ka is so small, the -x term after 0.0783 can be ignored, leaving

x^2 / 0.0797 = 2.27 x 10^-11
x^2 = 1.8 x 10^-12
x = 1.3 x 10^-6 = [H3O+]

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

d) 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 59.73 mL + 6.0 mL = 65.73 mL = 0.06573 L

[H3O+] = moles H3O+ / L
= 9.0*10^-4 moles / 0.06573 L
= 1.369*10^-2 M

pH = -log [H3O+]
= -log (1.422*10^-2 M)
= 1.86