Question

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.

Express your answer using three significant figures.

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.

Answer 1

CH3NH2 + HBr --> CH3NH3+ + Br-
weak base + strong acid
remember to add up the volumes to determine final molarity

Kb of CH3NH2 = 4.4x10^-4 so Ka = 2.27x10^-11
Ka = [CH3NH2][H+] / [CH3NH3+]
2.27x10^-11 = x^2 / 0.180M
initial [H+] = 2.02x10^-6M so initial pH = 5.69

at eq point, moles HBr = moles CH3NH2
0.023L x 0.180M CH3NH2 = 0.00414moles
0.00414moles HBr / 0.15L = 0.0276L or 27.6ml

pH at eq point:
0.00414moles CH3NH3+ / 0.0506L = 0.08182 M CH3NH3+
Ka = 2.27x10^-11 = x^2 / 0.08182
[H+] = 1.36x10^-6M
pH = 5.87

pH after 4ml HBr
0.004L x 0.15M HBr = 6x10^-4moles HBr used and 6x10^-4moles CH3NH3+ produced
6x10^-4moles CH3NH3+ / 0.027L = 2.22x10^-2M
2.27x10^-11 = x^2 / 2.22x10^-2M
[H+] = 5.04 x 10^-13
pH = 12.29

0.006L x 0.15M HBr = 9x10^-4moles HBr or H+
at eq point, [H+] from CH3NH3+ = 2.02x10^-6M
added [H+] from HBr = 9x10^-4moles / 0.0566L = 0.0159M
[H+] = 0.0159 + 2.02x10^-6 = 0.0159M
pH = 1.798