Question

Consider the titration of a 26.0-mL sample of 0.180 M CH3NH2 with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

Already found:

volume of added acid required to reach the equivalence point.=32.3 ml

pH at 6.0 mL of added acid=11.28

pH at one-half of the equivalence point.= 10.64

Part A

Determine the initial pH

Part B

Determine the pH at the equivalence point.

Part C

Determine the pH after adding 5.0 mL of acid beyond the equivalence point.

Answer 1

part A )

Kb = 4.4 x 10^-4

CH3NH2 + H2O --------------------CH3NH3+ + OH-

0.180 -x                                            x                 x

Kb = x^2 / 0.180-x

4.4 x 10^-4 = x^2 / 0.180-x

x = 8.68 x 10^-3

[OH-] = 8.68 x 10^-3 M

pOH = -log [OH-] = 2.06

pH + pOH = 14

pH = 11.94

part B)

at the equivalence point salt is only found its concentration = millimoles / total volume

                                                                                             = (26 x 0.180 ) / (26 +32.3)

                                                                                             = 0.080 M

pH = 7 - 1/2 [pKb +logC] = 7 - 1/2 [3.36 +log 0.080]

pH = 5.87

part C)

HBr added = 32.3 +5 = 37.3 mL

[HBr] remaining = ( 37.3 x 0.145 - 26 x 0.180 ) / (26 +37.3) = 0.0115 M

pH = -log[H+] = -log (0.0115) = 1.94

pH = 1.94