Question

Consider the titration of a 27.0 −mL sample of 0.175 MCH3NH2 with 0.150 M HBr. Express answers using two decimal places. Determine each of the following.

A) the initial pH

B) the volume of added acid required to reach the equivalence point

C) the pH at 6.0 mL of added acid

D) the pH at one-half of the equivalence point

E) the pH at the equivalence point

F) the pH after adding 4.0 mL of acid beyond the equivalence point

Answer 1

A) pOH = 1/2[pKb-logC]= 1/2[3.36-log(0.00472)]= 2.84

pH = 14-2.84 = 11.16

B) M1V1 = M2V2

0.175*27 = 0.15*V2

V2 = 31.5 mL

C) After adding 6.0 mL of acid

[base] = 27*0.175/33 =0.143 ; [acid]= 6*0.15/33 = 0.027

base left = 0.143-0.027 = 0.116

pOH = pKb+log(salt/base) = 3.36+log(0.027/0.116) = 2.73

pH = 14-2.73 = 11.27

D) [base]= 27*.175/42.75 = 0.11 ; [acid]=15.75*.15/42.75 = 0.055

base left = 0.11-0.055 = 0.055

pOH = 3.36+log(0.055/0.055) = 3.36

pH = 14-3.36 = 10.64

E) equivalence point, [salt] = 31.5*0.15/58.5 =0.081

pH = 7-1/2(pKb+logC) = 7-1/2(3.36+log(0.081)) = 5.86

F) 31.5+4 = 35.5 ml of acid

[base] = 27*.175/62.5 =0.0756; [acid] = 35.5*.15/62.5 = 0.0852

acid left = 0.0852-0.0756 =0.0096

pH = -log(0.0096) = 2.02