Question

Consider the titration of a 23.0 −mL sample of 0.180 M CH3NH2 with 0.145 M HBr. Determine each of the following.

Part A the initial pH Express your answer using two decimal places. pH = 11.95 SubmitMy AnswersGive Up Correct Part B the volume of added acid required to reach the equivalence point V = 28.6   mL   SubmitMy AnswersGive Up Correct Part C the pH at 5.0 mL of added acid Express your answer using two decimal places. pH = Part D the pH at one-half of the equivalence point Express your answer using two decimal places. pH = Part E the pH at the equivalence point Express your answer using two decimal places. pH = Part F the pH after adding 5.0 mL of acid beyond the equivalence point Express your answer using two decimal places.

Answer 1

millimoles of base = 23 x 0.18 = 4.14

c) millimoles of acid added = 5.0 x 0.145 = 0.725

4.14 - 0.725 = 3.415 millimoles base left.

0.725 millimoles salt formed.

[salt] = 0.725 / 23 + 5 = 0.026 M

[base] = 3.415 / 23 + 5 = 0.122 M

pKb of methyl amine = 3.36

pOH = pKb + log [salt] / [base]

pOH = 3.36 + log [0.026] / [0.122]

pOH = 2.69

pH = 14 - 2.69

pH = 11.31

d) at half equivalence point

pOH = pKb

pOH = 3.36

pH = 14 - 3.36

pH = 10.64

e) at equivalence point 4.14 millimoles acid must be added

4.14 = v x 0.145

V = 28.55

total volume = 28.55 + 23 = 51.55

[salt] = 4.14 / 51.55 = 0.08 m

pOH = 1/2 [pKw + pKb + log C]

pOH = 1/2 [14 +3.36 + log 0.08]

pOH = 8.13

pH = 14 - 8.13

pH = 5.87

f) total volume of acid added = 28.55 + 5 = 33.55 mL

millimoles of acid = 33.55 x 0.145 = 4.865

4.865 - 4.14 = 0.725 millimoles acid left

[acid] = 0.725 / 33.55 + 23 = 0.725 / 56.55 = 0.0128 M

pH = - log [H⁺]

pH = - log [0.0128]

pH = 1.89