Question

Consider the titration of a 26.0 −mL sample of 0.180 M CH3NH2 with 0.155 M HBr. Determine each of the following.

1. the pH at 4.0 mL of added acid

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

3. the pH at the equivalence point

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

Answer 1

millimoles of CH3NH2 = 26.0 x 0.180 = 4.68

pKb of CH3NH2 = 3.38 standard value

1) millimoles of HBr added = 4.0 x 0.155 = 0.62

4.68 - 0.62 = 4.0 millimoles base left

0.62 millimoles salt formed

[base] = [4.0 / 30 ] = 0.13 M

[salt] = 0.62 / 30 = 0.021 M

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

pOH = 3.38 + log [0.021] / [0.13]

pOH = 2.60

pH = 14 - 2.60

pH = 11.40

2) at one-half of the equivalence point

pOH = pKb

pOH = 3.38

pH = 14 - 3.38

pH = 10.62

3) at equivalence point

all base becomes salt

4.68 millimoles HBr must be added

4.68 = V x 0.155

V = 30.1 mL HBr added

[salt] = 4.68 / 26 + 30.1 = 4.68 / 56.1 = 0.083 M

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

pOH = 1/2 [14 + 3.38 + log 0.083]

pOH = 8.15

pH = 14 - 8.15

pH = 5.85

4) millimoles of exess HBr = 5.0 x 0.155 = 0.775

total volume = 26 + 30.1 + 5.0 = 61.1 mL

.[HBr] = 0.775 / 61.1 = 0.013 M

as HBr strong base

[HBr] = [H⁺] = 0.013 M

pH = - log [H⁺]

pH = - log [0.013]

pH = 1.89