Question

Determine the pH during the titration of 36.0 mL of 0.235 M trimethylamine ((CH₃)₃N, K_b = 6.3×10^-5) by 0.235 M HBr at the following points. (Assume the titration is done at 25 °C.)
Note that state symbols are not shown for species in this problem.

a.) Before the addition of any HBr

b.) After the addition of 15.3 mL of HBr

c.) At the titration midpoint

d.) At the equivalence point

e.) After adding 56.2 mL of HBr

Answer 1

millimoles of (CH₃)₃N = 36 x 0.235 = 8.46

kb= 6.3×10^-5

pKb = -logKb = -log (6.3×10^-5) = 4.2

a) before the addition of any HBr

pOH = 1/2 [pKb -logC] = 1/2 [4.2 -log0.235] = 2.41

pH + pOH = 14

pH = 11.59

b) after the addition of 15.3 mL HBr

millimoles of HBr = 15.3 x0.235 = 3.6

(CH₃)₃N + HBr ---------------------->(CH₃)₃NH+Br-

8.46         3.6                                0

4.86            0                                   3.6

pOH = pKb + log (3.6 /4.86)

pOH = 4.07

pH = 9.93

c) At the titration midpoint

it is half equivalence point . so

pOH = pKb

pOH = 4.2

pH +pOH =14

pH = 9.80

d) At the equivalence point

volume at the equivalence point = 36 mL

here salt only remains .so

[salt] = 36 x 0.235 / (36 + 36) = 0.1175

pH = 7 - 1/2 (pKb + log C)

      = 7 - 1/2 (4.2 + log 0.1175)

      = 5.36

pH = 5.36

e) after the addition of 56.2 mL HBr

millimoles of acid = 56.2 x 0.235 = 13.207

(CH₃)₃N   + HBr ----------------------> (CH₃)₃NH+Br-

8.46         13.207                                0

0               4.747                                  8.46

[H+] = 4.747 / (36 + 56.2) = 0.0514

pH = -log [H+] = 1.29

pH = 1.29