Question

Consider the titration of 40.0 mL of 0.0600 M (CH₃)₂NH (a weak base; K_b = 0.000540) with 0.100 M HClO₄. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 6.0 mL pH =	(c) 12.0 mL pH =
(d) 18.0 mL pH =	(e) 24.0 mL pH =	(f) 26.4 mL pH =

Answer 1

(a)

(CH3)2NH + H2O <----> (CH3)2NH 2+ + OH-; Kb = 0.000540
Kb = [OH-]*[(CH3)2NH2+]/[(CH3)2NH]

0.000540 = X*X/ 0.0600

X = [OH-] ≈ √[0.060*0.000540] = 5.69*10^-3
pOH = -log [OH-]= 2.24

pH = 14-pOH=14.00-2.24

= 11.76.

(b) 6.0 mL of 0.100 M HClO₄ added

6.0 mL 0.100 M HClO₄ adds 0.0006 mol of acid or H+.

Here the original (CH3)2NH solution contained 0.0600 M*40 ml/1000 mL=

=0.0024 mol (CH3)2NH
After the addition of acid, 0.0018 mol (CH3)2NH remains, and

Molrity = 0.0018/ 0.046 = 0.039 M

(CH3)2NH2+ = 0.0024 mol + 0.0006 mol = 0.0003 mol (CH3)2NH2+

Molarity = 0.003/0.046=0.065

pKb + log[salt]/[base]

pOH= 3.27 + log(0.065/ 0.039) = 3.49
pH = 14-pOH=14.00-3.49

= 10.51

(c)

12.0 mL of 0.100 M HClO₄ added

12.0 mL 0.100 M HClO₄ adds 0.0012 mol of acid or H+.

Here the original (CH3)2NH solution contained 0.0600 M*40 ml/1000 mL=

=0.0024 mol (CH3)2NH
After the addition of acid, 0.0012 mol (CH3)2NH remains, and

Molrity = 0.0012/ 0.052 = 0.023 M

(CH3)2NH2+ = 0.0024 mol + 0.0012 mol = 0.0036 mol (CH3)2NH2+

Molarity = 0.0036/0.052=0.069

pKb + log[salt]/[base]

pOH= 3.27 + log(0.069/ 0.023) = 3.75
pH = 14-pOH=14.00-3.75

= 10.25

(d)

18.0 mL of 0.100 M HClO₄ added

18.0 mL 0.100 M HClO₄ adds 0.0018 mol of acid or H+.

Here the original (CH3)2NH solution contained 0.0600 M*40 ml/1000 mL=

=0.0024 mol (CH3)2NH
After the addition of acid, 0.0006 mol (CH3)2NH remains, and

Molrity = 0.0006/ 0.058 = 0.0103M

(CH3)2NH2+ = 0.0024 mol + 0.0018 mol = 0.0042 mol (CH3)2NH2+

Molarity = 0.0042/0.058=0.072

pKb + log[salt]/[base]

pOH= 3.27 + log(0.072/ 0.0103) = 4.11
pH = 14-pOH=14.00-4.11

= 9.89