Consider the titration of 40.0 mL of 0.0600 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HCI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (b) 6.0 mL (c) 12.0 mL (d) 18.0 mL
(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
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