(a)
(CH3)2NH is a weak base.
Ka of (CH3)2NH = 5.9 * 10^-4
So, Kb = Kw / Ka
= (1.0 * 10^-14) / (5.9 * 10^-4)
= 1.7 * 10^-11
Now,
(CH3)2NH
+ H2O
<------->
(CH3)2NH2+
+ OH-
IC:
0.383
0
0
C:
-
x
+ x
+
x
EC: 0.383 – x
x
x
Kb = [(CH3)2NH2+]
[OH-] / [(CH3)2NH]
or, 1.7 * 10^-11 = (x) (x) / (0.383 – x)
or, 1.7 * 10^-11 = x2 / (0.383 – x)
or, 1.7 * 10^-11 = x2 /
(0.383) [Kb
is very small, so x term in denominator is discarded]
or, x2 = (1.7 * 10^-11) * (0.383)
or, x2 = 6.5 * 10^-12
or, x = 2.55 * 10^-6
So, [OH-] = x = 2.55 * 10^-6
pOH = - log [OH-]
= - log (2.55 *
10^-6)
= 5.59
Now, pH = 14 – pOH
= 14 – 5.59
= 8.41
(2)
Moles of (CH3)2NH = 0.383 M x 0.0215 L =
0.0082345 mol
Moles of perchloric acid = 0.279 M x 0.0118 L = 0.0032922 mol
So, 0.0032922 mol of perchloric acid will react with 0.0032922 mol of (CH3)2NH to produce 0.0032922 mol of (CH3)2NH2+.
So, mole of (CH3)2NH left = 0.0082345 mol - 0.0032922 mol = 0.0049423 mol
Total volume = 21.5 mL + 11.8 mL = 33.3 mL = 0.033 L
So, the concentration terms are
[(CH3)2NH] = 0.0049423 mol / 0.033 L = 0.1498
M
[(CH3)2NH2+] =
0.0032922 mol / 0.033 L = 0.0998 M
Using the Henderson - Hasselbalch equation
pOH = pKb + log ([salt] / [base])
pOH = pKb + log {
[(CH3)2NH2+] /
[(CH3)2NH] }
Some preliminary calculations :
Kb = 1.7 * 10^-11
So, = pKb = - log Kb = - log (1.7*10^-11) = 10.77
Now,
pOH = pKb + log {
[CH3)2NH2+] /
[CH3)2NH] }
pOH = 10.77 + log (0.0998 / 0.1498)
pOH = 10.77 + log (0.666)
pOH = 10.77 - 0.18
pOH = 10.59
Now, pH = 14 – pOH
= 14 – 10.59
= 3.41
(3)
At mid point;
Half of the [(CH3)2NH is neutralized by the
acid. Therefore, the moles of (CH3)2NH =
moles of
(CH3)2NH2+.
So, [(CH3)2NH] = [(CH3)2NH2+]
Using the Henderson - Hasselbalch equation
pOH = pKb + log {
[(CH3)2NH2+] /
[(CH3)2NH] }
pOH = pKb + log 1
pOH = 10.77 +
pOH = 10.77
Now, pH = 14 – pOH
= 14 – 10.77
= 3.23
4.
At equivalence point;
At equivalence point, all the base [(CH3)2NH] is neutralized with acid.
Moles of (CH3)2NH = 0.0082345 mol
So, moles of perchloric acid = 0.0082345 mol
molarity of perchloric acid = 0.279 M
So, volume of perchloric acid = 0.0082345 mol / 0.279 M = 0.0295 L
= 29.5 mL
Total volume = 21.5 mL + 29.5 mL = 51 mL = 0.051 L
Now, 0.0082345 mol of perchloric acid will react with 0.0082345 mol of (CH3)2NH to produce 0.0082345 mol of (CH3)2NH2+.
[(CH3)2NH2+] = (0.0082345 mol) / (0.051 L) = 0.16 M
Now,
(CH3)2NH2+
<------->
(CH3)2NH
+
H+
IC:
0.16
0
0
C:
- x
+
x
+
x
EC: 0.16 – x
x
x
Ka = [(CH3)2NH] [H+] /
[(CH3)2NH2+]
or, 5.9*10^-4 = (x) (x) / (0.16 – x)
or, 5.9*10^-4 = x2 / (0.16 – x)
or, 5.9*10^-4 = x2 /
(0.16) [Ka is
very small, so x term in denominator is discarded]
or, x2 = (5.9*10^-4) * (0.16)
or, x2 = 9.44 * 10^-5
or, x = 0.0097
So, [H+] = x = 0.0097
pH = - log [H+]
= - log (0.0097)
= 2.01
(5)
Moles of (CH3)2NH = 0.383 M x 0.0215 L =
0.0082345 mol
Moles of perchloric acid = 0.279 M x 0.0422 L = 0.0117738 mol
So, 0.0082345 mol of CH3)2NH will react with 0.0082345 mol of perchloric acid.
So, mole of perchloric acid left = 0.0117738 mol - 0.0082345 mol = 0.0035393 mol
Total volume = 21.5 mL + 42.2 mL = 63.7 mL = 0.0637 L
So, [perchloric acid] = 0.0035393 mol / 0.0637 L = 0.056 M
Perchloric acid is a strong acid and hence will dissociate completely.
So, [H+] = 0.056 M
pH = - log [H+]
= - log (0.056)
= 1.25
Ka of dimethylamine = 5.9*10^-4 A 21.5 mL sample of 0.383 M dimethylamine, (CH32NH, is titrated...
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