CHM514Leçon6KaCorr

Capsule
d’études
secondaires
2020
CHM 514
La constante d'acidité
Corrigé
Par S. Lavallé
e
Prob. 6

14. a)
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
2
.
8
𝑥
10
]
=
7
.
6
b)
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
3
.
6
𝑥
10
]
=
7
.
4
15.
[
𝐻
]
=
10
=
10
.
=
[
2
.
5
𝑥
10
]
16.
1
𝑚𝑜𝑙𝑒
𝐻𝐶𝑙
36
.
5
𝑔
=
𝑥
𝑚𝑜𝑙𝑒
𝐻𝐶𝑙
2
.
00
𝑔
→
𝑥
=
1
𝑚𝑜𝑙𝑒
𝐻𝐶𝑙
∙
2
.
00
𝑔
3
6
.
5
𝑔
=
5
.
48
𝑥
10
𝑚𝑜𝑙𝑒
𝑐
=
𝑛
𝑣
=
5
.
48
𝑥
10
𝑚𝑜𝑙𝑒
2
.
5
𝐿
=
[
2
.
19
𝑥
10
]
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
2
.
19
𝑥
10
]
=
1
.
7
17. a)
𝐻𝑁𝑂
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝑁𝑂
(
)
b)
𝐻
𝑆
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝑂𝐶𝑙
(
)
c)
𝐻𝑂𝐶𝑙
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝐻𝑆
(
)
d)
𝐻𝑆𝑂
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝑆𝑂
(
)
18. a)
𝐶𝐻
𝑁𝐻
(
 +
𝐻
𝑂
(
)
𝐶𝐻
𝑁𝐻
(
)
+
𝑂𝐻
(
)
b)
𝐶
𝐻
𝑁
(
)
+
𝐻
𝑂
(
)
𝐶
𝐻
𝑁𝐻
(
)
+
𝑂𝐻
(
)
c)
(
𝐶𝐻
)
𝑁
(
 +
𝐻
𝑂
(
)
(
𝐶𝐻
)
𝑁
𝐻
(
)
+
𝑂𝐻
(
)
d)
𝐻𝑆𝑒
(
)
+
𝐻
𝑂
(
)
𝐻
𝑆𝑒
(
)
+
𝑂𝐻
(
)
19. a)
𝐻
𝑆𝑂
(
)
+
(
𝐶𝐻
)
𝑁𝐻
(
)
𝐻𝑆𝑂
(
)
+
(
𝐶𝐻
)
𝑁𝐻
(
)
A B BC AC
b)
𝑁𝐻
𝑂𝐻
(
)
+
𝐻
𝐶
𝑂
(
)
𝐻
𝐶
𝑂
(
)
+
𝑁𝐻
𝑂𝐻
(
)
B A BC AC
c)
𝐻
𝑃𝑂
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝐻
𝑃𝑂
(
)
A B AC BC
d)
𝐻𝐶𝑂
(
)
+
𝐻𝐶𝑂𝑂𝐻
(
)
𝐻𝐶𝑂𝑂
(
)
+
𝐻
𝐶𝑂
(
)
B A BC AC
20. a)
𝐻𝐵𝑟
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝐵𝑟
(
)
b)
𝑁𝐻
𝑁𝐻
(
)
+
𝐻
𝑂
(
)
𝑁𝐻
𝑁𝐻
(
)
+
𝑂𝐻
(
)
c)
𝐻𝑁𝑂
(
)
+
𝐻
𝑂
(
)
𝐻
𝑂
(
)
+
𝑁𝑂
(
)
d)
𝐶
𝐻
𝑁𝐻
(
)
+
𝐻
𝑂
(
)
𝐶
𝐻
𝑁𝐻
(
)
+
𝑂𝐻
(
)

21.
𝑝𝑂𝐻
=
−
log
[
𝑂𝐻
]
=
−
log
[
2
.
3
𝑥
10
]
=
4
.
6
𝑝𝐻
+
𝑝𝑂𝐻
=
14
→
𝑝𝐻
=
14
−
𝑝𝑂𝐻
=
14
−
4
.
6
=
9
.
4
22.
𝑝𝐻
+
𝑝𝑂𝐻
=
14
→
𝑝𝑂𝐻
=
14
−
𝑝𝐻
=
14
−
3
.
7
=
10
.
3
[
𝑂𝐻
]
=
10
=
10
.
=
5
.
0
𝑥
10
23. a)
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
7
.
8
𝑥
10
]
=
7
.
1
b)
𝑝𝑂𝐻
=
−
log
[
𝑂𝐻
]
=
−
log
[
0
.
0001
]
=
4
→
𝑝𝐻
=
14
−
𝑝𝑂𝐻
=
14
−
4
=
10
c)
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
0
.
0065
]
=
2
.
2
d)
𝑝𝑂𝐻
=
−
log
[
𝑂𝐻
]
=
−
log
[
0
.
3
]
=
5
.
2
→
𝑝𝐻
=
14
−
5
.
2
=
14
−
4
.
6
=
8
.
8
c<a<d<b
24.
𝟏
𝑵𝑯
𝟑
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝟏
𝑵𝑯
𝟒
(
𝒂𝒒
)
+
𝟏
𝑶𝑯
(
𝒂𝒒
)
D
[
0
.
25
]
-
[
0
]
[
0
]
R
-
[
𝑥
]
- +
[
𝑥
]
+
[
𝑥
]
É
[
0
.
25
−
𝑥
]
-
[
𝑥
]
[
𝑥
]
𝐾
=
𝑁𝐻
(
)
∙
𝑂𝐻
(
)
𝑁𝐻
(
)
=
[
𝑥
]
∙
[
𝑥
]
[
0
.
25
−
𝑥
]
=
𝑥
0
.
25
−
𝑥
=
1
.
79
𝑥
10
𝑥
+
1
.
79
𝑥
10
𝑥
−
4
.
48
𝑥
10
𝑥
=
−
𝑏
±
√
𝑏
−
4
𝑎𝑐
2
𝑎
=
−
1
.
79
𝑥
10
±
(
1
.
79
𝑥
10
)
−
(
4
∙
1
∙
−
4
.
48
𝑥
10
)
2
∙
1
→
𝑥
=
−
[
0
.
021
]
𝑟𝑒𝑗𝑒𝑡
é
,
𝑥
=
[
0
.
021
]
𝑂𝐻
(
)
É
=
[
0
.
021
]
→
𝑝𝑂𝐻
=
−
log
[
0
.
021
]
=
1
.
68
𝑝𝐻
+
𝑝𝑂𝐻
=
14
→
𝑝𝐻
=
14
−
𝑝𝑂𝐻
=
14
−
1
.
68
=
1
.
23
𝟏
(
𝑪
𝟐
𝑯
𝟓
)
𝟐
𝑵𝑯
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
(
𝑪
𝟐
𝑯
𝟓
)
𝟐
𝑵𝑯
𝟐
(
𝒂𝒒
)
+
𝟏
𝑶𝑯
(
𝒂𝒒
)
D
[
0
.
25
]
-
[
0
]
[
0
]
R
-
[
𝑥
]
- +
[
𝑥
]
+
[
𝑥
]
É
[
0
.
25
−
𝑥
]
-
[
𝑥
]
[
𝑥
]
𝐾
=
(
𝐶
𝐻
)
𝑁𝐻
(
)
∙
𝑂𝐻
(
)
(
𝐶
𝐻
)
𝑁𝐻
(
)
=
[
𝑥
]
∙
[
𝑥
]
[
0
.
25
−
𝑥
]
=
𝑥
0
.
25
−
𝑥
=
3
.
09
𝑥
10
𝑥
+
3
.
09
𝑥
10
𝑥
−
7
.
76
𝑥
10

𝑥
=
−
𝑏
±
√
𝑏
−
4
𝑎𝑐
2
𝑎
=
−
3
.
09
𝑥
10
±
(
3
.
09
𝑥
10
)
−
(
4
∙
1
∙
−
7
.
76
𝑥
10
)
2
∙
1
→
𝑥
=
−
[
0
.
0089
]
𝑟𝑒𝑗𝑒𝑡
é
,
𝑥
=
[
0
.
0086
]
𝑂𝐻
(
)
É
=
[
0
.
0086
]
→
𝑝𝑂𝐻
=
−
log
[
0
.
0086
]
=
2
.
07
𝑝𝐻
+
𝑝𝑂𝐻
=
14
→
𝑝𝐻
=
14
−
𝑝𝑂𝐻
=
14
−
2
.
07
=
11
.
9
Le
(
𝐶
𝐻
)
𝑁𝐻
(
)
possède le plus grand pH.
25.
𝐾
=
[
1
𝑥
10
]
[
1
𝑥
10
]
∙
[
0
.
1
−
1
𝑥
10
]
=
1
𝑥
10
L'acide ayant un K
a
est le H
2
S.
26.
𝟏
𝑪
𝟓
𝑯
𝟓
𝑪𝑶𝑶𝑯
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝟏
𝑪
𝟓
𝑯
𝟓
𝑪𝑶𝑶
(
𝒂𝒒
)
+
𝟏
𝑯
𝟑
𝑶
(
𝒂𝒒
)
D
[
0
.
012
]
-
[
0
]
[
0
]
R
-
[
𝑥
]
- +
[
𝑥
]
+
[
𝑥
]
É
[
0
.
012
−
𝑥
]
-
[
𝑥
]
[
𝑥
]
𝐾
=
𝐶
𝐻
𝐶𝑂𝑂
(
)
∙
𝐻
𝑂
(
)
𝐶
𝐻
𝐶𝑂𝑂𝐻
(
)
=
[
𝑥
]
∙
[
𝑥
]
[
0
.
012
−
𝑥
]
=
𝑥
0
.
012
−
𝑥
=
6
.
4
𝑥
10
𝑥
+
6
.
4
𝑥
10
𝑥
−
7
.
68
𝑥
10
𝑥
=
−
𝑏
±
√
𝑏
−
4
𝑎𝑐
2
𝑎
=
−
6
.
4
𝑥
10
±
(
6
.
4
𝑥
10
)
−
(
4
∙
1
∙
7
.
68
𝑥
10
)
2
∙
1
→
𝑥
=
−
[
0
.
00091
]
𝑟𝑒𝑗𝑒𝑡
é
,
𝑥
=
[
0
.
00084
]
𝑝𝐻
=
−
log
[
𝐻
]
=
−
log
[
0
.
00084
]
=
3
.
1
27.
𝑝𝐻
+
𝑝𝑂𝐻
=
14
→
𝑝𝑂𝐻
=
14
−
𝑝𝐻
=
14
−
8
.
2
=
5
.
8
[
𝑂𝐻
]
=
10
=
10
.
=
1
.
58
𝑥
10
𝑹𝑵𝑯
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝑹𝑵𝑯
𝟐
(
𝒂𝒒
)
+
𝟏
𝑶𝑯
(
𝒂𝒒
)
D
[
0
.
2
]
-
[
0
]
[
0
]
R
-
[
1
.
58
𝑥
10
]
- +
[
1
.
58
𝑥
10
]
+
[
1
.
58
𝑥
10
]
É
[
0
.
2
−
1
.
58
𝑥
10
]
-
[
1
.
58
𝑥
10
]
[
1
.
58
𝑥
10
]

𝐾
=
𝑅𝑁𝐻
2
(
𝑎𝑞
)
+
∙
𝑂𝐻
(
)
𝑅𝑁𝐻
(
𝑎𝑞
)
=
[
1
.
58
𝑥
10
−
6
]
∙
[
1
.
58
𝑥
10
−
6
]
[
0
.
20
−
1
.
58
𝑥
10
−
6
]
=
1
.
26
𝑥
10
28. a)
Chang
.
Réac
.
𝟏
𝑯𝑵𝑶
𝟐
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝟏
𝑯
𝟑
𝑶
(
𝒂𝒒
)
+
𝟏
𝑵𝑶
𝟐
(
𝒂𝒒
)
Rép.
-
Si on ajoute du
𝑁𝑂
(
)
on favorise la réaction inverse. et on descend la
concentration de
𝐻
𝑂
(
)
donc le pH augmente (plus basique).
b)
Chang
. 
Réac
.
𝟏
𝑯𝑵𝑶
𝟐
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝟏
𝑯
𝟑
𝑶
(
𝒂𝒒
)
+
𝟏
𝑵𝑶
𝟐
(
𝒂𝒒
)
Rép.
 -  
Si on ajoute du
𝐻𝑁𝑂
(
)
on favorise la réaction inverse. et on descend la
concentration de
𝐻
𝑂
(
)
donc le pH augmente (plus basique).
c)
Chang
.  
Réac
.
𝟏
𝑯𝑵𝑶
𝟐
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝟏
𝑯
𝟑
𝑶
(
𝒂𝒒
)
+
𝟏
𝑵𝑶
𝟐
(
𝒂𝒒
)
Rép.
 -  
Si on ajoute du
𝑂𝐻
(
)
on descend la concentration du
𝐻
𝑂
(
)
et le système tant
à le reformer en augmentant sa concentration car on favorise la réaction directe.
Le pH diminue (devient plus acide).
29.
[
𝐻
]
=
10
=
10
.
=
[
5
.
01
𝑥
10
]
5
.
01
𝑥
10
𝑚𝑜𝑙𝑒
1000
𝑚𝑙
=
𝑥
𝑚𝑜𝑙𝑒
100
𝑚𝑙
→
𝑥
=
5
.
01
𝑥
10
𝑚𝑜𝑙𝑒
∙
100
𝑚𝑙
1000
𝑚𝑙
=
5
.
01
𝑥
10
𝑚𝑜𝑙𝑒
5
.
01
𝑥
10
𝑚𝑜𝑙𝑒
100
%
=
𝑥
𝑚𝑜𝑙𝑒
95
%
→
𝑥
=
5
.
01
𝑥
10
𝑚𝑜𝑙𝑒
∙
95
%
100
%
=
4
.
76
𝑥
10
𝑚𝑜𝑙𝑒
30.
𝑪
𝟔
𝑯
𝟓
𝑵𝑯
𝟐
(
𝒂𝒒
)
+
𝟏
𝑯
𝟐
𝑶
(
𝒍
)
𝑪
𝟔
𝑯
𝟓
𝑵𝑯
𝟑
(
𝒂𝒒
)
+
𝟏
𝑶𝑯
(
𝒂𝒒
)
D
[
0
.
012
]
-
[
0
]
[
0
]
R
-
[
𝑥
]
- +
[
𝑥
]
+
[
𝑥
]
É
[
0
.
012
−
𝑥
]
-
[
𝑥
]
[
𝑥
]