CHM514Leçon6KaCorr

Capsule
d’études
secondaires
2020
CHM 514
La constante d'acidité
Prob. 6
Corrigé
Par S. Lavallée
14.
a)
b)
𝑝𝐻 = − log [𝐻 ] = − log [2.8𝑥10 ] = 7.6
𝑝𝐻 = − log [𝐻 ] = − log [3.6𝑥10 ] = 7.4
15.
[𝐻 ] = 10
16.
1 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙 𝑥 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙
1 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙 ∙ 2.00 𝑔
=
→𝑥=
= 5.48𝑥10 𝑚𝑜𝑙𝑒
36.5 𝑔
2.00 𝑔
36.5 𝑔
𝑐=
= 10
.
= [2.5𝑥10 ]
𝑛 5.48𝑥10 𝑚𝑜𝑙𝑒
=
= [2.19𝑥10 ]
𝑣
2.5𝐿
𝑝𝐻 = − log [𝐻 ] = − log [2.19𝑥10 ] = 1.7
17.
18.
19.
20.
a)
b)
c)
d)
𝐻𝑁𝑂 (
𝐻 𝑆(
𝐻𝑂𝐶𝑙 (
𝐻𝑆𝑂 (
)
+
+
+
+
a)
b)
c)
d)
𝐶𝐻 𝑁𝐻 (
𝐶 𝐻 𝑁( )
(𝐶𝐻 ) 𝑁(
𝐻𝑆𝑒( )
+
+
+
+
a)
𝐻 𝑆𝑂
A
+
b)
𝑁𝐻 𝑂𝐻(
B
c)
𝐻 𝑃𝑂
A
d)
𝐻𝐶𝑂 (
B
a)
b)
c)
d)
𝐻𝐵𝑟( )
𝑁𝐻 𝑁𝐻 ( )
𝐻𝑁𝑂 ( )
𝐶 𝐻 𝑁𝐻 ( )
)
)
(
(
)
)
)
)
)
𝐻
𝐻
𝐻
𝐻
𝑂( )
𝑂( )
𝑂( )
𝑂( )
𝐻
𝐻
𝐻
𝐻




𝐻
𝐻
𝐻
𝐻




𝑂( )
𝑂( )
𝑂( )
𝑂( )
𝐻 𝐶𝑂
A
(
+
𝐻 𝑂( )
B
+
𝐻𝐶𝑂𝑂𝐻(
A
+
+
+
+
𝐻
𝐻
𝐻
𝐻
)
)
)
)
+
+
+
+
𝐶𝐻 𝑁𝐻 (
𝐶 𝐻 𝑁𝐻(
(𝐶𝐻 ) 𝑁𝐻(
𝐻 𝑆𝑒( )
𝑁𝑂 (
𝑂𝐶𝑙(
𝐻𝑆(
𝑆𝑂 (
)
)
)
+
+
+
+

𝐻𝑆𝑂 (
BC

𝐻𝐶 𝑂 (
BC

𝐻 𝑂(
AC
)

𝐻𝐶𝑂𝑂(
BC
𝑂( )
𝑂( )
𝑂( )
𝑂( )




𝐻 𝑂(
𝑁𝐻 𝑁𝐻
𝐻 𝑂(
𝐶 𝐻 𝑁𝐻
(𝐶𝐻 ) 𝑁𝐻(
B
+
𝑂(
𝑂(
𝑂(
𝑂(
)
)
𝑂𝐻(
𝑂𝐻(
𝑂𝐻(
𝑂𝐻(
)
)
)
)
)
)
+ 𝑁𝐻 𝑂𝐻(
AC
)
)
)
)
)
)
(
)
+ (𝐶𝐻 ) 𝑁𝐻(
AC
)
(
)
)
+
𝐻 𝑃𝑂 (
BC
+
𝐻 𝐶𝑂 (
AC
+ 𝐵𝑟( )
+ 𝑂𝐻( )
+ 𝑁𝑂 ( )
+ 𝑂𝐻( )
)
)
)
)
21.
𝑝𝑂𝐻 = − log [𝑂𝐻 ] = − log [2.3𝑥10 ] = 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 . = 5.0𝑥10
23.
a)
b)
c)
d)
𝑝𝐻 = − log [𝐻 ] = − log [7.8𝑥10 ] = 7.1
𝑝𝑂𝐻 = − log [𝑂𝐻 ] = − log [0.0001] = 4 → 𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 14 − 4 = 10
𝑝𝐻 = − log [𝐻 ] = − log [0.0065] = 2.2
𝑝𝑂𝐻 = − log [𝑂𝐻 ] = − log [0.3] = 5.2 → 𝑝𝐻 = 14 − 5.2 = 14 − 4.6 = 8.8
c<a<d<b
24.
D
R
É
𝐾 =
+ 𝟏 𝑯𝟐 𝑶(𝒍)
-
𝟏 𝑵𝑯𝟑 (𝒂𝒒)
[0.25]
- [𝑥]
[0.25 − 𝑥]
𝑁𝐻
(
)
𝑁𝐻
∙ 𝑂𝐻(
(
)
=
)

𝟏 𝑵𝑯𝟒 (𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
+
𝟏 𝑶𝑯(𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
[𝑥] ∙ [𝑥]
𝑥
=
= 1.79𝑥10
[0.25 − 𝑥] 0.25 − 𝑥
𝑥 + 1.79𝑥10 𝑥 − 4.48𝑥10
𝑥=
± (1.79𝑥10 ) − (4 ∙ 1 ∙ −4.48𝑥10 )
→
2∙1
−𝑏 ± √𝑏 − 4𝑎𝑐 −1.79𝑥10
=
2𝑎
𝑥 = −[0.021] 𝑟𝑒𝑗𝑒𝑡é, 𝑥 = [0.021]
𝑂𝐻(
)É
= [0.021] → 𝑝𝑂𝐻 = − log [0.021] = 1.68
𝑝𝐻 + 𝑝𝑂𝐻 = 14 → 𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 14 − 1.68 = 1.23
D
R
É
𝐾 =
+ 𝟏 𝑯𝟐 𝑶(𝒍 
-
𝟏 (𝑪𝟐 𝑯𝟓 )𝟐 𝑵𝑯(𝒂𝒒)
[0.25]
- [𝑥]
[0.25 − 𝑥]
(𝐶 𝐻 ) 𝑁𝐻
(
)
∙ 𝑂𝐻(
(𝐶 𝐻 ) 𝑁𝐻(
)
𝑥 + 3.09𝑥10 𝑥 − 7.76𝑥10
)
=
(𝑪𝟐 𝑯𝟓 )𝟐 𝑵𝑯𝟐 (𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
+
[𝑥] ∙ [𝑥]
𝑥
=
= 3.09𝑥10
[0.25 − 𝑥] 0.25 − 𝑥
𝟏 𝑶𝑯(𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
𝑥=
−𝑏 ± √𝑏 − 4𝑎𝑐 −3.09𝑥10
=
2𝑎
± (3.09𝑥10 ) − (4 ∙ 1 ∙ −7.76𝑥10 )
→
2∙1
𝑥 = −[0.0089] 𝑟𝑒𝑗𝑒𝑡é, 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(
25.
𝐾 =
)
possède le plus grand pH.
[1𝑥10 ][1𝑥10 ] ∙
= 1𝑥10
[0.1 − 1𝑥10 ]
L'acide ayant un Ka est le H2S.
26.
𝟏 𝑪𝟓 𝑯𝟓 𝑪𝑶𝑶𝑯(𝒂𝒒) + 𝟏 𝑯𝟐 𝑶(𝒍)
[0.012]
D
R
- [𝑥]
[0.012 − 𝑥]
É
𝐾 =
𝐶 𝐻 𝐶𝑂𝑂(
)
∙ 𝐻 𝑂(
𝐶 𝐻 𝐶𝑂𝑂𝐻(
)
=
)

+
𝟏 𝑪𝟓 𝑯𝟓 𝑪𝑶𝑶(𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
𝟏 𝑯𝟑 𝑶(𝒂𝒒)
[0]
+ [𝑥]
[𝑥]
[𝑥] ∙ [𝑥]
𝑥
=
= 6.4𝑥10
[0.012 − 𝑥] 0.012 − 𝑥
𝑥 + 6.4𝑥10 𝑥 − 7.68𝑥10
𝑥=
−𝑏 ± √𝑏 − 4𝑎𝑐 −6.4𝑥10
=
2𝑎
± (6.4𝑥10 ) − (4 ∙ 1 ∙ 7.68𝑥10 )
→
2∙1
𝑥 = −[0.00091] 𝑟𝑒𝑗𝑒𝑡é, 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 . = 1.58𝑥10
D
R
É
𝑹𝑵𝑯(𝒂𝒒)
[0.2]
- [1.58𝑥10 ]
[0.2 − 1.58𝑥10 ]
+
𝟏 𝑯𝟐 𝑶(𝒍)
-

𝑹𝑵𝑯𝟐 (𝒂𝒒)
[0]
+ [1.58𝑥10 ]
[1.58𝑥10 ]
+
𝟏 𝑶𝑯(𝒂𝒒)
[0]
+ [1.58𝑥10 ]
[1.58𝑥10 ]
𝐾 =
28.
a)
𝑅𝑁𝐻+
2 (𝑎𝑞) ∙ 𝑂𝐻(
)
=
𝑅𝑁𝐻(𝑎𝑞)
Chang.
Réac.
Rép.
[1.58𝑥10−6 ] ∙ [1.58𝑥10−6 ]
= 1.26𝑥10
[0.20 − 1.58𝑥10−6 ]
+ 𝟏 𝑯𝟐 𝑶(𝒍)
-
𝟏 𝑯𝑵𝑶𝟐 (𝒂𝒒)

𝟏 𝑯𝟑 𝑶(𝒂𝒒)

+

𝟏 𝑵𝑶𝟐 (𝒂𝒒)

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.
Rép.

𝟏 𝑯𝑵𝑶𝟐 (𝒂𝒒)

Chang.
Réac.
Rép.
𝟏 𝑯𝟑 𝑶(𝒂𝒒)
+
𝟏 𝑵𝑶𝟐 (𝒂𝒒)

-
Si on ajoute du 𝐻𝑁𝑂 (
concentration de 𝐻 𝑂(
c)
+ 𝟏 𝑯𝟐 𝑶(𝒍)

on favorise la réaction inverse. et on descend la
) donc le pH augmente (plus basique).
)

𝟏 𝑯𝑵𝑶𝟐 (𝒂𝒒)

+ 𝟏 𝑯𝟐 𝑶(𝒍)
𝟏 𝑯𝟑 𝑶(𝒂𝒒)

+
𝟏 𝑵𝑶𝟐 (𝒂𝒒)

-

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
.
= [5.01𝑥10 ]
5.01𝑥10 𝑚𝑜𝑙𝑒 𝑥 𝑚𝑜𝑙𝑒
5.01𝑥10 𝑚𝑜𝑙𝑒 ∙ 100 𝑚𝑙
=
→𝑥=
= 5.01𝑥10
1000 𝑚𝑙
100 𝑚𝑙
1000 𝑚𝑙
5.01𝑥10 𝑚𝑜𝑙𝑒 𝑥 𝑚𝑜𝑙𝑒
5.01𝑥10 𝑚𝑜𝑙𝑒 ∙ 95 %
=
→𝑥=
= 4.76𝑥10
100 %
95 %
100 %
30.
D
R
É
𝑪𝟔 𝑯𝟓 𝑵𝑯𝟐 (𝒂𝒒)
[0.012]
- [𝑥]
[0.012 − 𝑥]
+ 𝟏 𝑯𝟐 𝑶(𝒍)
-

𝑪𝟔 𝑯𝟓 𝑵𝑯𝟑 (𝒂𝒒) + 𝟏 𝑶𝑯(𝒂𝒒)
[0]
[0]
+ [𝑥]
+ [𝑥]
[𝑥]
[𝑥]
𝑚𝑜𝑙𝑒
𝑚𝑜𝑙𝑒
𝐾 =
𝐶6 𝐻5 𝑁𝐻+
3 (𝑎𝑞) ∙ 𝑂𝐻(
𝐶6 𝐻5 𝑁𝐻2 (𝑎𝑞)
)
=
[𝑥] ∙ [𝑥]
𝑥
=
= 3.8𝑥10
[0.012 − 𝑥] 0.012 − 𝑥
𝑥 + 3.8𝑥10 𝑥 − 4.56𝑥10
𝑥=
−𝑏 ± √𝑏 − 4𝑎𝑐 −3.8𝑥10
=
2𝑎
± (3.8𝑥10 ) − (4 ∙ 1 ∙ −4.56𝑥10
2∙1
𝑥 = −[2.14𝑥10 ] 𝑟𝑒𝑗𝑒𝑡é, 𝑥 = [2.14𝑥10 ]
𝑂𝐻(
)É
= [2.14𝑥10 ] → 𝑝𝑂𝐻 = − log [2.14𝑥10 ] = 5.7
𝑝𝐻 + 𝑝𝑂𝐻 = 14 → 𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 14 − 5.7 = 8.3
)
→