Acid-base Equilibria Chem 17 Handout New

A�� B�� II: R�� R�� A�� B�� D�� • A�� D�� +

H (aq) + OH

•

•

(aq)

→ H 2 O (l ) ←

ACID AC IDS S – dona donate tes s H+ HNO3, H3PO4, H2�O4, HC�, HI, HB�, CH3COOH, ��COOH, H2�O3

•

•

BASES – donates OHN�OH, KOH, L�OH, C�OH, M�(OH)2,��(OH)2,B�(OH)2 , C�(OH)2

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• B��L�� D�� + + NH 4(aq) H 2O(l) + NH3(g) ⇔ OH (aq)

•

ACID AC IDS S – dona donate tes s H+ (proton donor)

•

•

•

•

HNO3, H3PO4, H2�O4, HC�, HI, HB�,

BASES – accepts H+ (proton acceptor)

•

NH3, ��NH2, N�OH, KOH, L�OH, C�OH, M�(OH) ,��(OH) ,B�(OH) , C�(OH)















• B��L�� D��: C�� L��















• A��: – �� . – A�� A��:: �� H��42� + H2�  H2��4� + �H� H��42� + H2�  ��43� + H3�+















• L�� D�� NH base

3

+ H 2O → ← NH

•

4

+ OH

-

acid

NaF + BF base

+

3

→ Na ←

+

+ BF

4

acid

ACID AC IDS S – elect electro ronn-pa pair ir acce accept ptor or

•

BASES BASES – electron electron-pai -pairr donor donor















A�� B��: C�� 16  17 • �� A��

+ → H 2O(l) + H 2O(l) ← H3O(aq) + OH(aq)

E�� E�� :

B�� (�� ) �� 25

K c =

[ H 3O + ][OH - ] [H 2O]2 [

O + ][OH ]

















�H = �� H3O+� �� H+� �OH = ��OH�� K� = 1.00 � 10�14 = �H+��OH�� 25�C ∴ �OH + �H = 14.00

















• C�� H3O+�, �H, �OH��, �� OH �� 0.020 � HNO3 ��. – I� HNO3 � �� ? – �� H3O+� ?

















• �� H �� ? I� �� ? – G��: ∆H� �H2O(�) = �285.83 �J/�� ∆H� � OH�(��) = �230.0 �J/��

















R�� R�� A�� B�� C�� A��B�� P�� 

�� , �� . ��.



�� , ��

















I�� C�� M�� A�� B�� , ��, • C�� CH3COOH. �� ?























































I�� C�� M�� A��





































←H O + + CH COO CH COOH + H O →









































































































































































































+ H 2 CO3(aq) + H 2 O (l) ↔ HCO3(aq) + H 3 O (aq) 2+ + H 2 O (l) ↔ CO3(aq) + H 3 O (aq) HCO3(aq) 2− + H 2 O(l) ↔ HCO3(aq) CO3(aq) + HO(aq) − + H 2 O (l) ↔ H 2 CO3(aq) + HO(aq) HCO3(aq)

K a1

K b1 K

K a1 = 4.4 x 10-7 K a2 = 4.7 x 10 -11 K b1 = 2.1 x 10-4 K b2 = 2.3 x 10-8















+ H3PO4(aq) + H2O(l) → H2 PO-4(aq) + H3O(aq) 4(aq)

H2 PO

+ H2O(l) → HPO

24(aq)

+

+ H3O(aq)

2+ + H2O(l) → PO34(aq) + H3O(aq) HPO4(aq)

K a1 = 7.50 x 10-3 K a2 = 6.20 x 10

-8

K a3 = 3.60 x 10-13

− 2+ H 2 O (l) → HPO 4(aq) PO 34(aq) + HO (aq)

K b1

− 2HPO 4(aq) + H 2 O (l) → H 2 PO -4(aq) + HO (aq)

K b2

− H 2 PO -4(aq) + H 2 O (l) → H 3 PO 4(aq) + HO (aq)

K

K

K b3















• �� (�� ) �� − HCO3(aq) + H 2 O(l) → H 2 CO3(aq) + OH(aq) 2+ + H 2 O(l) → CO3(aq) + H3O(aq) HCO3(aq)

K b2 = 2.38 x 10-8 K a2 = 4.80 x 10-11

2+ + H3O(aq) H 2 PO-4(aq) + H 2 O(l) → HPO4(aq)

K a2 = 6.20 x 10-8

− H 2 PO-4(aq) + H 2 O(l) → H3 PO4(aq) + OH(aq)

K b3 = 1.33 x 10-12















• �� H �� 1.52 � �� N�H 2PO4�2H2O �� 50.00 �L ��? I� �� N� 2HPO4�2H2O ��, �� ?

• �� (2.0 M HC�) �� . �� , �� , �� .















I�� L�� A�� H�� M�� I�� • B�� , �� . �� – D�� , �� H+ Fe(H2 O)

3+ 6 ( aq )

→ + 2+ Fe(H2 O)5 (OH)(aq) + H (aq) ←

K a = 2.0 x 10-3

















�� A�� E��: N�C� (�� HC� �� N�OH) K2�O4 (�� KOH �� H 2�O4) 

















�� • A�� , �� . �� K �� K� �� H.

















C�� I�� E��

















C�� I�� E�� A �� L� C�� C�� • C�� 0.05 M �� , CH CH3COOH (50.00 �L)

→ − + CH3COOH(aq) + H 2 O( l) CH3COO(aq) + H 3O(aq) ←

K a = 1.8 x 10-5

















C�� I�� E�� A �� L� C�� C��

��, • C�� 0.05 M �� , CH3COOH (50.00 �L)























































H��H�� E�� E��













































































































































































































































�� C�� D�� A�� B��















• A 0.1044�� 0.1044�� 22.10 �L �� 0.0500 M N�OH �� . (�) �� ? (�) A� �� , �� H �� 11.05 �L �� 4.89. �� K� �� ?



A �� 750 750 �L �� H 4.50. �� , N�C 2H3O2, �� , HC 2H3O2, �� . G�� 99% �� 1.05 �/�L. I� ��

















�� H �� 0.30 �� N�OH, 0.25 �� N� 2HPO4 , �� 0.20 �� H 3PO4 �� 1.00 L? H3PO4 �K�1 = 2.12 �K�2 = 7.21 �K�3 = 12.38

Acid-base Equilibria Chem 17 Handout New

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