This document provides an overview of acid-base concepts including definitions, examples, and calculations. It defines acids as producing H+ ions and bases as producing OH- ions. Strong acids and bases fully dissociate in water while weak acids and bases partially dissociate. Acids are proton donors and bases are proton acceptors. The document also discusses using pH and pOH scales to calculate hydrogen and hydroxide ion concentrations in solutions and determine whether they are acidic or basic.

2. Peta konsep
ASAM BASA
PENGERTIAN
ARRENIUS
BROWNSTED LOWRY
LEWIS
ASAM
ASAM KUAT
ASAM LEMAH
BASA
BASA KUAT
BASA LEMAH
MANFAAT
2

3. ASAM : Menghasilkan H+ atau (H3O+)
Exp : HCl(aq) → H+
(aq) + Cl-
(aq)
BASA : Menghasilkan OH-
Exp : NaOH(aq) → Na+
(aq) + OH-
(aq)
Menghasilkan
H+
Menghasilkan
OH-
3

4. ASAM : donor H+ atau proton
Exp : HCl(aq) + H2O(aq) → Cl-
(aq) + H3O+
(aq)
BASA : aseptor H+ atau proton
Exp : NH3(aq) + H2O(aq) → NH4
+
(aq) + OH-
(aq)
Asam Basa Konjugasi
Basa Asam Konjugasi
4By : Ika Septiana_ReSaint

5. ASAM : aseptor e-
BASA : donor e-
Exp : H F
H-N̈-H + F-B-F → H-N : B-F
H F H F
Memiliki PEB untuk donor
Tidak Memiliki PEB,maka
Aseptor
5By : Ika Septiana_ReSaint

6.  Larutan Elektrolit
 pH < 7
 rasanya asam
 ABIMA (Asam Biru
Merah)
 Asam Kuat → ἁ = 1
 Asam Lemah → ἁ < 1
6By : Ika Septiana_ReSaint

7. Exp : HCl, HBr, H2SO4, HNO3, Dll.
Hitung pH
Contoh : Berapa pH larutan HCl 0.01 M ?
Jawab : HCl(aq) → H+
(aq) + Cl-
(aq)
0.01 M 0.01 M
Maka [H+]=0.01 M → pH= -Log[H+]
= - log 10-2
= 2
pH= -Log [H+]
7By : Ika Septiana_ReSaint

8. Exp : HF, H2SO3, H3PO4, HNO2,
CH3COOH,HCOOH, H2CO2 Dll.
Hitung pH: [H+]=
Contoh:Berapa pH larutan HF 0.01 M dg Ka=10-4 ?
Jawab : [H+]= = = 10-3
Maka [H+]= 10-3 M → pH= -Log[H+]
= - log 10-3
= 3
pH= -Log [H+]
8
By : Ika Septiana_ReSaint

9.  Larutan Elektrolit
 pH > 7
 rasanya pahit
 BAMBU (Basa Merah
Biru)
 Basa Kuat → ἁ = 1
 Basa Lemah → ἁ < 1
9By : Ika Septiana_ReSaint

10. Exp : NaOH, Ca(OH)2,KOH, Ba(OH)2, Dll.
Hitung pH:
Contoh : Berapa pH larutan Ca(OH)2 0.005 M ?
Jawab : Ca(OH)2(aq) → Ca+2
(aq) + 2OH-
(aq)
0.005 M 0.01 M
Maka [OH-]=0.01 M → pOH= -Log[OH-]
= - log 10-2
pH=14-pH = 2
pH= 14-2 =12
pOH= -Log [OH-]
pH=14 - pOH
10By : Ika Septiana_ReSaint

11. Exp : NH3, Al(OH)3,Fe(OH)3 Dll.
Hitung pH: [OH-]=
Contoh : Berapa pH larutan NH3 0.01 M dg Kb=10-4 ?
Jawab : [NH3]=10-2
[0H-] = = = 10-3
Maka [H+]= 10-3 M → pOH = -Log[0H-]
= - log 10-3
pH=14-pH = 3
pH= 14-3 =11
pOH= -Log [OH-]
pH=14 - pOH
11By : Ika Septiana_ReSaint

12. 12By : Ika Septiana_ReSaint

13. 13