AP Chemistry: Aqueous Equilibrium
Guide
1. Acid-Base Fundamentals Example: Weak Acid pH
The pH Square Find pH of 0.20 M HC2H3O2 (Ka = 1.8 × 10−5 ).
Step 1: Setup K expression
Kw = [H+ ][OH− ]
[H3O+] [OH–] [x][x] x2
Ka = ≈
[0.20 − x] 0.20
− log[H+ ] 10−pH 10−pOH − log[OH− ] Step 2: Solve for x
p
pH + pOH = 14 x = 1.8 × 10−5 × 0.20 = 0.0019 M
pH pOH
Step 3: Convert to pH
pH = − log(0.0019) = 2.72
2. Identifying Acids & Bases
MUST MEMORIZE: The Strong Species
4. Buffers
Strong Acids (Dissociate 100%): Key Formulas
• HCl, HBr, HI Henderson-Hasselbalch Equation:
• HNO3, H2SO4, HClO4
Strong Bases (Dissociate 100%):
[Base]
• Group 1 Hydroxides (e.g., NaOH, KOH) pH = pKa + log
[Acid]
• Heavy Group 2 (Ca2+, Sr2+, Ba2+) Hydroxides
Best capacity when [Base] ≈ [Acid].
Tip: How to Spot Weak Species
Weak Acids: 5. Titrations
• Any acid not on the Strong List.
• HF is WEAK. Key Points on the Curve:
• -COOH group (Carboxylic acids).
• Acids are generally identified in chemical formulas • Half-Equiv: pH = pKa .
by having hydrogen at the beginning, followed by
a nonmetal or polyatomic anion • Equivalence: Moles Acid = Moles Base.
Weak Bases:
• Ammonia (NH3) & Amines (CH3NH2). Example: Titration Calculation (Hard!)
• Anions of Weak Acids (e.g., F–, CN–).
• Bases can be identified from their formula by look- Problem: Titrate 50mL of 0.1M CH3COOH (Ka =
1.8e-5) with 0.1M NaOH. Find pH at Equivalence Pt.
ing for compounds that release hydroxide (OH–)
1. Stoichiometry (Moles): nacid = 0.05L × 0.1M =
ions, such as metal hydroxides (MOH), contain 0.005 mol. At Eq. Pt, added 0.005 mol NaOH. All
basic anions like carbonate or bicarbonate, or Acid converts to Conj. Base (CH3COO–). Total Vol
are nitrogen-containing compounds like ammonia = 50mL + 50mL = 100mL (0.10L).
(NH3). A common, easy indicator is finding a 2. New Concentration: [A–] = 0.005 mol
0.10 L
= 0.05 M.
metal combined with (OH–) 3. Equilibrium (Base Hydrolysis): A– +
H2O HA + OH– We need Kb : Kb = K w
Ka
=
10−14
= 5.6e-10.
1.8e-5
4. Solve for [OH-]:
3. Weak Acid Math x2
5.6e-10 = ⇒ x = 5.3e-6 M (OH-)
0.05
Use RICE tables. Assume x is small if K < 10−4 .
5. Finish: pOH = − log(5.3e-6) = 5.28. pH =
14 − 5.28 = 8.72.
HA + H2O A– + H3O+
1
Guide
1. Acid-Base Fundamentals Example: Weak Acid pH
The pH Square Find pH of 0.20 M HC2H3O2 (Ka = 1.8 × 10−5 ).
Step 1: Setup K expression
Kw = [H+ ][OH− ]
[H3O+] [OH–] [x][x] x2
Ka = ≈
[0.20 − x] 0.20
− log[H+ ] 10−pH 10−pOH − log[OH− ] Step 2: Solve for x
p
pH + pOH = 14 x = 1.8 × 10−5 × 0.20 = 0.0019 M
pH pOH
Step 3: Convert to pH
pH = − log(0.0019) = 2.72
2. Identifying Acids & Bases
MUST MEMORIZE: The Strong Species
4. Buffers
Strong Acids (Dissociate 100%): Key Formulas
• HCl, HBr, HI Henderson-Hasselbalch Equation:
• HNO3, H2SO4, HClO4
Strong Bases (Dissociate 100%):
[Base]
• Group 1 Hydroxides (e.g., NaOH, KOH) pH = pKa + log
[Acid]
• Heavy Group 2 (Ca2+, Sr2+, Ba2+) Hydroxides
Best capacity when [Base] ≈ [Acid].
Tip: How to Spot Weak Species
Weak Acids: 5. Titrations
• Any acid not on the Strong List.
• HF is WEAK. Key Points on the Curve:
• -COOH group (Carboxylic acids).
• Acids are generally identified in chemical formulas • Half-Equiv: pH = pKa .
by having hydrogen at the beginning, followed by
a nonmetal or polyatomic anion • Equivalence: Moles Acid = Moles Base.
Weak Bases:
• Ammonia (NH3) & Amines (CH3NH2). Example: Titration Calculation (Hard!)
• Anions of Weak Acids (e.g., F–, CN–).
• Bases can be identified from their formula by look- Problem: Titrate 50mL of 0.1M CH3COOH (Ka =
1.8e-5) with 0.1M NaOH. Find pH at Equivalence Pt.
ing for compounds that release hydroxide (OH–)
1. Stoichiometry (Moles): nacid = 0.05L × 0.1M =
ions, such as metal hydroxides (MOH), contain 0.005 mol. At Eq. Pt, added 0.005 mol NaOH. All
basic anions like carbonate or bicarbonate, or Acid converts to Conj. Base (CH3COO–). Total Vol
are nitrogen-containing compounds like ammonia = 50mL + 50mL = 100mL (0.10L).
(NH3). A common, easy indicator is finding a 2. New Concentration: [A–] = 0.005 mol
0.10 L
= 0.05 M.
metal combined with (OH–) 3. Equilibrium (Base Hydrolysis): A– +
H2O HA + OH– We need Kb : Kb = K w
Ka
=
10−14
= 5.6e-10.
1.8e-5
4. Solve for [OH-]:
3. Weak Acid Math x2
5.6e-10 = ⇒ x = 5.3e-6 M (OH-)
0.05
Use RICE tables. Assume x is small if K < 10−4 .
5. Finish: pOH = − log(5.3e-6) = 5.28. pH =
14 − 5.28 = 8.72.
HA + H2O A– + H3O+
1
