Dilution prep pH / pOH basics 8.2 Ka from pH 8.3 Conjugate pairs & Keq 8.3
FORMULA CORE How to solve: pH → [H⁺] → ICE table GOLDEN RULE
M₁V₁ = M₂V₂ pH = −log[H⁺] | pOH = −log[OH⁻] → Ka Ka × Kb = Kw = 1.0×10⁻¹⁴
How to solve: plug in 3 knowns, solve ALWAYS TRUE AT 25°C 1 [H⁺] = 10⁻ᵖᴴ = x Kb = Kw / Ka
for 4th. Always check units — mL or L, pH + pOH = 14 2 Ka = x² / (C₀ − x) Larger Ka → smaller Kb (conjugate)
be consistent.
Find [H⁺] [H⁺] = 10⁻ᵖᴴ 3 If x << C₀, use Ka ≈ x²/C₀ Keq > 1 stronger species on LEFT
Q: Make 100 mL of 0.115 M from 2.000 Strong acid pH pH = −log[HA] Keq < 1 stronger species on RIGHT
Q: 0.115 M acid, pH 2.92
M stock
Strong base pH pOH=−log[MOH]→pH=14−pOH x = 1.20×10⁻³
V₁ = (0.115×100)/2.000 = 5.75 mL Q: Ka acetic = 1.8×10⁻⁵
Ka = (1.20×10⁻³)² / (0.115−0.0012)
Kb acetate = 10⁻¹⁴/1.8×10⁻⁵ =
Q: pH=2.92 → [H⁺]=10⁻²·⁹²=1.20×10⁻³ = 1.27×10⁻⁵
Procedure tip: use volumetric flask (not 5.6×10⁻¹⁰
M
beaker!), add water to the line, use eye % DISSOCIATION
[OH⁻]=10⁻¹⁴/1.20×10⁻³=8.33×10⁻¹² M Particulate diagrams: mostly HA = weak
dropper for last drops, stopper and invert
([H⁺] / C₀) × 100%
to mix. acid (small Ka). Mostly ions = strong
acid.
Net ionic equations — write these every time 8.4 Excess reagent → pH 8.4
NET IONIC 1 mol = M × V(L) for each
REACTION TYPE AP TIP
EQUATION
2 subtract → find excess mol
Strong acid + Strong H⁺ + OH⁻ → H₂O drop ALL spectator ions (Na⁺, K⁺, Cl⁻)
base 3 divide by total volume
Weak acid + Strong base HA + OH⁻ → A⁻ + keep HA molecular (weak = not 4 excess H⁺: pH=−log[H⁺]
H₂O dissociated)
5 excess OH⁻: pOH→pH=14−pOH
Weak base + Strong acid B + H⁺ → BH⁺ keep B molecular
Weak acid + Weak base HA + B ⇌ A⁻ + BH⁺ equilibrium arrow, not one-way Q: 200mL 0.40M HClO₄ + 300mL
0.30M KOH
Brønsted-Lowry pair: identify the proton donor (acid) and acceptor (base). Then name the
mol H⁺=0.080, mol OH⁻=0.090
conjugate: acid loses H⁺ → conjugate base; base gains H⁺ → conjugate acid.
excess OH⁻=0.010mol/0.500L=0.020M
pOH=1.70 → pH=12.30
Equivalence point pH — the most tested concept 8.4/8.5 Molarity from titration 8.5 Half-equiv → Ka from graph 8.5
SPECIES AT EQ. 1:1 RATIO KEY FACT
TITRATION PH HOW TO SOLVE
PT.
Ma × Va = Mb × Vb at half-equiv: pH = pKa
Strong + Strong neutral salt =7 done — it's neutral
1 mol titrant = M × V(L) 1 find equiv. vol on curve
Weak acid + Strong A⁻ (weak base) >7 Kb=Kw/Ka → ICE → [OH⁻] → pOH
base → pH 2 mol analyte = mol titrant 2 halve it → half-equiv vol
Weak base + Strong BH⁺ (weak acid) <7 Ka=Kw/Kb → ICE → [H⁺] → pH 3 M = mol / V(L) analyte 3 read pH at that volume
acid
4 pKa = that pH
Magic phrase AP wants: "The conjugate base/acid hydrolyzes water, producing OH⁻/H⁺ Q: 20.52mL of 0.173M NaOH to eq. pt.
mol=0.02052×0.173=3.55×10⁻³ 5 Ka = 10⁻ᵖᴷᵃ
which makes the solution basic/acidic."
[acid]=3.55×10⁻³/0.02500=0.142M
pKb from graph: same idea but read
Q: weak acid eq. pt. → [A⁻]=0.20M, Kb=5.6×10⁻¹⁰ → x=√(5.6×10⁻¹⁰×0.20)=1.06×10⁻⁵ →
Mixture tip: if NaCl + weak acid mixed pOH at half-equiv → pKb = pOH. Or
pOH=4.97 → pH=9.03
— NaCl does NOT react with NaOH. pKb = 14 − pKa.
Only weak acid reacts. Use only weak
acid moles. Q: equiv at 20mL → half-equiv at 10mL
→ pH=6.03 → pKa=6.03 →
Ka=10⁻⁶·⁰³=9.3×10⁻⁷
Buffers & H-H 8.9 Titration curve shapes — sketch these from memory 8.5 Ksp & solubility 8.6
HENDERSON-HASSELBALCH STRONG + WEAK ACID + STRONG WEAK BASE + STRONG AG₂C₂O₄ EXAMPLE
FEATURE
STRONG BASE ACID
pH = pKa + log([A⁻]/[HA]) Ksp = [Ag⁺]²[C₂O₄²⁻]
Start pH very low (~1) low but higher (~3–4) high (~10–11)
[A⁻]=[HA] pH = pKa exactly Buffer none — steep flat gradual region flat gradual region How to solve: let s = molar solubility.
Add H⁺ A⁻+H⁺→HA (A⁻ drops) region Set up ion concentrations in terms of s,
Half-equiv no meaning = pKa = 14−pKb
substitute into Ksp, solve.
Add OH⁻ HA+OH⁻→A⁻ (HA drops)
pH
Dilute buffer pH stays same (ratio Q: Ksp=5.4×10⁻¹² → 4s³=5.4×10⁻¹² →
Equiv pt. pH =7 >7 <7
s=1.10×10⁻⁴M
unchanged)
Why? neutral salt A⁻+H₂O⇌HA+OH⁻ BH⁺+H₂O⇌B+H₃O⁺
Buffer capacity ↑ when both [HA] and Solubility ↑ in acid anion reacts with H⁺
[A⁻] are large. Best range: pKa ± 1. Reversed curve: if you swap beaker/buret, the curve flips — starts high, ends low (or vice Common ion ↓ solubility (shifts left)
versa). Same equiv. vol. Larger sample: same shape, same equiv. pH, just shifted RIGHT
(more titrant needed). Write net ionic for why solubility
increases in acid: anion + H⁺ →
conjugate acid
Lab glassware precision lab Indicators & oil/water 8.5 AP exam tips — things that lose points all
GLASSWARE PRECISION USE FOR Strong+Strong bromothymol blue (pH~7) MISTAKE WHAT TO DO INSTEAD
Buret ±0.05 mL delivering Weak acid+Strong base phenolphthalein Writing Ka with [H₂O] in denominator Omit pure liquids/solids from equilibrium
titrant (pH 8–10) expressions always
Volumetric ±0.05 mL making Saying strong+strong equiv pt "has no Say "neutral salt forms, neither ion hydrolyzes —
Weak base+Strong acid methyl red (pH 4–
flask exact species" pH=7"
volume 6)
Using mL instead of L in molarity calc Always divide mL by 1000 before using in M×V
Graduated ±0.5 mL rough Pick indicator whose transition range Saying weak acid+strong base equiv pt = Always >7 — conjugate base hydrolyzes, say why
cylinder measurement includes the equivalence point pH. 7
Beaker/flask ±5 mL mixing only
Explore our developer-friendly HTML to PDF API Printed using PDFCrowd HTML to PDF
FORMULA CORE How to solve: pH → [H⁺] → ICE table GOLDEN RULE
M₁V₁ = M₂V₂ pH = −log[H⁺] | pOH = −log[OH⁻] → Ka Ka × Kb = Kw = 1.0×10⁻¹⁴
How to solve: plug in 3 knowns, solve ALWAYS TRUE AT 25°C 1 [H⁺] = 10⁻ᵖᴴ = x Kb = Kw / Ka
for 4th. Always check units — mL or L, pH + pOH = 14 2 Ka = x² / (C₀ − x) Larger Ka → smaller Kb (conjugate)
be consistent.
Find [H⁺] [H⁺] = 10⁻ᵖᴴ 3 If x << C₀, use Ka ≈ x²/C₀ Keq > 1 stronger species on LEFT
Q: Make 100 mL of 0.115 M from 2.000 Strong acid pH pH = −log[HA] Keq < 1 stronger species on RIGHT
Q: 0.115 M acid, pH 2.92
M stock
Strong base pH pOH=−log[MOH]→pH=14−pOH x = 1.20×10⁻³
V₁ = (0.115×100)/2.000 = 5.75 mL Q: Ka acetic = 1.8×10⁻⁵
Ka = (1.20×10⁻³)² / (0.115−0.0012)
Kb acetate = 10⁻¹⁴/1.8×10⁻⁵ =
Q: pH=2.92 → [H⁺]=10⁻²·⁹²=1.20×10⁻³ = 1.27×10⁻⁵
Procedure tip: use volumetric flask (not 5.6×10⁻¹⁰
M
beaker!), add water to the line, use eye % DISSOCIATION
[OH⁻]=10⁻¹⁴/1.20×10⁻³=8.33×10⁻¹² M Particulate diagrams: mostly HA = weak
dropper for last drops, stopper and invert
([H⁺] / C₀) × 100%
to mix. acid (small Ka). Mostly ions = strong
acid.
Net ionic equations — write these every time 8.4 Excess reagent → pH 8.4
NET IONIC 1 mol = M × V(L) for each
REACTION TYPE AP TIP
EQUATION
2 subtract → find excess mol
Strong acid + Strong H⁺ + OH⁻ → H₂O drop ALL spectator ions (Na⁺, K⁺, Cl⁻)
base 3 divide by total volume
Weak acid + Strong base HA + OH⁻ → A⁻ + keep HA molecular (weak = not 4 excess H⁺: pH=−log[H⁺]
H₂O dissociated)
5 excess OH⁻: pOH→pH=14−pOH
Weak base + Strong acid B + H⁺ → BH⁺ keep B molecular
Weak acid + Weak base HA + B ⇌ A⁻ + BH⁺ equilibrium arrow, not one-way Q: 200mL 0.40M HClO₄ + 300mL
0.30M KOH
Brønsted-Lowry pair: identify the proton donor (acid) and acceptor (base). Then name the
mol H⁺=0.080, mol OH⁻=0.090
conjugate: acid loses H⁺ → conjugate base; base gains H⁺ → conjugate acid.
excess OH⁻=0.010mol/0.500L=0.020M
pOH=1.70 → pH=12.30
Equivalence point pH — the most tested concept 8.4/8.5 Molarity from titration 8.5 Half-equiv → Ka from graph 8.5
SPECIES AT EQ. 1:1 RATIO KEY FACT
TITRATION PH HOW TO SOLVE
PT.
Ma × Va = Mb × Vb at half-equiv: pH = pKa
Strong + Strong neutral salt =7 done — it's neutral
1 mol titrant = M × V(L) 1 find equiv. vol on curve
Weak acid + Strong A⁻ (weak base) >7 Kb=Kw/Ka → ICE → [OH⁻] → pOH
base → pH 2 mol analyte = mol titrant 2 halve it → half-equiv vol
Weak base + Strong BH⁺ (weak acid) <7 Ka=Kw/Kb → ICE → [H⁺] → pH 3 M = mol / V(L) analyte 3 read pH at that volume
acid
4 pKa = that pH
Magic phrase AP wants: "The conjugate base/acid hydrolyzes water, producing OH⁻/H⁺ Q: 20.52mL of 0.173M NaOH to eq. pt.
mol=0.02052×0.173=3.55×10⁻³ 5 Ka = 10⁻ᵖᴷᵃ
which makes the solution basic/acidic."
[acid]=3.55×10⁻³/0.02500=0.142M
pKb from graph: same idea but read
Q: weak acid eq. pt. → [A⁻]=0.20M, Kb=5.6×10⁻¹⁰ → x=√(5.6×10⁻¹⁰×0.20)=1.06×10⁻⁵ →
Mixture tip: if NaCl + weak acid mixed pOH at half-equiv → pKb = pOH. Or
pOH=4.97 → pH=9.03
— NaCl does NOT react with NaOH. pKb = 14 − pKa.
Only weak acid reacts. Use only weak
acid moles. Q: equiv at 20mL → half-equiv at 10mL
→ pH=6.03 → pKa=6.03 →
Ka=10⁻⁶·⁰³=9.3×10⁻⁷
Buffers & H-H 8.9 Titration curve shapes — sketch these from memory 8.5 Ksp & solubility 8.6
HENDERSON-HASSELBALCH STRONG + WEAK ACID + STRONG WEAK BASE + STRONG AG₂C₂O₄ EXAMPLE
FEATURE
STRONG BASE ACID
pH = pKa + log([A⁻]/[HA]) Ksp = [Ag⁺]²[C₂O₄²⁻]
Start pH very low (~1) low but higher (~3–4) high (~10–11)
[A⁻]=[HA] pH = pKa exactly Buffer none — steep flat gradual region flat gradual region How to solve: let s = molar solubility.
Add H⁺ A⁻+H⁺→HA (A⁻ drops) region Set up ion concentrations in terms of s,
Half-equiv no meaning = pKa = 14−pKb
substitute into Ksp, solve.
Add OH⁻ HA+OH⁻→A⁻ (HA drops)
pH
Dilute buffer pH stays same (ratio Q: Ksp=5.4×10⁻¹² → 4s³=5.4×10⁻¹² →
Equiv pt. pH =7 >7 <7
s=1.10×10⁻⁴M
unchanged)
Why? neutral salt A⁻+H₂O⇌HA+OH⁻ BH⁺+H₂O⇌B+H₃O⁺
Buffer capacity ↑ when both [HA] and Solubility ↑ in acid anion reacts with H⁺
[A⁻] are large. Best range: pKa ± 1. Reversed curve: if you swap beaker/buret, the curve flips — starts high, ends low (or vice Common ion ↓ solubility (shifts left)
versa). Same equiv. vol. Larger sample: same shape, same equiv. pH, just shifted RIGHT
(more titrant needed). Write net ionic for why solubility
increases in acid: anion + H⁺ →
conjugate acid
Lab glassware precision lab Indicators & oil/water 8.5 AP exam tips — things that lose points all
GLASSWARE PRECISION USE FOR Strong+Strong bromothymol blue (pH~7) MISTAKE WHAT TO DO INSTEAD
Buret ±0.05 mL delivering Weak acid+Strong base phenolphthalein Writing Ka with [H₂O] in denominator Omit pure liquids/solids from equilibrium
titrant (pH 8–10) expressions always
Volumetric ±0.05 mL making Saying strong+strong equiv pt "has no Say "neutral salt forms, neither ion hydrolyzes —
Weak base+Strong acid methyl red (pH 4–
flask exact species" pH=7"
volume 6)
Using mL instead of L in molarity calc Always divide mL by 1000 before using in M×V
Graduated ±0.5 mL rough Pick indicator whose transition range Saying weak acid+strong base equiv pt = Always >7 — conjugate base hydrolyzes, say why
cylinder measurement includes the equivalence point pH. 7
Beaker/flask ±5 mL mixing only
Explore our developer-friendly HTML to PDF API Printed using PDFCrowd HTML to PDF
