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AP Chemistry > Acid - Base Equilibrium > Flashcards

Acid - Base Equilibrium Flashcards

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1
Q

pKw, pH, pOH

Equations

A
  • pH= -log([H3O+])
  • Also H+ = 10^-pH
  • pOH= -log([OH-])
  • OH- = 10^-pOH
  • pKw = pH + pOH = -log([OH-] [H3O+])
  • Ka * Kb = Kw
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2
Q

pH, pOH

How to decide concentration of H3O+ or OH-

A
  • if in strong acid just that from the acid because self-ionization of water is negligible
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3
Q

Conjugate Base/Acid

Hydrolysis Rxns

A
  • Acid - base rxns between ions(from salts) and H2O molecules
  • May react with water in a/b rxn
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4
Q

Conjugate Base/Acid

pH vs H3O+ / OH- concentration

A
  • 7: [H30+] = [OH-]
  • <7: [H30+] > [OH-]
  • > 7: [H30+] < [OH-]
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5
Q

Conjugate Base/Acid

Determining Conjugate Base/Acid

A
  • If acid, then find with extra electron (goes from donator to acceptor) ⇒conj base
  • If base, then find with less electron (goes from acceptor to donator) ⇒ conj acid
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6
Q

Acid, Bases, and Acid-Base Equilibria

Acid

A
  • Proton Donor
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7
Q

Acid, Bases, and Acid-Base Equilibria

Base

A
  • Proton Acceptor
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8
Q

Acid, Bases, and Acid-Base Equilibria

Strong Acids

A
  1. HCl
  2. HBr
  3. HI
  4. HNO3
  5. H2SO4
  6. HCLO4
    * Complete Ionization: →
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9
Q

Acid, Bases, and Acid-Base Equilibria

Weak Acids

A
  1. CH3COOH
  2. NH4+
    * Partial Ionization: ⇌
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10
Q

Acid, Bases, and Acid-Base Equilibria

Strong Bases

A
  • Group 1A and 2A with Hydroxides
  • Complete Ionization: →
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11
Q

Acid, Bases, and Acid-Base Equilibria

Weak Base

A
  • NH3
  • Partial Ionization: ⇌
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12
Q

Acid, Bases, and Acid-Base Equilibria

Ka/ Kb

A
  • Acid/Base Ionization Constant
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13
Q

Acid, Bases, and Acid-Base Equilibria

Relation bet a/b and conj a/b

A
  • Stronger the original, weaker the conjugate and vice versa
  • Favored in the direction = stronger to weaker
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14
Q

Acid, Bases, and Acid-Base Equilibria

Kw

A
  • Ion product of H2O
  • = [H3O+][OH-] = 1.0 * 10^-14 M
  • both H3O+ and OH- have concentrations of 1.0 * 10^-7
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15
Q

Acid, Bases, and Acid-Base Equilibria

How to deal with weak Acid/ Base equilibria

A
  • Use ICE Table
  • If M(a/b)/K(a/b) > 100, then [A/B]≈ Eq. value
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16
Q

Acid, Bases, and Acid-Base Equilibria

Which ions hydrolyze

A
  • Weak acids or bases hydrolyze appreciably
  • Strong a/b form neutral solutions
17
Q

Acid, Bases, and Acid-Base Equilibria

Relationship between weak and strong a/b

A
  • Weak Acids and Strong Bases = basic
  • Strong Acids and Weak Bases = acidic
  • Weak a + b = any of them
18
Q

Common Ion Effect

Common Ion Effect

ASK ABT THIS

A
  • Supression of ionization of weak acids or weak bases by presence of common ion from a strong electrolyte
19
Q

Buffer Solutions

Definition

A

Solution that changes pH only slightly when small amounts of strong acid or base added
* Usually contain: weak acid/base with salt

20
Q

Buffer Solutions

Buffer Rules

A
  1. ratio of [conjugate base/ acid] / [weak acid/ base] between .10 and 10
  2. both [conjugate base/ acid] and [weak acid/ base] exceed Ka by factor of 100+
21
Q

Buffer Solutions

Henderson–Hasselbalch equation

A

pH = pKa + log([conj base]/ [weak acid])

22
Q

Buffer Solutions

Heylman equation

A

pOH = pKb + log([conj acid]/ [weak base])

23
Q

Buffer Solutions

Buffer Action

A
  • best buffer is when [conj acid / base] = [weak base / acid] → best buffer!
24
Q

Buffer Solutions

Effective Buffer Range

A
  • Approximately when pH = pKa ± 1 pH-unit
25
Q

Acid-Base Titrations

Analyte

A
  • solution of unknown concetration
26
Q

Acid-Base Titrations

Titrant

A

Solution with known concentration

27
Q

Acid-Base Titrations

Equivalence point

AKA Stoichiometric pt

A
  • # of moles of H3O+ = # of moles of OH-
28
Q

Indicators

What are they

A

A weak organic acid that has a different color than conjugate base

29
Q

Indicators

Relation bet. [HIn] and [In-]

A
  • equal means intermediate color
  • [In-]/[HIn] >10: solution will be color of In-
  • [In-]/[HIn] < .10: solution will be color of HIn
30
Q

Indicators

Endpoint

A
  • Point where indicator changes color
31
Q

Indicators

Equation

A
  • Ka/[H3O+] = [In-]/[HIn]

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