acids & bases chapter 16 (& part of chapter 17)

Acids & BasesCHAPTER 16

(& part of CHAPTER 17)

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

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CHAPTER 16: Acids & Bases

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives:

Define Brønsted-Lowry Acid/Base Define Lewis Acid/Base Evaluate the strength of acids/bases

Strong vs weak acids/bases Periodic trends Conjugate acids/bases

Identify likely compounds that will form acids and bases from the periodic table Acidic metal ions

Acid/Base equilibrium: pH, pOH Ka, Kb, pKa, pKb Kw of water

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CHAPTER 16: Acids & Bases


Lecture Road Map:

① Brønsted-Lowry Acids/Bases

② Trends in acid strength

③ Lewis Acids & Bases

④ Acidity of hydrated metal ions

⑤ Acid/Base equilibrium

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Acid/Base Equilibrium

CHAPTER 16 Acids & Bases

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Acid/Base Equilibrium Weak Acids & Bases


• Incompletely ionized• Molecules and ions exist in equilibrium• HA = any weak acid; B = any weak base HA(aq) + H2O A–(aq) + H3O+(aq) B (aq) + H2O B H+(aq) + OH–(aq)

CH3COOH(aq) + H2O CH3COO–(aq) + H3O+(aq)

HSO3–(aq) + H2O SO3

2–(aq) + H3O+(aq)

NH4+(aq) + H2O NH3 (aq) + H3O+ (aq)

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• Often simplify as• HA (aq) A –(aq) + H+(aq)

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Or generallyHA(aq) + H2O A–(aq) + H3O+(aq)

• But [H2O] = constant (55.6 M ) so rewrite as

• Where Ka = acid ionization constant

Acid + Water Conjugate Base + Hydronium Ion

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But [H2O] = constant so rewrite as Where Kb = base ionization constant

CH3COO–(aq) + H2O CH3COOH(aq) + OH–(aq)

NH3(aq) + H2O NH4+(aq) + OH–(aq)

• Or generallyB (aq) + H2O B H+(aq) + OH–(aq)

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Acid/Base Equilibrium pH


• Lots of weak acids and bases– How can we quantify their relative strengths?

• Need reference– Choose H2O

• Water under right voltage – Slight conductivity– Where does conductivity come from?

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• Trace ionization self-ionization of water• H2O + H2O H3O+(aq) + OH–(aq) acid base acid base

• Equilibrium law is:

• But [H2O]pure = = 55.6 M

[H2O] = constant even for dilute solutions

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H2O + H2O H3O+(aq) + OH–(aq)• Since [H2O] = constant, equilibrium law is

• K w = is called the ion product of water

• Often omit second H2O molecule and write• H2O H+(aq) + OH–(aq)

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Acid/Base Equilibrium Kw


H2O H+(aq) + OH–(aq) • For pure H2O at 25 °C

– [H+] = [OH–] = 1.0 × 10–7 M – Kw = (1.0 × 10–7)(1.0 × 10–7) = 1.0 × 10–14

– See Table 17.1 for K w at various temperatures

• H2O auto-ionization occurs in all solutions– When other ions present

• [H+] is usually NOT equal to [OH–] • But Kw = [H+][OH–] = 1.0 × 10–14

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Acid/Base Equilibrium Definition of Acidic & Basic


• In aqueous solution, – Product of [H3O+] and [OH–] equals K w

– [H3O+] and [OH–] may not actually equal each other– Solutions are classified on the relative

concentrations of [H3O+] and [OH–] Solution Classification

Neutral [H3O+] = [OH–]Acidic [H3O+] > [OH–]Basic [H3O+] < [OH–]

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Acid/Base Equilibrium Weak Acids & Bases: Example


Ex. 1 In a sample of blood at 25 °C, [H+] = 4.6 10–8 M. Find [OH–] and determine if the solution is acidic, basic or neutral.

•So 2.2 × 10–7 M > 4.6 × 10–8 M•[OH–] > [H3O+] so the solution is basic

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Jesperson, Brady, Hyslop.

Chemistry: The Molecular

Nature of Matter, 6E

• Arrhenius (of kinetics fame) – Sought an easy way to write the very small

numbers associated with [H+] and [OH–] – Developed the “p” notation where p stands for

the –log mathematical operation

– Result is a simple number• pH is defined as:

– Define pOH as:

– Define pKw as:• Take anti-log to obtain [H+], [OH–] or Kw

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Acid/Base Equilibrium General Properties of Logarithms


Using Logarithms• Start with • Taking –log of both sides of eqn. gives

• So at 25 °C:

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Acid/Base Equilibrium Definition of Acidic and Basic


• As pH increases, [H+] decreases; pOH decreases, and [OH–] increases

• As pH decreases, [H+] increases; pOH increases, and [OH–] decreases

Neutral pH = 7.00Acidic pH < 7.00Basic pH > 7.00

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Acid/Base Equilibrium Measuring pH


1. pH meter– Most accurate– Calibrate with solutions of known pH before use– Electrode sensitive to [H+]– Accurate to 0.01 pH unit

2. Acid-base indicator– Dyes, change color depending on [H+] in solution– Used in pH paper and titrations– Give pH to 1 pH unit

3. Litmus paper– Red pH 4.7 acidic– Blue pH 4.7 basic– Strictly acidic vs. basic

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Acid/Base Equilibrium


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Acid/Base Equilibrium Example pH Calculations


Calculate pH and pOH of blood in Ex. 1.We found [H+] = 4.6 × 10–8 M [OH–] = 2.2 × 10–7 M pH = –log(4.6 × 10–8) = 7.34 pOH = –log(2.2 × 10–7) = 6.66 14.00 = pKw Or pOH = 14.00 – pH = 14.00 – 7.34 = 6.66

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What is the pH of NaOH solution at 25 °C in which the OH– concentration is 0.0026 M? [OH–] = 0.0026 M pOH = –log(0.0026) = 2.59pH = 14.00 – pOH = 14.00 – 2.59 = 11.41

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Acid/Base Equilibrium Strong Acids


• Assume 100% dissociated in solution– Good ~ if dilute

• Makes calculating [H+] and [OH] easier• 1 mole H+ for every 1 mole HX

– So [H+] = CHX for strong acids • Thus, if 0.040 M HClO4 • [H+] = 0.040 M• And pH = –log (0.040) = 1.40

Strong Acids

HClHBrHI

HNO3

H2SO4

HClO3

HClO4

HX (general term for a

strong acid)

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Acid/Base Equilibrium Strong Bases


• 1 mole OH– for every 1 mole M OH • [OH–] = CMOH for strong bases

• 2 mole OH– for 1 mole M(OH)2 • [OH–] = for strong bases

Strong BasesNaOHKOHLiOHCa(OH)2

Ba(OH)2 Sr(OH)2

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Acid/Base Equilibrium Effect of Auto ionization of Water with

Strong Bases


• The auto-ionization of H2O will always add to [H+] and [OH–] of an acid or base. Does this have an effect on the last answer?– The previous problem had 0.00022 M [OH–] from the

Ca(OH)2 but the [H+] must have come from water. If it came from water an equal amount of [OH–] comes from water and the total [OH–] is

– [OH–]total = [OH–]from Ca(OH)2 + [OH–]from H2O

– [OH–]total = 0.00022 M + 4.6 × 10–11 M = 0.00022 M (when properly

rounded)

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– So [H+] from H2O must also be 5.0 10–13 M

• [H+]total = 0.020 M + (5.0 10–13 M)= 0.020 M (when properly rounded)

• So we see that [H+]from H2O will be negligible except in very dilute solutions of acids and bases

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Acid/Base Equilibrium Looking at Weak Acids Again


What is the pKa of HF if Ka = 3.5 × 10–4?HF(aq) + H2O F–(aq) + H3O+(aq)orHF(aq) F–(aq) + H+(aq)

= 3.5 × 10–4

pKa = –log Ka = –log(3.5 × 10–4) = 3.46bb logp KK

bpb 10 KK

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Acid/Base Equilibrium Conjugate Acid-Base Pairs


1. Consider ionization reaction of generic acid and waterHA(aq) + H2O A–(aq) + H3O+(aq)

2. Consider reaction of a salt containing anion of this acid (its conjugate base) with water

A–(aq) + H2O HA(aq) + OH–(aq)

HA(aq) + H2O A–(aq) + H3O+(aq)

A–(aq) + H2O HA(aq) + OH–(aq)

2H2O H3O+(aq) + OH–(aq)

wba ]][OH[H][A

][HA][OH[HA]

]][H[A KKK

For any conjugate acid base pair:14

wba 100.1 KKK (at 25 °C)

Acid/Base Equilibrium Conjugate Acid-Base Pairs


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Acid/Base Equilibrium More About Logarithms


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• Then taking –log of both sides of equation gives:

• Earlier we learned the inverse relationship of conjugate acid-base strengths, now we have numbers to illustrate this.

• The stronger the conjugate acid, the weaker the conjugate base.

)100.1log(log)log( 14wba

KKK

)00.14(logloglog wba KKKSo

(at 25 °C)

acids & bases chapter 16 (& part of chapter 17)

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