1 ch. 8: acids and bases chem 20 el camino college

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Ch. 8: Acids and Bases

Chem 20

El Camino College

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Two Acid-Base Theories

Arrhenius TheoryAn acid solution contains more H+ ions than

OH- ionsA base solution contains more OH- ions

than H+ ions

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Two Acid-Base Theories

Note--H+ is a proton Bronsted-Lowry Theory

An acid is a proton donorA base is a proton acceptor

HBr(aq) + H2O(l) H3O+(aq) + Br- (aq)

NH3(aq) + H2O(l) NH4+(aq) + OH- (aq)

acid base

acidbase

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Conjugate Acid-Base Pairs

Conjugate Acid-Base Pairs differ by one H+

The reactants side has the acid and the base The products side has the conjugate acid and the

conjugate base

HBr(aq) + H2O(l) H3O+(aq) + Br- (aq)conjugate

acidconjugate

baseacid base

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Conjugate Acid-Base Pairs

Acid: H+ donor on left side Conjugate base: missing H+ on right side Base: H+ acceptor on left side Conjugate acid: received H+ on right side

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Conjugate Acid-Base Pairs

Label and link conjugate acid-base pairs.

NH3(aq) + CH3CO2H(aq) NH4+(aq) + CH3CO2

- (aq)conjugate

acidconjugate

baseacidbase

H2SO4(aq) + HSO3-(aq) HSO4

-(aq) + H2SO3 (aq)

conjugateacid

conjugatebase

acid base

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The Water Equilibrium

H2O(l) H+(aq) + OH-(aq)

Kw = (concentration H+ in M)* (conc. OH- in M)Kw = [H+][OH- ] = 1.0 x 10-14

In aq. soln, if [H+]=[OH- ], the soln is neutral.

In aq. soln, if [H+]>[OH- ], the soln is acidic.

In aq. soln, if [H+]<[OH- ], the soln is basic.

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Examples

Use this equation: [H+][OH- ]=1.0 x 10-14

Ex. If the conc. of H+ is 3.5 x 10-3 M, find [OH- ]. Is the solution acidic or basic?

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

3.5x10-3M= 2.9 x 10-12 M

acidic

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Examples

Ex. If the conc. of H+ is 9.9 x 10-11 M, find [OH- ]. Is the solution acidic or basic?

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

9.9x10-11M= 1.0 x 10-4 M

Ex. If [OH- ] is 1.7 x 10-10 M, find [H+]. Is the solution acidic or basic?

[H+] = 1 x 10-14

[OH- ]

= 1 x 10-14

1.7x10-10M= 5.9 x 10-5 M

basic

acidic

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pH Values

pH = 7 is a neutral solution pH < 7 is an acidic solution pH > 7 is a basic solution

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Table 18-2, p. 516

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pH = -log [H+]

Ex. If [H+] = 1.5 x 10-6 M, find [OH- ] and pH

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

1.5x10-6M= 6.7 x 10-9 M

pH = -log [H+] = -log(1.5 x 10-6) = 5.82

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pH = -log [H+]

Ex. If [OH-] = 3.3 x 10-4 M, find [H+ ] and pH

[H+ ] = 1 x 10-14

[OH-]

= 1 x 10-14

3.3x10-4M= 3.0 x 10-11 M

pH = -log [H+] = -log(3.0 x 10-11) = 10.52

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pH = -log [H+]

Ex. If [H+] = 8.5 x 10-1 M, find [OH- ] and pH

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

8.5x10-1M= 1.2 x 10-14 M

pH = -log [H+] = -log(8.5 x 10-1) = 0.07

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pH to [H+]

[H+] = 10-pH or antilog(-pH)

The minus sign goes on the pH value first

Ex. If the pH = 5.55, find [H+]

[H+] = 10-pH = 10-5.55 = 2.8 x 10-6 M

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pH to [H+]

Ex. If the pH = 8.88, find [H+]

[H+] = 10-pH = 10-8.88 = 1.3 x 10-9 M

Ex. If the pH = 13.00, find [H+] and [OH-]

[H+] = 10-pH = 10-13.00 = 1.0 x 10-13 M

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

1.0x10-13M= 1.0 x 10-1 M

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Titration

In the acid-base titration we’ll do in lab, a flask contains a mixture of acid and phenolphthalein (a ccs)

Base is added by buret When the soln turns pale pink for 30 seconds,

moles acid moles base.

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Fig. 16-8, p. 457

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p. 459

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p. 459

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p. 459

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Titration: How many mL? Ex. The flask is filled with 25.00 mL of a 0.500 M HCl soln. How many mL of 0.300 M NaOH soln will neutralize the acid?

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Start with volume of acid, convert to L

Use molarity of acid as a conversion factor

Use a mole ratio to convert mol acid to mol base

Use molarity of base as a conversion factor, convert to mL

.500 mol HCl 1 L

= 41.7 mL25.00mL

HCl1 mol HCl

1 mol NaOH .300 mol

NaOH

1 L1000 mL

1 L 1 L

1000 mL

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Titration: How many mL? Ex. The flask is filled with 75.00 mL of a 0.200 M HCl soln.

How many mL of 0.150 M NaOH soln will neutralize the acid?

Ex. The flask is filled with 55.00 mL of a 1.5 M HCl soln. How many mL of 0.90 M NaOH soln will neutralize the acid?

.200 mol HCl 1 L

= 100. mL75.00mL

HCl1 mol HCl

1 mol NaOH .150 mol

NaOH

1 L1000 mL

1 L 1 L

1000 mL

1.5 mol HCl 1 L

= 91.7 mL55.00mL

HCl1 mol HCl

1 mol NaOH 0.90 mol

NaOH

1 L1000 mL

1 L 1 L

1000 mL

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Titration: Find Molarity Ex. The flask is filled with 30.00 mL of a 0.100 M HCl soln. What is the molarity of the NaOH soln if it takes 23.45 mL to neutralize the acid?

Start with volume of acid, convert to L

Use molarity of acid as a conversion factor

Use a mole ratio to convert mol acid to mol base

Solve for mol base

***Divide mol base by mL base to find molarity. Convert to L.

.100 mol HCl 1 L

30.00mL

HCl1 mol HCl

1 mol NaOH1000 mL

1 L= 0.00300 mol NaOH

= 0.128 M 1 L 1000 mL0.00300 mol NaOH

23.45 mL

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Titration: Find Molarity Ex. The flask is filled with 45.00 mL of a 0.996 M HCl soln.

What is the molarity of the NaOH soln if it takes 52.33 mL to neutralize the acid?

.996 mol HCl 1 L

45.00mL

HCl1 mol HCl

1 mol NaOH1000 mL

1 L= 0.0448 mol NaOH

= .856 M 1 L 1000 mL0.0448 mol NaOH

52.33 mL

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Titration: Find Molarity Ex. The flask is filled with 80.00 mL of a 2.30 M

H2SO4 soln. What is the molarity of the NaOH soln if it takes 70.00 mL to neutralize the acid?

H2SO4(aq) + 2 NaOH(aq) Na2SO4 (aq) + 2 H2O(l)

2.30 mol H2SO4

1 L

80.00mL

H2SO4 1 mol H2SO4

2 mol NaOH1000 mL

1 L= 0.184 mol NaOH

= 2.63 M 1 L 1000 mL0.184 mol NaOH

70.00 mL

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Strong and Weak Acids

Strong acids completely break down into ions when dissolved in water Weak acids only break down into a few ions in water. Most of the

weak acid molecules stay together in water.

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Strong and Weak Acids These are the only strong acids

HCl hydrochloric acidHBr hydrobromic acidHI hydroiodic acidHNO3 nitric acid

H2SO4 sulfuric acid

HClO3 chloric acid

HClO4 perchloric acid.

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Strong and Weak Acids

All other acids are weak acids. Here are some examplesHF hydrofluoric acidH3PO4 phosphoric acid

CH3CO2H acetic acid

H2CO3 carbonic acid.

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Buffers

When a small amount of acid or base is added to pure water, pH changes dramatically A buffer resists change in pH when small amounts of acids or bases are added Blood is buffered in the body to a pH of 7.4 A buffer is a combination of a weak acid and its conjugate base (found in an ionic cmpd) A strong acid cannot make a buffer.

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How Buffers Work

This is a buffer made of CH3COOH and CH3COONa

When acid is added, the extra H+ reacts with CH3COO- to form more CH3COOH

When base is added, the extra OH- reacts with CH3COOH to form more CH3COO-

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Buffers

Which of the following represents a buffer system?HCl and NaClHF and NaFHNO3 and NaNO3

CH3COOH and CH3COONa

noyes

noyes

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Practice: Conjugate Acid-Base Pairs

Label and link conjugate acid-base pairs.

HNO3(aq) + CH3CO2- (aq) CH3CO2H (aq) + NO3

- (aq)conjugate

acidconjugate

basebaseacid

HCl(aq) + SO42-(aq) Cl-(aq) + HSO4

- (aq)conjugate

acidconjugate

baseacid base

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Practice: pH to [H+]

Ex. If the pH = 3.68, find [H+] and [OH-] in scientific notation.

[H+] = 10-pH = 10-3.68 = 2.1 x 10-4 M

[OH- ] = 1 x 10-14

[H+]

= 1 x 10-14

2.1x10-4M= 4.8 x 10-11 M

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Practice: Titration Ex. The flask is filled with 20.50 mL of a 0.0996 M HCl soln.

How many mL of 0.194 M NaOH soln will neutralize the acid?

.0996 mol HCl 1 L

= 10.5 mL20.50mL

HCl1 mol HCl

1 mol NaOH .194 mol

NaOH

1 L1000 mL

1 L 1 L

1000 mL

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Practice: Titration

Ex. The flask is filled with 25.00 mL of a 2.5 M HCl soln. How many mL of 0.50 M Ca(OH)2 soln will neutralize the acid?

2 HCl(aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2 H2O(l)

= 62.5 mL2.5 mol HCl 1 L

25.00mL

HCl2 mol HCl

1 mol Ca(OH)2

0.50 mol

Ca(OH)2

1 L1000 mL

1 L 1 L

1000 mL

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Practice: Titration Ex. The flask is filled with 20.60 mL of a 0.09662 M HCl soln.

What is the molarity of the NaOH soln if it takes 20.92 mL to neutralize the acid?

.09662 mol HCl 1 L

20.60mL

HCl1 mol HCl

1 mol NaOH1000 mL

1 L= 0.001990 mol NaOH

= 0.09512 M 1 L

1000 mL0.001990 mol NaOH20.92 mL