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Chapter 19 - Chapter 19 - Reactions of Acids Reactions of Acids and Bases and Bases

Chapter 19 - Reactions of Acids and Bases

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Chapter 19 - Reactions of Acids and Bases. Water is amphoteric - can act as either an acid or base. H 2 O H + + OH - (acting as acid) H 2 O + H + H 3 O + (acting as base) Water forms an equilibrium with itself H 2 O + H 2 O H 3 O + + OH - - PowerPoint PPT Presentation

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Page 1: Chapter 19 - Reactions of Acids and Bases

Chapter 19 - Chapter 19 - Reactions of Acids Reactions of Acids and Basesand Bases

Page 2: Chapter 19 - Reactions of Acids and Bases

Water is amphoteric - Water is amphoteric - can act as either an acid can act as either an acid or baseor base HH22O HO H++ + OH + OH-- (acting as acid) (acting as acid)

HH22O + HO + H++ H H33OO++ (acting as base) (acting as base) Water forms an equilibrium with itselfWater forms an equilibrium with itself HH22O + HO + H22O HO H33OO++ + OH + OH--

[H[H33OO++] [OH] [OH--] ]

Keq =Keq = --------------------------------

[H[H22O]O]22

Page 3: Chapter 19 - Reactions of Acids and Bases

KKeqeq [H [H22O]O]22 = [H = [H33OO++] [OH] [OH--]]

HH22O is a constant (the amount O is a constant (the amount does not change much on percent does not change much on percent basis)basis)

KKww = K = Keqeq [H [H22O]O]22 = [H = [H33OO++] [OH] [OH--] ]

KKww = [H = [H33OO++] [OH] [OH--] = 1.0 x 10] = 1.0 x 10-14-14

KKw w is the ion product constant for is the ion product constant for waterwater

KKww = x = x22 = 1.0 x 10 = 1.0 x 10-14-14

x = 1.0 x 10x = 1.0 x 10-7-7

Page 4: Chapter 19 - Reactions of Acids and Bases

If [HIf [H33OO++] > 1.0 x 10] > 1.0 x 10-7-7 the solution the solution is acidicis acidic

If [HIf [H33OO++] < 1.0 x 10] < 1.0 x 10-7-7 the solution the solution is basicis basic

See sample problem #1See sample problem #1 Assignment – practice problems 1 Assignment – practice problems 1

& 2 on page 628& 2 on page 628

Page 5: Chapter 19 - Reactions of Acids and Bases

pH scale – a more pH scale – a more compact way to express compact way to express [H[H33OO++] ] pH = -log [HpH = -log [H33OO++] ] If pH < 7 the solution is acidic If pH < 7 the solution is acidic If pH > 7 the solution is basicIf pH > 7 the solution is basic Example problem: What is the pH if Example problem: What is the pH if

[H[H33OO++] = 7.3 x 10] = 7.3 x 10-5-5 M? M?

pH = -log (7.3 x 10pH = -log (7.3 x 10-5-5 ) = 4.14 ) = 4.14

Do practice problems 3 & 4 on page Do practice problems 3 & 4 on page 630630

Page 6: Chapter 19 - Reactions of Acids and Bases

More on pHMore on pH

Measuring pH – use acid-base Measuring pH – use acid-base indicators (will change color with pH)indicators (will change color with pH)

Strong acids dissociate completely.Strong acids dissociate completely.

What is the pH of a .1 M solution of What is the pH of a .1 M solution of HCl?HCl?

What is the pH of a 1 M solution of HCl?What is the pH of a 1 M solution of HCl?

What is the pH of a 6 M solution of HCl?What is the pH of a 6 M solution of HCl?

One is not the lowest pH we can have.One is not the lowest pH we can have.

Page 7: Chapter 19 - Reactions of Acids and Bases

Buffers – resist change in Buffers – resist change in pH by absorbing or pH by absorbing or releasing Hreleasing H++ ions. ions. The most common buffers are The most common buffers are

mixtures of weak acids and their mixtures of weak acids and their conjugate bases.conjugate bases.

Example buffers: HCExample buffers: HC22HH33OO22 + NaC + NaC22HH33OO22

Acetic acid and its conjugate base the Acetic acid and its conjugate base the CC22HH33OO22

-- ion ion

NHNH44Cl and NHCl and NH33 (acid and conj. Base) (acid and conj. Base)

Page 8: Chapter 19 - Reactions of Acids and Bases

How a buffer worksHow a buffer works

show diagramshow diagram

Page 9: Chapter 19 - Reactions of Acids and Bases

The buffering capacity is limited The buffering capacity is limited by the amount of HCby the amount of HC22HH33OO22 and and NaCNaC22HH33O2. If we run out of either O2. If we run out of either the buffering capacity is exceeded.the buffering capacity is exceeded.

Equal amount of acetic acid and Equal amount of acetic acid and acetate ion stablize the pH near acetate ion stablize the pH near 4.74.7

We can derive this from the Ka of We can derive this from the Ka of acetic acid, which is 1.8 x 10acetic acid, which is 1.8 x 10-5-5

Page 10: Chapter 19 - Reactions of Acids and Bases

[H[H33OO++] [C] [C22HH33OO22--] ]

Ka =Ka = ---------------- = 1.8 x 10---------------- = 1.8 x 10-5-5

[HC[HC22HH33OO22] ]

If [CIf [C22HH33OO22--] and [HC] and [HC22HH33OO22] are equal ] are equal

they cancel out of the equationthey cancel out of the equation

Ka = [HKa = [H33OO++] = 1.8 x 10] = 1.8 x 10-5-5 M M

pH = -log 1.8 x 10pH = -log 1.8 x 10-5-5 = 4.74 = 4.74

Page 11: Chapter 19 - Reactions of Acids and Bases

Experimental Experimental Determination of acid Determination of acid concentration concentration The concentration of a The concentration of a very very

strongstrong acid can be determined by acid can be determined by pH. pH.

The concentration of a weak acid The concentration of a weak acid is more accurately determined by is more accurately determined by an acid/base titration. an acid/base titration.

An acid-base titration is a An acid-base titration is a carefully controlled neutralization carefully controlled neutralization reaction. (see page 637)reaction. (see page 637)

Page 12: Chapter 19 - Reactions of Acids and Bases

Acid – Base Titration Acid – Base Titration

Equivalence point – when we Equivalence point – when we have added just enough base to have added just enough base to neutralize all of the acid.neutralize all of the acid.

End point – when we are at the End point – when we are at the color change point of a titrationcolor change point of a titration

Phenolphthalein indicator – Phenolphthalein indicator – changes color over a pH range of changes color over a pH range of 8 – 10 (clear to red)8 – 10 (clear to red)

Page 13: Chapter 19 - Reactions of Acids and Bases
Page 14: Chapter 19 - Reactions of Acids and Bases

Do example problem on page 638 Do example problem on page 638 together in classtogether in class

Assignment – problems 5 & 6 on Assignment – problems 5 & 6 on page 638page 638

Questions page 646-647 (1-19, Questions page 646-647 (1-19, 32, 33)32, 33)

Problem bank 37, 39, 41, 43, 45, Problem bank 37, 39, 41, 43, 45, 47, 49, 51, 53, 54, 56, 58, 60 47, 49, 51, 53, 54, 56, 58, 60 (answers on page 945)(answers on page 945)