1 Acid-Base Reactions Chapter 17. 2 3 Acid-Base Reactions Reactions always go from the stronger A-B...

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1 Acid-Base Reactions Chapter 17

Transcript of 1 Acid-Base Reactions Chapter 17. 2 3 Acid-Base Reactions Reactions always go from the stronger A-B...

Page 1: 1 Acid-Base Reactions Chapter 17. 2 3 Acid-Base Reactions Reactions always go from the stronger A-B pair (larger K) to the weaker A-B pair (smaller K).

11Acid-Base ReactionsAcid-Base Reactions

Chapter 17

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Acid-Base Reactions

Reactions always go from the stronger A-B pair

(larger K) to the weaker A-B pair (smaller K).

ACIDS CONJUGATE BASESACIDS CONJUGATE BASES

STRONGSTRONG

weakweak

weakweak

STRONGSTRONG

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Acid-Base ReactionsAcid-Base Reactions

A strong acid is 100% dissociated.

STRONG ACID = a good H+ donor—must have a WEAK CONJUGATE BASE

Or a poor H+ acceptor.

HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq)

STRONG A base acid weak B

•Every A-B reaction has two acids and two bases.

•Equilibrium always lies toward the weaker pair.•Here K is very large.

•Every A-B reaction has two acids and two bases.

•Equilibrium always lies toward the weaker pair.•Here K is very large.

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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions

We know from experiment that HNO3 is a strong acid.

1. It is a stronger acid than H3O+

2. H2O is a stronger base than NO3-

3. K for this reaction is large

WEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2OWEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2O

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K and Acid-Base ReactionsK and Acid-Base ReactionsK and Acid-Base ReactionsK and Acid-Base Reactions

Acetic acid is only 0.42% ionized.

It is a WEAK ACID

HOAc + H2O H3O+ + OAc-

WEAK A base acid STRONG B

[H3O+] is small, this means

1. H3O+ is a stronger acid than HOAc

2. OAc- is a stronger base than H2O

3. K for this reaction is small

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Acid/Base ReactionsAcid/Base Reactions

Strong acid + Strong base

H+ + Cl- + Na+ + OH- H2O + Na+ + Cl-

Net ionic equation

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

K = 1/Kw = 1 x 1014

Mixing equal molar quantities of a strong acid and strong

base produces a neutral solution.

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Acid/Base ReactionsAcid/Base Reactions

Weak acid + Strong base

CH3CO2H + OH- H2O + CH3CO2-

This is the reverse of the reaction of CH3CO2

- (conjugate base) with H2O.

OH- stronger base than CH3CO2-

K = 1/Kb = 5.6 x 104

Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate

base. The solution is basic.

Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate

base. The solution is basic.

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Acid/Base ReactionsAcid/Base Reactions

Strong acid + Weak base

H3O+ + NH3 H2O + NH4+

This is the reverse of the reaction of NH4+ (conjugate

acid of NH3) with H2O.

H3O+ stronger acid than NH4+

K = 1/Ka = 5.6 x 104

Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate

acid. The solution is acid.

Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate

acid. The solution is acid.

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Acid/Base ReactionsAcid/Base Reactions

Weak acid + Weak baseWeak acid + Weak base

•Product cation = CA of weak base.•Product anion = CB of weak acid.

pH of solution depends

on relative strengths of

cation and anion.

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Summary

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MX + H2O ----> acidic or basic??

Consider NH4Cl

NH4Cl(aq) NH4+(aq) + Cl-(aq)

(a) Reaction of Cl- with H2O

Cl- + H2O ----> HCl + OH-

base acid acid base

Cl- ion is a VERY weak base because its conjugate acid is strong.

Therefore, Cl- = neutral solution

SaltsSalts

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(b) Reaction of NH4+ with H2O

NH4+ + H2O NH3 + H3O+

acid base base acid

NH4+ ion is a moderate acid because its

conjugate base is weak.

Therefore, NH4+ = acidic solution

See TABLE 17.4 for a summary of acid-base properties of ions.

SaltsSalts

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Calculate the pH of a 0.10 M Na2CO3.

A)

Na+ + H2O neutralB)

CO32- + H2O HCO3

- + OH-

base acid acid base

Kb = 2.1 x 10-4

Salts

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Calculate the pH of a 0.10 M solution of Na2CO3.

Kb = 2.1 x 10-4

Step 1. Set up ICE table

[CO32-] [HCO3

-] [OH-]

I 0.10 0 0

C -x +x +x

E 0.10 - x x x

SaltsSalts

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Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x

Assume 0.10 - x ≈ 0.10,

because 100•Kb < Co

x = [HCO3-] = [OH-] = 0.0046 M

Step 2.Step 2. Solve the equilibrium expression

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Step 3. Calculate the pH

[OH-] = 0.0046 M

pOH = - log [OH-] = 2.34

pH + pOH = 14,

so pH = 11.6

the solution is ________.

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