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Chapter 10 Acids and Bases

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3

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Arrhenius acids Produce H+ ions in water.

H2OHCl H+(aq) + Cl– (aq)

Are electrolytes. Have a sour taste. Corrode metals. React with bases to form salts

and water.

10.1 Acids and Bases in Aqueous Solution

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Arrhenius bases Produce OH– ions in

water. Taste bitter or chalky. Are electrolytes. Feel soapy and slippery. React with acids to

form salts and water.

Bases

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Identify each as a characteristic of an

A) acid or B) base

____1. Has a sour taste.

____2. Produces OH- in aqueous solutions.

____3. Has a chalky taste.

____4. Is an electrolyte.

____5. Produces H+ in aqueous solutions.

Learning Check

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Identify each as a characteristic of an

A) acid or B) base

A 1. Has a sour taste.

B 2. Produces OH– in aqueous solutions.

B 3. Has a chalky taste.

A, B 4. Is an electrolyte.

A 5. Produces H+ in aqueous solutions.

Solution

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Names of Acids

Acids with H and one nonmetal are named with the prefix hydro- and end with -ic acid.HCl hydrochloric acid

Acids with H and a polyatomic ion are named by changing the end of an –ate ion to -ic acid and an –ite ion to -ous acid.

HClO3 chloric acid

HClO2 chlorous acid

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Name each of the following as acids: A. HBr 1. bromic acid

2. bromous acid 3. hydrobromic acid

B. H2CO3 1. carbonic acid

2. hydrocarbonic acid3. carbonous acid

Learning Check

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A. HBr 3. hydrobromic acid

The name of an acid with H and one nonmetal begins with the prefix hydro- and ends with -ic acid.

B. H2CO3 1. carbonic acid

An acid with H and a polyatomic ion is named by changing the end of an –ate ion to -ic acid.

Solution

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Bases with OH- ions are named as the hydroxide of the metal in the formula.

NaOH sodium hydroxide

KOH potassium hydroxide

Ba(OH)2 barium hydroxide

Al(OH)3 aluminum hydroxide

Fe(OH)3 iron (III) hydroxide

Some Common Bases

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Match the formulas with the names:

A. ___ HNO2 1) hydrochloric acid

B. ___Ca(OH)2 2) sulfuric acid

C. ___H2SO4 3) sodium hydroxide

D. ___HCl 4) nitrous acid

E. ___NaOH 5) calcium hydroxide

Learning Check

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Match the formulas with the names:A. 4 HNO2 4) nitrous acid

B. 5 Ca(OH)2 5) calcium hydroxide

C. 2 H2SO4 2) sulfuric acid

D. 1 HCl 1) hydrochloric acid

E. 3 NaOH 3) sodium hydroxide

Solution

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According to the Brønsted-Lowry theory, Acids are hydrogen ion (H+) donors. Bases are hydrogen ion (H+) acceptors.

donor acceptor hydronium ion

HCl + H2O H3O+ + Cl-

+ -

+ +

BrØnsted-Lowry Acids and Bases

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When NH3 dissolves in water, a few NH3 molecules react with water to form ammonium ion NH4

+ and a hydroxide ion.

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

acceptor donor

+ - + +

NH3, A Bronsted-Lowry Base

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Conjugate Acids and Bases An acid that donates H+ forms a conjugate base. A base that accepts a H+ forms a conjugate acid. In an acid-base reaction, there are two conjugate

acid-base pairs.

acid 1 conjugate base 1 + +

base 2 conjugate acid 2

HF H2O H3O+ F-

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Conjugate Acid-Base Pairs

A conjugate acid-base pair is two substances related by a loss or gain of H+.

+ +

acid 1– conjugate base 1

base 2– conjugate acid 2

HF H2O H3O+ F-

HF, F-

H2O, H3O+

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Learning Check

A. Write the conjugate base of the following:

1. HBr

2. H2S

3. H2CO3

B. Write the conjugate acid of the following:

1. NO2–

2. NH3

3. OH–

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Solution

A. Write the conjugate base of the following:

1. HBr Br-

2. H2S HS–

3. H2CO3 HCO3–

B. Write the conjugate acid of the following:

1. NO2– HNO2

2. NH3 NH4+

3. OH– H2O

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Learning Check

Identify the following that are acid-base conjugate pairs.

1. HNO2, NO2–

2. H2CO3, CO32 –

3. HCl, ClO4 –

4. HS–, H2S

5. NH3, NH4+

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Solution

Identify the following that are acid-base conjugate pairs.

1. HNO2, NO2–

4. HS–, H2S

5. NH3, NH4+

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Learning Check

A. The conjugate base of HCO3– is

1. CO32– 2. HCO3

– 3. H2CO3

B. The conjugate acid of HCO3– is

1. CO32– 2. HCO3

– 3. H2CO3

C. The conjugate base of H2O is

1. OH– 2. H2O 3. H3O+

D. The conjugate acid of H2O is

1. OH– 2. H2O 3. H3O+

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Solution

A. The conjugate base of HCO3 – is

1. CO32–

B. The conjugate acid of HCO3– is

3. H2CO3

C. The conjugate base of H2O is

1. OH–

D. The conjugate acid of H2O is 3. H3O+

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Strong acids completely ionize (100%) in aqueous solutions.HCl + H2O H3O+ (aq) + Cl– (aq)

(100 % ions) Strong bases completely (100%) dissociate

into ions in aqueous solutions. H2O

NaOH Na+ (aq) + OH–(aq) (100 % ions)

10.6 Acid and Base Strength

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Strong and Weak Acids

In an HCl solution, the strong acid HCl dissociates 100%.

A solution of the weak acid CH3COOH contains mostly molecules and a few ions.

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Only a few acids are strong acids. The conjugate bases of strong acids are weak

bases.

Strong Acids

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Most acids are weak acids. The conjugate bases of weak acids are strong bases.

Weak Acids

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Most bases in Groups 1A and 2A are strong bases. They include

LiOH, NaOH, KOH, and Mg(OH)2, Ca(OH)2

Most other bases are weak bases.

Strong Bases

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Learning Check

Identify each of the following as a strong or weak acid or base.

A. HBr

B. HNO2

C. NaOH

D. H2SO4

E. Cu(OH)2

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Solution

Identify each of the following as a strong or weak acid or base.

A. HBr strong acid

B. HNO2 weak acid

C. NaOH strong base

D. H2SO4 strong acid

E. Cu(OH)2 weak base

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Table 10.1 Relative strengths of acids and Table 10.1 Relative strengths of acids and basesbases

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Learning Check

A. Identify the stronger acid in each pair.

1. HNO2 or H2S

2. HCO3– or HBr

3. H3PO4 or H3O+

B. Identify the strong base in each pair.

1. NO3– or F-

2. CO32– or NO2

–

3. OH– or H2O

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Solution

A. Identify the stronger acid in each pair.

1. HNO2 2. HBr

3. H3O+

B. Identify the stronger base in each pair. 1. F-

2. CO32–

3. OH–

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Strong Acids

In water, the dissolved molecules of a strong acid are essentially all separated into ions.

The concentrations of H3O+ and the anion (A–) are large.

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Weak Acids

In weak acids, The equilibrium

favors the undissociated (molecular) form of the acid.

The concentrations of the H3O+ and the anion (A–) are small.

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In water, H+ is transferred from one H2O molecule to another.

One water molecule acts as an acid, while another acts as a base.H2O + H2O H3O+ + OH –

.. .. .. .. :O: H + :O:H H:O:H + + :O:H–

.. .. .. .. H H H water molecules hydronium hydroxide

ion (+) ion (–)

10.8 Dissociation of Water

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In pure water, the ionization of molecules produces small, but equal quantities of H3O+ and OH-

ions.H2O + H2O H3O+ + OH-

Molar concentrations are indicated as [H3O+] and [OH-].[H3O+] = 1.0 x 10-7 M[OH-] = 1.0 x 10-7 M

Pure Water is Neutral

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Adding an acid to pure water increases the [H3O+].

In acids, [H3O+] exceeds 1.0 x 10–7 M.

As [H3O+] increases, [OH–] decreases.

Acidic Solutions

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Adding a base to pure water increases the [OH–].

In bases, [OH–] exceeds 1.0 x 10–7M.

As [OH–] increases, [H3O+] decreases.

Basic Solutions

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Comparison of [H3O+] and [OH–]

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The ion product constant, Kw, for water is the product of the concentrations of the hydronium and hydroxide ions.

Kw = [H3O+][OH–] We can obtain the value of Kw using the

concentrations in pure water.

Kw = [1.0 x 10–7][1.0 x 10–7]

= 1.0 x 10–14

Ion Product of Water, Kw

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Kw in Acids and Bases

In neutral, acidic, and basic solutions, the Kw is equal to 1.0 x 10–14.

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What is the [H3O+] of a solution if [OH–] is 1.0 x 10-8

M?

Rearrange the Kw expression for [H3O+ ].

Kw = [H3O+][OH–] = 1.0 x 10-14

[H3O+] = 1.0 x 10-14

[OH–]

[H3O+] = 1.0 x 10-14 = 1.0 x 10-6 M 1.0 x 10- 8

Calculating [H3O+]

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The [H3O+] of lemon juice is 1.0 x 10–3 M.

What is the [OH–] of the solution?

1) 1.0 x 103 M

2) 1.0 x 10–11 M

3) 1.0 x 1011 M

Learning Check

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The [H3O+] of lemon juice is 1.0 x 10–3 M. What is the [OH–] of the solution?

2) 1.0 x 10–11 M

Rearrange the Kw to solve for [OH–]

Kw = [H3O+ ][OH–] = 1.0 x 10–14

[OH- ] = 1.0 x 10 -14 = 1.0 x 10–11 M 1.0 x 10 - 3

Solution

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10.9 Measuring Acidity in Aqueous Solution: pH

The pH scale: Is used to indicate the acidity of a solution. Has values that usually range from 0 to 14. Indicates an acidic solution when the values

are less than 7. Indicates a neutral solution with a pH of 7. Indicates a basic solution when the values

are greater than 7.

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Fig 10.2 The pH scale and pH ofsome common substances

Fig 10.3 relationship of pH to H+ and OH- ion concentrations

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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Neutral

[H3O+]>[OH-] [H3O+] = [OH-] [H3O+]<[OH-]

Acidic Basic

pH Range

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Identify each solution as1. acidic 2. basic 3. neutralA. ___ HCl with a pH = 1.5 B. ___ Pancreatic fluid [H3O+] = 1 x 10–8 M

C. ___ Sprite soft drink pH = 3.0D. ___ pH = 7.0E. ___ [OH– ] = 3 x 10–10 MF. ___ [H3O+ ] = 5 x 10–12 M

Learning Check

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Identify each solution as

1. acidic 2. basic 3. neutral

A. 1 HCl with a pH = 1.5

B. 2 Pancreatic fluid [H3O+] = 1 x 10–8 M

C. 1 Sprite soft drink pH = 3.0

D. 3 pH = 7.0

E. 1 [OH–] = 3 x 10–10 M

F. 2 [H3O+] = 5 x 10–12 M

Solution

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Testing the pH of Solutions

The pH of solutions can be determined using a a) pH meter, b) pH paper, and c) indicators that have specific colors at different pHs.

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pH is: Defined as the negative log of the hydrogen ion

concentration.

pH = - log [H3O+]

The value of the negative exponent of [H3O+] for concentrations with a coefficient of 1.

[H3O+] = 1 x 10-4 pH = 4.0

[H3O+] = 1 x 10-11 pH = 11.0

pH Scale

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A calculator can be used to find the pH of a

solution with a [H3O+] of 1 x 10-3 as follows. 1. Enter 1 x 10-3 by pressing 1 (EE) 3 (+/–). The EE key gives an exponent of 10 and the

+/– key changes the sign.2. Press the log key to obtain log 1 x 10-3. log (1 x 10-3) = –33. Press the +/– key to multiply by –1. (-3) +/– = 3

Calculations of pH

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Significant Figures in pH

When expressing log values, the number of decimal places in the pH is equal to the number of significant figures in the coefficient of [H3O+].

[H3O+] = 1 x 10-4pH = 4.0

[H3O+] = 1.0 x 10-6 pH = 6.00

[H3O+] = 2.4 x 10-8 pH = 7.62

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A. The [H3O+] of tomato juice is 2 x 10-4 M.

What is the pH of the solution?

1) 4.0 2) 3.7 3) 10.3

B. The [OH-] of a solution is 1.0 x 10-3 M.

What is the pH of the solution?

1) 3.00 2) 11.00 3) –11.00

Learning Check

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A. 2) 3.7 pH = – log [ 2 x 10-4] = 3.7 2 (EE) 4 (+/–) log (+/–)

B. 2) 11.00 Use the Kw to obtain [H3O+] = 1.0 x 10-11

pH = – log [1.0 x 10- 11] = – 11.00 (+/–) = 11.00

Solution

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Calculating [H3O+] from pH

The [H3O+] can be expressed with the pH as the negative power of 10.

[H3O+] = 1 x 10-pH

If the pH is 3.0, the [H3O+] = 1 x 10-3

On a calculator

1. Enter the pH value 3.0

2. Use (+/–) to change sign 3.0 (+/–) = –3.0

3. Use the inverse log key (or 10x) to obtain

the [H30+]. = 1 x 10-3 M

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A. What is the [H3O+] of a solution with a pH of 10.0?

1) 1 x 10 -4 M 2) 1 x 1010 M

3) 1 x 10 -10 MB. What is the [H3O+] of a solution with a

pH of 5.7?

1) 2 x 10- 6 M 2) 5 x 105 M

3) 1 x 10-5 M

Learning Check

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A. What is the [H3O+] of a solution with a pH of 10.0? 3) 1 x 10-10 M 1 x 10-pH

B. What is the [H3O+] of a solution with a pH of 5.7? 1) 2 x 10- 6 M

5.7 (+/–) inv log (or 10x) = 2 x 10-6

Solution

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pOH

The pOH of a solution: Is related to the [OH-]. Is similar to pH.

pOH = – log [OH-] Added to the pH value is equal to 14.0.

pH + pOH = 14.0

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Learning Check

What is the pH and the pOH of coffee if the [H3O+] is 1 x 10-5 M?

1) pH = 5.0 pOH =7.0

2) pH = 7.0 pOH = 7.0

3) pH = 5.0 pOH = 9.0

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Solution

What is the pH and the pOH of coffee if the [H3O+] is 1 x 10-5 M?

3) pH = 5.0 pOH = 9.0

pH = – log [1 x 10-5] = –(– 5.0) = 5.0

pH + pOH = 14.0

pOH = 14.0 – pH = 14.0 – 5.0 = 9.0

or [OH-] = 1 x 10-9

pOH = – log [1 x 10-9] = –(– 9.0) = 9.0

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An acid such as HCl produces H3O+ and a base such as NaOH provides OH-.HCl + H2O H3O+ + Cl-

NaOH Na+ + OH-

In a neutralization reaction, the H3O+ and the OH- combine to form water.

H3O+ + OH- 2 H2O The positive ion (metal) and the negative ion

(nonmetal) are the ions of a salt.

Neutralization Reactions

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In the equation for neutralization, an acid and a base produce a salt and water.

NaOH + HCl NaCl + H2O

base acid salt water

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

base acid salt water

Neutralization Equations

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Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid.

1. Write the formulas of the acid and base.Mg(OH)2 + HNO3

2. Balance to give equal OH- and H+.

Mg(OH)2 + 2 HNO3

3. Write the products (a salt and water) and balance: Mg(OH)2 + 2HNO3 Mg(NO3)2 + 2H2O

Balancing Neutralization Reactions

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Write strong acids, bases, and salts as ions.

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

Cross out matched ions.

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

Write a net ionic reaction.

H+ + OH- H2O

Net Ionic Equations for NeutralizationNaOH + HCl NaCl + H2O

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Select the correct group of coefficients for the

following neutralization equations.

A. __ HCl + __ Al(OH)3 __AlCl3 + __ H2O

1) 1, 3, 3, 1 2) 3, 1, 1, 1 3) 3, 1, 1, 3

B. __ Ba(OH)2 + __H3PO4 __Ba3(PO4)2 + __ H2O

1) 3, 2, 2, 2 2) 3, 2, 1, 6 3) 2, 3, 1, 1

Learning Check

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A. 3) 3, 1, 1 3

3HCl + 1Al(OH)3 1AlCl3 + 3H2O

B. 2) 3, 2, 1, 6

3Ba(OH)2 + 2H3PO4 1Ba3(PO4)2 + 6H2O

Solution

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Antacids neutralize stomach acid (HCl).Alka-Seltzer NaHCO3, citric acid, aspirin

Chooz, Tums CaCO3

Di-gel CaCO3 and Mg(OH)2

Gelusil, Maalox, Al(OH)3 and Mg(OH)2

Mylanta Al(OH)3 and Mg(OH)2

Milk of Magnesia Mg(OH)2

Antacids

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Learning Check

Write the neutralization reactions for stomach acid HCl and Mylanta.

Mylanta Al(OH)3 and Mg(OH)2

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Solution

Write the neutralization reactions for stomach acid HCl and Mylanta.

Mylanta: Al(OH)3 and Mg(OH)2

3HCl + Al(OH)3 AlCl3 + 3H2O

2HCl + Mg(OH)2 MgCl2 + 2H2O

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Some Salt Solutions are Neutral

A salt containing the ions of a strong acid and a strong base forms a neutral solution.

The metal ion does not produce H+ and the negative ion does not attract H+ from water.

For example, KNO3 contains ions from a strong base (KOH) and a strong acid (HNO3). A solution of KNO3 would be neutral.

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Some Salt Solutions are Basic

A salt containing the ions of a weak acid and a strong base forms a basic solution.

The ion of the weak acid attracts H+ from water to form a basic solution.

For example, KHCO3 contains ions from a strong base (KOH) and a weak acid (H2CO3). A solution of KHCO3 is basic.

HCO3- + H2O H2CO3 + OH-

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Some Salt Solutions are Acidic

A salt containing the ions of a strong acid and a weak base forms an acidic solution.

The ion of the weak base produces H+ in water to form an acidic solution.

For example, NH4Cl contains ions from a weak base (NH4OH) and a strong acid (HCl). A solution of NH4Cl is acidic.

NH4+ + H2O H3O+

+ NH3

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10.16 Acidity and basicity of Salt Solution

Salt solutions can be neutral, acidic, or basic depending on ions present, because some ions react

with water to produce H3O+ and some other ions react with water to produce OH- ions.

The general rule for predicting acidity or basicity of a salt solution is that the stronger partner from which the salt was formed dominates:

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A salt from strong acid and weak base will be acidic.

A salt from weak acid and strong base will be basic.

A salt from strong acid and strong base will be neutral.

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Learning Check

Predict whether a solution of each salt will be

1) acidic 2) basic or 3) neutral

A. Li2CO3

B. Mg(NO3)2

C. NH4Cl

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Solution

Predict whether a solution of each salt will be

1) acidic 2) basic or 3) neutral

A. Li2CO3 2) basic; CO3-2 forms HCO3

-

and OH-

B. Mg(NO3)2 3) neutral

C. NH4Cl 1) acidic; NH4+ produces

H3O+

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When an acid or base is added to water, the pH changes drastically.

A buffer solution resists a change in pH when an acid or base is added.

10.12 Buffer Solutions

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Buffers Absorb H3O+ or OH- from foods and

cellular processes to maintain pH. Are important in the proper functioning of

cells and blood. In blood maintain a pH close to 7.4. A

change in the pH of the blood affects the uptake of oxygen and cellular processes.

Buffers Animation

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A buffer solution Typically has equal concentrations of a weak

acid and its salt. Or equal concentration of a weak base and its

salt . Contains a combination of acid-base conjugate

pairs. Contains a weak acid and a salt of the

conjugate base of that acid.

Components of a Buffer

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The acetic acid/acetate buffer contains acetic acid (CH3COOH) and sodium acetate (CH3COONa).

The salt produces sodium and acetate ions.

CH3COONa CH3COO- + Na+

The salt provides a higher concentration of the conjugate base CH3COO- than the weak acid.

CH3COOH + H2O CH3COO- + H3O+ Large amount Large amount

Buffer Action

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The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate.

CH3COOH + OH– CH3COO– + H2O

Function of the Weak Acid

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The function of the acetate ion CH3COO- (conjugate base) is to neutralize H3O+ from acids. The weak acid product adds to the weak acid available.

CH3COO- + H3O+ CH3COOH + H2O

Function of the Conjugate Base

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The weak acid in a buffer neutralizes base. The conjugate base in the buffer neutralizes acid. The pH of the solution is maintained.

Summary of Buffer Action

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Learning Check

Does each combination make a buffer solution? 1) yes 2) no Explain.

A. HCl and KCl

B. H2CO3 and NaHCO3

C. H3PO4 and NaCl

D. CH3COOH and CH3COOK

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Solution

Does each combination make a buffer solution?

1) yes 2) no Explain.

A. HCl + KCl no; HCl is a strong acid.

B. H2CO3 + NaHCO3 yes; weak acid and its salt.

C. H3PO4 + NaCl no; NaCl does not contain a conjugate base of H3PO4.

D. CH3COOH + CH3COOK

yes; weak acid and its salt.

90

pH of a Buffer The [H3O+] in the Ka expression is used to

determine the pH of a buffer.

Weak acid + H2O H3O+ + Conjugate base

Ka = [H3O+][conjugate base] [weak acid]

For values of Ka see Page 303 of your book.[H3O+] = Ka x [weak acid]

[conjugate base]

pH = –log [H3O+]

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Calculation of Buffer pH

The weak acid H2PO4- in a blood buffer

H2PO4-/HPO4

2- has Ka = 6.2 x 10-8. What is the pH of the buffer if it is 0.20 M in both H2PO4

- and HPO42-?

[H3O+] = Ka x [H2PO4-]

[HPO42-]

[H3O+] = 6.2 x 10-8 x [0.20 M] = 6.2 x 10-8

[0.20 M]pH = –log [6.2 x 10-8] = 7.21

92

Learning Check

What is the pH of a H2CO3 buffer that is 0.20 M H2CO3 and 0.10 M HCO3

-?

Ka = 4.3 x 10-7

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Solution

What is the pH of a H2CO3 buffer that is 0.20 M H2CO3 and 0.10 M HCO3

- ?

Ka (H2CO3) = 4.3 x 10-7

[H3O+] = Ka x [H2CO3] [HCO3

-]

[H3O+] = 4.3 x 10-7 x [0.20 M] = 8.6 x 10-7

[0.10 M]

pH = –log [8.6 x 10-7] = 6.07

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10. 13 Buffers in the Body

The pH of body fluid is maintained by the following three major buffer systems: Carbonic acid – bicarbonate system Dihydrogen phosphate – hydrogen phosphate

system Proteins acting as either proton acceptor or proton donors at different pH values.

Relationships of the bicarbonate buffer system to the lungs and kidneys shown in the following figure 10.7.

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Fig 10.7 Lungs and kidneys relation with the bicarbonate Fig 10.7 Lungs and kidneys relation with the bicarbonate buffer systembuffer system

96

10.15 Titration Titration is a

laboratory procedure used to determine the molarity of an acid.

In a titration, a base such as NaOH is added to a specific volume of an acid.

Base (NaOH)

Acid

solution

97

Indicator

A few drops of an indicator is added to the acid in the flask.

The indicator changes color when the base (NaOH) has neutralized the acid.

98

End Point of Titration At the end point, the

indicator has a permanent color.

The volume of the base used to reach the end point is measured.

The molarity of the acid is calculated using the neutralization equation for the reaction.

99

Fig 10.9 Flow diagram for acid-base titrationFig 10.9 Flow diagram for acid-base titration

100

Calculating Molarity

What is the molarity of an HCl solution, if 18.0 mL of a 0.25 M NaOH, are required to neutralize 10.0 mL of the acid?

1. Write the neutralization equation. HCl + NaOH NaCl + H2O

2. Calculate the moles of NaOH. 18.0 mL NaOH x 1 L x 0.25 mole NaOH

1000 mL 1 L= 0.0045 moles NaOH

101

Calculating Molarity (continued)

3. Calculate the moles of HCl. HCl + NaOH NaCl + H2O 0.0045 mole NaOH x 1 mole HCl

1 mole NaOH= 0.0045 moles HCl

4. Calculate the molarity of HCl. 10.0 mL HCl = 0.010 L HCl 0.0045 mole HCl = 0.45 M HCl 0.0100 L HCl

102

Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH.

H2SO4 + 2KOH K2SO4 + 2H2O

1) 12.5 mL

2) 50.0 mL

3) 200. mL

Learning Check

103

Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH.

H2SO4 + 2KOH K2SO4 + 2H2O1) 12.5 mL 0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x

1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL2 mole H2SO4 1 L

Solution

104

Chapter Summary Arrhenius acid is a substance that gives

hydrogen ions, H+, when dissolved in water. Arrhenius base is a substance that gives

hydroxide ions, OH-, when dissolved in water. The neutralization reaction of an acid with a

base yields water plus a salt. Bronsted-Lowry Acids: Any substance that is

able to give a hydrogen ion, H+, to another ion or molecule.

Bronsted-Lowry bases: Any substance that is able to accept a hydrogen ion, H+, from an acid.

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Chapter Summary Contd.

Conjugate acid-base pair: Two substances whose formula differ by only a hydrogen ion, H+.

According to Bronsted-Lowry acid-base theory, water is both an acid and a base, property known as amphoteric.

Strong acid: An acid that gives up H+ easily and is essentially 100% dissociated (splits to produce H+ and an anion) in water.

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Chapter Summary Contd.

Weak acid: An acid that gives up H+ with difficulty and is less than 100% dissociated in water.

Strong base: A base that has a high affinity for H+ and holds it tightly.

Weak base: A base that has only a slight affinity for H+ and holds it weakly.

The stronger the acid, the weaker its conjugate base; the weaker the acid, the stronger its conjugate base.

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Chapter Summary Contd.

Strong acids have ka value much greater than 1; Weak acids have ka value much less than 1.

Ion product constant for water, kw:

kw = ka[H2O] = [H3O+][OH-] = [1.00 x 10-7][1.00 x 10-7] = 1.00 x 10-14 at 25oC. Product of [H3O+] and [OH-] is a constant.

Therefore, in an acidic solution where [H3O+] is large and [OH-] must be small.

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Chapter Summary Contd.Acidic solution: pH < 7 [H3O+ ] > 1.00 x 10-7 M

Neutral solution: pH = 7 [H3O+ ] = 1.00 x 10-7 M Basic solution: pH > 7 [H3O+ ] < 1.00 x 10-7 MBuffer: A combination of substances that act together to prevent a drastic change in pH.

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End of Chapter 10