Arrhenius Theory Acids release hydrogen ions (H + ) Acids release hydrogen ions (H + ) HCl H + + Cl...

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Arrhenius TheoryArrhenius Theory AcidsAcids release hydrogen release hydrogen

ions ions (H(H++)) HCl → HCl → HH++ + Cl + Cl--

BasesBases release hydroxide release hydroxide ions ions (OH(OH--))

NaOH → NaNaOH → Na++ + + OHOH--

ACID-BASE THEORIESACID-BASE THEORIES• The most general theory for

common aqueous acids and bases is the BRØNSTED - LOWRY

theory

• ACIDS DONATE H+ IONS

• BASES ACCEPT H+ IONS

You PracticeYou PracticeNHNH33 + H + H22O → NHO → NH44

++ + OH + OH--

HNOHNO33 + NaOH → H + NaOH → H22O + NaNOO + NaNO33

NaHCONaHCO33 + HCl → NaCl + H + HCl → NaCl + H22COCO33

Lewis acid = a substance that accepts

an electron pair

Lewis Acids & BasesLewis Acids & Bases

Lewis base = a substance that

donates an electron pair

Formation of hydronium ion is also an excellent example.

Lewis Acids & BasesLewis Acids & Bases

•Electron pair of the new O-H bond originates on the Lewis

••••••

O—HO—HH+

1111Lewis Acid/Base Reaction

Why is Water Neutral?Why is Water Neutral?

When one water gains, When one water gains, another loses a H+another loses a H+

[H[H33O+ ] = [OH-]O+ ] = [OH-]

Relative ion Relative ion concentrationsconcentrations

ppHH is a relative measure of is a relative measure of the the hydrogen ionhydrogen ion concentrationconcentration

ppHH is a is a ratingrating; ; ranges fromranges from 0 – 14 0 – 14

0 = most0 = most,, 7 equal, 7 equal, 14 = least14 = least

Any pAny pXX Scales ScalesIn generalIn generalpX = -log XpX = -log X

pOH = - log [OHpOH = - log [OH--]] pH = - log [HpH = - log [H++]]

Determining pOHDetermining pOHpH + pOH = 14

If know one can determine the other.

If pH = 13, what is the pOH?13 + pOH = 14

pOH = 14 – 13 = 1

Why at a pH = 7 ?Why at a pH = 7 ? Determined by concentration [ X ] of Determined by concentration [ X ] of

each ioneach ion

[H+ ]= [OH-] = 10[H+ ]= [OH-] = 10--77MM

[H+ ]= 10[H+ ]= 10-pH-pHMM

[OH-] = 10[OH-] = 10-pOH-pOHMM

2020Water

H2O can function as both an ACID and a BASE.

Equilibrium constant for water = Kw

Kw = [H3O+] [OH-] =

1.00 x 10-14 at 25 oC

AUTOIONIZATION

2121Water

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

In a neutral solution [H3O+] = [OH-]

so [H3O+] = [OH-] = 1.00 x 10-7 M

Autoionization

2222Acid-Base Reactions

Chapter 17

ACIDSACIDSNonmetal oxides can be acids

CO2(aq) + H2O(liq) ---> H2CO3(aq)

SO3(aq) + H2O(liq) ---> H2SO4(aq)

and can come from burning coal and oil.

24BASESMetal oxides are

CaO in water. Indicator shows solution is basic.

CaO(s) + H2O(liq) --> Ca(OH)2(aq)

2525Hydrolysis

27Acid-Base ReactionsAcid-Base Reactions• sometimes called NEUTRALIZATIONS

because the solution is neither acidic nor basic at the end.

• The other product of the A-B reaction is a SALT, MX.

HX + MOH ---> MX + H2O Mn+ comes from base & Xn- comes from acid

This is one way to make compounds!

Acid-Base ReactionsAcid-Base Reactions The “driving force” is the formation of water.

NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(liq)

Net ionic equation

OH-(aq) + H+(aq) ---> H2O(liq)

This applies to ALL reactions of

STRONG acids and bases.

Weak Acids and Weak Acids and BasesBases

Acid Conjugate Baseacetic, CH3CO2H CH3CO2

-, acetateammonium, NH4

+ NH3, ammoniabicarbonate, HCO3

- CO32-, carbonate

A weak acid (or base) is one that ionizes to a VERY small extent (<

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7

AcidsAcids

ConjugatConjugatee

BasesBases

Kw = Ka* Kb

3636Weak Acids and Weak Acids and BasesBases

acetic acid, CH3CO2H (HOAc)HOAc + H2O H3O+ + OAc-

Acid Conj. base

Ka [H3O+][OAc- ][HOAc]

1.8 x 10-5

(K is designated Ka for ACID)

[H3O+] and [OAc-] are SMALL, Ka << 1.

3737Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid

You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH

Step 1. ICE table.

[HOAc] [H3O+] [OAc-]

[HOAc] [H3O+] [OAc-]I 1.00 0 0C -x +x +xE 1.00-x x x

Note that we neglect [H3O+] from H2O.

Ka 1.8 x 10-5 = [H3O+][OAc- ][HOAc]

1.00 - x

Assume x is very small because Ka is so small.

Ka 1.8 x 10-5 = x2

1.00Now we can more easily solve this Now we can more easily solve this

approximate expression.approximate expression.

4040Equilibria Involving A Weak Acid

Step 3. Solve Ka approximate expression

Ka 1.8 x 10-5 = x2

x = [H3O+] = [OAc-] = [Ka • 1.00]1/2

x = [H3O+] = [OAc-] = 4.2 x 10-3 M

pH = -log (4.2 x 10-3) = 2.37

Chapter 17 – Acids and Bases

What is the [H+] of a 1.0 M acetic acid if Ka = 1.8 x 10-5 for the solution:

A. 1.0 M B. 0.3 M C. 0.0042 M D. 1.8 x 10-5 M

Chapter 17 – Acids and Bases

A 0.100 M solution of HX has a pH of is 4.5 Determine the Ka for the acid, HX.

A. 1.0 x 10-8

B. 3.0 x 10-7

C. 2.5 x 10-8

D. 4.5 x 10-9

Weak BasesWeak Bases

4444Equilibria Involving A Weak Equilibria Involving A Weak BaseBase

You have 0.010 M NH3. Calc. the pH.

NH3 + H2O NH4+ + OH-

Kb = 1.8 x 10-5

Step 1; ICE table

[NH3] [NH4+] [OH-]

Weak BaseWeak BaseStep 1. ICE table

[NH3] [NH4+] [OH-]

I 0.010 0 0

C -x +x +x

E 0.010 - x x x

Weak BaseWeak BaseStep 2. Solve the equilibrium expression

Kb 1.8 x 10-5 = [NH4+][OH- ]

[NH3 ] = x2

0.010 - x

Assume x is small (100•Kb < Co), so

x = [OH-] = [NH4+] = 4.2 x 10-4 M

[NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M

The approximation is valid !

Neutralization Neutralization ReactionsReactions

NaOH + HCl NaOH + HCl → NaCl + H→ NaCl + H2200

REACT a base with an acid.REACT a base with an acid.

forms salt and waterforms salt and water

19.4 Neutralization Reactions >19.4 Neutralization Reactions >

A reaction between an acid and a base will go to completion when the solutions contain equal moles of hydrogen ions and hydroxide ions.

Acid-Base ReactionsAcid-Base Reactions

• The balanced equation provides the correct ratio of acid to base.

Neutralization occurs when the number of moles of

hydrogen ions is EQUALEQUAL to the number of moles of hydroxide

TitrationTitration

For hydrochloric acid and sodium hydroxide, the mole ratio is 1:1.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)1 mol 1 mol 1 mol 1 mol

For sulfuric acid and sodium hydroxide, the ratio is 1:2.

1 mol 2 mol 1 mol 2 mol

• Two moles of the base are required to neutralize one mole of the acid.

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

Similarly, hydrochloric acid and calcium hydroxide react in a 2:1 ratio.

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)2 mol 1 mol 1 mol 2 mol

Sample Problem 19.7

Finding the Moles Needed for Neutralization

How many moles of sulfuric acid are required to neutralize 0.50 mol of sodium hydroxide? The equation for the reaction is

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O.

KNOWNS

mol NaOH = 0.50 mol

1 mol H2SO4/2 mol NaOH (from balanced equation)

Analyze List the knowns and the unknown.1

UNKNOWN

mol H2SO4 = ? mol

To determine the number of moles of acid, you need to know the number of moles of base and the mole ratio of acid to base.

Sample Problem 19.7

Calculate Solve for the unknown.2

Use the mole ratio of acid to base to determine the number of moles of acid.

Sample Problem 19.7

0.50 mol NaOH × 1 mol H2SO4

2 mol NaOH= 0.25 mol H2SO4

Evaluate Does the result make sense?3

Because the mole ratio of H2SO4 to NaOH is 1:2, the number of moles of H2SO4 should be half the number of the moles of NaOH.

Sample Problem 19.7

You can use a neutralization reaction to determine the concentration of an acid or base.

• The process of adding a measured amount of a solution of known concentration to a solution of unknown concentration is called a

titration.

TitrationTitration

A flask with a known volume of acids (and an indicator) is placed beneath a buret that is filled with a base of known concentration.

The base is slowly added from the buret to the acid.

A change in the color of the solution is the signal that neutralization has occurred.

The indicator that is chosen for a titration must change color at or near the pH of the equivalence point.

TitrationTitration

• The point at which the indicator changes color is the end point of the titration.

19.2 Hydrogen Ions and Acidity >19.2 Hydrogen Ions and Acidity >

An indicator (HIn) is a weak acid or base that dissociates in a known pH range.

• Indicators work because their acid form and base form have different colors in solution.

Acid-Base Indicators

• The acid form of the indicator (HIn) is dominant at low pH and high [H+].

• The base form (In−) is dominant at high pH and high [OH−].

The pH change of a solution during the titration of a strong acid (HCl) with a strong base (NaOH).

• The equivalence point for this reaction occurs at a pH of 7.

• As the titration nears the equivalence point, the pH rises dramatically because hydrogen ions are being used up.

Sample Problem 19.8

Determining Concentration by Titration

A 25-mL solution of H2SO4 is neutralized by 18 mL of 1.0M NaOH.

What is the concentration of the H2SO4 solution?

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

KNOWNS

[NaOH] = 1.0M

VNaOH = 18 mL = 0.018 L

VH2SO4 = 18 mL = 0.018 L

Analyze List the knowns and the unknown.1

UNKNOWN

[H2SO4] = ?M

The conversion steps are as follows: L NaOH → mol NaOH → mol H2SO4 → M H2SO4.

Sample Problem 19.8

Convert volume to liters because molarity is in moles per liter.

Use the molarity to convert the volume of base to moles of base.

Sample Problem 19.8

0.018 L NaOH × 1.0 mol NaOH

1 L NaOH= 0.018 mol NaOH

Use the mole ratio to find the moles of acid.

Sample Problem 19.8

0.018 mol NaOH × 1.0 mol H2SO4

2 mol NaOH= 0.0090 mol H2SO4

Calculate the molarity by dividing moles of acid by liters of solution.

molarity = mol of solute

L of solution

0.0090 mol

0.025 L=

= 0.36M H2SO4

Sample Problem 19.8

19.5 Salts in Solution >19.5 Salts in Solution >

Interpret Graphs

• One curve is for the addition of sodium hydroxide, a strong base, to ethanoic acid, a weak acid.

• An aqueous solution of sodium ethanoate exists at the equivalence point.

CH3COOH(aq) + NaOH(aq) →

CH3COONa(aq) + H2O(l)

69Acetic acid titrated with NaOHAcetic acid titrated with NaOH

Figure 18.5Figure 18.5

Weak acid titrated with a strong base

70Acid-Base TitrationSection 18.3

You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?

HBz + NaOH ---> Na+ + Bz- + H2O

CC66HH55COCO22H = HBzH = HBz Benzoate ion = Bz-

Kb = 1.6 x 10-10

71QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.

pH at pH at equivalence equivalence point?point?

pH of solution of pH of solution of benzoic acid, a weak benzoic acid, a weak acidacid

Benzoic acid Benzoic acid + NaOH+ NaOH

pH at pH at half-way half-way point?point?

72Acid-Base ReactionsAcid-Base Reactions

Strategy — find the conc. of the conjugate base Bz- in the solution AFTER

the titration, then calculate pH.This is a two-step problem

1.stoichiometry of acid-base reaction2.equilibrium calculation

QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the

equivalence point. What is the pH of the final solution?

73STOICHIOMETRY PORTION

M * V = mol1. Calc. moles of NaOH req’d

(0.100 L)(0.025 M) = 0.0025 mol HBz

This requires 0.0025 mol NaOH

2. Calc. volume of NaOH req’d

0.0025 mol (1 L / 0.100 mol) = 0.025 L

25 mL of NaOH req’d

74STOICHIOMETRY PORTION3. Moles Bz- produced = moles HBz = 0.0025 mol

4. Calc. conc. of Bz-

There are 0.0025 mol of Bz- in a TOTAL SOLUTION VOLUME of

125 mL125 mL

[Bz[Bz--] ] = 0.0025 mol / 0.125 L= 0.0025 mol / 0.125 L = = 0.020 M0.020 M

75Equivalence PointMost important species in solution is benzoate ion,

Bz-. It will react to form the weak conjugate base, benzoic acid, HBz.

Bz- + H2O HBz + OH- Kb = 1.6 x 10-10

[Bz-] [HBz] [OH-]I 0.020 0 0

C - x +x +x

E 0.020 - x x x

x = [OH-] = 1.8 x 10-6

pOH = 5.75 -----> pH = 8.25

Kb 1.6 x 10-10 = x2

0.020 - x

77QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?

pH at pH at half-way half-way point?point?

Equivalence Equivalence point point pH = 8.25pH = 8.25

78Acid-Base ReactionsAcid-Base ReactionsYou titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH. What is the pH at the half-way point?

[H3O ] = [HBz][Bz- ]

• Ka

At the half-way point, [HBz] = [Bz-]Therefore, [H3O+] = Ka = 6.3 x 10-5

pH = 4.20 = pKa of the acid

Both HBz and BzBoth HBz and Bz-- are present.are present.

This is a BUFFER!This is a BUFFER!

Figure 18.7Figure 18.7

Weak base (NH3) titrated with a

strong acid (HCl)

• Strong acid + strong base HCl + NaOH ----> SALT WATER

• Strong acid + weak base HCl + NH3 ---> ACID

• Weak acid + strong base HOAc + NaOH ---> BASIC• Weak acid + weak base HOAc + NH3 ---> Ka / Kb

Arrhenius Theory Acids release hydrogen ions (H + ) Acids release hydrogen ions (H + ) HCl H + + Cl...

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