Acids and Bases

http://www.shodor.org/unchem/basic/ab/

http://www.chemtutor.com/acid.htm

Strong Acids

Strong acids completely dissociate in water,

HCl(aq) =>H+1 (aq) + Cl-1 (aq) 1moldm-3 1moldm-3 1moldm-3

HCl - hydrochloric acid

HNO3 - nitric acid

H2SO4 - sulfuric acid

HBr - hydrobromic acid

HI - hydroiodic acid

HClO4 - perchloric acid

Weak Acids (Carboxylic Acids)

A weak acid only partially dissociates in water

CH3COOH(aq) =>H+1 (aq) + CH3COO-

(aq) 1moldm-3 x moldm-3 x

moldm-3

Examples of weak acids include hydrofluoric acid, HF, and

acetic acid, CH3COOH.

2

Strong Bases

The hydroxides of the Group I and Group II metals usually are considered to be strong bases.

Ionize completely

NaOH(aq) => Na+(aq) + OH- (aq)

LiOH - lithium hydroxide NaOH - sodium hydroxide KOH - potassium hydroxide RbOH - rubidium hydroxide Ca(OH)2 - calcium hydroxide

Ba(OH)2 - barium hydroxide

Weak Bases ( AMINES)

Examples of weak bases include ammonia, NH3,

Ionize partially

NH3 + H2 O=> NH4+1 + OH-

Methylamine and diethylamine, (CH3CH2)2NH.

3

Which is not a strong acid?

A. Nitric acid

B. Sulfuric acid

C. Carbonic acid

D. Hydrochloric acid

4

5

Properties of Acids (page 145)

1.Produce H+ (as H3O+) ions in water

HCl (aq) => H+ (aq) + Cl+ (aq)

2.Taste sour

3.Corrode metals:

Zn(s) + HCl(aq)=> ZnCl2 (aq) + H2

(g)

4.Electrolytes

5.React with bases to form a salt and water:

HCl + NaOH => NaCl + H2 O

6.pH is less than 7

7.Turns blue litmus paper to red

8. React with carbonates and bicarbonates to produce carbon dioxide.

CaCO3 (s) + 2HCl(aq) => CaCl2 (aq) + H2 O(l) + CO2

(g)

6

Properties of Bases (page 146)

Generally produce OHGenerally produce OH-- ions in water: ions in water:

NaOH => NaNaOH => Na+ + + OH + OH- -

Taste bitter, chalky,soapy,slipperyTaste bitter, chalky,soapy,slippery

Are electrolytesAre electrolytes

Displacement of ammonia from ammonium saltsDisplacement of ammonia from ammonium salts

React with acids to form salts and waterReact with acids to form salts and water

pH greater than 7pH greater than 7

Turns Turns redred litmus paper to litmus paper to blueblue

Which substance, when dissolved in water, to give a 0.1 mol dm− solution, has the highest pH?

A. HCl B. NaCl C. NH3

D. NaOH

7

Definitions

Arhenius Bronsted Lowry Lewis

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Arrhenius DefinitionArrhenius

Acid - Substances in water that increase the concentration of hydrogen ions (H+).

HCl(g) => H+1 (aq) + Cl-1 (aq)

Base - Substances in water that increase concentration of hydroxide ions (OH-).

NaOH(s) => Na+1 (aq) + OH-1 (aq)

10

Bronsted-Lowry Definition

Acids are species that donate a proton (H+).

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

(aq)

NO3- is called the conjugate base of the acid HNO3, and H3O+ is the conjugate

acid of the base H2O

Bases are species that accept a proton.

NH3 (aq) + H2O(l) => NH4+

(aq) + OH-(aq)

NH3 is a base and H2O is acting as an acid. NH4+ is the

conjugate acid of the base NH3, and OH- is the conjugate base of the acid H2O.

A compound that can act as either an acid or a base, such

as the H2O in the above examples, is called amphoteric 11

12

Conjugate Base - The species remaining after an acid has transferred its proton.

Conjugate Acid - The species produced after base has accepted a proton.

Conjugate Acid Base Pairs

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Acid Base Conjugate Acid Conjugate Base

HCl + H2O H3O+ + Cl-

H2PO4- + H2O

H3O+ + HPO4

2-

NH4+ + H2O

H3O+ + NH3

Base Acid Conjugate Acid Conjugate Base :NH3 + H2O

NH4+ + OH-

PO43- + H2O

HPO42- + OH-

Bronsted-Lowry Acid Base Systems

Lewis Definition A Lewis acid is defined to be any species that accepts

lone pair electrons. A Lewis base is any species that donates lone pair

electrons.

H+ + : OH-1 => H2O

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Auto Ionization of Water http://www.wwnorton.com/college/chemistry/gilbert2/tutorials/interface.asp?chapter=chapter_16&folder=self_ionization self-ionization of water

H2O (l) => H+ (aq) + OH− (aq)

or

2 H2O (l) => H3O+ (aq) + OH− (aq)

hydronium ion

It is an example of autoprotolysis, and relies on the amphoteric nature of water.

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Kw

The autoionization of water in equilibrium:

H2O(l) H+(aq) + OH-(aq) ΔH> 0

The equilibrium constant expression for this reaction is given by:

Keq = [H+] [OH-]/[H2 O] Kc x [H2O]= Kw = 1.0 x 10-14

ion product for water

The value for Kw is for room temperature, 25 °C, and 1.0 atm

The equilibrium constant expression applies not only to pure (distilled) water but to any aqueous solution. It can be used to calculate either [H+] or [OH-] provided one of them is known. 16

Kw and T: http://mmsphyschem.com/autoIon.htm

H2O(l) H+(aq) + OH-(aq) ΔH> 0

T Kw

0 1.14 x 10-15 5 1.85 x 10-15

10 2.92 x 10-15

15 4.53 x 10-15

20 6.81 x 10-15

25 1.01 x 10-14

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The ionization of water is endo so,

as the temperature increases,so does the Kw

FormulaspH = -log [H] pKa = -log Ka

pOH = -log [OH] pKb = -log Kb

pH + pOH = 14 pKb + pKa = 14

[H] [OH] = 1 x 10 -14 = Ka x Kb = Kw

Kw = 1.0 x 10-14 pKw = 14 pKw = pH + pOH

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pH Formulas

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20

Lime(calcium hydroxide) is added to a lake to neutralize the effects of acid rain. The pH value of the lake water rises from 4 to 7. What is the change in concentration of H+ in the lake water?

A. An increase by a factor of 3 B. An increase by a factor of 1000 C. A decrease by a factor of 3 D. A decrease by a factor of 1000

21

The pH Scale

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Calculating the pHpH = - log [H3O+]

Example 1: If [H3O+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)

pH = 10

Example 2: If [H3O+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)

pH = 4.74

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pH and acidity

The pH values of several common substances are shown at the right.

Many common foods are weak acids

Some medicines and many household cleaners are bases.

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Indicators: Substances that change color when the concentration of hydrogen changes.

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Neutralization An acid will neutralize a base,

giving a salt and water as products

Examples:

HCl + NaOH NaCl + H2O

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

H3PO4 + 3 KOH K3PO4 + 3 H2O

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

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Neutralization Problems

# moles acid = # moles base

If an acid and a base combine in a 1 to 1 ratio, then the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base

Vacid M acid = V base M base

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Neutralization Problems Example 1:

HCl + KOH KCl + H2O

If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of KOH solution, what is the concentration of the KOH solution?

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Neutralization Problems

Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation

n Vacid C acid = V base C base

The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa)

Example

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

moles base = 2 x moles acid

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Neutralization Problems Example 3:

H3PO4 + 3 KOH K3PO4 + 3 H2O If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of

H3PO4 solution, what is the concentration of the H3PO4 solution?

Solution:

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Neutralization Problems Example 2: Sulfuric acid reacts with sodium hydroxide according to the following reaction:

H2SO4 + 2 NaOH Na2SO4 + 2 H2O If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of

NaOH solution, what is the concentration of the NaOH solution?

Solution:In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore

2 Vacid Cacid = Vbase Cbase

2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase

Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 )

Cbase = 0.500 M

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Neutralization Problems Example 4:

2 HCl + Ca(OH)2 CaCl2 + 2 H2O If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of

Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution?

YEAR 2

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Acid Base DissociationAcid-base reactions are equilibrium processes.

The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant: Ka and Kb

The stronger the acid or base, the larger the value of the

dissociation constant.

]B[:

][OH [HB] K O][H K

[HA]

][H ]A[: K O][H K

constant. is solutions dilute in O][H

][H OH

:Note

O][H ]B:[

]OH][HB[ K

O][H HA][

]OH][ A[: K

waterin base a For waterin acid an For

-

-

b2eq

-

a2eq

2

3

2-

-

eq2

3-

eq

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Acid Strength Strong Acid - Transfers all of its protons to water;

- Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated.

Weak Acid - Transfers only a fraction of its protons to water;

- Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water

As acid strength decreases, base strength increases. The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base

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Acid Dissociation ConstantsDissociation constants for some weak acids

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Base Strength Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons.

Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons.

As base strength decreases, acid strength increases. The stronger the base, the weaker its conjugate acid. The weaker the base the stronger its conjugate acid.

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Weak Acid EquilibriaA weak acid is only partially ionized.Both the ion form and the unionized form exist at equilibrium HA + H2O H3O+ + A-

The acid equilibrium constant is

Ka = [H3O+ ] [A-] [HA]

Ka values are relatively small for most weak acids. The greatest part of the weak acid is in the unionized form

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Weak Acid Equilibrium Constants

Sample problem . A certain weak acid dissociates in water as follows: HF + H2O H3O+ + F-

If the initial concentration of HF is 1.5 M and the equilibrium concentration of H3O+ is 0.0014 M. Calculate Ka for this acid

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Weak Base EquilibriaWeak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constantGeneral form: B + H2O BH+ + OH-

Kb = [BH+][OH-] [B]

ExampleNH3 + H2O NH4

+ + OH-

Kb = [NH4+][ OH-]

[NH3]

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Weak Base Equilibrium Constants

Sample problem . A certain weak base dissociates in water as follows: B + H2O BH+ + OH-

If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is 0.0011 M. Calculate Kb for this baseSolutionKb = [BH+ ] [OH-]

[B]

I C E Substituting[B] 1.2 -x 1.2-x Kb = (0.0011)2 = 1.01 x 10-6 [OH-] 0 +x x 1.1989[BH+ ] 0 +x x

x = 0.0011 1.2-x = 1.1989

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Weak Acid Equilibria Concentration Problems

Problem 1. A certain weak acid dissociates in water as follows: HA + H2O H3O+ + A-

The Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+ ] and pH of a 2.0 M solutionSolutionKa = [H3O+ ] [A-] = 2.0 x 10-6

[HA] I C E Substituting

[HA] 2.0 -x 2.0-x Ka = x2 = 2.0 x 10-6

[A-] 0 +x x 2.0-x[H3O+ ] 0 +x x If x <<< 2.0 it can be dropped

from the denominator

The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3

[A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998

pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7

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Weak Acid Equilibria Concentration Problems

Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O H3O+ + CH3COO-

The Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-] [H3O+ ] and pH of a 0.100 M solutionSolution

Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5

[CH3COOH] I C E Substituting

[CH3COOH] 0.100 -x 0.100-x Ka = x2 = 1.8 x 10-5

[CH3COO- ] 0 +x x 0.100-x[H3O+] 0 +x x If x <<< 0.100 it can be dropped

from the denominatorThe x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3

[CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH ] = 0.100 - 0.0013 = 0.0987

pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88

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Weak Base EquilibriaExample1. Ammonia dissociates in water according to the following equilibrium

NH3 + H2O NH4+ + OH-

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

[NH3]Calculate the concentration of [NH4

+][ OH-] [NH3 ]and the pH of a 2.0M solution.

I C E Substituting[NH3] 2.0 -x 2.0-x Kb = x2 = 1.8x 10-5

[OH-] 0 +x x 2.0-x[NH4

+] 0 +x x If x <<< 2.0 it can be dropped from the denominator

The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3 [OH-] = [NH4

+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994pOH = - log [OH-] =-log (6.0 x10-3) = 2.22pH = 14-pOH = 14-2.22 = 11.78

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Amphoteric Solutions A chemical compound able to react with both an

acid or a base is amphoteric.    Water is amphoteric. The two acid-base couples

of water are H3O+/H2O and H2O/OH-

It behaves sometimes like an acid, for example

And sometimes like a base :

Hydrogen carbonate ion HCO3- is also amphoteric,

it belongs to the two acid-base couples H2CO3/HCO3

- and HCO3-/CO3

2-

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Common Ion EffectThe common ion effect is a consequence of Le Chatelier’s Principle When the salt with the anion (i.e. the conjugate base) of a weak acid is added to that acid,

It reverses the dissociation of the acid.Lowers the percent dissociation of the

acid.A similar process happens when the salt with the cation (i.e, conjugate acid) is added to a weak base.These solutions are known as Buffer Solutions.

Buffers A buffer solution is na aqueous solution that resists a

change in pH when a small amount of acid, base or water is added to it.

Acidic Buffers: Weak acid and its salt

Can be prepared by using excess of a weak acid and a strong base .

NaOH(aq) + CH3 COOH(aq) => CH3 COONa(aq) + CH3 COOH(aq) + H2 O (l)

salt excess weak acid

CH3COOH (aq) => CH3COO- (aq) + H+ (aq) Ka = 1.74 x 10-5

A CB

pH = pKa + log [ CB] / [ A ] Henderson Hasselbach Equation

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FormulaspH = -log [H] pKa = -log Ka

pOH = -log [OH] pKb = -log Kb

pH + pOH = 14 pKb + pKa = 14

[H] [OH] = 1 x 10 -14 = Ka x Kb

Kw = 1.0 x 10-14 pKw = 14

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1. Given 30 cm3 of 0.1 M ethanoic acid and 10 cm3 of 0.1 M NaOH, find the pH of the buffer solution.

CH3COOH (aq) => CH3COO- (aq) + H+ (aq) Ka = 1.74 x 10-5

A CB pKa = -log (1.74 x 10-5 )=4.76

n CH3COOH = 0.1 x 0.03 = 0.003

n NaOH = 0.1 x 0.01 = 0.001

excess CH3COOH after titration = 0.002

pH = pKa + log [ CB] / [ A ]

pH = 4.76 + log(0.001/0.002)= 4.5

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2. Calculate the pH of a solution that is 0.50 M CH3COOH and 0.25 M NaCH3COO.

CH3COOH + H2O H3O+ + CH3COO- (Ka = 1.8 x 10-5)

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Buffer Solution CalculationsCalculate the pH of a solution that is 0.50 M CH3COOH and 0.25 M NaCH3COO.

CH3COOH + H2O H3O+ + CH3COO- (Ka = 1.8 x 10-5)

Solution

Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5

[CH3COOH] I C E . Substituting

[CH3COOH] 0.50 -x 0.50-x Ka = x (0.25+x) = 1.8 x 10-5

[CH3COO-] 0.25 +x 0.25+x (0.50-x)[H3O+] 0 +x x If x <<< 0.25 it can be dropped from both

expressions in ( ) since adding or subtracting a small amount will not significantly change the value of the ratio

Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5

x = 3.6 x 10-5 = [H3O+]

pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44

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Using the Henderson -Hasselbach Equation

pH = pKa + log([A-]/[HA]) Example

Calculate the pH of the following of a mixture that contains 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 + H2O H3O+ + C3H5O3-

Solution

Using the Henderson-Hasselbach equation

pH = - log (1.4 x 10-4) + log ( 0.25/0.75 )

= 3.85 + (-0.477) = 3.37

Basic Buffers: Weak base and its salt

Can be prepared by using excess of a weak base and a strong acid .

HCl(aq) + CH3 NH2(aq) => CH3 NH2Cl(aq) + CH3 NH3 +(aq) + H2 O (l)

salt excess base

CH3NH2 (aq) + H2 O => CH3NH3+ (aq) + OH- (aq) Kb = 4.37 x 10-4

B CA

pOH = pKb + log [ CA] / [ B ] Henderson Hasselbach Equation

54

Find the pH of a buffer made with 0.025 M methylamine and 0.010 M HCl.

55

Methyl amine is a weak base with a Kb or 4.38 x 10-4

CH3NH2 + H2O CH3NH3+ + OH-

Calculate the pH of a solution that is 0.10 M in methyl amine and 0.20 M in methylamine hydrochloride.

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Henderson-Hasselbach Equation & Base Buffers Methyl amine is a weak base with a Kb or 4.38 x 10-4

CH3NH2 + H2O CH3NH3+ + OH-

Calculate the pH of a solution that is 0.10 M in methyl amine and 0.20 M in methylamine hydrochloride.

pOH = pKb + log ([BH+] / [B])

Solution pOH = -log (4.38 x 10-4) + log (0.20 / 0.10)

= 3.36 + 0.30 = 3.66

pH = 14- 3.66 = 10.34

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Additional Buffer ProblemsHow many grams of sodium formate, NaCHOO, would have to

be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to

make the solution a buffer of pH 3.80? Ka= 1.78 x 10-4

pH = pKa + Log ([A-]/[HA])

Solution 3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12]

3.80 = 3.75 + Log [A-] - (-0.92)

Log [A-] = 3.80 - 3.75 - 0.92 = - 0.87

[A-] = 10-0.87 = 0.135 mol dm-3

The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1

So (0.135 mol dm-3)(58.0 gmol-1 ) = 7.8 grams per dm-3

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Relationship of Ka, Kb & Kw HA weak acid. Its acid ionization is

A- is the conjugate base Its base ionization is

Multiplying Ka and Kb and canceling like terms

60

Titration Curves Simulations:

http://chem-ilp.net/labTechniques/AcidBaseIdicatorSimulation.htm

A graph showing pH vs volume of acid or base added

The pH shows a sudden change near the equivalence point

The equivalence point is the point at which the moles of OH- are equal to the moles of H3O+

The end point is when the indicatorchanges color acurate to the addition of one dro.

Equivalence Point x End Pointhttp://www.chem.ubc.ca/courseware/pH/section15/index.html

The endpoint of a titration is NOT the same thing as the equivalence point.

The equivalence point is a single point defined by the reaction stoichiometry as the point at which the base (or acid) added exactly neutralizes the acid (or base) being titrated.

The endpoint is defined by the choice of indicator as the point at which the colour changes. Depending on how quickly the colour changes, the endpoint can occur almost instantaneously or be quite wide.

Intuition may suggest that the endpoint of the titration will occur at the equivalence point if we choose an indicator whose pKa is equal to the pH of the equivalence point

61

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Indicators: Substances that change color when the concentration of hydrogen changes. The choice of an indicator is determined by the pH of the solution at the equivalence point http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm

An indicator is a solution of a weak acid in which the the conjugate base has a different color from that of the undissociated acid

Given na aqueous solution of na indicator HIn(aq)

HIn(aq) H+(aq) + In-(aq)

Color A Color B

(in acid solution) ( in basic solution)

The end point of the indicator will be when pH = pkIN

At low pH we see color A At high pH we see color B 63

64

Strong acid-strong base At equivalence point, Veq:

Moles of H3O+ = Moles of OH-

There is a sharp rise in the pH as one approaches the equivalence point

With a strong acid and a strong base, the equivalence point is at

pH =7

Indicators: methyl red and phenophtalein

Change between 4 and 10

65

a strong acid-strong base has a very sharp equivalence point, meaning that a very large change in pH occurs due to the addition of a very small amount of titrant, often a single drop.

. This means that any indicator that starts to change colour in this range will signal equally well that the equivalence point has been reached.

Any acid-base indicator that changes colour between pH 4 and pH 10 is suitable to detect the end-point for a strong acid - strong base titration. Both methyl orange and phenolphthalein could be used. Just one drop of the added base will bring about a change in colour of the indicator.

66

67

68

Weak acid-strong base The increase in pH is more gradual

as one approaches the equivalence point

With a weak acid and a strong base, the equivalence point is higher than pH = 7

CH3COOH + NaOH

Ka

Indicator : phenophtalein

NaOH(aq) + HF(aq) NaF(aq) + H2O(l)

WEAK ACID - STRONG BASEWEAK ACID - STRONG BASE

Strong Acid X Weak Base

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The equivalence point is below 7 because the salt (NH4Cl) formed at the neutralization reacts with water to give H+ ions. The equivalence point lies at about pH 5.3. It is, therefore necessary to use an indicator with pH range slightly on the acidic side. Methyl orange can be used. Phenolphthalein is not suitable because its colour change occurs away from the equivalence point.

Weak Acid Weak Base

There will hardly be a pH change around the equivalence point

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Buffered Weak Acid-Strong Base Titration Curve

The initial pH is higher than the unbuffered acid

As with a weak acid and a strong base, the equivalence point for a buffered weak acid is higher than pH =7

The conjugate base is strong enough to affect the pH

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Polyprotic Weak Acid-Strong Base Titration Curve

Phosphoric Acid has three hydrogen ions.

There are three equivalence points

H3P04 + H2O H3O+ + H2PO4-

H2PO4- +H2O H3O+ + HPO4

2-

HPO42- +H2O H3O+ + PO4

3-

SALTS & Hydrolysis Salts are ionic and already completely

dissociated. They are also strong electrolytes.

I. When a salt comes from a strong acid and a strong base, they form neutral solutions when they dissolve in water.

NaCl(aq) => Na+(aq) + Cl-(aq)

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II. When a salt comes from a weak acid and a strong base, it forms na alkaline solution when it dissolves in water.

CH3COOH(aq) + NaOH(aq) => CH3COONa(aq) + H2O(l)

CH3COOH(aq) => Na+(aq) + CH3COO-(aq)

+

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

↨

CH3COOH(aq)

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III. When a salt comes from a strong acid and a weak base, it will be acidic in solution.

HCl(aq) + NH3(aq) => NH4Cl(aq)

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

+

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

↨

NH3(aq)

76

http://home.clara.net/rod.beavon/AlCl3_and_water.htm

The acidity of a salt also depends on the size and charge of the anion.

Aluminum chloride reacts vigorously with water to produce a strong acidic solution.

AlCl3(s) + 3H2O(l) → Al(OH )3 (s) + 3HCl(g) amphoteric nature

The charge of aluminum is spread over the small ion giving it a high charge density. It will then attract six pair of electrons forming the hexahydrated ion CC page 155

http://en.wikipedia.org/wiki/Hydrolysis

The greater the charge of nthe ion or 77