Concepts of Acid Base Neutralization

7/30/2019 Concepts of Acid Base Neutralization

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An electrolyteis a substance that, when dissolved in water, results

in a solution that can conduct electricity.

A nonelectrolyteis a substance that, when dissolved, results in a

solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte

ELECTROLYTIC PROPERTIES


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METHOD OF DISTINGUISHING BETWEENELECTROLYTES AND NONELECTROLYTES

A pair of inert electrodes (Cu or Pt) is immersed in a beaker

of water.

To light the bulb, electric current must flow from one

electrode to the other, thus completing the circuit.

By adding NaCl (ionic compound), the bulb will glow.

NaCl breaks up into Na+ and Cl- ions when dissolves in water.

Na+ are attracted to the negative electrode.

Cl- are attracted to the positive electrode.

The movement sets up an electric current that is equivalent to

the flow of electrons along a metal wire.


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Strong Electrolyte

100% dissociation (breaking up of compound intocations and anions

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte

not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

A reversiblereaction. The reaction can occur in

both directions.


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Hydrationis the process in which an ion is surrounded by

water molecules arranged in a specific manner.

Hydration helps to stabilize ions and prevents cations fromcombining with anions.

d+d-

H2O

d+


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Nonelectrolyte does not conduct electricity?

No cations (+) and anions (-) in solution

C6H12O6 (s) C6H12O6 (aq)H2O


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Have a sour taste.- Vinegar owes its taste to acetic acid.

- Citrus fruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon

dioxide gas

Cause color changes in plant dyes.

2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g)

2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l)

Aqueous acid solutions conduct electricity.

PROPERTIES ACIDS


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Have a bitter taste.

Feel slippery. Many soaps contain bases.

Cause color changes in plant dyes.

Aqueous base solutions conduct electricity.

Examples:

PROPERTIES OF BASES


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ROLE OF WATER TO SHOW PROPERTIES OF

ACIDS

Anhydrous pure acid (without water) does not show acidicproperties.

In dry form, acids exist as neutral covalent molecules.

Dry acids do not dissociate to form hydrogen ion (H+).

When a pure acid is dissolved in water, it will show the

properties of acids.

This is because acids will dissociate in water to form H+ or

hydroxonium/hydronium ion, H3O+ which are free to move.

For example:

i) HCl in liquid methylbenzene (organic solvent) - does not

show acidic properties.

ii) HCl in watershow acidic properties


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ROLE OF WATER TO SHOW PROPERTIES OF

ALKALI

Dry base does not show alkaline properties.

A base in dry form, contains hydroxide ions (OH-) that are

not free to move. Thus, the alkaline properties cannot be

shown. In the presence of water, bases can dissociate in water to

form hydroxide ions, OH-, which are free to move. Thus,

alkaline properties are shown.

For example:i) ammonia in tetrachlomethane (organic solvent)do not

show alkaline properties

ii) ammonia in watershow alkaline properties


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DEFINITION OF ACID AND BASE

Arrhenius Brnsted-

LowryLewis


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Arrhenius acid is a substance that produces H+ (hydrogen ion)

or hydronium ion(H3O+) in water

Arrhenius base is a substance that produces OH- in water

DEFINITION OF ACID AND BASE BY ARRHENIUS


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Examples of acid:

CO2

(g) + H2

O (l) H2

CO3

(aq)

H2CO3 (aq) + H2O(l) H3O+ (aq) + HCO3

- (aq)

nonmetal oxides + H2O acid

Examples of bases:

NaOH (s) Na+ (aq) + OH- (aq)

N2H4 (aq) + H2O N2H5+ (aq) + OH- (aq)

metal oxides + H2O bases

Na2O (s) + H2O (l) 2NaOH (aq)

* Limited only to aqueous solutions


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A Brnsted acid is a proton donor

A Brnsted base is a proton acceptor

Example:

acid base acid base

A Brnsted acid must contain at least one ionizable

proton!

DEFINITION OF ACID AND BASE BY BRNSTED-LOWRY

HCl (aq) +H2O (l) H3O+ (aq) + Cl- (aq)

HCl is a acid because it donates proton to H2O

H2O is a base because it accepts proton from HCl


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Brnsted acids and bases

Conjugate acid-base pair:

i) Conjugate base of a Brnsted acid

- the species that remains when one proton has beenremoved from the acid

ii) Conjugate acid

- addition of a proton to a Brnsted base


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Examples:

HCl (aq) +H2O (l) H3O+ (aq) + Cl- (aq)

acid1 base2 acid2 base1

Cl- is a conjugate base of HCl and HCl is a conjugate acid of Cl-

H2O is a base conjugate of H3O+ and H3O

+ is a acid conjugate of

H2O

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

base1 acid2 acid1 base2

subscripts 1 and 2 = two conjugate acid-base pair


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When a strong acid react with a strong base in Brnsted acid-base

reaction, it will give a weak conjugate acid and conjugate base.

Examples:

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

strong acid strong base weak conjugate weak conjugateacid base


Weak base weak acid strong conjugate strong conjugate

acid base

H2O can function as acid or base which called amphoteric

Amphoteric or amphiprotic substance is one that can react as

either an acid or base


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Identify each of the following species as a Brnsted acid, base, or

both. (a) HI, (b) CH3COO-, (c) H2PO4

-

HI (aq) H+ (aq) + I- (aq) Brnsted acid

CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brnsted base

H2PO4- (aq) H+ (aq) + HPO4

2- (aq)

H2PO4- (aq) + H+ (aq) H3PO4 (aq)

Brnsted acid

Brnsted base


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A Lewis acidis a substance that can accept a pair of electrons

A Lewis baseis a substance that can donate a pair of electrons

H+ H O H

+ OH-

acid base

N HH

H

H+ +

acid base

N H

H

H

H+

DEFINITION OF ACID AND BASE BY LEWIS


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Examples of Lewis Acids and Bases reactions:

N HH

H

acid base

F B

F

F+ F B

F

F

N H

H

H

b) Ag+ (aq) + 2NH3 (aq) Ag(NH3)2+ (aq)

acid base

c) Cd+ (aq) + 4I- (aq) CdI2-4 (aq)

acid base

d) Ni (s) + 4CO (g) Ni(CO)4 (g)

acid base

a)


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Acidsi) Strong acids:

- Acids that completely ionized in solution.

- Example:

HCl (aq) H+ (aq) + Cl- (aq)

ii) Weak acids

-Acids that incompletely ionized in solution

- Example:

CH3COOH (aq) CH3COO- (aq) + H+ (aq)

TYPES OF ACIDS-BASES


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Monoprotic acid:

- each unit of the acid yields one hydrogen ion upon

ionization

HCl H+ + Cl-

HNO3 H+ + NO3

-

CH3COOH H+ + CH3COO-

Strong electrolyte, strong acid


Weak electrolyte, weak acid

Diprotic acid:

- each unit of the acid gives up two H+

ions, in two separatesteps

H2SO4 H+ + HSO4

-

HSO4- H+ + SO4

2-




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Triprotic acids:

- yield three H+ ions

H3PO4 H+ + H2PO4

-

H2PO4- H+ + HPO4

2-

HPO42- H+ + PO4

3-





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Bases

i) Strong bases:

- Bases that completely ionized in solution.

- Example:

NaOH (s) Na+ (aq) + OH- (aq)

ii) Weak bases

- bases that incompletely ionized in solution

- Example:



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Acids and bases as electrolytes

Strong acids such as HCl and HNO3 are strongelectrolytes, while weak acid such as acetic acid(CH3COOH) is a weak electrolyte.


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

- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4

- (aq)


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Strong Acidsare strong electrolytes



- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4

- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4

- (aq)

Acids and bases as electrolytes

HF (aq) + H2O (l) H3O+

(aq) + F-

(aq)

Weak Acidsare weak electrolytes


- (aq)

HSO4- (aq) + H2O (l) H3O

+ (aq) + SO42- (aq)

H2O (l) + H2O (l) H3O+

(aq) + OH-

(aq)


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Strong Basesare strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H

2O

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

H2O

F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Basesare weak electrolytes

NO2- (aq) + H2O (l) OH

- (aq) + HNO2 (aq)


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Conjugate acid-base pairs:

The conjugate base of a strong acid has no measurable

strength.

H3

O+ is the strongest acid that can exist in aqueous solution.

The OH- ion is the strongest base that can exist in aqueous

solution.


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H2O (l) H

+

(aq) + OH

-

(aq)

autoionizationof water

Can act either as a acid or as a base.

Water functions as a base with acids such as HCl and

CH3COOH and function as acid in reaction with bases. Water is a very weak electrolyte and undergo ionization to a

small extent:

ACID-BASE PROPERTIES OF WATER

h d f


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

The Ion Product of Water

Kc=[H+][OH-]

[H2O]

[H2O] = constant

Kc

= equilibrium constant

Kc[H2O] = Kw= [H+

][OH-

]

The ion-product constant(Kw) is the product of the molar

concentrations of H+ and OH- ions at a particular temperature.

At 250C

Kw= [H+][OH-] = 1.0 x 10-14

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Isneutral

acidic

basic


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What is the concentration of OH- ions in a HCl solution whosehydrogen ion concentration is 1.3 M?

Kw= [H+][OH-] = 1.0 x 10-14

[H+] = 1.3 M

[OH-] =Kw

[H+]

1 x 10-14

1.3= = 7.7 x 10-15M


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pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 250C

pH [H+]

pH-A MEASURE OF ACIDITY

pHthe negative logarithm of the hydrogen in

concentration (in mol/L)

Oth i t t l ti hi


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pOH = -log [OH-]

[H+][OH-] = Kw= 1.0 x 10-14

-log [H+]log [OH-] = 14.00

pH + pOH = 14.00

Other important relationships

pH Meter


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1) The pH of rainwater collected in a certain region of the

northeastern United States on a particular day was 4.82. What is

the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH= 10-4.82 = 1.5 x 10-5M

2) The OH- ion concentration of a blood sample is 2.5 x 10-7 M.

What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00pOH = 14.006.60 = 7.40


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CALCULATION OF pH FORSOLUTION CONTAINING A

STRONG ACID AND ASOLUTION OF A STRONG

BASE

1) i f 2 10 3 O i ?


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1) What is the pH of a 2 x 10-3 MHNO3 solution?

HNO3 is a strong acid100% dissociation.


- (aq)

pH = -log [H+

] = -log [H3O+

] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

2) What is the pH of a 1.8 x 10-2MBa(OH)2 solution?

Ba(OH)2 is a strong base100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00pOH = 14.00 + log(0.036) = 12.6


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HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka= [H+

][A-

][HA]

Kais the acid ionization constant

Kaweak acid

strength


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1) Wh t i th H f 0 5M HF l ti ( t 250C)?


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1) What is the pH of a 0.5MHFsolution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka=[H+][F-]

[HF]= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka=x2

0.50 - x= 7.1 x 10-4

Ka x20.50 = 7.1 x 10-40.50x 0.50K

a


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When can I use the approximation?

0.50x 0.50Ka


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Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

In most cases, you can ignore the autoionization of water.

Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of

single unknown x.

3. Write Kain terms of equilibrium concentrations. Solve for x

by the approximation method. If approximation is not valid,

solve for xexactly.

4. Calculate concentrations of all species and/or pH of the

solution.


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1) What is the pH of a 0.122Mmonoprotic acid whose

Kais 5.7 x 10-4?


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka=x2

0.122 - x= 5.7 x 10-4

Ka

x2

0.122 = 5.7 x 10-4

0.122x 0.122Ka


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Ka=x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x6.95 x 10-5 = 0

ax2 + bx+ c=0-b b24ac

2ax=

x= 0.0081 x= - 0.0081


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x= 0.0081 M

pH = -log[H+] = 2.09

W k B d B I i ti C t t


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NH3

(aq) + H2O (l) NH

4

+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb=[NH4

+][OH-]

[NH3]

Kbis the base ionization constant

Kbweak base

strength

Solve weak base problems like weak acids exceptsolve for

[OH-] instead of [H+].


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CONCENTRATION OF SOLUTION


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The concentrationof a solution is the amount of solute

present in a given quantity of solvent or solution.

M= molarity=moles of solute

liters of solution

1) What mass of KI is required to make 500. mL of a 2.80

MKI solution?

volume of KI solution moles KI grams KIMKI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

CONCENTRATION OF SOLUTION


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Preparing a Solution of Known

Concentration


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DILUTION OF SOLUTIONS


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Dilutionis the procedure for preparing a less concentrated solution

from a more concentrated solution.

DilutionAdd Solvent

Moles of solute

before dilution (i)

Moles of solute

after dilution (f)=

MiVi MfVf=

DILUTION OF SOLUTIONS

EXAMPLE


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EXAMPLE:

1) How would you prepare 60.0 mL of 0.200 MHNO3

from a stock solution of 4.00 MHNO3?

MiVi = MfVf

Mi = 4.00 M Mf= 0.200 M Vf= 0.0600 L Vi = ? L

Vi =MfVf

Mi=

0.200 Mx 0.0600 L

4.00 M= 0.00300 L = 3.00 mL

Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

Concentration Units


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Concentration Units

The concentrationof a solution is the amount of solute present

in a given quantity of solvent or solution.

Percent by Mass (%w/w)

% by mass = x 100%mass of solute

mass of solute + mass of solvent

= x 100%mass of solute

mass of solution


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Percent by Volume (%v/v)

% by volume = x 100%

Volume of solute

Volume of solution

Mole Fraction (X)

XA =moles of A

sum of moles of all components


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M=moles of solute

liters of solution

Molarity (M )

Molality (m)

m=moles of solute

mass of solvent (kg)


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Quantitative analytical process based on measuringvolumes.

The most common form of VA is the titration, aprocess whereby a standard solution of knownconcentration is chemically reacted with a solutionof unknown concentration in order to determine theconcentration of the unknown.

VOLUMETRIC ANALYSIS (VA)


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In a titrationa solution of accurately known concentration(standard solution) is added gradually added to another solution

of unknown concentration until the chemical reaction between the

two solutions is complete.

Equivalence pointthe point at which the reaction is complete

Indicatorsubstance that changes color at (or near) the

equivalence point

TITRATIONS

Titrations can be used in the analysis of acid-base reactions

H2SO4 + 2NaOH 2H2O + Na2SO4


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Slowly add base

to unknown acid

UNTIL

the indicator

changes color

PROCEDURE FOR THE TITRATION


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EXAMPLE


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EXAMPLE:

1) What volume of a 1.420 MNaOH solution is required to titrate

25.00 mL of a 4.50 MH2SO4 solution?

WRITE THE CHEMICAL EQUATION!

volume acid moles red moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4x

1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

M

acid

rxn

coef.

M

base

ACID-BASE TITRATIONS


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Strong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H2

O (l) + NaCl (aq)

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

pH PROFILE OF THE TITRATION (TITRATION CURVE)


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Before addition of NaOH

- pH = 1.00

When the NaOH added

- pH increase slowly at first

Near the equivalence point (the point which equimolar

amounts of acid and base have reacted)

- the curve rises almost vertically

Beyond the equivalence point

- pH increases slowly

pH PROFILE OF THE TITRATION(TITRATION CURVE)

CALCULATION OF Ph AT EVERY STAGE OF


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TITRATION

1) After addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M

HCl

Total volume = 35.0 mL

Moles of NaOH in 10.0 mL

= 10.0 mL x 0.100 mol NaOH x 1L

1L NaOH 1000 mL

= 1.00 x 10-3 mol

Moles of HCl in 25.0 mL

= 25.0 mL x 0.100 mol HCl x 1L

1 L HCl 1000 mL

= 2.50 x 10-3 mol


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Amount of HCl left after partial neutralization

= (2.50 x 10-3)-(1.00 x 10-3)

= 1.50 x 10-3 mol

Concentration of H+ ions in 35.0 mL

1.50 x 10-3 mol HCl x 1000 mL = 0.0429 M HCl

35.0 mL 1L

[H+] = 0.0429 M,

pH = -log 0.0429 = 1.37

2) After addition of 25.0 mL of 0.100 M NaOH to 25.0 mL 0f 0.100 MHCl

[H+] = [OH-] = 1.00 x 10-7

pH = 7.00

3) After addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 mL


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of HCl

Total volume = 60.0 mL

Moles of NaOH added

= 35.0 mL x 0.100 mol NaOH x 1L1 L NaOH 000 mL

= 3.50 x 10-3 mol

Moles of HCl in 25.0 mL solution = 2.50 x 10-3 mol

After complete neutralization of HCl, no of moles of NaOH left= (3.50 x 10-3)-(2.50x10-3)

= 1.00 x 10-3 mol

Concentration of NaOH in 60.0 mL solution

= 1.00 x 10

-3

mol NaOH x 1000 mL60.0 mL 1L

= 0.0167 M NaOH

[OH-] = 0.0167 M

pOH = -log 0.0167 = 1.78

pH = 14.001.78

= 12.22

Exactly 100 mL of 0.10 MHNO2 are titrated with a 0.10 M


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NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.01

0.200= 0.05 MFinal volume = 200 mL

Kb=[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05x

0.05 x

1.05 x 10

-6

= [OH

-

]

pOH = 5.98

pH = 14pOH = 8.02

Acid-Base Indicators


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

10[HIn][In

-

]

Color of acid (HIn) predominates

10[HIn][In-]

Color of conjugate base (In-) predominates

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Concepts of Acid Base Neutralization