Chapter 3

50
Chapter 3 ACID AND BASES

description

semoga bermanfaat

Transcript of Chapter 3

Page 1: Chapter 3

Chapter 3

ACID AND BASES

Page 2: Chapter 3

• Electrolyte - substance that, when dissolved in water, results in a solution that can conduct electricity.

• Nonelectrolyte - substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte

ELECTROLYTIC PROPERTIES

Page 3: Chapter 3

• Strong Electrolyte – 100% dissociation (breaking up of compound into cations and anions

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

• Weak Electrolyte – not completely dissociated

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

A reversible reaction. The reaction can occur in both directions.

Page 4: Chapter 3

Nonelectrolyte does not conduct electricity?

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

C6H12O6 (s) C6H12O6 (aq)H2O

Page 5: Chapter 3
Page 6: Chapter 3

• Hydration - process in which an ion is surrounded by water molecules arranged in a specific manner.

- helps to stabilize ions and prevents cations from combining with anions.

H2O

Page 7: Chapter 3

• Have a sour taste.

• Turn litmus paper from blue to red. • React with certain metals to produce hydrogen gas.

• React with carbonates and bicarbonates to produce carbon dioxide gas

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

Page 8: Chapter 3

• Have a bitter taste.

• Turn litmus paper from blue to red.• Feel slippery. Many soaps contain bases.

• Aqueous base solutions conduct electricity.

• Examples:

PROPERTIES OF BASES

Page 9: Chapter 3

ROLE OF WATER TO SHOW PROPERTIES ROLE OF WATER TO SHOW PROPERTIES OF ACIDSOF ACIDS

Anhydrous pure acid does not show acidic properties

do not dissociate to form hydrogen ion (H+). When dissolved in water show the properties of acids. Why? - acids will dissociate in water to form H+ / H3O+ ions

which are free to move. E.g:

i) HCl in liquid methylbenzene (organic solvent) - does not show acidic properties.

ii) HCl in water – show acidic properties

Page 10: Chapter 3

ROLE OF WATER TO SHOW PROPERTIES ROLE OF WATER TO SHOW PROPERTIES OF ALKALIOF ALKALI

• Dry base does not show alkaline properties

hydroxide ions (OH-) are not free to move

• In water, bases dissociate form hydroxide ions, OH-, which are free to move

• For example:

i) ammonia in tetrachlomethane (organic solvent) – do not show alkaline properties

ii) ammonia in water – show alkaline properties

Page 11: Chapter 3

DEFINITION OF ACID AND BASE

Arrhenius Brønsted-Lowry

Lewis

Page 12: Chapter 3

Arrhenius acid - substance that produces H+ (hydrogen ion) or hydronium ion (H3O+) in water

Arrhenius base - substance that produces OH- in water

DEFINITION OF ACID AND BASE BY ARRHENIUS

Page 13: Chapter 3

Examples of acid:

CO2 (g) + H2O (l) H2CO3 (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

Page 14: Chapter 3

• A Brønsted acid is a proton donor• A Brønsted base is a proton acceptor• Example:

acid base acid base

• A Brønsted acid must contain at least one ionizable proton!

DEFINITION OF ACID AND BASE BY BRØNSTED-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

Page 15: Chapter 3

Conjugate acid-base pair:

i) Conjugate base of a Brønsted acid - the species that remains when one proton has been removed from

the acid

ii) Conjugate acid - addition of a proton to a Brønsted base

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

Page 16: Chapter 3

• When a strong acid react with a strong base in Brønsted 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 conjugate acid base NH3 (aq) + H2O (l) NH4

+ (aq) + OH- (aq) 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

Page 17: Chapter 3

Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4

-

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

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

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

2- (aq)

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

Brønsted acid

Brønsted base

Page 18: Chapter 3

• Lewis acid - substance that can accept a pair of electrons

• Lewis base - substance that can donate a pair of electrons

H+ H O H••••

+ OH-••••••

acid base

N H••

H

H

H+ +

acid base

N H

H

H

H+

DEFINITION OF ACID AND BASE BY LEWIS

Page 19: Chapter 3

Examples of Lewis Acids and Bases reactions:

N H••

H

H

acid base

F B

F

F

+ F B

F

F

N H

H

Ha)

Page 20: Chapter 3
Page 21: Chapter 3
Page 22: Chapter 3

ACIDS

i) 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

Page 23: Chapter 3
Page 24: Chapter 3

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

Strong electrolyte, strong acid

Weak electrolyte, weak acid

Diprotic acid:

- each unit of the acid gives up two H+ ions, in two separate steps

H2SO4 H+ + HSO4-

HSO4- H+ + SO4

2-

Strong electrolyte, strong acid

Weak electrolyte, weak acid

Page 25: Chapter 3

Triprotic acids: - yield three H+ ions

H3PO4 H+ + H2PO4-

H2PO4- H+ + HPO4

2-

HPO42- H+ + PO4

3-

Weak electrolyte, weak acidWeak electrolyte, weak acidWeak electrolyte, weak acid

Page 26: Chapter 3

• 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:

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

Page 27: Chapter 3

ACIDS AND BASES AS ACIDS AND BASES AS ELECTROLYTESELECTROLYTES

Strong acids (HCl,HNO3 ) - strong electrolytes

Weak acid (CH3COOH) - weak electrolyte

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

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

Page 28: Chapter 3

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O

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 Bases are weak electrolytes

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

Page 29: Chapter 3
Page 30: Chapter 3

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

autoionization of water

• Can act either as a acid or as a base.

• Water is a very weak electrolyte and undergo ionization to a small extent:

ACID-BASE PROPERTIES OF WATER

Page 31: Chapter 3

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-A MEASURE OF ACIDITY

• pH – the negative logarithm of the hydrogen in

concentration (in mol/L)

Page 32: Chapter 3

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

Page 33: Chapter 3

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 10-5 M

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 = 14.00 – pOH = 14.00 – 6.60 = 7.40

Page 34: Chapter 3

CALCULATION OF PH FOR CALCULATION OF PH FOR SOLUTION CONTAINING A SOLUTION CONTAINING A

STRONG ACID AND A STRONG ACID AND A SOLUTION OF A STRONG SOLUTION OF A STRONG

BASEBASE

Page 35: Chapter 3

1) What is the pH of a 0.002 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (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 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% 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 = 14.00 – pOH = 14.00 + log(0.036) = 12.6

Page 36: Chapter 3

•Concentration - 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 M KI solution?

volume of KI solution moles KI grams KIM KI 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

Page 37: Chapter 3

Preparing a Solution of Known Concentration

Page 38: Chapter 3
Page 39: Chapter 3

Dilution - procedure for preparing a less concentrated solution from a more concentrated solution.

DilutionAdd Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=

DILUTION OF DILUTION OF SOLUTIONSSOLUTIONS

Page 40: Chapter 3

EXAMPLE:

1) How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3?

M1V1 = M2V2

M1 = 4.00 M M2 = 0.200 M V2 = 0.0600 L V1 = ? L

V1 =M2V2

M1

=0.200 M x 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.

Page 41: Chapter 3

Concentration UnitsConcentration Units

Percent by Mass (%w/w)

% by mass = x 100%mass of solute

mass of solute + mass of solvent

= x 100%mass of solutemass of solution

Percent by Volume (%v/v)

% by volume = x 100%Volume of soluteVolume of solution

Page 42: Chapter 3

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

Page 43: Chapter 3

• In a titration a 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 point – the point at which the reaction is complete

• Indicator – substance that changes color at (or near) the equivalence point

TITRATIONS

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

Page 44: Chapter 3

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

PROCEDURE FOR THE TITRATION PROCEDURE FOR THE TITRATION

Page 45: Chapter 3

EXAMPLE:1) What volume of a 1.420 M NaOH solution is required to titrate 25.00 mL of a 4.50 M H2SO4 solution?

WRITE THE CHEMICAL EQUATION!

H2SO4 + 2NaOH 2H2O + Na2SO4

MaVa

MbVb

=a

b

Ma = concentration of acidMb = concentration of baseVa = volume of acidVb = volume of basea = coefficient of acidb = coefficient of base

(4.50 M) (25 mL)

(1.420 M) (Vb)=

1

2

Vb = 158 mL

Page 46: Chapter 3

ACID-BASE TITRATIONSStrong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

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

pH PROFILE OF THE TITRATION (TITRATION CURVE)pH PROFILE OF THE TITRATION (TITRATION CURVE)

Page 47: Chapter 3

• Before addition of NaOH

- pH = 1.00

• When the NaOH added

- pH increase slowly at first

• Near the equivalence point

- the curve rises almost vertically

• Beyond the equivalence point

- pH increases slowly

pH PROFILE OF THE TITRATION pH PROFILE OF THE TITRATION (TITRATION CURVE) (TITRATION CURVE)

Page 48: Chapter 3

CALCULATION OF PH AT EVERY STAGE OF CALCULATION OF PH AT EVERY STAGE OF TITRATIONTITRATION

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

Page 49: Chapter 3

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

pH = 7.00

Page 50: Chapter 3

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

Total volume = 60.0 mLMoles of NaOH added= 35.0 mL x 0.100 mol NaOH x 1L 1 L NaOH 1000 mL= 3.50 x 10-3 mol

Moles of HCl in 25.0 mL solution = 2.50 x 10-3 molAfter 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 mL 60.0 mL 1L= 0.0167 M NaOH[OH-] = 0.0167 MpOH = -log 0.0167 = 1.78pH = 14.00 – 1.78 = 12.22