CHM 160 (CHAPTER 2)

ACID AND BASES

AND

PREPARATION AND DILUTION OF STOCK SOLUTION

• An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

• A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte

ELECTROLYTIC PROPERTIES

METHOD OF DISTINGUISHING BETWEEN ELECTROLYTES 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.

• 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.

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

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

H2O

Nonelectrolyte does not conduct electricity?

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

C6H12O6 (s) C6H12O6 (aq)H2O

• 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

• 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

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

• Anhydrous pure acid (without water) does not show acidic properties.

• 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 water – show acidic properties

ROLE OF WATER TO SHOW PROPERTIES ROLE OF WATER TO SHOW PROPERTIES OF ALKALIOF 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 water – show alkaline properties

DEFINITION OF ACID AND BASE

Arrhenius Brønsted-Lowry

Lewis

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

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

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

* Limited only to aqueous solutions

• 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

Brønsted acids and bases

• 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

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

base1 acid2 acid1 base2

• subscripts 1 and 2 = two conjugate acid-base pair

• 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

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

• A Lewis acid is a substance that can accept a pair of electrons

• A Lewis base is a 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

Examples of Lewis Acids and Bases reactions:

N H••

H

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)

• 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

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

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

• 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)

Acids and bases as electrolytesAcids and bases as electrolytes

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

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

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

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

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

Strong Acids are strong electrolytes

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

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

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

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

Acids and bases as electrolytesAcids and bases as electrolytes

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

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)

Conjugate acid-base pairs:

• The conjugate base of a strong acid has no measurable strength.

• H3O+ is the strongest acid that can exist in aqueous solution.

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

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

autoionization of 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

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

• pH – the negative logarithm of the hydrogen in

concentration (in mol/L)

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

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

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

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

• The concentration of 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 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

Preparing a Solution of Known Concentration

Dilution is the 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

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

M2

=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.

Concentration UnitsConcentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass

% 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

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

• Quantitative analytical process based on measuring volumes.

• The most common form of VA is the titration, a process whereby a standard solution of known concentration is chemically reacted with a solution of unknown concentration in order to determine the concentration of the unknown.

VOLUMETRIC ANALYSIS (VA)

• 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

H2SO4 + 2NaOH 2H2O + Na2SO4

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

PROCEDURE FOR THE TITRATION PROCEDURE FOR THE TITRATION

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

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)

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

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

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 10-7pH = 7.00

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 000 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