CHAPTER 1

CHAPTER 1

Objectives :

At the end of the lesson, the students should be able to:

1. Understand the concept of acid and base2. Define strong weak acids/bases and the

meaning of % of ionization or degree of dissociation

3. Measure acidity and alkalinity4. Understand the methods of preparing

soluble and insoluble salts

1.1 Concept of acids and bases

1.1.1 Definition: Arrhennius.

Acid : An acid is something that produceshydrogen ions, H+ (H3O+).

Example: HCl in water ionizes to H+ + Cl-

HCl → H+ + Cl-

Hence an aqueous solution of HCl is acidic

Base : a base is something that produces hydroxyl ions, OH- in water.

Example : NaOH in water ionizes to Na+ + OH-

NaOH → Na+ + OH-

Hence, an aqueous solution of NaOH is basic.

1.1.2 Examples of Acids and Bases

Acids BasesHCI (hydrochloric acid)H2SO4 (sulfuric acid)HNO3 (nitric acid)CH3COOH (acetic acid)HNO2 (nitrous acid)

NaOH (sodium hydroxide)KOH (potassium hydroxide)Ba(OH)2 (barium hydroxide)Sr(OH)2 (strontium hydroxide)NH3 (ammonia)

1.2 Properties of Acids and Bases

Acids:• React with zinc, magnesium or aluminum

and form H2 (g)• React with compounds containing CO3

2-

and form carbon dioxide and water.• Turn blue litmus to red.• Taste sour

Bases :• They react with most cations to precipitate

hydroxides• They turn red litmus to blue• Feel soapy and slippery• Taste bitter

1.3 Strength of Acids and Bases

Acids1. Strong Acids• Completely dissociate in water (ionizes 100%),

forming H+ and an anion.• Degree of dissociation, α = 1• Example: 0.1 M HCl ionizes to 0.1 M [H+] and

0.1 M [Cl-].HCl (aq) →H+(aq) + Cl-(aq)0.1M 0.00M 0.00M initial0.00M 0.1M 0.1M after ionization

Example of strong acids:• There are only 6 strong acids. The

remainder of the acids therefore are considered weak acids.

HClH2SO4

HNO3

HClO4

HBrHI

2.Weak acids• A weak acid only partially dissociates in water to

give H+ and the anion.• Example: HF dissociates in water to give H+ and

F-. It is a weak acid with a dissociation equation :HF ↔ H+ + F-

Note the use of the double arrow with the weak acid. This is because an equilibrium exists between the dissociated and the undissociated molecule.

Example of weak acids:

Acid FormulaFormicAceticTrichloroaceticHydrofluoricHydrocyanicHydrogen sulfideWaterConjugate acids of weak bases

HCOOHCH3COOHCCl3COOHHFHCNH2SH2O

NH4+

Bases1. Strong bases

• They dissociate 100 % into the cation and OH-

(hydroxide ion).Example: 0.3 M NaOH ionizes to 0.3 M [Na+]

and 0.3 M [OH-].NaOH (aq) → Na+ (aq) + OH- (aq)

0.3 M 0.00M 0.00 M initial 0.00M 0.3M 0.3M after ionization

What are the strong bases?

The hydroxides of Group I and II

BASES Vs ALKALIS• A base is a substance that neutralizes

an acid• An Alkali is a soluble base• All Alkalis are bases, but not all bases

are alkalis.• Exp. of Bases (insoluble): CuO, Al2O3,

Zn(OH)2, Fe2O3 and Pb(OH)2.• Exp. Of Alkalis: KOH, NaOH, NH3(aq),

Ca(OH)2 and Ba(OH)2.

2. Weak bases

• Partially ionized in the solutions.• Most weak bases are anions of weak acids.General reaction:Weak base + H2O → weak acid + OH-

Examples:NH3 + H2O ↔ NH4

+ + OH-

CH3NH2 + H2O ↔ CH3NH3+ + OH-

NO2- + H2O OH- + HNO2

Examples of Weak basesBase Formula

AmmoniaTrimethyl ammoniaPyridineAmmonium hydroxideWaterHS-

Conjugate bases weak acids

NH3

N(CH3)3

C5H5NNH4OHH2OHS-

e.g.: HCOO-

1.4 measurement of acidity and alkalinity1.4.1 The ion product of Water

H2O ↔ H+ + OH-

Kc = [H+][OH-] [H2O] = constant[H2O]

Kc = [H+][OH-] = Kw = [H+][OH-]The ion-product constant (Kw) is the product of the molar concentration of H+ and OH- ions at a particular temperature.

[H+] = [OH-] = neutral[H+] > [OH-] = acidic[H+] < [OH-] = basic

At 250CKw = [H+][OH-]= 1.0 x 10-14

Example:What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?

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

[H+] = 1.3 M[OH-] = Kw = 1 x 10-14 = 7.7 x 10-15M

[H+] 1.3

pH- A Measure of Acidity

• The pH is defined as the negative logarithm of the H+ concentration or the H3O+ concentration.

pH = - log [ H+ ]

• For pure water and neutral solution[ H+ ] = 1 x 10-7

• pH = -log [H+ ] = - log (1 x 10- 7) = 7

pH Scale

Neutral [H+] = [OH-] ; [H+] = 1 x 10-7 pH = 7Acidic [H+] >[OH-] ; [H+] > 1 x 10-7 pH < 7Basic [H+] <[OH-] ; [H+] < 1 x 10-7 pH > 7

pH ↑ ; [H+] ↓

pH indicators

An indicator is a chemical which indicates the nature of the solution, by means of a sharp change in color

Universal indicator Acid-base indicator

• Is a mixture of organic dyes

• Indicates strength of solution

•Change in color is sharp at any pH

•Is a chemical such as Methyl orange, phenolphthalein

•Does not indicate strength of solution

•Change in its color is abrupt at a particular pH

related• The [H+] and [OH-] are as follows:• The function pOH is defined in a manner parallel

to that of pH. Thus,

• [H+] x [OH-] = Kw = 1.0 x 10-14 (25°C)• log[H+] + log[OH-] = logKw

• -log[H+] - log[OH-] = -logKw

• pH + pOH = pKw ; Kw = 1.0 x 10-14, pKw = 14Therefore,

pH + pOH = 14

pOH = - log [ OH- ]

pH

[H+] pH =-log[H+] pH value

1.0 x 10-3 M = -log [1.0 x 10-3] 3.00

4.55 x 10-7 M = -log [4.55 x 10-7] 6.34

2.93 x 10-10 M = -log[2.93 x 10-10] 9.53

Examples

1. What is the [H+] of a sample of lake water with [OH-] of 5.0 x 10-9 M? Is the lake acidic, basic or neutral?Sol:Kw = [H+][OH-][H+] = 1 x 10-14 = 2 x 10-6 M.

5.0 x 10-9

Therefore, the lake is slightly acidic.

2. Calculate the [OH-] of a solution of baking soda with a pH of 8.5.Solution:if pH = 8.5, then antilog of -8.5 = 3.2 x 10-9.thus, [H+] = 3.2 x 10-9 M.[OH-] = Kw = 1.0 x 10-14 = 3.1 x 10-6 M

[H+] 3.2 x 10-9

3. Calculate the pH of a solution of household ammonia whose [OH-] is 7.93 x 10-3 M.

Solution:This time you first calculate the [H+] from the

[OH-]

7.93 x 10-3 M = 1.26 x 10-12 M [H+]

Then find the pH pH = -log[1.26 x 10-12] = 11.9

• Practice #1. What is the pH of a solution of NaOH that has a [OH-] of 3.5 x 10-3 M?

• Practice #2. The [H+] of vinegar that has a pH of 3.2 is what?

• Practice #3. What is the pH of a 0.001 M HCl solution?

Some Problems

• Determine the [H+], [OH-], pH and pOH for each of the following solutions

(a) 0.0010M HBr, (b) 0.010M NaOH,(c) 1.0M HClO4, (d) 10M HCl, (e) 1.0M KOH

Degree of dissociation• It is a fraction of the total number of moles

of an acid or base or electrolyte that dissociates into ions in an aqueous solution when equilibrium is reached. It is represented by α

Thus, greater the degree of dissociation (α) stronger the acids or bases and vice versa.

HA ↔ H+ + A−

• Degree of dissociation, α =

• % of ionization =

]HA[]H[ +

100x]HA[]H[ +

Example

What is the percent of ionization if the pH for a solution of ammonia is 1.50 M is 12.20?

(March’04)Solution:

% ionization = x100%; pH = -log[H+] = ?

percent of ionization= 1.05%

]NH[]OH[

3

−

1.5 Salts

Definition

an ionic compound made of a cation and an anion, other than hydroxide.The product besides water of a neutralization reaction.

1.5.1 Salts-preparations

• neutralization (i.e., the reaction between an acid and a base),

• displacement (i.e., when a more reactive element displaces a less reactive element from its compound),

• and precipitation (i.e., when two soluble ionic compounds react to form a precipitate).

Examples of three reaction types; neutralization (1 - 4), displacement (5 & 6), and precipitation (7).

(1) NiO(s) + 2HCl(aq) → NiCl2(aq) + H2O(l) (2) Mg(OH)2(s) + H2SO4(aq) → MgSO4(aq) + 2H2O(l)(3) PbCO3(s) + 2HNO3(aq) → Pb(NO3)2(aq) + H2O(l)

+ CO2(g) (4) 2NH3(aq) + H3PO4(aq) → (NH4)2HPO4(aq) (5) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) (6) 2Al(s) + 3Zn(NO3)2(aq) → 2Al(NO3)3(aq) + 3Zn(s) (7) 2KOH(aq) + FeSO4(s) → 2Fe(OH)2(s) + K2SO4(aq)

Table of SolubilityType of Compound Solubility

Group 1 compound and all ammonium compounds

Soluble

Nitrates Soluble

Chlorides Soluble (except lead (II) chloride and silicone chloride

Sulfates Soluble (except calcium sulfate, lead sulfate and barium sulfate)

Carbonates Insolubles (except group 1 and ammonium carbonates)

1.5.2 Salt Preparation

• The method used depends on two factors:

The solubility of the metal/base usedThe solubility of the salt to be made.

Preparation of soluble Salts

1. Acid + Baseexp: 2 KOH (ak) + H2SO4 (ak) → K2SO4 (ak) + 2 H2O (l)

2. Acid + Metalexp: Mg (s) + 2 HCl (ak) → MgCl2 (ak) + H2 (g)

3. Acid + Carbonateexp: ZnCO3(s) + 2HNO3 → Zn(NO3) + CO2(g) + H2O(l)

4. Acid + Oxide Baseexp: ZnO (s) + H2SO4 (ak) → ZnSO4 (ak) + H2O (l)

Soluble Salts-PreparationsTwo main methods:• (i) for a metal or insoluble base reacting

with an acid to produce a soluble salt filtration is used, e.g. reacting copper(II) oxide with sulphuric acid to make copper(II) sulphate.

• (ii) for a soluble base reacting with an acid to produce a soluble salt titration is used, e.g. reacting sodium hydroxide with hydrochloric acid to make sodium chloride.

Preparation of Copper(II) sulfate

• Stage 1: The addition of black copper(II) oxide to sulphuric acid. Mild heating is required for a full reaction to occur; however, make sure that the acid does not boil as this would be a great safety hazard.

• The copper(II) oxide is added until no more visible reaction can be seen, i.e. the base no longer dissolves and a black solid is seen in the blue solution.

• Stage 2 : Filteration to remove the excess black solid and leave a clear blue solution in the evaporating bowl. If the blue solution is heated gently, to remove some of the water and allowed to cool down slowly, crystals will appear. The slower this crystallization is allowed to occur, the larger the crystals that will be produced.

Preparation of sodium chloride

• N.B.: This method of preparation is used when any and all sodium or potassium salts are made.

• The problem with the reaction of a soluble base reaction with acid is that once all the acid has reacted any excess base will not be visible. This problem is overcome by adding a third chemical into the reaction mixture called an indicator.

• Indicators are chemicals that change colour with a change in pH. So, if the indicator is added to an acid it will be one colour. As base is added the pH of the solution is raised until, once all the acid has reacted, i.e.been neutralized, and the base is now in excess, the indicator changes colour.

• Example indicators -• Indicator colour in acid colour in base

phenolphthalein colourless purplemethyl orange red yellow

The base is added from the burette slowly until the indicator changes colour. The volume of base required can be read from the markings on the side of the burette.

Insoluble Salts-Preparation

• for an insoluble base reacting with an acid to produce an insoluble salt a two stage process involving filtration and then precipitation is used, e.g. reacting lead(II) oxide firstly with nitric acid to produce lead(II) nitrate and then reacting the lead(II) nitrate with aqueous iodide ions to produce lead(II) iodide.

Preparation of lead(II) iodide• There is no direct method of preparing an insoluble salt

from an insoluble base. The problem us that as the base is added the salt is produced so filtration would yield a mixture of two solids. Titration won't work either as the base is not soluble in water.

• Using the preparation of lead(II) iodide as an example, the solution is a two stage process. The first stage is the same as the method for making copper(II) sulphate(shown above). Lead(II) oxide is added to hot nitric acid and when an excess of oxide is present the mixture is filtered.

• This gives a filtrate of lead(II) nitrate solution. An equal volume of potassium iodide solution is then added and the resulting mixture filtered to give solid lead(II) iodide as the residue.