CHAPTER 8 – ACIDS & BASES
Section 8.1-Understanding Acids & Bases
Properties of Pure & Aqueous Substances
Substance
Most molecular
compounds
Most Ionic
compounds
Acids
Bases
Conductivity
Liquid
Solid
Liquid
Aqueous
Aqueous
No
No
No
No Effect
No
No
No
Yes
No
Yes
Yes
Yes
Yes
No Effect
Blue-Red
Red-Blue
 NaOH(s)-Na+(aq) + OH-(aq) …..similar to ionic compounds
 acids in liquids are still showing acid’s molecular nature
 something happens when a solid acid is dissolved in water i.e.
HCl(g) -- H+(aq) + Cl-(aq)
 Arrhenius hypothesized that bases will dissociate into individual positive &
negative ions
e.g. Ba(OH)2(s) ---> Ba+2(aq) + 2OH-1(aq)
 Arrhenius also theorized that acids first dissolve as individual molecules and
then ionize into H+ and X- in solution
e.g. HCl(g) -- H+(aq) + Cl-(aq)
Sample Problem 1
Q: Write dissociation or ionization equation equations (as appropriate) for
the following chemicals dissolving in water. Label each equation as either
dissociation or ionization;
a) potassium chloride (a salt substitute)
b) hydroiodic acid ( a strong acid)
A: a) KCl(s) --K+(aq) + Cl-(aq) =dissociation
b) HI(aq) -- H+(aq) + I-(aq) =ionization
REMEMBER: Ionic substances dissociate, acids ionize!
 acids as pure substances are molecular compounds and may be solid, liquid,
or gas
 examples;
HCl(g) or HCl(aq)-- H+(aq) + Cl-(aq)
= gaseous acid
HC2H3O2(l) or HC2H3O2(aq) -- H+(aq) + C2H3O2-(aq) =liquid acid
H2C2O4(l) or H2C2O4(aq) -- H+(aq) + HC2O4-(aq)
=solid acid
Strong and Weak Acids
 acids can be classified as strong or weak
Strong acids – have high electrical conductivity which is due to their
High percentage ionization >99%
e.g. H2SO4(aq), HNO3(aq), HCl(aq)
 most other acids have low electrical conductivity due to low percentage
ionization and are called weak acids
 to communicate the percentage ionization and strength of an acid, we can
write the percentage ionization over the chemical reaction arrow
i.e. strong acids are usually >99%
weak acids ionize <50%
>99%
HNO3(aq) ------- H+(aq) + NO3-(aq)
=strong acid
<50%
HC2H3O2(aq) ------ H+(aq) + C2H3O2-(aq) =weak acid
Sample Problem 2
Q: The following acidic solutions were tested (at equal concentrations and
temperature) for electrical conductivity. Write ionization equations to explain
the relative conductivity of each acid.
a) hydrobromic acid (aqueous hydrogen bromide): high conductivity
b) hydrofluoric acid (aqueous hydrogen fluoride): low conductivity
>99%
A: HBr(aq) ------- H+(aq) + Br-(aq)
<50%
HF(aq) ------- H+(aq) + F-(aq)
Strong Bases: XOH(s) -- X+(aq) + OH-(aq)
(100% dissociated as ions)
*assume all bases are ionic hydroxides, all of which are strong bases
Finish questions 1-6 on page 367.
8.2 pH of a Solution
 pH is a way of indicating the concentration of hydrogen ions [H+] in a
solution
 according to Arrhenius’ theory H+ are responsible for the properties of acids
 the higher the concentration of H+ the more acidic the solution will be
 pure water would not be expected to contain any H+ or OH- at all
 careful testing has shown that 2 out of every billion water molecules will
break down into;
H2O(l)---- H+(aq) + OH-(aq)
 this will give a H+ concentration of 1 x 10-7mol/L
 therefore in a neutral solution [H+(aq)] = 1 x 10-7mol/L
in an acidic solution [H+(aq)] > 1 x 10-7mol/L
in a basic solution [H+(aq)] < 1 x 10-7mol/L
this lead to the development of the pH scale
 the pH of a solution is defined as the negative of the exponent to the base
10 of the H+(aq) concentration (expressed as mol/L)
 examples: if [H+(aq)] is 10-7mol/L, the pH is 7
or
if pH =2, the [H+(aq)] is 10-2mol/L
Sample Problem 1
Q: What is the pH of
(a) 1 x 10-2 mol/L [H+(aq)] in vinegar
(b) [H+(aq)] =1.0 x 10-12mol/L in ammonia
A: (a) pH=2
(b) pH=12.00
Do questions 1-6 on page 371.
Sample Problem 2
Q: What is the [H+(aq)] of;
(a) a carbonated beverage with a pH of 3.0
(b) an antacid solution for which the pH is 10.00
A: (a) [H+(aq)] = 1 x 10-3mol/L
(b) [H+(aq)] = 1.0 x 10-10mol/L
 measuring pH of a solution can be done using pH indicators or pH paper to
estimate the value
 for more precise measurements one would normally use a pH meter
 these meters work like small electric cells in which the electricity produced
depends on the acidity of the solution
Calculating pH and [H+(aq)]
 the relationship between [H+(aq)] and pH is easy to calculate if the
concentration is only a power of 10
i.e. 10-3mol/L where pH=3 or [H+(aq)] =1 x 10-7mol/L where pH=7.0
But what if [H+(aq)] =2.7 x 10-3mol/L?
The logarithm of a number is the exponent when the number is written in
exponential form
100=102
0.001=10-3
log10 (102)=2
log10 (10-3)=-3
 in general, if Y=10X, then log10(Y)=X
 fortunately we can use calculators to solve these problems.
 because [H+(aq)] are usually <0.1mol/L we define pH as the negative
logarithm to avoid having almost all pH’s as negative numbers. i.e. pH=log10 [H+(aq)]
 the digits preceding the decimal point in a pH value are determined by the
digits in the exponent of the [H+(aq)]
 these digits locate the position of the decimal point in the concentration
value not the certainty of the value
 the number of digits following the decimal point in the pH value is equal to
the number of significant digits expressing the certainty of the [H+(aq)]
Example:
[H+(aq)] =2.7 x 10-3mol/L
pH=2.57
a) note 2 significant digits-----2 decimal places
b) 3 in 2.7 x 10-3 only indicates where the decimal point goes
c) .57 in pH indicates a certainty of 2 decimal places
d) 2 in pH indicates where the decimal place will go
Sample Problem 3
Q: An antacid solution has a hydrogen concentration of
4.7 x 10-11 mol/L. What is its pH?
A: pH=-log10 [H+(aq)]
=-log(4.7 x 10-11)
pH=10.33
4
1
1
.
7
+/-
Exp
=
log
+/-
OR
-
Exp
+/-
log
4
1
1
.
=
 we may have to convert a pH reading to the molar concentration of
[H+(aq)]
 therefore pH becomes the exponent
i.e. [H+(aq)] =10-pH
7
Sample Problem 4
Q: The pH reading of a solution is 10.33. What is its hydrogen ion
concentration? Be sure to indicate your answer with the correct certainty.
A:
[H+(aq)] =10-pH
=10-10.33mol/L
= 4.7 x 10-11mol/L
1
0
+/-
.
3
Either
OR
3
Inv
2nd
log
log
OR depending on your calculator
10
YX
10.33
+/-
XY
^
Summary
 pH is the negative power of 10 of the [H+(aq)]
 pH=-log[H+(aq)] or [H+(aq)] =10-pH
 solution:
[H+(aq)] :
pH
:
Acidic
Neutral
>10-7
<7
107
7
Basic
<10-7
>7
* the higher the hydrogen ion molar concentration, the lower the pH.
Do questions 8-10 on page 374.
=
Read section 8.3 on page 376.
8.4 Acid-Base Theories
Arrhenius Concept of Acids & Bases
To summarize Arrhenius’ definitions;
Acid-- H+(aq) + an anion
Base-- a cation + OH-
Revision of Arrhenius’ Definitions
The following cannot be explained by our original definitions;
1. compounds of hydrogen polyatomic ions
e.g. NaHCO3(aq) & NaHSO4(aq)
2. oxides of metals & nonmetals
e.g. CaO(aq) & CO2(g)
3. compounds that are neither oxides nor hydroxides e.g. NH3(aq) &
Na2CO3(aq) but are still bases
4. compounds that contain no hydrogen but are acids e.g. Al(NO3)3(aq)
 revised theory looks at collisions with water molecules and the nature of
the hydrogen ion
 Canadian scientist, Paul Giguere, has shown that there are hydrated protons
e.g. hydronium ion H3O+
 using H3O+(aq) to explain the formation of acidic solutions by strong acids
such as HCl(g) is represented as;
>99%
HCl(g) + H2O(l) ------- H3O+(aq) + Cl-1(aq)
 for weak acids;
1.3%
HC2H3O2(aq) + H2O(l) ------ H3O+(aq) + C2H3O2-(aq)
(acetic acid)
* in general, acidic solutions form when substances react with water to form
hydronium ions
Strong & Weak Bases
 strong bases dissociate to increase the concentration of OH-(aq) in an
aqueous solution
 evidence indicates that all ionic hydroxides are all strong bases
 a strong base, like Ba(OH)2(s) will dissociate completely i.e.
Ba(OH)2(s) -----Ba2+(aq) + 2OH-1(aq) and we don’t have to worry about
water being present or needed
 for a weak base, like ammonia (NH3) we would need water i.e.
<50%
NH3(aq) +H2O(l) ------OH-(aq) + NH4-(aq)
 what about Na2CO3?
 Na2CO3(s) ---- 2Na+(aq) + CO3-2(aq) ( no H+(aq), no H3O+(aq), no OH-(aq))
<50%
 but
CO32-(aq)
+ H2O(l) ------ OH-(aq) + HCO3-1(aq)
Sample Problem 1
Q: A forensic technician tested the pH of a sodium cyanide solution and found
that it had a pH greater than 7. Explain this evidence using chemical
equations.
A: NaCN(s) ---Na+(aq) + CN-(aq)
<50%
CN-(aq)
+ H2O(l) ----- OH-(aq) + HCN(aq)
Questions 12-16 on page 386.
The BrØnsted-Lowry Concept
 BrØnsted (Denmark) & Lowry (England) focused on the role of acids & bases
in a reaction
ACID: HCl(aq) + H2O(l) ------ H3O+(aq) + Cl-(aq)
BASE: NH3(aq) + H2O(l) ------ OH-(aq) + NH4+(aq)
 water does not have to be a reactant
e.g. H3O+(aq) + NH3(aq) ---- H2O(l) + NH4+(aq)
 water does not even have to be present.
e.g. HCl(g) + NH3(g) ---- NH4Cl(s)
 acids and bases using this theory are called BrØnsted – Lowry acids or
BrØnsted – Lowry bases BUT only for specific reactions
i.e. a substance can be a BrØnsted – Lowry acid in one reaction and a
BrØnsted – Lowry base in another
 a substance that appears as a BrØnsted – Lowry acid in one reaction and a
BrØnsted – Lowry base in another is called amphiprotic.
e.g. HCO3-(aq) in baking soda
 all hydrogen polyatomic ions are amphiprotic (sometimes called amphoteric)
e.g. HCO3-(aq) + H3O+(aq) ----H2CO3(aq) + H2O(l) (neutralizes a strong acid)
HCO3-(aq) + OH-(aq) ---- CO32-(aq) + H2O(l) (neutralizes a strong base)
 according to the BrØnsted – Lowry concept acid-base reactions involve the
transfer of a proton so that the products formed in these reactions must
differ from the reactants by a proton.
e.g.
HC2H3O2(aq) + H2O(l) ---- H3O+(aq) + C2H3O2-(aq)
- a product formed as a result of an acid losing a proton is called a
conjugate base. e.g. C2H3O2-(aq)
- a product formed as a result of a base gaining a proton is called a
conjugate acid. e.g. H3O+(aq)
- a pair of substances that differ only by a proton is called a
conjugate acid-base pair.
HC2H3O2(aq) + H2O(l) ---- H3O+(aq) + C2H3O2-(aq)
Questions 1-11 on page 392.
8.5 Acid-Base Reactions
 active metal + acid---- H2(g) + ionic compound
e.g. Mg(s) + 2HCl(aq) --- H2(g) + MgCl2(aq)
 to clean up spills, emergency response teams use the knowledge that all
acids can react with carbonates (Na2CO3(aq) and CaCO3(s)) to form carbonic
acid, H2CO3(aq)
 carbonic acid is unstable and can quickly form CO2 and H2O so therefore an
acid + a carbonate---- H2CO3(aq) + an ionic compound
--- CO2(g) + H2O(l) + an ionic compound
e.g. 2HNO3(aq) + Na2CO3(aq) ---- CO2(g) + H2O(l) + 2NaNO3(aq)
 lastly, acids can react with bases in a neutralization reaction
i.e. HCl(aq) + KOH(aq) ----- H2O(l) + KCl(aq)
H+(aq) + Cl-(aq) + K+(aq) + OH-(aq) -- H2O(l) + K+(aq) + Cl-(aq)
(total ionic equation)
H+(aq) + OH-(aq) --- H2O(l)
(net ionic equation)
or
H3O+(aq) + OH-(aq) --- 2H2O(l)
(net ionic equation)
 precipitation reaction
i.e. acid + ionic compound -- precipitate + acid
e.g. 2HI(aq) + Pb(NO3)2(aq) -- PbI2(s) + 2HNO3(aq)
 can be used to determine the initial concentration of HI(aq)
Acid-Base Titration
Titration – is an important and common technique used to determine the
concentration of substances in solution
 known volume of a sample is put in an Erlenmeyer flask
Titrant – (solution in the buret) is added slowly to the sample in the flask
 an indicator is used to tell us when the reaction ends
 the point at which the indicator changes colour is called the end point
 you must know the concentration of one of the reactants (standard solution)
Titration Requirements
1.
2.
3.
4.
spontaneous-chemicals should react without adding extra energy
fast- should react quickly
quantitative- more than 99% complete
stoichiometric-single whole number mole ratio of reactants &
products
Example: Titration of Hydrochloric Acid
-Take 1.59g Na2CO3(s) to make 100.0mL of stock solution
-Samples (10.00mL)of stock solution are taken and titrated with HCl(aq) which
has been diluted by a factor of 10
Results:
TRIAL
Final Buret Reading (mL)
Initial Buret Reading (mL)
Volume of HCl Added (mL)
Calculate C of Na2CO3(s)
VNa2CO3= 100mL
MNa2CO3=105.99g/mol
mNa2CO3=1.59g
1
13.3
0.2
13.1
2
26.0
13.3
12.7
3
38.8
26.0
12.8
4
13.4
0.6
12.8
1st step: Calculate n of Na2CO3(s) using n=m/M
n=1.59g/105.99g/mol
=0.0150mol
2nd step: CNa2CO3=n/v
=0.0150mol/0.10000L
=0.150mol/L
step 3: use balanced equation
2 HCl(aq) +
Na2CO3(aq)-----H2CO3(aq) + 2NaCl(aq)
VAV=12.8mL V=10.00mL
C=?
C=0.150mol/L
Note: we are using 10.00mL not 100.00mL
NNa2CO3=10.00mL x 0.150mol/L
=1.50mmol
nHCl=1.50mmol x 2/1 = 3.00mmol
CHCl=n/v
3.00mmol/12.8mL
=0.234mol/L
Therefore the molar concentration of the HCl was 0.234mol/L
Sample Problem 1
Q: An acid rain sample containing sulphurous acid was analyzed in a laboratory
using a titration with a standard solution of sodium hydroxide. Use the
evidence given to determine the concentration of the sulphurous acid.
Titration of 25.0mL of H2SO3(aq) with 0.105mol/L NaOH(aq)
TRIAL
Final Buret Reading (mL)
Initial Buret Reading (mL)
Volume of HCl Added (mL)
1
11.1
0.3
10.8
2
21.7
11.1
10.6
3
32.4
21.7
10.7
A: H2SO3(aq) + 2 NaOH(aq) --- Na2SO3(aq) + 2H2O(l)
V=25.0mL
V=10.7mL
C=?
C=0.105mol/L
nNaOH = 10.7mL x 0.105mol/L
= 1.12mmol
nH2SO3 = 1.12mmol x 1/2 = 0.562mmol
CH2SO3 = n/V
= 0.562mmol/25.0mL
= 0.0225mol/L
Therefore the molar concentration of the H2SO3(aq) was 0.0225mol/L