Acids, Bases and Salts
IV.1 Arrhenius Theory
acids• donates H+(a proton) when in aqueous solution
• normally written HX
bases • donates OH- when in aqueous solution
• normally written XOH
salts • neutralization product which is ionic and neither acid nor base
• should be easy to spot based on recognition of ions from Chem.
11.
Generally:
Acid + Base → Salt + Water
Balancing neutralization reactions:
• balance H’s and OH’s on the reactant side of the equation
• on product side write formula for water and formula for resulting
salt.
___HCl + ____Ca(OH)2 →
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Hebden p.110 Exercises 1 & 2
Properties of acids and bases
Acids
• react with bases
• are electrolytes (carry an electric current)
• react with some metals to produce H2 gas
• turns litmus red
• tastes sour
Bases
• react with acids (duh)
• are electrolytes (carry an electric current)
• feel slippery
• turns litmus blue
• tastes bitter
IV.2 Common Acids
Read Hebden p.112 to 114 an do exercises
IV.3 True Nature of H+(aq): Background Theory
2
Hebden p.115 Exercises
IV.4 The Bronsted-Lowry Theory
an ACID is a proton DONOR
a BASE is a proton ACCEPTOR
E.g.
NH3 + H2O ↔ NH4+ + OHWhat is happening here?
Which is acting like an acid and which a base?
HCl
Mono-
(1)
H2SO4
Di-
(2) protic (proton containing) acid )polyprotic
H3PO4
Tri-
(3)
)
amphiprotic - when a substance can either be a proton donor or
acceptor (acid or a base)
Examples: water and polyprotic acids
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NH3 + H2O ↔ NH4+ + OHCH3COOH + H2O ↔ CH3COO- + H3O+
H3PO4 ← H2PO4- →HPO42Amphiprotics must
a) possess a negative charge
b) still have an easily removable hydrogen. Other than organic
substances, it is safe to assume all hydrogens on a negatively
charged ion are ‘easily removed’
Since reactions can be reversed, in every Bronsted-Lowry
reaction there is an acid and base on both sides of the equation.
CH3COOH + H2O ↔CH3COO- + H3O+
acid
base
base
acid
IV.5 Conjugate Acids and Bases
conjugate pair- a pair of chemical species that differ by only on
proton
conjugate acid - member of the pair that HAS the extra proton
conjugate base - member of the pair that LACKS the extra proton
Hebden p. 119 Examples and Exercises
When asked to give the conjugate acid of a species, you must
assume the species is a base ∴ conjugate acid will have one more
proton. Conversely if asked to give the conjugate BASE of a species,
you must assume the species is an acid ∴ conjugate base will have
one less proton.
E.g.
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Conjugate base of NH3?
Conjugate acid of NH3?
Writing Acid-Base Equilibria.
Example
Write the acid-base equilibrium which occurs when H2S and CO32are mixed in solution.
H2S + CO32- ↔
Hebden p.121 Exercises
IV.6 “Strong and Weak” Acids and Bases
Get the terminology right!
• a STRONG acid or base dissociates 100% (completely, ⇒ NO
EQUILIBRIUM)
• a WEAK acid or base dissociates <100% ( only partial
dissociation, ⇒ EQUILIBRIUM SITUATION)
• a CONCENTRATED acid or base ⇒ high molarity
• a DILUTE acid or base ⇒ low molarity
e.g.
10.0 M HF (aq) is concentrated and weak
0.1 M HCl (aq) is dilute and strong
THINK ABOUT THIS!!!
Introducing Table:
Relative Strengths of Bronsted-Lowry Acids and Bases
Read Hebden p. 121 - 125 Exercises p. 125-6
IV.7 The Equilibrium Constant for The Ionization of Water - Kw
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By definition:
Neutral solution: [H3O+] = [OH-]
Acid:
[H3O+] > [OH-]
Base:
[H3O+] < [OH-]
the net ionic equation for a neutralization reaction between a strong
acid and a strong base is:
H+ + OH- ↔ H2O + 59 kJ
This is equal but opposite to the self-ionization of water:
H2O + 59 kJ ↔ H+ + OHThe equilibrium expression for this reaction is
Kw = [H+][OH-] = [H3O+][OH-] = 1.00 X10-14 @ 25oC
Hebden p.127 exercises
Finding [H3O+] given the [OH-] & finding [OH-] given the [H3O+].
Remember in strong acids [H3O+] = [HX]
Remember in strong bases [OH-] = [XOH]
Hebden p.127 exercises
IV.8 Ka and Kb
The acid ionization constant Ka represent the equilibrium reached in
a reaction of a WEAK acid, such that in
CH3COOH (aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
Ka = ___[CH3COO-][ H3O+]___ = 1.76 X 10-5
[CH3COOH]
6
The base ionization constant Kb represent the equilibrium reached in
a reaction of a WEAK base, such that in:
NH3 (aq) + H2O(l) ↔ NH4+ (aq) + OH-(aq)
Kb = ___[NH4+][ OH-]___ = 1.79 X 10-5
[NH3]
Hebden p. 128 Exercises
IV.9 Ka and Kb for the Conjugate Pair
Ka(conjugate acid) X Kb (conjugate base) = Kw
Find Ka from Table
Calculate Kb from Kw and Ka
Hebden p. 129-30 Exercises
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IV.10 The Relative Strengths of Acids and Bases
Read Hebden p.130- 132
In a Bronsted-Lowry acid-base equilibrium, where there is a
competition between two acids to donate a proton, thee stronger acid
(as determined by their respective Ka’s) wins out. This means simply
that the side of the equilibrium equation that has the weaker acid will
be favored.
Given the following:
H2CO3 + SO32- ↔ HCO3- + HSO3There are two conjugate pairs in solution, what are they?
-
2-
-
(H2CO3, HCO3 and SO3 , HSO3 )
What competition is happening?
H2CO3 → HCO3- + H+
SO32-+ H+← HSO3-
(Ka = 4.3 X10-7)
(Ka = 1.0 X 10-7)
Which is stronger?
Which side will be favored?
How much will it be favored?…
…The Equilibrium Expression
Keq = ___[products]___ = ___Ka(reactant acid)___
[reactants]
Ka(product acid)
Hebden p.133 Exercises
IV.11 pH and pOH
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Aside: logs and anti-logs
Hebden p134,5 Exercises
pH = -log[H+]
pOH = -log[OH-]
Kw = [H3O+][OH-]
[H3O+]←⎯⎯⎯⎯⎯⎯⎯⎯→[ OH-]
pH = -log[H3O+]
pOH= -log[ OH-]
pH←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ pOH
pH + pOH = 14
NOTE; Sig Figs; in pH and pOH only the numbers after the decimal
are significant.
Hebden p139 – 140 Exercises
The pH Scale (potential of Hydrogen)
• copies of scale to be included with Relative Strengths of B-L Acids
and Bases
Read Hebden p140; Exercises p141
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IV.12 Mixtures of Strong Acids and Bases
Two points to remember:
• strong acids and bases dissociate completely. ( 1 mole HCl → 1
mole [H30+]; 1 mole NaOH → 1mole [OH-])
• mixing an acid and a base produces a solution which may be
either acidic, basic or neutral depending on the relative amonts of
reactants involved.
Example :
If 10.0 mL of 0.100 HCl is added to 90.0 mL of 0.100 NaOH, what is
the pH of the mixture?
Hebden p143 Exercises
10
Strong Acids and Bases Revisited
Points to remember
1. Following neutralization, pH/pOH depend on EXCESS of either
H3O+ or OH-2. Strong acids and bases dissociate 100% so whatever
concentration of acid and base is mixed: that is the
concentration of H3O+ and OH-.
3. Must perform dilution calculations to determine new
concentrations of acid and base using M1V1 = M2V2 IF USING
THE CONCENTRATION METHOD.
Or
Must change to moles IF USING mole METHOD.
***I SUGGEST USING mole METHOD
1. Use –log (remaining ion) to find pH or pOH.
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Solving Problems
Steps:
2. State neutralization equation determined by predicted ion in
excess. (don’t worry both versions mean the same thing).
E.g.
mol H3O+F = mol H3O+ST – mol OH-ST
2. Using :
Volume (L) X M ( mol / L) → Find mol for two
unknowns.
3. Solve for mol of unknown by subtraction.
4.Divide answer (#3) by TOTAL VOLUME to give the final [OH-.] or
[H3O+]
3. Use –log to find pH/pOH.
4. Calculate other (pH or pOH) : 14 -__________ = __________
5. Calculate [H3O+] / [OH-.]: antilog (-pH or pOH)
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IV.13 Hydrolysis
A few points to ponder:
• Hydrolysis OF A SALT is a reaction between water and the
anion, cation or both contained in the salt so as to produce an
acidic or basic solution
• Since salts dissociate to be 100% ionized in water; we pay
attention ONLY TO THE IONS.
• We pay attention only to the reactions between water and the
ions. Inter-ion reactions are ignored.
• The conjugates of strong acids and bases are SPECTATOR
IONS.
NaOH → Na+ + OH- (100% dissociation)
Na+ + H2O → NaOH + H+
∴ Na+ is a spectator
HCl → H+ + Cl- (100% dissociation)
Cl- + H2O → HCl + OH∴ Cl- is a spectator
• Spectator cations: Group 1 and Group 2 ions
• Spectator anions: ClO4-, I-, Br-, Cl- and NO3To determine the behavior of a salt in water
a) Determine ions produced during dissociation.
b) Discard spectators
c) Any remaining ions will act as acids if they are on the acid side
(left) of the Table (of Relative Strengths of B-L Acids and
Bases) and/or as bases if they are on the base side (right).
How will the following salts behave:
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NaCl
NH4Cl
NaF
NaHC2O4
NH4NO2
Fe(H2O)6Cl3
Al(H2O)5(OH)(NO3)2
Hebden p148 Exercises
14
IV.14 Calculations Involving Ka
• Problems involving weak acids are similar to those involving
Ksp and salts.
• There are three majour pieces of information which a
completely solved problem contains:
o [HA], the original concentration of the weak acid
o Ka for the acid
o [H3O+] or pH of the acid solution
• Three types of problems, all variations of:
Given two of the above, find the third
• Acid ionization expression is the only equation used to solve
the problems
• Must state simplifying assumption
“Assume 0.50 – X ≅ 0.50” (more on this in a minute)
• A calculation requiring the use of a Ka value from the Table of
Relative Strengths of Acids is restricted to a maximum of 2 Sig
Figs since all Ka’s are only good to 2 Sig Figs.
Work through examples
Hebden p149 – 151
Hebden p152 Exercises
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IV.16 Acid-Base Titrations (see other transparencies)
IV.17 Indicators
Indicator – a week organic acid or base with different colors for its
conjugate acid and base forms.
HIn + H2O ↔ In- + H3O+
Yellow
Red
HIn + H2O ↔ In- + H3O+
Yellow
Red
HIn + H2O ↔ In- + H3O+
Yellow
Red
An indicator changes colour at its equilibrium point. Since it is a weak
acid this color change is also it’s Ka.
Similar to pKw (-log Kw = -log (1.00 X 10-14) = 14)
pKa = -log(Ka)
∴ pKa of an indicator is the pH of its endpoint . The point where
[Hind] = [Ind-] (a.k.a. transition point )
Equivalence point- point at which chemically equivalent amounts of
acid and base have reacted.
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Calculating Ka for an indicator in the Provincial Exam.
Since you are given a RANGE of values; you need to find an
AVERAGE prior to taking the antilog(- pH/pKa).
Read Hebden p.160 Example and p161-162
p.162-3 Exercises
IV.18 Practical Aspects of Titration
In order for a titration to be accurate, you must titrate with a solution
of accurately known concentration. This is not quite as easy as
mixing coffee.
Two ways:
a.) A primary standard solution.
Must be of known molecular mass, pure, stable (unreactive) and
does not absorb H2O or CO2 from the air…this will change it’s
molecular mass.
Acidic
- potassium hydrogen phthalate (KHC8H4O4)
o monoprotic with NaOH
o phenophthalein or thymol blue indicator
- oxalic acid dehydrate H2C2O4.2H2O
o diprotic with NaOH
o phenolphthalein
Basic
- sodium carbonate (Na2CO3)
o diprotic with HCl
o methyl orange
a.) A standard-ized solution
A solution titrated against a primary standard in order to determine
the solution’s exact concentration.
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Read Hebden p.164-5 Do Exercise #122
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Titration Curves – STRONG ACIDS and STRONG BASES
Indicator – something close to pH = 7
Hebden p. 166-7 DO
Titration Curves – weak acids and STRONG BASES
Indicator – something close to pH = 8 - 10
Titration Curves – STRONG ACIDS and weak bases
Indicator – something close to pH = 4 - 6
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IV.19 Buffers
♦ a solution containing BOTH large amounts of a weak acid and
large amounts of it’s conjugate base.
♦ normally made by mixing a weak acid + a Na___________ of the
weak acid…
e.g. CH3COOH and CH3COONa
…and then diluting the system to a volume (usually 1.0 L)
so
CH3COOH + H2O ↔ CH3COO- + H3O+
1.0 M
1.0 M
Aside:
Can this be made by a solution of CH3COOH alone?
Why not?
-5
Ka = 1.8 X10
CH3COOH + H2O ↔ CH3COO- + H3O+
1.0 M
Addition of a small amount of acid:
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CH3COOH + H2O ↔ CH3COO- + H3O+
1.0 M
1.0 M
Addition of a small amount of base:
CH3COOH + H2O ↔ CH3COO- + H3O+
1.0 M
1.0 M
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Not only that but…
• Since [CH3COOH]=[ CH3COO-]
then Ka = ____[CH3COO-][H3O+]_____ = __1 [H3O+]__
[CH3COOH]
1
• Diluting a buffer has no effect on the pH since [CA] and [CB] are
diluted equally.
How a buffer works:
CH3COOH + H2O ↔ CH3COO- + H3O+
A buffer prevents the addition of either an acid or base from
changing the pH of a solution to any great extent.
Acidic buffer - buffers in the acidic region.
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CH3COOH + H2O ↔ CH3COO- + H3O+
Ka = 1.8 X 10 -5
pH = pKa = -log (1.8 X 10-5) = 4.74
Basic buffer - buffers in the basic region.
NH4+ + H2O ↔ NH3 + H3O+
Ka = 5.6 X10-10
pH = pKa = -log (5.6 X10-10) = 9.25
Read Hebden p 181
Do Exercises
Ka Calculations
Steps:
1. Write balanced Dissociation Equation
2. Set up ST, ∆, F table
3. Using Bronsted-Lowry Table and information given: fill in table
with values and inferred dissociation quantities (x’s
usually).
4. Set Ka = B-L Table value = Ka Expression (Equation)
5. Plug in dissociation quantities
6. WRITE DOWN ASSUMPTION
7. Solve for x.
8. Ask yourself: What does this x value represent (Hint: it is
always a concentration)? Is this what the question is asking me
for? If not (sometimes they ask for pH/pOH/[other ion] )
Calculate using
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Kb Calculations
Steps:
1. Exactly the same as Ka Calculations EXCEPT Kb MUST BE
CALCULATED FROM:
Kb = Kw / Ka
Hint:
♦ Dealing with an acid ⇒ use Ka
♦ Dealing with a base ⇒ use Kb
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