4.1
The Arrhenius Theory Of Acids And Bases
There are several different definitions of acids and bases.
Arrhenius’ explanation is the one you have been exposed to so
far…it got him the Nobel Prize in 1903.
Acid
•
•
•
Base
•
•
•
Salt
•
Any substance that releases H+(aq) in water
Ionic and starts with an H
HCl, HNO3, H2SO4…
Any substance that released OH-(aq) in water
Ionic and ends with OH
NaOH, KOH, Ca(OH)2…
The neutralization product that results when an acid and a
base react
• Any ionic compound that is not an acid or a base
Assign 1
Acid base neutralization reactions have the following form:
• Acid + Base  Salt + Water
• The production of water is just the combination of
o H+ + OH-  H2O
To balance an acid/base neutralization:
• Count the H’s and OH’s in the formulae
• Put the smallest coefficients in that will balance the H’s and
OH’s
• The same number of H2O’s will be produced as OH’s were
used
• Write down the formula for the salt with the remaining ions
Try with:
____HCl + _____Ca(OH)2  _____H2O + _______________
Assign 2 a, c, e.
Descriptive definitions of acids and bases
Acids
• React with bases
• Are electrolytes
• Act on some metals (like magnesium) to produce H2(g)
• Turn litmus red
• Taste sour
Bases
• React with acids
• Are electrolytes
• Feel slippery
• Turn litmus blue
• Taste bitter
Assign 3-4
Assign reading of 4.2 and questions 5-9 but no notes.
4.3 The true nature of H+(aq)
The hydrogen atom is a proton with an electron. If the electron is
removed then a naked proton is left
• extremely concentrated positive charge that is strongly
attracted to any region of negative charge
../../Chemistry 11/Unit VIII Atoms & Periodic Tab/moleculepolarity_en.jar
• water molecules are bent because of lone pairs of electrons
• the proton is attracted to the negative lone pairs so that the
hydronium ion (hydrated proton) H3O+ is formed.
• H+ + H2O  H3O+
• There is no such thing as a bare proton in water
• We need to change our dissociation equations to adjust for
this reality
o HCl(g)  H+(aq) + Cl-(aq) becomes:
 HCl(g) + H2O(l)  H3O+(aq) + Cl-(aq)
Assign 10
4.4 The Bronsted-Lowery Theory of Acids and Bases
This is a more general theory of acids and bases and the one we
will use.
• Allows for equilibrium reactions
• Acids are substances that donate protons (gives away H+)
• Bases are substances that accept protons (receives an H+)
NH3 + H2O ↔ NH4+ + OHBase + Acid ↔ Acid + Base
CH3COOH + H2O ↔ CH3COO- + H3O+
•
Mono
Di
Tri
Poly
To decide if something is an acid or a base look on the
other side of the arrow for something that differs by
only one H. The acid loses the H and the base picks it
up.
protic acids can supply 1 hydrogen
protic acids can supply 2 hydrogens
protic acids can supply 3 hydrogens
protic acids can supply More than 1 hydrogen
Assign 11-12
Water can be both an acid and a base
• Amphiprotic
o Can donate or accept protons
o H2O, H2PO4-, HS-, HCO3-
o If a substance has a negative charge and still has an
easily removed hydrogen (attached to anything besides
carbon)
In every Bronsted-Lowery reaction there is an acid and a base on
both sides of the equation
• Look for conjugate acid-base pairs to determine which is
which
o Differ by only one H on opposite sides of the
equation
• You need an acid and a base on both sides
Assign 13-14
4.5
CONJUGATE ACIDS AND BASES
Conjugate acid-base pair (conjugate pair)
• a pair of chemical species that differ by only one proton
• the acid has the extra proton
• the base lacks the extra proton
NH4+ +
H2O ↔
NH3 +
H3O+
a
b
b
a
Assign 15
The conjugate acid of x has one more proton than x.
The conjugate base of y has one less proton than y.
A Bronsted-Lowery acid base reaction is just an equilibrium
proton transfer.
Do example page 120
Assign 16-19
4.6
STRONG AND WEAK ACIDS AND BASES
Strong acid or base
• 100% ionized in solution
 NaOH(s)  Na+(aq) + OH-(aq)
 HCl(g)  H+(aq) + Cl-(aq)
Weak acid or base
o Less than 100% ionized in solution (equilibrium)
 NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
 HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
• Equilibriums involve weak acids and bases, not strong acids and
bases
 99% ionized is weak by definition
• in practice they are less than 50% ionized
 100% ionized is strong
• Strong/weak and concentrated/dilute are not interchangeable
 Strong/weak refers to % ionization
 Concentrated/dilute refers to molarity
 You can have a dilute solution of a strong acid
Discuss 20
acid-base-solutions_en.jar
Pull out and discuss page 6 of the data book
Strong acids
• Left hand of table
• Top six are 100% ionized
 There is no equilibrium because the reverse reaction
doesn’t occur
 All six make H+
• Then the seventh reaction occurs
• The net result of adding any strong acid to water is
o H+ + H2O ↔ H3O+
Strong bases
• Right hand of table
• Bottom two are 100% ionized
 There is no equilibrium because the forward reaction
doesn’t occur
 Third is H2O ↔ H+ + OH• This is the net result of adding any strong base
(including those that release OH-) to water
Weak acids
• Left side of the table with double arrows
 Note that hydroxide and ammonia never act as acids in
aqueous solutions
Weak bases
o Right side of the table with double arrows
 Note that the six top species never act as bases in aqueous
solutions
All of the weak acids and bases reactions can be written either
way.
H2S ↔ H+ + HS-
You can work with H2S as an acid or HS- as a base.
H2S ↔ H+ + HS- (gives off H+)
Added to
H+ + H2O ↔ H3O+
(accepts the H+)
Gives
H2S + H2O ↔ H3O+ + HSStronger acids give more H3O+
Or…
H+ + HS- ↔ H2S (accepts a proton from water)
Added to H2O ↔ H+ + OH- (water donates a proton)
Gives HS-+ H2O ↔ H2S + OHStronger bases give more OHOther notes on the table
• The higher an acid is on the left, the stronger the acid
• The lower a base is on the right the stronger the base
• The stronger the acid the weaker its conjugate base and vice
versa
The leveling effect
• All strong acids (bases) are 100% dissociated in aqueous
solutions and are equivalent to solutions of H3O+ (OH-).
• Only the central portion of the table (the double headed arrows)
show reactions that actually occur in aqueous solution
Assign 21-27
4.7
THE EQUILIBRIUM CONSTANT FOR THE
IONIZATION OF WATER
Neutral solution
• [H3O+] = [OH-]
Acidic solution
• [H3O+] > [OH-]
Basic solution
• [H3O+] < [OH-]
The reaction of a strong acid and base is very exothermic
HCl(aq) + NaOH(aq) NaCl(aq) +H2O(l) + 59kJ
But in complete ionic form this is:
H+ + Cl- + Na+ + OH-  Na+ + Cl- +H2O(l) + 59kJ
Drop the spectator ions to get:
H+(aq)+ OH-(aq) ↔ H2O(l) + 59kJ
The reverse reaction is:
H2O(l) + 59kJ ↔ H+(aq)+ OH-(aq) or…
2H2O(l) + 59kJ ↔ H3O+(aq)+ OH-(aq)
Which is the self ionization of water which has a special Keq
expression called
• Kw=[H3O+] [OH-] =1.00 x 10-14 at 25oC
o Unless told otherwise assume that temperature is 25oC
• As [H3O+] increases [OH-] decreases and vice versa
Assign 28-29
Both [H3O+] and [OH-] are present at some concentration in water.
Using Kw you can figure out the concentration of one once you
have the other.
Ex What is the [H3O+] and [OH-] in 0.0010 M HCl(aq)?
Since HCl is a strong acid [H3O+]=[HCl]=1.0 x 10-3M (we can
ignore the contribution from the self ionization of water 10-7M)
Assign 30
4.8
Ka AND Kb
The acid ionization of acetic acid (a weak acid) is
CH3COOH(aq) + H2O(l) ↔ CH3COO- (aq) + H3O+(aq)
The Ka is the acid ionization constant.
The greater the Ka, the greater the ionization, and the stronger the
acid.
• for strong acids the denominator is zero and the
expression is infinitely large
The base ionization of ammonia (a weak base) is
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
The Kb is the base ionization constant
The greater the Kb, the greater the ionization, and the stronger the
base.
Assign 31-34
4.9
THE RELATIONSHIP BETWEEN Ka AND Kb FOR A
CONJUGATE PAIR
The acid ionization of ammonium is
NH4+ + H2O ↔NH3(aq) + H3O+
With
While the base ionization is
NH3(aq) + H2O ↔ NH4+ + OHWith
There is a relationship between Ka and Kb for a conjugate pair
• For a conjugate pair Ka(conjugate acid) x Kb(conjugate base) =Kw
o You can find Ka and Kb from the table
Ex:
Find the Ka and Kb of H2PO4-.
• Ka can be read directly from the table
o Find H2PO4- on the left and Ka is 6.2 E-8
• Kb can be calculated from the Ka of its conjugate acid
o H3PO4 ↔ H+ + H2PO4- ….Ka= 7.5E-3
o When H2PO4- acts as a base the equation is
o H2PO4- + H2O ↔ H3PO4 + OH-
Assign 35-37
4.10 THE RELATIVE STRENGTHS OF ACIDS AND BASES
If you mix solutions containing H2CO3 and SO32- the SO32- can
only act as a base.
H2CO3 + SO32- ↔ HCO3- + HSO3• There are 2 conjugate pairs in solution
• (We will not consider 2 proton transfers in our BronstedLowery reactions)
H2CO3 + SO32- ↔ CO32- + H2SO3
If you mix CO32- and H2PO4-:
CO32- + H2PO4- ↔ HCO3- + HPO42• There are two acids in equilibrium, both H2PO4- and HCO3want to donate their proton. So who wins?
o H2PO4-…Ka = 6.2E-8
o HCO3- …Ka = 5.6E-11
o H2PO4- is the stronger acid and is better at donating its
protons.
 The reaction shifts right
 The side of the equilibrium with the weaker acid
is favored
There is another way to figure out the favored side of the reaction
• See pages 131-132 if you are interested
Assign 38-46
4.11 pH AND pOH
pH = -log10[H3O+]
pOH = -log10[OH-]
So we need a math mini lesson:
log10(x) means
• the logarithm to the base 10
of x
• 10(what power) = x
• what is the equivalent
power of 10 that will give
me the same number?
Ex
• Log10 (1000) = 3 …….because (103) = 1000
• Log10 (0.01) = ?
• 0.01 is really 10-2 so…… Log10 (0.01) = -2
• Make sure you know how to use your calculator and try 47.
The reverse procedure is taking the antilog
• Antilog (x) = 10x
Ex what is the antilog of 4?
• Antilog(4) = 104= 10 000
Your calculator might have a invlog button, a 10x button or an
antilog button
• Make sure you know how to use your calculator and try 48.
Log and antilog cancel each other out, giving you the original
value.
Ex
• Antilog(log(3))= antilog(0.4771)=3
• Log(antilog(0.5))=log(3.162)=0.5
There is one other thing about logs you need to know:
• Log(10x 10y) = log (10x) + log(10y)
• Or if A = 10x and B = 10y
• Log(AB)=log(A) + log (B)
Back to Chemistry…
acid-base-solutions_en.jar
ph-scale_en.jar
If [H3O+] = 3.94 E-4M, what is pH?
pH = -log10[H3O+] = -log (3.94 E-4) = - ( - 3.405) = 3.405
If [OH-] = 9.51E-12 what is pOH?
pOH = -log10[OH-]= -log (9.51E-12) = -(-11.022)=11.022
Finding [H3O+] or [OH-] from pH and pOH
• Reverse the steps
• [H3O+] = antilog(-pH)
• [OH-] = antilog(-pOH)
Ex
• pH = 3.405, what is [H3O+]?
•
[H3O+] = antilog(-3.405)= 3.94E-4M
• pOH=11.682, what is [OH-]?
• [OH-] = antilog(-11.682)=2.08E-12M
Remember from Kw at 25o
• [H3O+][OH-]=1.00E-14
• take the log of both sides
• log([H3O+][OH-])=log (1.00E-14) = -14
•
we can use Log(AB)=log(A) + log (B) so that
• log([H3O+][OH-])= log([H3O+]) + log([OH-])
• log([H3O+]) + log([OH-])=-14
•
multiply everything by -1
-log([H3O+]) + -log([OH-])=14
•
pH + pOH = 14 @ 25o
Now we can move between pH, pOH, [H3O+] and [OH-]
• copy the diagram on page 138
Ex if pH = 9.355 what is pOH?
pH + pOH = 14.000 so pOH= 14.000-9.355=4.645
Ex if pH = 6.330, what is [OH-]?
pOH = 14.000-pH = 7.670
[OH-] = antilog(-pOH)= antilog (-7.670) = 2.14E-8M
Notes
• Kw=[H3O+][OH-]=1.00E-14 so
• pKw=pH + pOH = 14
• In a pH only the digits after the decimal are significant.
Assign 49-54 ace
pH scale:
look at the diagram on page 140
ph-scale_en.jar
•
•
•
•
As pH goes up, pOH goes down (they add to 14)
pH < 7 or pOH > 7 is acidic
pH >7 or pOH < 7 is basic
the scale is logarithmic
• an increase of 3 pH units is a 1000X decrease in [H3O+]
Assign 55-57 all
4.12 MIXTURES OF STRONG ACIDS AND BASES
Mixing an acid and base can give you an acidic, basic or neutral
solution depending on the amount of reactants involved.
Example If 10.0 mL of 0.100M HCl is added to 90.0mL of 0.100M
NaOH, what is the pH of the mixture?
• First perform dilution calculations to get the starting
concentrations
• Since the reaction is
• H3O+ + OH-  2H2O
• and they react in a 1:1 ratio, there is excess OH- and all the
H3O+ will be used up.
[OH-] = [OH-]start - [H3O+]start = 0.0900-0.0100=0.0800M
pOH=-log [OH-] = -log (0.0800) = 1.097
pH=14.000-pOH = 14.000-1.097=12.903
Note you can get moles of H+ and moles of OH- at the beginning
instead of doing the dilution calculation and get to the same result.
Ex How many moles of HCl must be added to 40.0mL of 0.180M
NaOH to produce a solution with pH=12.50? Assume no change
in volume when HCl is added.
The pH indicates a basic solution so calculate the [OH-]
• pOH = 14.000-pH= 14.000-12.500=1.500
• [OH-] = antilog(-pOH) = antilog (-1.500)= 0.03162M
We need 0.03162M in excess of the HCl added.
[OH-]xs=[OH-]start- [H3O+]start
We are solving for [H3O+]start= [OH-]start - [OH-]xs
[H3O+]start= 0.180M – 0.0316M =0.1484M
So moles H3O+ (HCl) = (0.1484 mol/L)x0.0400L= 5.9E-3 mol
Assign 58-68 even
4.13 HYDROLYSIS
Hydrolysis of a salt
• a reaction between water and the cation or anion or both
contained in a salt so as to produce an acidic or basic solution
Note
• All salts are considered to be 100% ionized in water so it is the
ions that count, not the salt itself
Spectator ions do not react with water.
• Conjugates of strong acids and bases are spectator ions
•
HCl  H+ + Cl•
Cl- + H2O
HCl + OH•
Cl- cannot form HCl again so it will not react with
water.
• Column one and two give spectator cations
• Spectator anions are the first six anions found at the top right of
the strength of acids table
To determine the behaviour of a salt in water
1. Determine the ions produced
2. Discard any spectators
3. If an ion appears on the left of the strengths table it acts as an
acid and vice versa
Ex
NaCl Na+ + Cl• Both are spectators (conjugates of NaOH and HCl)
• Neutral solution (no hydrolysis)
NH4Cl  NH4+ + Cl• Ignore Cl• NH4+ is an acid
• NH4+ + H2O   H3O+ + NH3
• The solution is acidic
• Cationic hydrolysis
NaF  Na+ + F• Ignore Na+
• F- is a base
• F- + H2O   HF+ OH• Solution is basic
• Anionic hydrolysis
NaHC2O4 Na+ + HC2O4• Ignore Na+
• HC2O4- can be an acid or a base
• Check ka vs kb to see which occurs
• Ka(HC2O4)=6.4E-5
• Kb(HC2O4)= Kw/(ka(H2C2O4))= 1.0E-14/5.9E-2 = 1.7E-13
• Ka > Kb so it is an acid
• HC2O4- + H2O   H3O+ + C2O4-2
• The solution is acidic
• Further ionization
NH4NO2  NH4+ + NO2• Neither can be ignored (both undergo hydrolysis)
• NH4+ + H2O   H3O+ + NH3
• NO2- + H2O   OH- + HNO2
• To find out which reaction dominates you need to
compare Ka and Kb again
• Ka(NH4+)= 5.6E-10
• Kb(NO2-) = Kw/ (Ka(HNO2))= (1.0E-14) / (4.6E-4)= 2.2E-11
• Ka > Kb so more H3O+ is produced than OH• Solution is acidic
Fe(H2O)6Cl3  Fe(H2O)63+ + 3Cl• Ignore the Cl
• Keep the other ion intact (do not try to split away the
waters!)
• Iron hexaquoiron is on the left
• Fe(H2O)63+ + H2O   H3O+ + Fe(H2O)5(OH)2+
• The solution is acidic
Assign 69-73
4.14 Calculations involving Ka
When a weak acid HA is put in water, some of it ionized which
produces a certain amount of H3O+. Since none of the acid ionized
until it hit the water we can use an ICE box. Initial, change,
equilibrium.
A completely solved problem will have:
• [HA], the original concentration of the weak acid
• Ka for the acid
• [H3O+] or the pH of the acid solution
The only equation that applies to these questions is the Ka
expression so there are 3 possible forms the questions can take and
the other information must be given:
• find [H3O+] or pH given Ka and [HA]
Ex
If Ka = 1.8E-5 for CH3COOH what is the pH of 0.500M
CH3COOH?
• [H3O+] must be found to give us pH
CH3COOH + H2O  CH3COO- + H3O+
Initial
0.500
0
0
Change
-x
+x
+x
Equilibrium
0.500-x
x
x
•
• to solve this exactly you would need the quadratic formula
• in Chem 12 there is always a way around the quadratic
formula
• assume the acid is sufficiently weak that the amount
that ionizes will not significantly affect the
concentration of the acid
• assume 0.500-x ≈ 0.500
• you must write down your assumption or you will
get the question wrong!
•
• Check the assumption:
• 3.0E-3 << 0.500 so the assumption seems valid but
how do we know if it’s enough?
• If the percent dissociation is less than 5%
• pH = -log (3.0E-3) = 2.52
• This is 2 sig figs as only the digits after the decimal are
significant in a pH (has to do with the logging)
• find Ka given [HA] and [H3O+]or pH
Ex
If pH = 1.70 for a 0.100 M solution of an unknown weak acid
HA, what is the Ka for HA?
• pH must be converted to [H3O+]
• [H3O+] = antilog (-1.70) = 0.020M
• Now do the ICE box

HA
+ H2O
H3O+
+ AInitial
0.100
0
0
Change
-0.020
+0.020
+0.020
Equilibrium 0.080
0.020
0.020
• No assumption was needed because the calculation was
easy and besides
• 0.02/0.1 >5%
• find [HA] given [H3O+] or pH and Ka
Ex
What mass of NH4Cl will produce 1.50L of a solution having
a pH of 4.75?
• first write out the equation
• NH4Cl  NH4+ + Cl• Cl- is a spectator and NH4+ undergoes hydrolysis as a
weak acid
• NH4+ + H2O  NH3 + H3O+
• [NH4+] must be found
• Ka is from the table
• [H3O+] = antilog (-pH) = antilog (-4.75) = 1.78E-5M
• Once [H3O+] is found it can be converted to [NH4+]
which can then be converted to moles and then mass
of NH4Cl.
NH4+
+ H2O 
NH3
+ H3O+
Initial
X
0
0
Change
-1.78 E-5
+1.78 E-5 +1.78 E-5
Equilibrium X - 1.78 E-5
1.78 E-5 1.78 E-5
Assume X - 1.78 E-5 ≈ X
So
• Check the assumption
• X - 1.78 E-5 = 0.565 – 0.0000178 = 0.565
Mass NH4Cl = (0.565 mol/L)(1.50 L)(53.4g/mol) = 45g
• (limited to 2 sig figs if we used Ka from the table)
Notes
• for hydrolysis of polyprotic acids we will only consider the
ionization of the first hydrogen
• the second one can be ignored because it is so small
compared to the first
•
H2C2O4  HC2O4 – matters while HC2O4 – C2O4 2can be ignored
• Two of the acids on the table have never been observed but the
calculations are easier if we pretend they exist
• H2CO3 and H2SO4 can be assumed to be produced when
carbon dioxide or sulfur trioxide react with water.
Assign 74-82 even
4.15 Calculations involving Kb
When a weak base, A- is put in water, the base ionizes and a
certain amount of OH- is produced
• smaller Kb = less OHCalculations with Kb are similar to those with Ka (last day) with
the following changes
• Kb must be calculated
• The resulting solution will be basic so [OH-] and pOH
are used instead of [H3O+] and pH
Ex
What is the pH of a 0.10M solution of NaCN?
NaCN  Na+ + CNNa+ is a spectator and CN- acts as a weak base.
[NaCN] is given and Kb(CN-) can be calculated from Ka(HCN) on the
table. To find pH we need pOH which requires[OH-]
I
C
E
CN0.10
-x
0.10-x
+ H2O

HCN
0
+x
X
Assume that the weak base doesn’t dissociate much, so
• 0.10-x ≈ 0.10
then
+ OH0
+x
X
and
X= [OH-] = 1.43E-3
pOH = -log[OH-] = 2.845…. pH = 14.00-pOH = 11.15
Ex
The pOH of a 0.50M solution of the weak acid HA is 10.64.
What is Kb for A-?
• This is an acidic solution (high pOH) so we must work with
[H3O+] and not [OH-].
o Convert pOH to [H3O+]
 pH = 14.00 – pOH = 14.00-10.64 = 3.36
 [H3O+] = antilog (-3.36)= 4.37E-4M

HA
+ H2O
H3O+
+ AI
0.50
0
0
C -4.37E-4
+4.37E-4 +4.37E-4
E
0.504.37E-4
4.37E-4
4.37E-4
Assign 84-92 even
4.16 Acid-Base Titrations
Titration
• process in which a measured amount of a solution is reacted
with a known volume of a another solution until an equivalence
point (stoichiometric point) is reached.
• http://www.wfu.edu/~ylwong/chem/titrationsimulator/index.html
o One solution must be of known concentration and the
concentration of the other solution is calculated
In the reaction aA + bB  cC + dD
The equivalence point is reached when the
• When the mole ratio reacted is the same as the mole
ratio in the equation.
Remember that concentration =
There are 5 pieces of information in titration problems
[Acid]
[Base]
Base/acid mole ratio
Volume of acid
Volume of base
• The mole ratio can be easily calculated from the reaction
equation.
• Use c=n/V to calculate the moles of one substance
• Use the mole ratio to calculate the moles of the second
substance
• Use c=n/V to combine the moles of the second substance with
either the volume or concentration to find the missing value.
Sig figs are crucial because the whole purpose of a titration is
an accurate measure of concentration
Ex
H2SO4 + 2 NaOH  Na2SO4 + 2H2O
An equivalence point occurs when 23.10 mL of 0.2055M NaOH is
added to 25.00mL of H2SO4. What is [H2SO4]?
• Life can be made easy by using mmol
•
 1M =
moles NaOH = cV = 0.2055
x
23.10 mL = 4.7471mmol
moles H2SO4= 4.7471mmol NaOH x
=2.3735mmol
[H2SO4] =
=0.09494 M
Partial Neutralization:
• Polyprotic acids can donate more than one proton in a
neutralization reaction:
H4P2O7 + 1NaOH  NaH3P2O7 + 1H2O
H4P2O7 + 2NaOH  Na2H2P2O7 + 2H2O
H4P2O7 + 3NaOH  Na3HP2O7 + 3H2O
H4P2O7 + 4NaOH  Na4P2O7 + 4H2O
In partial neutralization questions the base/acid mole ratio can be
unknown because we don’t know how many of these reactions
actually occur.
Ex
An equivalence point is reached by reacting 25.00 mL of
0.1255M NaOH with 38.74mL of 0.02700M H4P2O7. How many
protons are removed on average and what is the balanced equation
for the reaction?
Moles NaOH used = 0.1255
x 25.00 mL = 3.1375 mmol
Moles H4P2O7 used=0.02700
x 38.74 mL = 1.0460 mmol
so 3 NaOH are required for every 1 H4P2O7:
H4P2O7 + 3NaOH  Na3HP2O7 + 3H2O;
• 3 protons removed on average
Percentage Purity
• Use the titration data to find the actual concentration
• Use the mass of the impure substance and its molar mass to
calculate the theoretical concentration
•
o % purity =
Ex
A 3.4786g sample of impure NaHSO4 is diluted to 250.0mL. A
25.00 mL sample of the solution is titrated with 26.77 mL of
0.09974 M NaOH. What is the percentage purity of the NaHSO4?
First find moles NaOH used
• 0.09974
x
26.77mL = 2.6700 mmol
NaHSO4 is monoprotic so the moles NaOH = moles NaHSO4
[NaHSO4] =
= 0.10680 M (actual)
[NaHSO4]expected =
= 0.11586 M(expected)
% purity =
= 92.18 %
Molar mass determination
• titration can determine the moles present in a solution of an
unknown acid or base
o the moles, along with the mass of the original sample
can be used to calculate the molar mass
Ex
A 3.2357g sample of an unknown monoprotic acid is diluted to
250mL. A 25.00 mL sample is titrated with 16.94mL of 0.1208M
KOH. What is the molar mass of the acid?
Moles KOH = 0.1208
x 16.94 mL = 2.0464 mmol
Since the acid is monoprotic, moles KOH = moles acid
[acid] =
= 0.081854 M
Moles in original container=0.081854
0.020464mol
Molar mass
x0.2500L=
= =158.1 g/mol
Experimental note
• titrations must be double checked for accuracy
o if not within 0.02mL then repeat
o take the average of the 2 that are close enough
Assign 95-107 odd
4.17 Indicators
Indicator
• weak organic acid or base with different colours for its
conjugate acid and base forms
o complex molecule that is usually abbreviated HIn or InHIn + H2O ↔ In- + H3O+
Yellow
Red
• When the indicator is put in an acid the equilibrium shifts to the
left and the yellow HIn dominates; [HIn] > [In-]
o Indicators are in their conjugate acid form in very
acidic solutions
• When the indicator is put in a base the equilibrium shifts to the
right and the red In- dominates [HIn] < [In-]
o Indicators are in their conjugate base form in very basic
solutions
If a base is added to an acid, at some point [HIn] = [In-] and an
orange solution will result.
• The point at which an indicator is half way through its colour
change is the end point or transition point.
• At the end point:
since [HIn] = [In-]
So the [H3O+] at which the indicator changes colour is the Ka value
of the indicator.
So
Ka=[H3O+]
Then
–log Ka = -log [H3O+]
Or
pKa = pH
An indicator is at the midpoint of its colour change when the pH of
the solution equals the pKa of the indicator
• The end point is the point where the colour changes
• The equivalence point is where the stoichiometry of the
reaction is satisfied
• A well chosen indicator will have an end point very
close to the equivalence point
 The colour actually changes over a range of about
2 pH units (see the bible)
Ex
Ethyl orange is red at pH < 3 and yellow at pH > 4.8 and
intermediate orange at pH 4.1. What is the approximate Ka of
ethyl orange?
• At the midpoint of the colour change pKa = pH = 4.1
• Ka = antilog (-4.1) = 8 E –5
Ex
Alizarin Yellow R changes from yellow to red at pH = 11.0.
Aliz ion is red. What colour is Alizarin Yellow R in 1E-5 M
NaOH?
HAliz + H2O ↔ Aliz- + H3O+
Yellow
Red
First establish the colour at the critical pH values
• pH = 11 [HAliz] = [Aliz-] orange
• pH < 11 [HAliz] > [Aliz-] yellow (acid form)
• pH > 11 [HAliz] < [Aliz-] red (base form)
-
If [OH-] = 1E-5 then [H+] = 1E-9 so the pH is 9.0
POH = -log [OH-] = 5 pH = 14.00-pOH = 9
• Alizarin Yellow R is in its acid form at pH 9.0 and is
yellow.
Universal Indicator
• Solution of several indicators that change colour several times
over a range of pH values
o Gives an estimate accurate within +/- 0.5 pH values
Look at the tables on page 162 for an example of a universal
indicator.
Assign 108-120
4.18 Practical aspects of titration
In order to carry out a titration you need a solution with an
accurately known concentration
• standard solution or standardized solution
o requires the use of a substance that is pure, stable, and
won’t absorb water from the air
o Oxalic acid dihydrate is a primary acid standard
 A primary acidic standard solution can be used to
standardize a basic solution for titrating acid
Primary standards
• Potassium hydrogen phthalate (acid)
o Use phenolpthalein or thymol blue
• Oxalic acid dihydrate
o Use phenolpthalein
• Sodium carbonate (base)
o Use methyl orange
Procedure for using primary standards
• Obtain a pure sample (analytical grade)
• Dry the sample in an oven and store with a dessicator
• Accurately weigh enough to make a 0.1000M solution
• Quantitatively transfer to a volumetric flask
• Dilute to the required volume
Assign 121-123 Calculate the [NaOH]
Types of titration curves
• Will help select appropriate indicators for titrations
Strong acid with a strong base
• Assign 124 (plot the data) titration curves (Q124).xls
• Vb is the volume of the base required to reach the
equivalence point
• pH rises almost vertically around the value of Vb
Choosing an indicator
• salt of a strong acid and a strong base is neutral (no
hydrolysis)
• Indicator should have pKa “around” 7
Titration of a weak acid with a strong base
• There is an initial up swing of pH
• Vb is the volume of base required to get to the equivalence point
o V1/2 = ½ Vb
o The pH at V ½ = pH ½
o The pH before any NaOH is added = pH init
You can calculate the [acid] using normal titration calculations if
you know [NaOH], vol NaOH and vol acid.
Optional
But you can also use the graph to calculate [acid] if you don’t
know the [NaOH]:
Step 1
Assume you start with 0.100M CH3COOH in 1.00L solution.
At the equivalence point 0.100 mol of NaOH is added and at the
half volume, V ½ , 0.050mol NaOH is added. Look at the ICE box
at the half volume point
CH3COOH
OHI
0.100
0.050
C
-0.050
-0.050
E
0.050
0
So [CH3COOH] = [CH3COO-]
↔
CH3COO0
+0.050
0.050
H2O
CH3COOH + H2O ↔ CH3COO- + H3O+
And
So pKa=pH ½
The Ka can be determined by using the pH value at point pH ½
Step 2
Combine the Ka value with pH init to determine the concentration of
the weak acid
Let [CH3COOH] = the initial concentration
Let X = the amount of dissociation
Let [CH3COOH]eq = the concentration at equilibrium (assume that
it is equivalent to [CH3COOH]
I
C
E
CH3COOH
[CH3COOH]
-x
[CH3COOH]eq
H2O
↔
CH3COO0
+x
x
H3O+
0
+x
x
The [H3O+] can be calculated from pHinit which was before any
base was added.
X= [H3O+]= [CH3COO-]
End optional
Choosing an indicator
• A weak acid and a strong base will produce a salt with
a basic pH (hydrolysis of the anion)
• CH3COOH + NaOH  Na+ + CH3COO- + H2O
• pKa of the indicator should be basic pKa 8-10 will
usually work.
The titration curve of a weak base with a strong acid
• exactly the same as a weak acid with a strong base but
reversed
• convert pH to pOH for the calculations
• pKb= pOH ½
• kb = [OH-]1/2
•
To remember whether a titration gives an acidic basic or neutral
solution and the indicator needed:
• Salt of STRONG ACID and STRONG BASE is
• a tie (neutral, no hydrolysis)
• indicator changes at “close” to pH = 7
• Salt of a STRONG ACID and weak base
• STRONG beats weak  acidic (cationic hydrolysis)
• indicator changes at close to pH = 4-6
• Salt of a weak acid and a STRONG BASE
• STRONG beats weak  base (anionic hydrolysis)
• indicator changes at close to pH = 8-10
Example
The following data was obtained when titrating ethylamine (weak
base) with HCl
25.00 mL = volume ethylamine solution titrated
19.22 mL = volume of HCl required to get to equivalence point
11.855
= initial pH of the ethylamine solution
10.807
= pH at 9.60 ml point of titrtation (half point)
a) Calculate Kb for ethylamine
Kb = [OH-] 1/2 (from the half volume point)
pOH1/2 = 14.000- pH1/2 = 14.000-10.807 = 3.193
Kb = [OH-] 1/2= antilog ( -3.193) = 6.412E-4
b) Calculate [ethylamine]
This equation refers to the situation before the titration starts so
we need [OH-] from the pHinitial
pOHinit= 14.000 - pHinitial =14.000 – 11.855 = 2.145
[OH-] = antilog (-2.145) = 7.161E-3 M
= 0.0800M
But we need [ethylamine]initial which includes the amount that
dissociated to give OH[ethylamine]initial = [ethylamine]eq + [OH-] = 0.0800 + 7.161E-3
= 0.0872 M
c) Calculate the [HCl] used
at the equivalence point Moles HCl = moles ethylamine
Moles ethylamine = (0.0872 mmol/mL)(25.00mL) = 2.18 mmol
[HCl] = (2.18mmol)/(19.22mL) = 0.113M
d) Suggest a suitable indicator
Weak base plus strong acid = acidic solution…we want the
change in the pH = 4-6 range (methyl red)
Assign 125, 126b
4.19 Buffers
Buffer
• A solution containing appreciable amounts of a weak acid and
its conjugate weak base
CH3COOH + H2O ↔ CH3COO- + H3O+
1M
1M
If [CH3COOH] = [CH3COO ] then
•
=
[H3O+]= 1.8E-5
• pH= pKa
When equal concentrations of a weak acid and its conjugate weak
base are added to water, the pH of the resulting buffer will equal
the pKa of the weak acid.
What happens when a buffer is made?
CH3COOH + H2O ↔ CH3COO- + H3O+; Ka = 1.8E-5
If a solution is made by adding CH3COOH to water there will be a
small amount of both CH3COO- + H3O+. (It is not a buffer yet)
• Adding extra CH3COO- causes the equilibrium to shift
left.
• There is a large relative decrease in [H3O+] but only a
very small amount of the CH3COO- added will be used
up because there wasn’t much H3O+ available in
solution.
• Overall, the buffer will have a very large concentration
of the weak acid and it’s conjugate base but a very
small concentration of H3O+
When acids and bases are diluted their concentrations and
therefore their pH’s are changed.
If a buffer is diluted both the acid and conjugate base are diluted
equally.
If [CH3COOH] = [CH3COO-]= 1M
Then
If the solution is diluted 10 times so
[CH3COOH] = [CH3COO-]= 0.1M
then
• Diluting a buffer does not affect its pH.
Why are buffers important?
• To maintain almost constant pH
How?
The Ka expression can be rearranged
The Henderson Hasselbalch Equation
Assume you have [CH3COOH] = [CH3COO-]=1M
CH3COOH + H2O ↔ CH3COO- + H3O+
The original ratio of
was
If you add 0.1 mol HCl, the equilibrium shifts left and more of the
conjugate acid is produced.
[CH3COOH] = 1.1 and [CH3COO-]= 0.9
Enough H3O+ was added to neutralize 10% of the conjugate base
but the pH only changed 0.08 units
Without the buffer, the pH would change 6 units!
The same effect occurs when adding OH-, except that [base]
increases while the [acid] decreases and the pH is virtually
unchanged.
A buffer prevents the addition of either an acid or a base from
changing the pH of a solution to any great extent
• Because the
ratio does not change
much when a buffer reacts with a small amount of acid
or base.
• CH3COOH + OH-  CH3COO- + H2O (neutralizes base)
• CH3COO- + H3O+  CH3COOH + H2O (neutralizes acid)
o The pH of a solution is stabilized near the pKa of the
conjugate acid of the buffer system
Ex
Acidic buffer
• Mix 1.0 mol acetic acid and 1.0 mol sodium acetate and
dilute to 1.0L
• pH will stay near pKa = -log(1.8E-5) = 4.74
Basic buffer
• Mix 1.0 mol ammonia with 1.0 mol ammonium nitrate
and dilute to 1.0 L
• pH will stay near pKa = - log (5.6E-10) = 9.25
There is a limit to how much acid or base a buffer can neutralize.
• If there was 1.0 mol of conjugate acid initially it can
only neutralize a maximum of 1.0 mol of base
• And vice versa
Look at page 181
Whenever a weak acid or base is titrated, a buffer solution will
occur in the middle portion of the titration curve (the flat area
where adding some acid or base has little effect on the pH.)
Assign 131-140
4.20 Buffers in Biological Systems
Life works on a principle of homeostasis
• Tries to maintain all chemical species within an acceptable
range
• Too much or too little can harm an organism
Buffers help keep the chemistry of the body from varying too
much
Hemoglobin carries oxygen in the body
HHb + O2 + H2O H3O+ + HbO2The optimum pH for blood is 7.35.
• If the pH is too low (acidosis)
• The equilibrium shifts left and oxygen is not absorbed properly
in the lungs
• If the pH is too high (alkalosis)
• The equilibrium shifts right and the hemoglobin won’t release
oxygen in the cells
• Buffers keep the [H3O+] in our bodies constant so that our
chemistry works properly
• Otherwise eating an orange could kill you!
The CO2/HCO3- system
• Main buffer in blood
• Two equilibria arise from the production of CO2 (aq) in our cells
• CO2 (aq) + 2 H2O  H3O+ + HCO3• CO2(g) CO2(aq)
As CO2 accumulates the pH drops
• Your body gives you a signal to breathe out CO2 when the pH
of your blood drops
• Had nothing to do with oxygen levels
• If you hyperventilate you loose too much CO2 and the pH of
your blood rises
• You pass out.
The H2PO4-/HPO42- system
• Stabilizes the pH in many cells
• Prevents a build up of acid from the byproducts of metabolism
• H2PO4- + H2O  H3O+ + HPO42Assign 141-143
4.21
Applied acid/base chemistry
Aqueous solutions of metal and nonmetal oxides (anhydrides)
• When an ionic metal oxide is added to water it initially
dissociates to form the strong base O2- ion
• Na2O (s)  2Na+ (aq) + O2-(aq)
• O2- + H2O  2OH-…basic solution
• When you combine the two reactions you get
• Na2O + H2O  2NaOH
• When a covalent nonmetal oxide reacts with water the water
bonds to the oxide to make and acidic solution
• SO3 + H20  H2SO4
• CO2 + H2O  H2CO3
Assign 144-145
Acid rain
• All rain is naturally acidic pH = 5.6 due to dissolved CO2 from
the atmosphere
• CO2 + H2O  H2CO3
• Any rain with pH < 5.6 (more acidic than normal) is acid rain
• Sources
• Sulphur contaminants in hydrocarbons
• S + O2  SO2
• SO2 + O2  SO3
• SO2 + H2O  H2SO3 (sulphurous acid)
• SO3 + H2O  H2SO4 (sulphuric acid)
• A mixture SO2 and SO3 is called “SOx”
• Nitrogen in internal combustion engines
• N2 + O2  2NO
• N2 + 2O2  2NO2
• 2NO + O2  2NO2
• 2NO2 + H2O  HNO2 + HNO3 (nitrous and nitric acids)
• Nitrogen oxides are referred to as “NOx”
Natural protection
• Most lakes have CO2 /HCO3- buffering ability
• Limestone deposits can neutralize the acid (Kal lake)
• Added limestone can help a lake recover
Problems with acid rain
• Fish and plant growth
• Some lakes in the east have lost all their fish
• Maple forests have been wiped out
• Some food crops are very sensitive (apples, tomatoes…)
• Dissolves minerals out of rocks and soils
• Many are poisonous and get into the water
• Affects general health of those in the region
• Other nutrients are lost from the soil
• Metal and stone structures get eaten away
• Marble and limestone statues
• Often falls far from the source
• The people who create it don’t suffer
• Less incentive to fix problems
• The cost of using materials that don’t produce acid rain is high
We need to find cheap technology that will not require the
production of these pollutants!
Assign 146-147