Acids & Bases and Titrations
Properties of Acids & Bases
• Acids
–
–
–
–
–
taste sour
feel like water
good electrolytes
turn blue litmus paper “pink”
pH less than 7 at 25°C
• Bases
–
–
–
–
–
taste bitter
feel slippery
good electrolytes
turn pink litmus paper “blue”
pH greater than 7 at 25°C
3 Definitions of Acids
• Arrhenius: acids are substances which
produce H+ ions in water.
• Brønsted-Lowry: acids are substances
which donate H+ ions to other substances.
• Lewis: acids are substances which accept
electron pairs from other substances.
3 Definitions of Bases
• Arrhenius: Bases are substances that
produce OH- ions in water.
• Brønsted-Lowry: Bases are substances that
accept H+ ions from other substances.
• Lewis: Bases are substances that donate a
pair of electrons to another substance.
Conjugate Acids-Base Pairs
HF (aq) + H2O (l)  H3O+(aq) + F- (aq)
acid
base
acid
base
SO42-(aq) + HCO3-(aq) HSO4-(aq) + CO32(aq)
base
acid
acid
base
Acid Strength
When an acid is placed in water, it dissociates as it
donates its H+ ions to water molecules:
HCl(aq) + H2O (l)  H3O+(aq) + Cl-(aq)
A strong acid is one which completely dissociates.
This means it has a LARGE ionization constant, Ka.
(the equilibrium constant for this reaction)
It is a very good electrolyte.
Relative Acid Strength
Common Strong Acids
• Hydrochloric acid, HCl
• Nitric acid, HNO3
• Hydrobromic acid, HBr
• Hydroiodic acid, HI
• Sulfuric acid, H2SO4
(only the first H+ dissociates strongly)
Common Weak Acids
•
•
•
•
Hydrofluoric acid, HF
Acetic acid, CH3COOH
Nitrous acid, HNO2
Benzoic acid, C6H5COOH
Ka = 6.8 x 10-4
Ka = 1.8 x 10-5
Ka = 4.5 x 10-4
Ka = 6.5 x 10-5
Rank these acids in order of increasing strength.
Write a dissociation reaction for each acid.
Write the Ka expression for each acid.
Weak Acids
Which of these
"molecular" pictures
best represents a
concentrated
solution of the weak
acid HA,
with Ka = 1 x 10-5?
Explain your choice.
Polyprotic Acids
• A polyprotic acid is
one that has more than
one H+ to donate:
Monoprotic acids:
•Polyprotic acids dissociate
“stepwise”:
e.g. Sulfuric Acid:
H2SO4 + H2O  H3O+ + HSO4-
HF, HCl, HNO3
Diprotic acids:
HSO4- + H2O  H3O+ + SO42-
H2SO4, H2CO3
Triprotic acid:
H3PO4
Write the 3 steps in the dissociation of
phosphoric acid! Write the Ka
expression for each step!
Dissociation Constants - Polyprotic Acids
Amphoteric Substances
• An amphoteric substance is a substance
which can behave as either an acid or a
base.
eg. HNO3 (aq) + H2O (l)  H3O+(aq) + NO3- (aq)
NH3 (aq) + H2O (aq)  NH4+ (aq) + OH- (aq)
To be amphoteric, a substance must have a hydrogen ion to donate.
Most amphoteric substances are negatively charged.
(e.g. HCO3- , H2PO4- etc…)
The pH Scale
• Acids produce H+ ions in water that immediately react
with water to form hydronium ions, H3O+.
• The concentration of H+ (or H3O+) in a solution is a
measure of the solution’s acidity.
• pH is one way to express the hydronium concentration.
pH = - log [H3O+]
When [H3O+] = 0.010 M in a solution, the pH = 2.00.
When the pH of a solution is 8.00, [H3O+] = 1.0 x 10-8 M.
What is [H3O+] when the pH = 3.25?
The pH Scale (continued)
Acidic, Basic, or Neutral?
Auto-Ionization of Water
• We’ve seen that water is amphoteric. In pure water, a very
small fraction of the water molecules undergo an acid-base
reaction with each other:
H2O + H2O  H3O+ + OH-
• At 25°C, the equilibrium constant (Kw) for this reaction is very
small.
Kw = 1.0 x 10-14 =[H3O+][OH-]
• A simple equilibrium calculation shows that in pure water at
25°C, [H3O+] = [OH-] = 1.0 x 10-7 M.
• Thus, at 25°C, the pH of pure water is 7.00 (and pOH = 7.00).
Ion product constant @ 25oC:
Kw=[H3O+][OH-]= 1.0 X 10-14
A Logarithm Lesson
Kw = 1.0 x 10-14 =[H3O+][OH-]
- log (1.0 x 10-14) = - log ([H3O+][OH-])
14.00 = -log [H3O+] + -log[OH-]
14.00 = pH + pOH
pH Practice
[H3O+]
[OH-]
pH
pOH
0.015 M
8.41
2.16
3.7 x 10-8 M
Calculate the pH of 0.10-M HNO3 solution.
HNO3 + H2O  H3O+ + NO3HNO3 is a strong
acid - Ka is large!
It dissociates
completely.
0.10 M
- 0.10
----
---+ 0.10
0.10 M
---+ 0.10
0.10 M
Thus, [H3O+] = 0.10 M
pH = - log (0.010) = 2.00
Why do we let [H3O+] = 0 at the start? Isn’t there some H3O+ in pure water??
Are there any HNO3 molecules in the solution??
Would this solution be a good electrolyte??
Calculate the pH of 0.15-M HF solution.
HF is a weak acid its Ka is 6.7 x 10-4.
It doesn’t
dissociate
completely.
K a  6.7 x10 4 
6.7 x10
4
HF + H2O  H3O+ + F0.15 M
-x
0.15 - x

---+x
x

[ H 3O ][ F ]
[ HF ]
2
x

0.15  x
Since Ka is small let’s assume
that x << 0.15 M
---+x
x
2
x
6.7 x10 4 
0.15
x1 = 0.010 M
x2 = 0.0097 M
x3 = 0.0097 M = [H3O+]
pH = - log (0.0097) = 2.01
Percent Dissocation (Ionization)
• Calculate the %-Dissociation of the HF
molecules in the 0.15-M solution of HF.
Of the 0.15 mol/L HF molecules, we know
that 0.0097 mol/L have dissociated. Thus…
0
.
0097
% Dissociation =
x100 = 6.5%
0.15
Does it make sense
that the %-dissociation
is so small for HF?
% Dissociation - Effect of Concentration
Calculate the pH of 0.20M H2SO4 (aq)
H2SO4 is a diprotic acid
- it dissociates in 2 steps.
Ka for the first
dissociation is LARGE.
Note that the second step
starts with both HSO4and H3O+ at 0.20 M.
The second dissociation
is weak - Ka2 = 0.012.
H2SO4 + H2O  H3O+ + HSO40.20 M
- 0.20
----
---+ 0.20
0.20 M
---+ 0.20
0.20 M
HSO4- + H2O  H3O+ + SO420.20 M
-x
0.20 - x
0.20 M
+x
0.20 + x
---+x
x
pH of 0.20M H2SO4 (aq) (cont’d)
Ka2  0.012 
Since the Ka is small for
the second dissociation,
we’ll assume that
x << 0.20 M
Note that the assumption
applies TWICE - in the
numerator and
denominator.
Find the TOTAL [H3O+]
from BOTH steps of the
dissociation in order to
find the pH.
0.012 
[ H 3O  ][ SO4 2 ]
[ HSO4  ]
(0.20  x )( x )
0.20  x
(0.20)( x )
0.012 
 x1
0.20
x2 = 0.011 M
This is the [H3O+] from the 2nd dissociation.
[H3O+] for both steps = 0.20 + 0.011 = 0.21 M
pH = - log (0.21) = 0.68
Other Polyprotic Acids
• For most polyprotic acids (H2CO3, H3PO4 etc…) the
first dissociation is the only one that contributes
significantly to the [H3O+] of the solution.
• This is because Ka1 >> Ka2 , Ka3
• Thus, the first step is the only one that needs to be
considered and the acids can be treated as weak
monoprotic acids.
Find the pH of 0.10 M H3PO4
H3PO4 is a triprotic acid
- it dissociates in 3 steps.
Ka for the first
dissociation is small.
Ka1 = 7.5 x 10 -3
H3PO4 + H2O  H3O+ + H2PO420.10 M
-x
0.10 - x


[
H
O
][
H
PO
]
3
3
2
4
K a  7.5 x10 
[ H 3 PO4 ]
2
x
7.5 x103 
0.10  x
2
x
7.5 x10 3 
0.10
Since Ka is small
assume that
x << 0.10 M
---+x
x
---+x
x
x1 = 0.027 M
x2 = 0.023 M
x3 = 0.024 M = [H3O+]
Since Ka1 >> Ka2, Ka3
assume these steps are negligible
pH = - log (0.024) = 1.62
Strong Bases
• Strong bases are metal
hydroxides:
•
•
•
•
•
NaOH  Na+ + OHKOH  K+ + OHLiOH  Li+ + OHCa(OH)2  Ca2+ + 2 OHSr(OH)2  Sr2+ + 2 OH-
• “Strong” refers to the
dissociation of the bases.
• Strong bases dissociate
completely into ions.
• They are good electrolytes
• Notice that hydroxide ions
are produced in a basic
solution
pH of Strong Base Solutions
• Calculating the pH of a strong base solution is relatively
easy because strong bases completely dissociate in water.
• For example, calculate the pH of 0.010-M NaOH solution.
NaOH  Na+ + OH-
Strong Bases
dissociate 100 %
In a 0.010-M solution, [Na+] = [OH-] = 0.010 M
Thus, pOH = - log [OH-] = - log (0.010) = 2.00
And pH = 14.00 - pOH = 14.00 - 2.00 = 12.00
Weak Bases
As you can see, most of the weak bases that are commonly
encountered are derived from ammonia, NH3.
Note that in addition to the weak bases listed below, every weak
acid has a corresponding weak base.
Calculate the pH of 0.25-M NH3 solution.
NH3 is a weak base its Kb is 1.8 x 10 -5.
It doesn’t dissociate
completely.
K b  1.8 x10 5 
NH3 + H2O  NH4+ + OH0.25 M
-x
0.25 - x

---+x
x

[ NH 4 ][OH ]
[ NH 3 ]
---+x
x
2
x
1.8 x10 5 
0.25
2
x
1.8 x10 5 
0.25  x
x1 = 0.0021 M
x2 = 0.0021 M = [OH -]
Since Kb is small let’s assume
that x << 0.25 M
pOH = - log (0.0021) = 2.68
pH = 14.00 - pOH = 11.32
Relationship between Ka and Kb
Consider the dissociation of HF and its conjugate base in water:
HF + H2O  H3O+ + F[ H 3O  ][ F  ]
Ka 
[ HF ]
F- + H2O  HF + OHKb 
[ HF ][OH  ]
[F  ]
[ H 3O  ][ F  ] [ HF ][OH  ]


K a ·K b 
·

[
H
O
][
OH
]  Kw
3

[ HF ]
[F ]
Ka · Kb = Kw
Relative Strengths of Conjugate Acid-Base Pairs
Acid-Base Properties of Ions
pH of Salt Solutions
• Salts are made of cations
and anions.
• Salt solutions will be acidic, basic
or neutral depending on the ions
that make up the salt.
• A few cations may be
acidic (NH4+)
e.g. Sodium Chloride
NaCl  Na+ + Cl-
• Many anions may be
basic (F-, SO42- etc…)
Na+ = neither acid nor base
= neutral
Cl- = conj. base of HCl (strong acid)
= Kb too small to affect pH
= neutral
Thus, NaCl is a neutral salt.
• Some ions are
amphoteric (HCO3-)
pH of Salt Solutions (cont’d)
e.g. Ammonium nitrate, NH4NO3
+
NH4 = conj. acid of ammonia,
NH3 (weak base)
= ACIDIC
NO3- = conj. base of nitric acid,
HNO3 (strong acid)
= Kb too small to affect
pH
= NEUTRAL
Thus, NH4NO3 is ACIDIC
e.g. Sodium Bicarbonate NaHCO3
Na+ = NEUTRAL
HCO3- = AMPHOTERIC
Acid: Ka = 5.6 x 10-11
Base: Kb = Kw ÷ Ka (of H2CO3)
= 2.3 x 10-8
Since its Kb > Ka , HCO3- will
be BASIC
Thus, NaHCO3 is BASIC
Acid-Base Titrations
• A titration is a volumetric technique often
involving acid-base neutralizations.
• It often involves reacting one solution of
known concentration, with another of
unknown concentration.
• Let’s take a look at how a titration is
performed.
• Then we’ll look at calculations.
Titration Description
• A titration is a procedure for quantitative analysis.
• In a titration two reagents are mixed, one with a
known concentration and one with an unknown
concentration.
• An acid-base indicator is added to determine when the
two reagents have reacted essentially completely
• At the end of the titration the unknown solution's
concentration (or molar mass) can be calculated.
Description
• One reagent is a solution and
is added from a buret.
• This solution is called the
titrant.
• The solution from the buret is
added to a flask that contains
either a measured volume of a
solution or a weighed quantity
of solid that has been
dissolved.
Preparing a Buret
• Rinse a clean buret three with
~ 5 mL portions of distilled
water.
• Repeat the rinse using the
titrant (the solution that will be
added to the flask).
• Allow the rinse to drain
through the buret stopcock to
rinse the tip of the buret.
• Discard the rinse portions.
Fill the Buret
• Clamp the buret into place
on the buret stand.
• Use a funnel to fill the buret
with the titrant.
• Allow some titrant to drain
out the tip. Remove any air
bubbles in the stopcock tip.
Reading the Volume in the Buret
• Make sure the volume
reading in the buret is at or
below the 0.00 mL mark.
• Record the volume reading
on the buret.
• Be sure to read the bottom
of the meniscus.
• Record all certain digits plus
one uncertain digit.
suggest a better way to
fill the buret!
Prepare the Sample
• If the sample to be titrated is a solution, pipet the desired
volume into an Erlenmeyer flask. Record the exact volume
transferred. Dilute the sample with a small portion of
distilled water (about 10 to 20 mL).
• If the sample is a solid, obtain the desired mass, and
dissolve it in about 25 mL of distilled water in an
Erlenmeyer flask. Record the exact mass of sample used.
• A few drops of chemical indicator is usually added to signal
the endpoint of the titration with a colour change.
• The choice of indicator depends on the pH at the end of the
neutralization reaction (consider the salt produced!)
Add the Titrant
• At the beginning of the titration,
titrant may be added quickly since
the indicator color disappears
rapidly.
• When the color persists for longer
periods of time, add titrant more
slowly (a drop or less at a time).
Adding the Titrant (cont’d)
• Be sure to mix the two reactant solutions thoroughly
by swirling the flask as the titrant is added.
• If solution splashes up to the side of the flask, you
can use distilled water to wash it back into the
solution.
• Placing the flask on a piece of white paper will
often help you observe the first appearance of color
change.
Determining the Endpoint
• Overshooting the endpoint
of the titration by adding too
much titrant is a common
error.
• The endpoint for this
titration is reached when
you reach a pale color that
persists throughout the
solution for several seconds.
• Record the volume in the
buret.
Endpoint Problems
Suggest what went wrong in each titration below!
Titration Calculations
Step 1: Write a balanced equation for the neutralization reaction.
Step 2: Write a mathematical equation using the mole-ratio for
the acid and base.
Step 3: For solutions, use “CV” to represent moles. For solids,
use “m/M” to represent moles.
Step 4: Substitute the given information into the equation.
Step 5: Solve for the unknown quantity.
Titration Calculations - example
A 25.00-mL sample of carbonic acid, H2CO3 , required 22.34 mL
of 0.126-M potassium hydroxide, KOH, for complete
neutralization.
Calculate the concentration of the carbonic acid solution.
H2CO3 + 2 KOH  K2CO3 + 2 H2O
moles KOH = 2 (moles H2CO3)
CbVb = 2 CaVa
CV
(0.126 M )(22.34 mL)
Ca  b b 
 0.0563 M H 2CO3
2 Va
2 (25.00 mL)
Titration Calculations - example
An unknown monoprotic acid, HA, was titrated using 0.0981-M
NaOH. A 0.214-g sample of the acid required 32.70 mL of the
base for complete neutralization.
Calculate the molar mass of the unknown acid.
HA + NaOH  NaA + H2O
moles HA = moles NaOH
ma
 CbVb
Ma
Ma 
ma
0.214 g

 66.7 g / mol
CbVb (0.0981 M )(0.03270 L)