Transcript Chapter 16 - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
www.cengage.com/chemistry/reger
Chapter 16
Reactions Between
Acids and Bases
Titration of Strong Acids and Bases
• Titration: a method used to determine
the concentration of a substance known
as the analyte by adding another
substance, the titrant, which reacts in a
known manner with the analyte.
•
analyte + titrant → products
Laboratory Titrations
(a) A known volume of acid is measured into a flask.
(b) Standard base is added from a buret.
(c) The endpoint is indicated by a color change.
(d) The volume of base is recorded.
Titration: Strong Acid and Base
• Titration Curve: a graph of pH of a
solution as titrant is added.
• For a titration of a strong acid with a
strong base, the pH will start at a very
low value and stay low as long as strong
acid is still present.
Titration: Strong Acid and Base
• The pH will rise sharply to 7 at the
equivalence point, where the acid and
base are present in stoichiometrically
equivalent amounts.
• After excess strong base has been
added, the pH levels off at a high value.
Titration Curve for a Strong Acid with a Strong
Base
The Equivalence Point in a Titration
• Calculate the equivalence point in the
titration of 20.00 mL of 0.1252 M HCl
with 0.1008 M NaOH.
HCl + NaOH  H2O + NaCl
Test Your Skill
• Calculate the equivalence point in the
titration of 40.00 mL of 0.2387 M HNO3
with 0.3255 M NaOH.
Units of Millimoles
• Millimole: one thousandth of a mole.
• If molarity is expressed in moles/liter (M)
and volume in milliliters (mL), n will be in
millimoles (mmol).
moles
• n  liter  milliliters  millimoles
Liters cancel but the milli- multiplier
remains.
.
Calculating a Titration Curve
• Calculate the pH in the titration of 20.0 mL of
0.125 M HCl with 0.250 M NaOH after 0,
2.00, 10.00, and 20.00 mL base are added.
Test Your Skill
• Calculate the pH in the titration of 20.0
mL of 0.125 M HCl with 0.250 M NaOH
after 5.00 mL and 12.00 mL base are
added.
Calculating a Titration Curve
Strong Base + Strong Acid Curve
• Titration curve of 50.00 mL 0.500 M KOH
with 1.00 M HCl.
Stoichiometry and Titration Curves
• 10 mL of two different 0.100 M acids
titrated with 0.100 M NaOH.
Estimating the pH of Mixtures
• Fill in first 3 three lines of sRf table.
• Look at the final solution (f-line).
• If a strong acid is present, the solution will be
strongly acidic.
• If a strong base is present the solution will be
strongly basic.
Solution
Estimate of pH
• If only water is
Strongly acidic
1
present, the
Neutral
7
solution will be
Strongly basic
13
neutral.
Buffers
• Buffer: a solution that resists changes in
pH.
• A buffer is a mixture of a weak acid or
base and its conjugate partner.
•
HA + OH- → H2O + AWeak acid reacts with any added OH-.
•
A- + H3O → HA + H2O
Weak base reacts with any added H3O+.
The pH of a Buffer System
• For the chemical reaction
HA + H2O → H3O+ + A
-
K a [HA]
[H3 O ][A ]

or [H3 O ] 
Ka 
[HA]
[A ]
take the - log (p - function) to obtain
-
[A ]
• pH  pK a  log
[HA]
The pH of a Buffer System
• Calculate the pH of a solution of 0.50 M
-10
HCN and 0.20 M NaCN, Ka = 4.9 x 10 .
The pH of a Buffer System
• Calculate the pH of a solution of 0.40 M
NH3 and 0.10 M NH4Cl. For NH3 Kb =
-5
1.8 x 10 .
Test Your Skill
• Calculate the pH of a buffer that is 0.25
M HCN and 0.15 M NaCN,
-10
Ka = 4.9 x 10 .
The Composition of a pH Buffer
• Calculate the amount of sodium acetate
that must be added to 250 mL of 0.16 M
acetic acid in order to prepare a pH 4.68
-5
buffer. Ka = 1.8 x 10
Test Your Skill
• How many moles of NaCN should be
added to 100 mL of 0.25 M HCN to
prepare a buffer with pH = 9.40?
-10
Ka = 4.9 x 10 .
Determining the Response of a Buffer to
Added Acid or Base
• Calculate the initial and final pH when
10 mL of 0.100 M HCl is added to (a)
100 mL of water, and (b) 100 mL of a
buffer which is 1.50 M CH3COOH and
1.20 M CH3COONa.
Test Your Skill
• Calculate the final pH when 10 mL of
0.100 M NaOH is added to 100 mL of a
buffer which is 1.50 M CH3COOH and
1.25 M CH3COONa.
Qualitative Aspects: Titration: Weak Acid +
Strong Base
Before any base added.
(a) Part way to equivalence point.
(b) Equivalence point.
(c) Beyond equivalence point.
Titration: Weak Acid + Strong Base
• HA + OH-  A- + H2O
• (a) Before any base is added the
solution is a weak acid has a low pH.
• Estimated pH = 3
pH 2-4 is typical
–Depends on the concentration of the acid.
–Depends on the value of Ka.
Titration: Weak Acid + Strong Base
• HA + OH-  A- + H2O
• (b) After some base is added, but before
the equivalence point is reached.
• The solution is a mixture of the weak
acid HA and its conjugate base A ;
therefore, the solution is a buffer.
• The estimated pH is equal to pKa.
Titration: Weak Acid + Strong Base
• HA + OH-  A- + H2O
• (c) At the equivalence point the solution
is salt of A , all the HA having been
consumed by the stoichiometric amount
of OH-.
• A- is the weak conjugate base of HA.
• The estimated pH is 10.
Titration: Weak Acid + Strong Base
• HA + OH-  A- + H2O
• (d) After excess strong base is added
OH- is in excess.
• The estimated pH is 13.
pH Estimates
Solution
Strongly acidic
Weakly acidic
Neutral
Weakly basic
Strongly basic
Buffer
(acidic to basic)
Estimate of pH
1
3
7
11
13
pKa
4-10 typical
Titration: Weak Acid + Strong Base
Titration Curves for Acids of Different
Strengths
Calculating the Titration Curve for a Weak
Acid
• Calculate the pH in the titration of 20.00
mL of 0.500 M formic acid (HCOOH
Ka=1.8 x 10-4) with 0.500 M NaOH after
0, 10.00, 20.00, and 30.00 mL of base
have been added.
• The titration reaction is
HCOOH + OH-  HCOO- + H2O
Titration of 25.00 mL of 0.500 M Formic Acid
with 0.500 M NaOH
Test Your Skill
• Calculate the pH in the titration of 12.00
mL of 0.100 M HOCl with 0.200 M NaOH
after 0, 3.00, 6.00, and 9.00 mL of base
have been added.
Titration of 20.00 mL of 0.500 M
Methylamine with 0.500 M HCl
pH Indicators
• Indicator: a substance that changes
color at the endpoint of a titration.
• pH indicators are weak acids or bases
whose conjugate species are a different
color.
pH Indicators
• HIn + H2O H3O+ + In
• KIn

[H3O ][In ]

[HIn]
• pKIn = -log(KIn)
pH Indicators
• When pH is lower than pKIn, the indicator
will be in the acid form.
• When pH is greater than pKIn, the
indicator will be in the base form.
• An indicator should be chosen which
changes at or just beyond the
equivalence point.
Properties of Indicators
Name
Acid Color Base Color pH Range pKIn
Thymol blue*
Red
Yellow
1.2–2.8
1.6
Methyl orange
Red
Yellow
3.1–4.4
3.5
Methyl red
Red
Yellow
4.2–6.3
5.0
Bromthymol blue Yellow
Blue
6.2–7.6
7.3
Phenolphthalein
Clear
Pink
8.3–10.0
8.7
Thymol blue*
Yellow
Blue
8.0–9.6
9.2
* Thymol blue is polyprotic and has three color forms.
Titration Curves for Strong and Weak Acids
Polyprotic Acids
• Polyprotic acids provide more than one
proton when they ionize.
• Polyprotic acids ionize in a stepwise
manner.
H2A + H2O ⇌ H3O+ + HA- Step 1
HA- + H2O ⇌ H3O+ + A2- Step 2
Polyprotic Acids
• There is a separate acid ionization
constant for each step
H2A + H2O ⇌ H3O+ + HA- Step 1


[H3 O ][HA ]
K a1 
[H2 A]
HA- + H2O ⇌ H3O+ + A2K a2 

2-
[H3 O ][A ]
[HA - ]
Step 2
Polyprotic Acids
-
• HA is the conjugate base of H2A, so it is
a weaker acid than H2A.
• Ka1 is always larger than Ka2.
• For triprotic acids (such as H3PO4), Ka2 is
always larger than Ka3.
Calculating Concentrations of Species in
Polyprotic Acid Solutions
• When successive Ka values differ by a
factor of 1000 or more, each step can be
assumed to be essentially unaffected by
the occurrence of the subsequent step.
Concentrations of Species in Polyprotic Acid
Solutions
• Calculate the concentrations of all
species in 0.250 M malonic acid,
-2
Ka = 1.6 x 10 .
• Consider the first ionization and solve
by usual approach.
H2C3H2O4 + H2O ⇌ HC3H2O4- + H3O+
Concentrations of Species in Polyprotic Acid
Solutions
• The second step is needed only to
calculate the concentration of C3H2O42because the concentration of H3O+ is
determined by the first step.
• You can ignore the effect of the second
step on the pH because the Ka1 is so
much larger than Ka2.
Test Your Skill
• Calculate the pH of a 0.040 M solution of
ascorbic acid. (Ka1 = 8.0 x 10-5,
Ka2 = 1.6 x 10-12)
Amphoteric Species
• Amphoteric: having both acidic and
basic properties.
• Conjugate bases of weak polyprotic
acids are amphoteric.
• The hydrogen oxalate ion, HC2O4-, is a
-4
weak acid (Ka2 = 1.6 ×10 ).
•
HC2O4- + H2O ⇌ C2O42- + H3O+
Amphoteric Species
• Weak Acid
• The hydrogen oxalate
ion, HC2O4-, is a weak
acid.
• Weak Base
• The hydrogen oxalate ion,
HC2O4-, can also act as a
weak base.
HC2O4- + H2O ⇌ C2O42- + H3O+ HC2O4- + H2O → H2C2O4 + OH-
• Ka2 = 1.6 x 10-4
• Kb = Kw/Ka1 = 1.0 x 10-14 /
5.6 x 10-2
• Kb = 1.9 x 10-13
• Since Ka > Kb, the ion will act as a weak acid in water.
• When comparing Ka to Kb note that Ka is Ka2 and
Kb is Kw/Ka1.
Test Your Skill
• Ka for the hydrogen malonate ion,
-6
HC3H2O4 , is 2.1 x 10 . Is a solution of
sodium hydrogen malonate acidic or
basic?
Factors That Influence Solubility
• The pH affects the solubility of salts
of weak acids.
• Complex ion formation affects the
solubility of salts of transition metal
cations.
Salts of Anions of Weak Acids
• The solubility of salts of anions of
weak acids is enhanced by lowering
the pH.
• Cd(CN)2(s) ⇌ Cd2+(aq) + 2CN-(aq)
-8
Ksp = 1.0 x 10
• Adding acid reduces [CN-] in solution,
by the reaction
• H3O+(aq) + CN-(aq) ⇌ HCN(aq) + H2O(l)
Salts of Transition Metal Cations
• Transition metal cations form complexes
with Lewis bases such as H2O, NH3, or
OH .
• Formation of a complex reduces the
concentration of metal ion and increases
the solubility of the salt.
Solubility of Amphoteric Species
• Amphoteric species, such as Be(OH)2,
Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3,
Ni(OH)2, Cu(OH)2, Zn(OH)2, and
Cd(OH)2 react with acid or base to form
the soluble metal ion or complex ions
+
x+
M(OH)x + xH  M + xH2O x = 2,3
y+
M(OH)x + yOH  M(OH)x x = 2,3,
y = 1,2