Transcript Buffers and Titration

Chapter 17
Sections 1-3
Common ions,
Buffers and Titration
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Sample Exercise 17.1 Calculating the pH When a Common Ion
is Involved
What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate
to enough water to make 1.0 L of solution?
Practice Exercise
Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2, Ka = 4.5  104) and 0.10 M
potassium nitrite (KNO2).
Aqueous
Equilibria
Sample Exercise 17.2 Calculating Ion Concentrations When a
Common Ion is Involved
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Practice Exercise
Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid
(HCOOH, Ka = 1.8  104) and 0.10 M in HNO3.
Aqueous
Equilibria
(a)
A.
B.
C.
D.
There are no spectator ions.
Cl–
Both NH4+ and Cl–
NH4+
(b)
A.
B.
C.
D.
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
NH4+(aq) + Cl-(aq)  NH3(aq) + HCl(l)
H3O+(aq) + OH–(aq)  2H2O(l)
NH4+(aq) + H2O(l)  H3O+(aq) + NH3(aq)
Aqueous
Equilibria
Buffers
• Buffers are solutions
of a weak conjugate
acid–base pair.
• They are particularly
resistant to pH
changes, even when
strong acid or base
is added.
Aqueous
Equilibria
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A. All listed pairs will not function as buffers.
–
B. HCHO2 and CHO2 will not work as a buffer
–
because HCHO2 is a weak acid and CHO2 is a
spectator ion.
C. HCO3– and CO32– will not work as a buffer
because HCO3– is a weak acid and HCO3– is a
spectator ion.
D. HNO3 and NO3– will not work as a buffer
because HNO3 is a strong acid and NO3– is a
spectator ion.
Aqueous
Equilibria
Buffers
If a small amount of hydroxide is added to an equimolar solution of HF
in NaF, for example, the
HF reacts with the OH to make F and water.
Similarly, if acid is added, the F reacts with it to form HF and water.
Aqueous
Equilibria
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A. There is no reaction because C2H3O2– and
HC2H3O2 form a buffer.
B. The NaOH reacts with C2H3O2– converting
some of it into HC2H3O2.
C. The NaOH reacts with HC2H3O2 converting
some of it into C2H3O2–.
D. The NaOH is neutralized and all concentrations
(HC2H3O2 and C2H3O2–) remain unchanged. Aqueous
Equilibria
A. There is no reaction because C2H3O2– and
HC2H3O2 form a buffer.
B. The HCl reacts with C2H3O2– converting some
of it into HC2H3O2.
C. The HCl is neutralized and all concentrations
(HC2H3O2 and C2H3O2–) remain unchanged.
D. The HCl reacts with HC2H3O2 converting some
of it into C2H3O2–.
Aqueous
Equilibria
Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12 M in lactic acid [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in
sodium lactate [CH3CH(OH)COONa, or NaC3H5O3]? For lactic acid, Ka = 1.4  104.
Aqueous
Equilibria
Buffer Calculations
Consider the equilibrium constant
expression for the dissociation of a
generic acid, HA:
H 3O + + A 
HA + H2O
[H3O+] [A]
Ka =
[HA]
Aqueous
Equilibria
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Buffer Calculations
Rearranging slightly, this becomes
]
[A
Ka = [H3O+]
[HA]
Taking the negative log of both side, we get
]
[A
log Ka = log [H3O+] + log
[HA]
pKa
pH
base
acid
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Aqueous
Equilibria
Buffer Calculations
• So
[base]
pKa = pH  log
[acid]
• Rearranging, this becomes
[base]
pH = pKa + log
[acid]
• This is the Henderson–Hasselbalch
equation.
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Aqueous
Equilibria
Sample Exercise 17.3 Calculating the pH of a Buffer
Continued
Practice Exercise
Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium
benzoate. (Refer to Appendix D.)
Aqueous
Equilibria
Sample Exercise 17.4 Preparing a Buffer
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00?
(Assume that the addition of NH4Cl does not change the volume of the solution.)
Practice Exercise
Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of
benzoic acid (C6H5COOH) to produce a pH of 4.00.
Aqueous
Equilibria
pH Range
• The pH range is the range of pH values
over which a buffer system works
effectively.
• It is best to choose an acid with a pKa
close to the desired pH.
Aqueous
Equilibria
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A.
B.
C.
D.
HClO because it is a stronger weak acid. A salt
containing ClO– is also needed.
HNO2 because it is a stronger weak acid. A salt
containing NO2– is also needed.
HClO because its pKa is closer to pH = 7.0. A salt
containing ClO– is also needed.
HNO2 because its pKa is closer to pH = 7.0. A salt
containing NO2– is also needed.
Aqueous
Equilibria
When Strong Acids or Bases Are
Added to a Buffer
When strong acids or bases are added to a buffer, it is
safe to assume that all of the strong acid or base is
consumed in the reaction.
Aqueous
Equilibria
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Addition of Strong Acid or Base
to a Buffer
1. Determine how the neutralization
reaction affects the amounts of
the weak acid and its conjugate
base in solution.
2. Use equilibrium to determine the
new pH of the solution.
Aqueous
Equilibria
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Sample Exercise 17.5 Calculating pH Changes in Buffers
A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make
1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this
solution after 5.0 mL of 4.0 M NaOH (aq) solution is added. (b) For comparison, calculate the pH of a
solution made by adding 5.0 mL of 4.0 M NaOH( aq) solution to 1.000 L of pure water.
Practice Exercise
Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of
0.020 mol HCl
(b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.000 L of pure
Aqueous
water.
Equilibria
Titration
In this technique, a
known concentration of
base (or acid) is slowly
added to a solution of
acid (or base).
Aqueous
Equilibria
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A.
B.
C.
D.
No change
Increase
Decrease
Need volumes and molarities
to answer the question
Aqueous
Equilibria
Titration
A pH meter or indicators are used to
determine when the solution has reached
the equivalence point, at which the
stoichiometric amount of acid equals
that of base.
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Aqueous
Equilibria
Titration of a Strong Acid with a
Strong Base
From the start of the titration to
near the equivalence point, the
pH goes up slowly.
Just before (and after) the
equivalence point, the pH
increases rapidly.
At the equivalence point, moles
acid = moles base, and the
solution contains only water and
the salt
from the cation of the base and
the anion of the acid.
As more base is added, the
increase in pH again levels off. Aqueous
Equilibria
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A.
B.
C.
D.
100.00 mL
50.00 mL
25.00 mL
12.50 mL
Aqueous
Equilibria
Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong
Base Titration
Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to
50.0 mL of 0.100 M HCl solution.
Practice Exercise
Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have
been added to 25.0 mL of 0.100 M KOH solution.
Aqueous
Equilibria
A.
B.
C.
D.
pH = 4.2
pH = 7.0
pH = 8.2
pH = 9.8
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
• Unlike in the previous
case, the conjugate
base of the acid
affects the pH when it
is formed.
• At the equivalence
point the pH is >7.
• Phenolphthalein is
commonly used as an
indicator in these
titrations.
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Aqueous
Equilibria
A. The volume of base does
not change but the equivalence
point pH increases.
B. The volume of base
increases but the equivalence
point pH does not change.
C. The volume of base
decreases and the equivalence
point pH does not change.
D. The volume of base does
not change but the equivalence
point pH decreases.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
At each point below the equivalence point, the
pH of the solution during titration is determined
from the amounts of the acid and its conjugate
base present at that particular time.
Aqueous
Equilibria
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Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong
Base Titration
Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100
M CH3COOH (Ka = 1.8  105).
Practice Exercise
(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M
benzoic acid (C6H5COOH, Ka = 6.3  105. (b) Calculate the pH in the solution formed by adding 10.0
mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.
Aqueous
Equilibria
Sample Exercise 17.8 Calculating the pH at the Equivalence Point
Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M
NaOH.
Practice Exercise
Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C6H5COOH, Ka =
6.3  105) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
With weaker acids,
the initial pH is
higher and pH
changes near the
equivalence point
are more subtle.
Aqueous
Equilibria
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A. The volume of base
does not change but the
equivalence point pH
increases.
B. The volume of base
increases but the
equivalence point pH
does not change.
C. The volume of base
decrease sand the
equivalence point does
not change.
D. The volume of base
does not change but the
Aqueous
equivalence point pH Equilibria
increases.
Titrations of Polyprotic Acids
When one
titrates a
polyprotic acid
with a base
there is an
equivalence
point for each
dissociation.
Aqueous
Equilibria
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Titration of a Weak Base with a
Strong Acid
• The pH at the equivalence point in these
titrations is <7, so using phenolphthalein
would not be a good idea.
• Methyl red is the indicator of choice.
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Aqueous
Equilibria
A. No, methyl red does not change
color at equivalence point.
B. No methyl red’s color change is not
sharp and gives only a rough
estimate of the equivalence point
pH.
C. Yes, methyl red changes color in the
pH region between 4 and 6 in a
steeply sloping region.
D. Yes, methyl red changes color in the
pH region between 8–10 in a
steeply sloping region.
Aqueous
Equilibria
A.
B.
C.
D.
The nearly vertical equivalence point portion of the titration curve
is large for a weak acid-strong base titration, and fewer indicators
undergo their color change so quickly because the change is
difficult to monitor.
The nearly vertical equivalence point portion of the titration curve
is smaller for a weak acid-strong base titration, and fewer
indicators undergo their color change within this narrow range.
Many indicators do not change colors at the equivalence points
of weak acid-strong base titrations.
Equivalence points at pH’s other than 7.00 are difficult to
determine.
Aqueous
Equilibria