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Buffers/Titration
Aqueous Equilibria - I
Review hydrolysis of salts
Slide 3 / 113
1
The hydrolysis of a salt of a weak base and a strong acid
should give a solution that is __________.
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2
A NaCl
A weakly basic
B
A solution of one of the following is acidic. The
compound is _________.
neutral
B CH3COONa
C strongly basic
C NH4Cl
D weakly acidic
D NH3
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3
Which salt would undergo hydrolyzes to form an acidic
solution?
A KCl
B
NaCl
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4
Which substance when dissolved in water will produce
a solution with a pH greater than 7?
A CH3COOH
B
NaCl
C NH4Cl
C NaC2H3O2
D LiCl
D HCl
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5
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A basic solution would result from the hydrolysis of one
of the ions in this compound. The compound
is__________.
A NaNO3
B
6
A water solution of which compound will turn blue
litmus red?
A K2CO3
B
NH4Cl
NH4Cl
C NaOH
C CH3COONa
D NaCl
D CaCl2
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The Common-Ion Effect
The Common-Ion Effect
· Consider an aqueous solution of acetic acid:
+
-
CH3COOH(aq) + H2O(l) ↔ H3O (aq) + CH3COO (aq)
· If acetate ion is added to the solution, Le
Châtelier 's Principle predicts that the equilibrium
will shift to the left.
“The extent of ionization of a weak electrolyte is
decreased by adding to the solution a strong
electrolyte that has an ion in common with the
weak electrolyte.”
In the following Sample Problem, compare the pH
of a 0.20 M solution of HF with the pH of the same
solution after some NaF is added to it.
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The Common-Ion Effect
The Common-Ion Effect
SAMPLE PROBLEM #1
a) Calculate the pH of a 0.20 M HFsolution. The Ka for HF is
6.8 ´ 10−4.
Ka =
[H3O+][F-]
= 6.8 x 10-4
[HF]
6.8 x 10-4 =
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
x2
(0.20)
x2 = (0.20) (6.8 ´ 10−4)
H3O+(aq)
F−(aq)
0.20 M
0
0
-x
+x
+x
0.20-x
x
x
HF(aq)
Initial
Change
At Equilibrium
H2O(l)
x = 0.012
So,
[H+] = [F-] = 0.012
and
pH = 1.93
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The Common-Ion Effect
The Common-Ion Effect
b) Calculate the pH of a for the same 0.20 M HF solution
that also contains 0.10 M NaF.
SAMPLE PROBLEM #1 (con't)
b) Calculate the pH of a for the same 0.20 M HF
solution that also contains 0.10 M NaF.
Ka for HF is 6.8 ´ 10−4.
· We use the same equation and the same Ka expression,
and set up a similar ICE chart.
· But because NaF is a soluble salt, it completely
dissociates, so F- is not initially zero.
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
HF(aq)
Ka =
Initial
Change
At Equilibrium
0
0.10 M
at Equilibrium
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The Common-Ion Effect
The Common-Ion Effect
Therefore,
x = [H3O+] = 1.4 x 10−3
+
H3O (aq)
F (aq)
0.20 M
0
0.10 M
-x
+x
+x
0.20-x
x
0.10 + x
H2O(l)
F−(aq)
Change
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HF(aq)
H3O+(aq)
0.20 M
Initial
[H3O+][F-]
= 6.8 x 10-4
[HF]
H2O(l)
−
and
pH = −log (1.4 x 10−3)
pH = 2.87
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The Common-Ion Effect
Ka =
6.8 x 10-4 =
(0.10)(x)
(0.20)
(0.20)(6.8 x 10-4)
(0.10)
=x
1.4 ´ 10−3 = x
Remember what the "x" is that you are solving for!
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The Common-Ion Effect
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
· Consider the two solutions we just examined in
Sample Problem #1.
· Compare the following:
Solution
Final [H3O+]
HF
0.12
HF and NaF
1.4 x 10-3
pH
· How do these results support Le Chatelier's Principle?
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The Common-Ion Effect
The Common-Ion Effect
The Ka for acetic acid, CH3COOH, is 1.8 x 10-5.
The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.
SAMPLE PROBLEM #2 - Answers
SAMPLE PROBLEM #2
a) Calculate the pH of a 0.30 M acetic acid solution.
a) Calculate the pH of a 0.30 M acetic acid, CH3COOH.
Ka =
b) Calculate the pH of a solution containing 0.30 M acetic
acid and 0.10 M sodium acetate.
1.8 x 10
-5
[H3O+][C2H3O2-]
[HC2H3O2]
=
= 1.8 x 10-5
x2
(0.30)
x = [H3O+] = ______________
and pH = ___________
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The Common-Ion Effect
The Common-Ion Effect
The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.
HC2H3O2 (aq) + H2O (l) ↔ C2H3O2 - (aq) + H3O+(aq)
SAMPLE PROBLEM #2 - Answers (con't)
b) Calculate the pH of a solution containing 0.30 M acetic acid
· Consider the two solutions we just examined in
Sample Problem #2.
and 0.10 M sodium acetate.
We use the same Ka expression, except now the acetate ion
concentration is not the same as [H3O+].
Ka =
1.8 x 10-5
+
-
[H3O ][C2H3O2 ]
[HC2H3O2]
· Compare the following:
Solution
= 1.8 x 10-5
[H3O+] = ______________
and
pH = ___________
HC2H3O2 and
NaC2H3O2
· How do these results support Le Chatelier's Principle?
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The Common-Ion Effect
SAMPLE PROBLEM #3
Calculate the pH of the following solutions:
a) 0.85 M nitrous acid, HNO2
b) 0.85 M HNO2 and 0.10 M potassium nitrite, KNO2
The Ka for nitrous acid is 4.5 x 10-4.
Answers:
a)
b)
pH
HC2H3O2
[H3O+] (0.10)
=
(0.30)
So now,
-
[OH ]
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7
The ionization of HF will be decreased by the
addition of:
A NaCl
B NaF
C HCl
D Both A and B
E Both B and C
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8
The dissociation of Al(OH)3 will be decreased by
the addition of:
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9
Which of the following substances will not
decrease the ionization of H3PO4?
A KOH
A K3PO4
B AlCl3
B HCl
C Mg(OH)2
C Na3PO4
D Both A and B
D None of them will decrease the ionization
E A, B and C
E All of them will decrease the ionization
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The Common-Ion Effect
The Common-ion effect can also be observed with weak
bases.
Consider the ionization of ammonia, NH3, which is a
weak base.
NH3 + H2O <---> NH4+ + OH-
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The Common-Ion Effect
Suppose a salt of the conjugate base is added to a
solution of ammonia. What effect would this have on
pH?
NH3 + H2O <---> NH4+ + OHAs with the previous Sample Problems, let us calculate
the pH of the following:
a) a 0.45 M solution of NH3
b) a 0.45 M solution of NH3 that also contains 0.15 M
NH4Cl, ammonium chloride -- a soluble salt that readily
yields NH4+ ions.
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The Common-Ion Effect
SAMPLE PROBLEM #4
Calculate the pH of the following solutions:
a) a 0.45 M solution of NH3
b) a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl
The Kb for NH3 is 1.8 x 10-5.
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The Common-Ion Effect
SAMPLE PROBLEM #4 - Answers
a) Calculate the pH of a 0.45 M solution of NH3.
The Kb for NH3 is 1.8 x 10-5.
NH3 + H2O <---> NH4+ + OH-
Kb = 1.8 x 10-5
=
1.8 x 10-5
=
[NH4+][OH- ]
[NH3]
x2
(0.45)
x = [OH-] = _____________
pH = _________________
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The Common-Ion Effect
The Common-Ion Effect
SAMPLE PROBLEM #4 - Answers (con't)
b) Calculate the pH of a 0.45 M solution of NH3 that also contains
-5
0.15 M NH4Cl. The Kb for NH3 is 1.8 x 10 .
NH3 + H2O <---> NH4+ + OH-
Kb = 1.8 x 10-5
=
1.8 x 10-5
=
[NH4+][OH- ]
[NH3]
SAMPLE PROBLEM #5
Calculate the pH of the following solutions:
a) 0.0750 M pyridine, C5H5N
b) 0.0750 M C5H5N and 0.0850 M pyridinium chloride, C5H5NHCl
Note that C5H5NHCl,( salt) dissociates into C5H5NH+ and Cl-.
The Kb for pyridine is 1.7 x 10-9.
(0.15) (x)
(0.45)
Answers:
a)
b)
-
So,
x = [OH ] = _____________
and
pH = _________________
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Buffers
Buffers
Buffers, or buffered
solutions, are special
mixtures that are
resistant to large pH
changes, even when
small amounts of strong
acid or strong base are
added.
· Buffers are able to resist large pH changes because they
contain both an acidic component and a basic component.
· Buffers are prepared by mixing either:
1) a weak acid and a salt containing its conjugate base
OR
2) a weak base and a salt containing its conjugate acid
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Buffers
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Buffers
Consider a buffer solution composed of HF and NaF.
Buffer after
addition of OH-
Consider a buffer solution composed of HF and NaF.
· The acidic component is HF. This component aids in the
neutralization of any strong base that is added to the buffer.
· The basic component is the fluoride ion, F-. This
component aids in the neutralization of any strong acid that is
added to the buffer.
HF
Buffer with equal
conc. of weak acid
and its conj. base
-
HF
F
OH-
F- + H2O <-- HF + OH-
HF
-
F
-
F
H+
H+ + F- --> HF
-
F
HF
If a small amount of KOH is added to this buffer, for
example, the HF reacts with the OH− to make __________.
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Buffers
Buffer with equal
conc. of weak acid
and its conj. base
Buffer after
addition of OH-
F-
HF
10
HF
OH
Buffer after
+
addition of H
HF
F-
-
H
-
-
H + F --> HF
F + H2O <-- HF + OH
A HA
B Na+
+
+
-
F-
If a buffer is made of HA (weak acid) and NaA,
which species will counteract the addition of a
strong base?
C A-
D Both HA and Na+
E Both HA and A-
Similarly, if a small amount of strong acid (H+) is
added, the F− reacts with it to form ____________.
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11
If a buffer is made of HNO2 and KNO2 , which
species will counteract the addition of a strong
acid?
A HNO2
B K+
D Both HNO2 and K+
Two important characteristics of buffers:
1- Its resistance to change in pH
2- Its buffer capacity
It depends on the amount of acid and base from which
the buffer is made.
E Both K+ and NO2-
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Buffer capacity and pH
Buffer A
0.1 M CH3COOH and
0.1 M CH3COONa
Total volume = 1.0 L
Total volume = 1.0 L
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12 Of the following solutions, which has the greatest buffering
capacity?
Buffer B
1.0 M CH3COOH and
1.0 M CH3COONa
Ka =
Buffer capacity and pH
Buffer capacity:
The amount of acid or base the buffer can neutralize
before the pH begins to change.
C NO2-
+
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A 0.1M NaF + 0.1MHF
B 0.5M NaF + 0.55M HF
C 0.8M NaF + 0.8M HF
-
[H3O ][C2H3O2 ]
[HC2H3O2]
Since
[C2H3O2- ]
[HC2H3O2]
D 0.2M NaF + 0.2M HF
=1
E 1M NaF + 1M HF
[ H3O+] = Ka
+
The above two combination of solution will have the same [H ].
The pH will depend on ka and the relative concentration of acid
and base only.
The buffering capacity will be higher for the first buffer. It
contains greater number of moles of CH3COOH and
NaCH3COO-.
It can neutralize more of the acid/base added.
The greater the amounts, greater resistance to pH change.
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Buffer Calculations
Buffer Calculations
Consider the equilibrium constant
expression for the dissociation of a generic
acid, HA:
HA + H2O ↔ H3O+ + A-
Ka =
Rearranging slightly, this becomes
Ka = [H3O+]
[A-]
[HA]
Taking the negative log of both sides, we get
[A-]
-log Ka = -log [H3O+] +(-log [HA] )
[H3O+][A-]
[HA]
pKa = pH - log [base]
[acid]
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Buffer Calculations
Buffer Calculations
pKa = pH - log [base]
[acid]
Since
Rearranging this:
pH = pKa + log [base]
[acid]
This is the Henderson-Hasselbalch equation
Buffer calculations typically require you to
calculate one or more of the following:
i) the pH of the buffer alone (Sample Prob #6)
ii) the pH of the buffer after a small amount of
strong base has been added (and neutralized)
(Sample Prob #7)
iii) the pH of the buffer after a small amount of
strong acid has been added (and neutralized)
(Sample Prob #8)
Because of the buffer system, the pH will not
change drastically; it is usually less than a
factor of 1.0 on the pH scale.
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Buffer Calculations
SAMPLE PROBLEM #6
What is the pH of a buffer that is 0.12 M in lactic
acid, CH3CH(OH)COOH, and 0.10 M in sodium
lactate? Ka for lactic acid is 1.4 ´ 10−4.
Method 1 - Traditional approach
Write the Ka expression. Solve for H+ and pH using ICE chart
HC3H5O3 (aq) + H2O(l) ↔ H3O+(aq) + C3H5O3-(aq)
0.12
0
0.10
-x
+x
+x
0.12-x
x
0.10+x
1.4 ´ 10
−4
= 0.10 x
0.12
x = 0.000168
PH =3.77
Buffer Calculations
SAMPLE PROBLEM #6
What is the pH of a buffer that is 0.12 M in lactic
acid, CH3CH(OH)COOH, and 0.10 M in sodium
lactate? Ka for lactic acid is 1.4 ´ 10−4.
Method 2 -Using the Henderson-Hasselbalch equation
Henderson–Hasselbalch Equation
pH = pKa + log [base]
[acid]
(0.10)
pH = -log (1.4 x 10-4) + log (0.12)
pH = 3.85 + (-0.08)
pH = 3.77
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13
The pH of a buffer solution that contains 0.818 M
acetic acid (Ka = 1.76x10-5) and 0.172 M sodium
acetate is __________.
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14
The pH of a buffer solution containing 0.145 M HF
and 0.183 M KF is _____. (Ka of HF is 3.5x10-4)
A 4.077
A 3.56
B 5.434
B 2.19
C 8.571
C 2.66
D 8.370
D 3.35
E 9.922
E 4.32
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Addition of Strong Acid or
Strong Base to a Buffer
Recall that buffers resist drastic changes in pH, even
when small amounts of either a strong acid or a strong
base are added to it.
When this happens, we assume that all of the strong
acid or strong base is consumed in the reaction. In other
words, all of the strong acid or base gets completely
neutralized by ONE of the components of the buffer
system.
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Addition of Strong Acid or
Strong Base to a Buffer
Remember that all of the strong acid or base is
consumed in the reaction.
-
+
X + H3O ⇒ HX + H2O
Add strong acid
Recalculate
HX and X-
-
Use Ka, [HX] and X
to calculate [H+]
Add strong base
HX + OH- ⇒ X- + H2O
stoichiometric calculation
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15
If a buffer is made of NH3 and NH4Cl, which
component of the buffer will neutralize a small
amount of HCl that is added?
A NH3
B NH4
+
equilibrium calculation
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16
If a buffer is made of HNO2 and KNO2 , which
component of the buffer will neutralize a small
amount of Ba(OH)2 that is added?
A HNO2
+
B K
C Cl-
C NO2-
D Both A and B
D Ba2+
E Both B and C
E Both K+ and NO2-
pH
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Addition of Strong Acid or
Strong Base to a Buffer
If a buffer is made of NH3 and NH4Cl, which
component will neutralize any KOH that is added?
OH -
17
Slide 56 / 113
.02
M
A NH3
pH =4.80
ad
d0
B NH4+
C Cl-
Buffer
D Both HNO2 and K+
0.3M HC2H3O2 pH= 4.74
0.3M NaC2H3O2
+
H
2M
0.0
add
E Both K+ and NO2-
· Determine how the
neutralization reaction affects
the amounts of the weak acid
and its conjugate base in
solution.
· Use the Henderson–
Hasselbalch equation to
determine the new pH of the
solution.
pH= 4.68
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Calculating pH Changes in Buffers
SAMPLE PROBLEM #7
A buffer is made by adding 0.300 mol HC2H3O2
and 0.300 mol NaC2H3O2 to enough water to make
1.00 L of solution. The pH of the buffer is 4.74.
Calculate the pH of this solution after 0.020 mol of
NaOH is added. Assume volume change is
negligible.
Calculating pH Changes in Buffers
Problem-solving strategy
Phase 1 - stoichiometric neutralization
· Identify the added substance.
· Identify the component of the buffer that will neutralize the
added substance.
· Write down the equation for nutralization.
· Solve for the new concentration of the buffer components.
Phase 2 - equilibrium
· Write the relevant dissociation equation.
· Create the ICE chart with the new [M] values(phase1) of acid
and base components of the buffer.
· Use Henderson-Hasselbalch equation to solve for pH.
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Calculating pH Changes in Buffers
SAMPLE PROBLEM #7 - Answer
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol
NaC2H3O2 to enough water to make 1.00 L of solution. The pH
of the buffer is 4.74. Calculate the pH of this solution after
0.020 mol of NaOH is added.
Phase 1 - Neutralization
· strong acid or a strong base added ?
· Which component of the buffer will neutralize the added
substance?
HA(aq)
acid
Before
0.3
OH-(aq)
H2O(aq)
added
substance
Calculating pH Changes in Buffers
SAMPLE PROBLEM #7 - Answer (con't)
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol
NaC2H3O2 to enough water to make 1.00 L of solution. The pH of
the buffer is 4.74. Calculate the pH of this solution after 0.020 mol
of NaOH is added.
Phase 2 - equilibrium
· Write the relevant dissociation equation.
· Make a ICE chart using the amounts from Phase 1.
A−(aq)
base
HA(aq)
H2O(l)
H3O+(aq)
A−(aq)
0.02
0.3
Initial
0.28 M
0
0.32 M
Neutralization -0.02
-0.02
0.3+0.02
Change
-x
+x
+x
After
0.0
0.32
At Equilibrium 0.28-x
x
0.32+x
0.28
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Calculating pH Changes in Buffers
Use the quantities from the ICE chart to calculate pH.
You will need to look up the Ka value.
Ka =
[H3O+][A−]
[HA]
1.8 x 10-5 =
(x) 0.32)
(0.28)
So,
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Calculating pH Changes in Buffers
Alternatively, you can use the Henderson-Hasselbalch equation to
calculate the new pH:
pH = pKa + log [base]
[acid]
pH = - log (1.8 x 10-5) + log
(0.320)
(0.280)
pH = 4.74 + 0.06
pH = 4.80
x = [H3O+] = _______________
and pH = ___________
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Calculating pH Changes in Buffers
pH Range for Buffers
SAMPLE PROBLEM #8
A buffer is made by adding 0.140 mol cyanic acid, HCNO, and
0.110 mol potassium cyanate, KCNO to enough water to make 1.00 L
of solution. Assume volume changes are negligible.
Ka for cyanic acid = 3.5 x 10-4
· The pH range is the range of pH values over which
a buffer system works effectively.
a) Calculate the pH of the buffer.
b) Calculate the pH of the buffer after the addition of 0.015 mol KOH.
c) Calculate the pH of the buffer after the addition of 0.018 mol HNO3.
Answers
· It is best to choose an acid with a pKa close to the
desired pH.
· After studying titration graphs, you will see why
buffers work best when pKa is close to the desired
pH.
a) 3.351
b) 3.46
c) 3.22
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Calculating pH Changes in Buffers
Slide 66 / 113
Calculating pH Changes in Buffers
A buffer is made by adding 15.0 g ammonia, NH3, and 55.0 g
ammonium chloride, NH4Cl. to enough water to make 1.00 L of solution.
Kb for NH3 = 1.8 x 10-5. Assume volume changes are negligible.
SAMPLE PROBLEM #10
A buffer is made by adding 25.0 g ammonia, NH3, and 45.0 g
ammonium chloride, NH4Cl. to enough water to make 1.75 L of
solution. Kb for NH3 = 1.8 x 10-5. Assume volume changes are
negligible.
a) Calculate the pH of the buffer.
b) Calculate the pH of the buffer after the addition of 0.013 mol HClO4.
c) Calculate the pH of the buffer after the addition of 0.015 mol KOH.
a) Calculate the pH of the buffer.
b) Calculate the pH of the buffer after the addition of 0.011 mol NaOH.
c) Calculate the pH of the buffer after the addition of 0.016 mol HNO3.
SAMPLE PROBLEM #9
Answers
a) 9.193
b) 9.181
c) 9.206
Answers
a) 9.50
b) 9.51
c) 9.49
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Titration
Slide 68 / 113
Titration
Important things to remember in getting
ready for titration:
In this technique a known
concentration of base (or
acid) is slowly added to a
solution of acid (or base).
The concentration of the
unknown will then be
determined.
Rinse the buret with distilled water
Rinse with the titrant
Fill the buret with the titrant
Drain a small portion of the titrant so
that air bubbles near the tip of the
burete is expelled and filled to the tip.
This is a quantitative analysis method.
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Titration
Important things to remember in getting
ready for titration:
Rinse the pipette with distilled water
Rinse with the solution to be pipeted
Pipette the solution , adjust the level and dispense
the fixed volume to a beaker or flask.
Slide 70 / 113
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.
At the equivalence point,
# of moles of acid = # of moles of base
MaVa = MbVb
By using an indicator you could visually see the
sudden change in color of the indicator at this point.
This point is also known as the "end point" where you
will stop adding the titrant.
The end point is the appearance of the first
permanent color change of the indicator.
Note that this point will have a slight excess of the
titrant in the solution
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Titration
Equivalence point and End point
At the equivalence point, ( stoichiometric)
# of moles of acid = # of moles of base
MaVa = MbVb
By using an indicator you could visually see the
sudden change in color of the indicator at this point.
This point is also known as the "end point" where you
will stop adding the titrant.
The end point is the appearance of the first
permanent color change of the indicator.
Note that this point will have a slight excess of the
titrant in the solution
This minute amount of the titrant is making the
indicator color visible.
Slide 72 / 113
Slide 73 / 113
Slide 74 / 113
Titration - Equivalence Point
Titration - Equivalence Point
MaVa = MbVb
At the equivalence point,
Here are some examples of reactions in which
the ratio of moles acid to moles base is 1:1.
# of moles of acid = # of moles of base
MaVa = MbVb
HCl + NaOH -->
HBr + KOH -->
HNO3 + LiOH -->
CH3COOH + NaOH -->
This equation is only applicable for reactions
in which the ratio of
moles acid to moles base
is 1:1.
So for reactions such as these, use the
equation above to calculate the molarity or
volume of acid or base at the equivalence
point.
Slide 75 / 113
Slide 76 / 113
Titration - Equivalence Point
Consider some examples of reactions in which
the ratio of moles acid to moles base is 2:1.
2 HCl + Ca(OH)2 --> CaCl2 + 2 H2O
2 HBr + Sr(OH)2 --> SrBr2 + 2 H2O
2 HNO3 + Ba(OH)2 --> Ba(NO3)2 + 2 H2O
moles acid
moles base
=
MaVa
MbVb
=
NaCl + H2O
KBr + H2O
LiNO3 + H2O
CH3COONa + H2O
Titration - Equivalence Point
Consider some examples of reactions in which
the ratio of moles acid to moles base is 1:2.
H2SO4 + 2 KOH --> K2SO4 + 2 H2O
H2SO3 + 2 LiOH --> Li2SO3 + 2 H2O
H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O
2
1
moles acid
moles base
Cross-multiplying, we obtain MaVa = 2 MbVb
=
MaVa
MbVb
=
1
2
Cross-multiplying, we obtain 2 MaVa = MbVb
Slide 77 / 113
Slide 78 / 113
Titration
18
Summary of Equations for
Equivalence Point
What is the concentration of hydrochloric acid
if 20.0 mL of it is neutralized by 40mL of 0.10M
sodium hydroxide?
A 0.025 M
Equation
Acid
Base
MaVa = MbVb
HCl, HNO3, HBr,
CH3COOH, HClO4
NaOH, KOH, LiOH
2 MaVa = MbVb
H2SO4, H2SO3,
H2C2O4
NaOH, KOH, LiOH
MaVa = 2 MbVb
HCl, HNO3, HBr,
CH3COOH, HClO4
Ca(OH)2, Sr(OH)2,
Ba(OH)2
B 0.050 M
C 0.10 M
D 0.20 M
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
Slide 79 / 113
19
Slide 80 / 113
What is the concentration of KOH if 60 mL of it is
neutralized by 20 mL 0.10M HCl?
20
A 0.005 M
What is the concentration of sulfuric acid if 50mL
of it is neutralized by 10mL of 0.1M sodium
hydroxide?
B 0.025 M
A 0.005M
C 0.033 M
B 0.01M
D 0.10 M
C 0.25M
D 0.5M
0.30 M
E
MaVa = MbVb
MaVa = MbVb
2 MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
MaVa = 2 MbVb
Slide 81 / 113
21
Slide 82 / 113
How much 0.5 M HNO3 is necessary to
titrate 25 mL of 0.05M Ca(OH)2 solution to
the endpoint?
How much 3.0 M NaOH is needed to exactly
neutralize 20.0 mL of 2.5 mL H2SO3?
A 8.3 mL
B 17 mL
A 2.5 mL
C 24 mL
B 5.0 mL
D 33 mL
C 10 mL
E
D 20 mL
E
22
25 mL
48 mL
MaVa = MbVb
MaVa = MbVb
2 MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
MaVa = 2 MbVb
Slide 83 / 113
23
How much 1.5 M NaOH is necessary to exactly
neutralize 20.0 mL of 2.5 M H3PO4?
Slide 84 / 113
24 What is the molarity of a NaOH solution if 15 mL is
exactly neutralized by 7.5 mL of a 0.02 M HC2H3O2
solution?
A 11 mL
B 12 mL
A 0.005 M
C 33 mL
B 0.010 M
D 36 mL
C 0.020 M
E 100 mL
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
D 0.040 M
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
Slide 85 / 113
Slide 86 / 113
Titration of a Strong Acid with a
Strong Base
Typically, there are 4 "zones" in which you may be asked to
calculate pH:
· Before any titrant is added from the buret
· After a small amount of titrant has been added
· At the equivalence point
· After the equivalence point
First, we will examine these zones on a titration graph.
Then, we will review the pH calculation for each region.
The strategy for calculating pH is different for each zone.
Slide 87 / 113
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.
Slide 88 / 113
Titration of a Strong Acid with a
Strong Base
Just before (and after)
the equivalence point, the
pH increases rapidly.
The low initial pH
indicates that the
substance being titrated
is a strong acid.
(2) After a
small amount
of titrant has
been added*
(1) Before any
titrant is
added from
the buret
Slide 89 / 113
Titration of a Strong Acid with a
Strong Base
Slide 90 / 113
Titration of a Strong Acid with a
Strong Base
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.
(3) At the
equivalence
point
As more base is added,
the increase in pH again
levels off.
(4) After the
equivalence
point
Slide 91 / 113
Titration of a Strong Acid with a
Strong Base
Slide 92 / 113
Solving Titration Problems:
Strong Acid - Strong Base
1) Before any titrant is added from the buret
The pH depends up on the concentration of the acid or base.
In summary, the four regions of any titration graph are:
· Before any titrant is added from the buret
· After a small amount of titrant has been added
· At the equivalence point
· After the equivalence point
The pH calculation differs for each "zone,"
so we will consider each one separately.
Slide 93 / 113
2) When some titrant is added from the buret, before the
equivalence point is reached
Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the
pH after 10 ml of base is added?
· It is recommended that you calculate the volume of the titrant
needed to neutralize the acid at the beginning to see which
segment of the titration we are in.
write down the neutralization reaction
+
-0.0025mol
NaOH
-->
0.0025mol
NaCl
+
H2O
-0.0025mol
0.0075mol
Remember to check the acid or base is polyprotic or
polyhydroxy.
For example:
· 20 ml 0.5M HCl is titrated with 0.25M NaOH
The original acid is 0.5M; The concentration of H+ = 0.5M
MaVa = MbVb
· 20 ml 0.5M H2SO4 is titrated with 0.25M NaOH
The original acid is 0.5M; The concentration of H+ = 0.5x2 M
2MaVa = MbVb
20 ml 0.5M HCl is titrated with 0.25M Ca(OH)2
The original acid is 0.5M; The concentration of H+ = 0.5 M
MaVa = 2MbVb2
Slide 94 / 113
Solving Titration Problems:
Strong Acid - Strong Base
Solving Titration Problems:
Strong Acid - Strong Base
HCl
0.01 mol
Use the molarity of the acid or base to determine the pH.
pH = - log [H3O+]
OR
pH = 14 - (-log [OH-]
0 mol left
[H+] after the 10 ml base had been neutralized by the acid is
= 0.0075 mols /0.030L = 0.25M
pH = 0.6
Slide 95 / 113
Solving Titration Problems:
Strong Acid - Strong Base
3) At the equivalence point:
Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH
at the equivalence point ?
· You will know you have reached the equivalence point when
MOLES of acid = MOLES of base
· Since there is no excess [H+] or [OH-], we must look to the salt that
is formed to see if it will affect pH.
· Since it is from a strong base and strong acid, will not undergo
hydrolysis to change the pH.
· In cases such as these, when a strong acid and strong base are
titrated, the pure water will have a pH = 7.0
Slide 96 / 113
Solving Titration Problems:
Strong Acid - Strong Base
4)
Beyond the equivalence point, when excess titrant has been
added from the buret
Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is
the pH after 45 ml of base had been added?
0.5 x 20ml = 0.25 x V2ml
V2 = 40 ml NaOH need to neutralize the 20 ml acid.
write down the neutralization reaction
HCl
0.01 mol
+
NaOH
-->
0.0125mol
NaCl
-0.01mol
-0.01mol
0 mol
0.0025mol excess
[OH-] after the 45 ml base added =
= 0.0025 mols /0.065L = 0.0385M
pOH = 1.415
pH = 12.585
+
H2O
Acid in the flask and titrant is the base.
The pH at eqequivalence point will be =7
Slide 97 / 113
Solving Titration Problems:
Strong Acid - Strong Base
Slide 98 / 113
25
The moles of acid equals the moles of base at the
equivalence point in a titration of _____ with
_____.
A strong acid, weak base
B strong base, weak acid
C strong acid, strong base
D weak acid, weak base
E All of the above
Acid in the flask and base is the titrant
Base in the flask and acid is the titrant
Slide 99 / 113
26
What is the pH of a titration between a weak acid
and a strong base at the equivalence point?
Slide 100 / 113
27
What is the pH of a titration between a weak base
and a strong acid at the equivalence point?
A Less than 7
A Less than 7
B Equal to 7
B Equal to 7
C Greater than 7
C Greater than 7
Slide 101 / 113
Solving Titration Problems:
Weak - Strong
1) Before any titrant is added from the buret
Consider the dissociation equation for the substance in the
flask:
For a weak acid in the flask
Ka = x2 / [acid]
pH = - log x
For a weak base in the flask
Kb = x2 / [base]
pH = 14 - (-log x)
Slide 102 / 113
Solving Titration Problems:
Weak - Strong
2) When some titrant is added from the buret, before the equivalence
point is reached
example# 11: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH;
10 ml NaOH is added: ka = 1.8 x 10-5
CH3COOH
0.01mol
-0.0025mol
0.0075mol
0.25M (acid)
+
NaOH
0.0025mol
-0.0025mol
0.0
-->
Na+CH3COO-
+
H2O
+ 0.0025mol
0.0025mol
0.083M ( c.base)
Remember you have a buffer situation now because of the salt
produced from the neutralizationof the weak acid.
Use the Henderson-Hasselbalch equation to solve for the pH of the
solution at this point.
pH = pKa + log ( [base] / [acid] )
Slide 103 / 113
Solving Titration Problems:
Weak - Strong
Sample # 12
calculate the pH of the solution after adding 10 ml of
0.05M KOH to 40 ml of 0.025M Benzoic acid. Ka for
benzoic acid = 6.3 x 10-5
Slide 105 / 113
Solving Titration Problems:
Weak - Strong
3) At the equivalence point
example # 14: Titration of 20 ml of 0.5M acetic acid with 0.25M
NaOH;
40 ml NaOH is added: ka = 1.8 x 10-5
CH3COOH
0.01mol
-0.01mol
0.0mol
+
NaOH
0.01mol
-0.01mol
0.0 mol
-->
Na+ CH3COO-
+
H2O
+ 0.01mol
0.01mol
Slide 104 / 113
Solving Titration Problems:
Weak - Strong
sample #13
Calculate the pH of the solution after adding 10 ml of
0.1M HCl to 20 ml of 0.1 M NH3
Kb of NH3 = 1.8 x 10-5
Slide 106 / 113
Solving Titration Problems:
Weak - Strong
sample # 15
Calculate the pH at equivalence point when 40 ml of
0.025M benzoic acid is titrated with 0.05M KOH. Ka of
benzoic acid is 6.3 10-5
0.166M (base)
The salt Na+ CH3COO- has a strong conjugate base CH3COOThe anion, CH3COO- will undergo hyrolysis as follows:
CH3COO-
+ H2O
-->
CH3COOH +
OH-
0.166M
-X
0.166-X
0
+x
x
0
+x
x
Kb = X2 / 0.166 ; Solve for x
Calculate pH
Will be slightly basic !
Remember Kb = Kw / Ka
Slide 107 / 113
Solving Titration Problems:
Weak - Strong
Slide 108 / 113
Titration of a Weak Acid with a
Strong Base
Sample # 16
calculate the pH at the equivalence point when 40 ml of
0.1M NH3 is titrated with 0.2 M HCl.
Kb of NH3 is 1.8 10-5
The pH at the equivalence point is above 7
strong acid vs strong base
weak acid vs strong base
Slide 109 / 113
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.
Slide 110 / 113
Titration of a Weak Base with a
Strong Acid
· The pH at the equivalence point in these titrations is < 7.
· Methyl red is the indicator of choice.
strong base with strong acid , pH =7
weak base NH3 with strong acid , pH = 5.5
Slide 111 / 113
Solving Titration Problems:
Weak - Strong
4) Beyond the equivalence point, when excess titrant
has been added from the buret
example: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH;
50 ml NaOH is added:
This is exactly the same as for strong acid-strong base titrations.
The strong base added after the end point will increase the OHin the solution making it strongly basic.
Calculate the [OH-] from the excess base in the new volume and
determine the pH.
Slide 113 / 113
Slide 112 / 113
Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is
an equivalence point for each dissociation.