Transcript Acids and Bases - University of California, Santa Barbara

Applications of Aqueous
Equilibria
Chapter 8
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Applications of Aqueous Equilibria– ch. 8
1. What is the pH at the equivalence point for the following titrations?
a. NaOH with HBr
pH = 7 pH > 7 pH < 7
b. HCl with NaCH3COO
pH = 7 pH > 7 pH < 7
c. KOH with HCN
pH = 7 pH > 7 pH < 7
Applications of Aqueous Equilibria– ch. 8
2. Which of the following combinations will result in a buffer solution
upon dissolving in 1 L of water? Calculate the pH of the solutions
that are buffers?
a. 0.1 mol HClO3 and 0.1 mol NaClO3
b. 0.5 mol H2S and 0.8 mol NaHS
c. 0.03 mol NH4I and 0.02 mol NH3
Applications of Aqueous Equilibria– ch. 8
Buffers ⇒ Solutions that can resist change in pH
Composed of a weak acid and it’s conjugate base
3 ways to make a buffer:
1. Mix a weak acid and it’s conjugate base
2. Mix a limiting amount of strong base with a weak acid
3. Mix a limiting amount of strong acid with a weak base
A–
pH = pKa + log
HA
−
or pH = pKa + log nnA
HA
Applications of Aqueous Equilibria– ch. 8
Buffers are classic examples of Le Chatlier’s Principle
Applications of Aqueous Equilibria– ch. 8
3. Which of the following buffer solutions can “absorb” the most acid
without changing the pH? (All solutions have the same volume)
a. 0.4 M HF /0.5 M NaF
b. 0.8 M HF /0.7 M NaF
c. 0.3 M HF/0.7 M NaF
Applications of Aqueous Equilibria– ch. 8
4. A 65.0-mL sample of 0.12 M HNO2 (Ka = 4.0 x 10–4 ) is titrated with
0.11 M NaOH. What is the pH after 28.4 mL of NaOH has been
added?
A) 10.43
B) 7.00
C) 3.57
D) 3.00
E) 3.22
Applications of Aqueous Equilibria– ch. 8
3 possible scenarios when titrating a weak acid with a strong base or
titrating a weak base with a strong acid
1. Buffer ⇒ nweak > nstrong
−
use pH = pKa + log nnA
HA
2. Equivalence point ⇒ nweak = nstrong
pH only depends on the conjugate of the weak acid or weak base
H3O+ A–
use Ka =
HA
OH− HA
or Kb =
A–
to solve for unknown
3. Beyond the equivalence point ⇒ nweak < nstrong
pH depends on the [xs strong acid] or [xs strong base]
Applications of Aqueous Equilibria– ch. 8
Titration Curves
Titration of a weak acid
with a strong base
Titration of a weak base
with a strong acid
>7
pH
<7
pH
Volume of strong base
Buffer
Zone
Volume of strong acid
Applications of Aqueous Equilibria– ch. 8
5. A 84.0-mL sample of 0.04 M KNO2 (Ka for HNO2 = 4.0 x 10–4 ) is
titrated with 0.2 M HI. What is the pH after 13.6 mL of HI has
been added?
Applications of Aqueous Equilibria– ch. 8
6. A 63.0 mL sample of 0.11 M HCN (Ka = 6.2 x 10–10) is titrated with
0.06 M KOH. What is the pH at the equivalence point?
Applications of Aqueous Equilibria– ch. 8
7. A 75.0 mL of 0.22M NH3 (Kb = 1.8 x 10–5) is titrated with 0.50M
HCl. What is the pH at the equivalence point?
Applications of Aqueous Equilibria– ch. 8
8. A 15.0 mL of 0.40M HF (Ka = 7.2 x 10–4) is titrated with 0.85M
LiOH. What is the pH after 8 mL of LiOH has been added?
Applications of Aqueous Equilibria– ch. 8
9. A 120 mL of 0.14M NaCH3COO is titrated with 0.36 M HCl. What
is the pH after 50 mL of HCl has been added?
Applications of Aqueous Equilibria– ch. 8
10. Draw a titration curve for a weak acid being titrated by a strong
base and label the following points:
a. the equivalence point
b. the region with maximum buffering
c. where pH = pKa
d. the buffer region
e. where the pH only depends on [HA]
f. where the pH only depends on [A-]
Applications of Aqueous Equilibria– ch. 8
11. Draw a titration curve for a weak base being titrated by a strong
acid and label the following points:
a. the equivalence point
b. the region with maximum buffering
c. where pH = pKa
d. the buffer region
e. where the pH only depends on [HA]
f. where the pH only depends on [A-]
Applications of Aqueous Equilibria– ch. 8
12. A solution contains 0.36 M HA (Ka = 2.0 x 10-7) and 0.24 M NaA.
Calculate the pH after 0.04 mol of NaOH is added to 1.00 L of
this solution.
a. 6.76
b. 6.52
c. 6.64
d. 6.40
e. 6.70
Applications of Aqueous Equilibria– ch. 8
13. Determine the solubility in mol/L and g/L for the following
compounds:
a. BaCO3 (Ksp = 1.6 x 10-9)
b. Ag2S (Ksp = 2.8 x 10-49)
Applications of Aqueous Equilibria– ch. 8
Solubility ⇒ the maximum amount of solute that can dissolve into a
given amount of solvent at any one temperature at this point the
solution is said to be saturated and in equilibrium
Applications of Aqueous Equilibria– ch. 8
14. Determine the Ksp values for the following compounds:
a. Ag3PO4 (solubility = 1.8 x 10-18 mol/L)
b. MgF2
(solubility = 0.0735 g/L)
Applications of Aqueous Equilibria– ch. 8
15. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M
solution of Na3PO4?
Applications of Aqueous Equilibria– ch. 8
16. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution
with pH of 11? Is Zn(OH)2 more soluble in acidic or basic
solutions?
Applications of Aqueous Equilibria– ch. 8
17. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5
M Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?
Applications of Aqueous Equilibria– ch. 8
You have completed ch. 8
Answer Key – ch. 8
1. What is the pH at the equivalence point for the following titrations?
a. NaOH (strong base) with HBr (strong acid)
pH = 7 pH > 7 pH < 7
b. HCl (strong acid) with NaCH3COO (weak base)
pH = 7 pH > 7 pH < 7
c. KOH (strong base) with HCN (weak acid)
pH = 7 pH > 7 pH < 7
Answer Key – ch. 8
2. Which of the following will result in a buffer solution upon mixing?
a. 0.1 mol HClO3 and 0.1 mol NaClO3 are put into 1L of water
Strong acid and it’s conjugate base is not a buffer
b. 0.5 mol H2S and 0.8 mol NaHS are put into 1L of water
weak acid and it’s conjugate base is a buffer
c. 0.03 mol KHC2O4 and 0.02 mol K2C2O4 are put into 1L of water
weak acid and it’s conjugate base is a buffer
d. 0.1 mol LiOH and 0.2 mol H3PO4 are put into 1L of water strong base
and weak acid will be a buffer as long as the strong base is the limiting reagent
e. 0.1 mol HNO3 and 0.04 mol NH3 are put into 1 L of water
strong acid and weak base could be a buffer if the strong as is limiting – however
in this case the weak base is limiting so no buffer
Answer Key – ch. 8
3. Which of the following 50 mL solutions can absorb the most acid
without changing the pH?
a. 0.4 M HF /0.5 M NaF
b. 0.8 M HF /0.7 M NaF
c. 0.3 M HF/0.7 M NaF
In order for a buffer to absorb an acid you want
[base] to be high and the [acid] to be low and vice
versa if absorbing base
Answer Key – ch. 8
4. Calculate the pH of the following.
a. 0.2 M HN3 (Ka = 1.9 x 10–5) /0.1 M NaN3
HN3/N3 – ⇒ weak acid/conjugate base ⇒ buffer
–]
[A
pH = pKa + log
[HA]
pH = - log (1.9 x 10–5) + log 0.1
0.2
pH = 4.4
b. 30 mL of 0.2 M HN3/ 80 mL of 0.1 M NaN3
mixing solutions causes dilution so we can use
M1V1 = M2V2 to get the new concentrations or we can alter the
Henderson Hasselbalch eqn (HH)⇒
pH = pKa + log nnA−
HA
… continue to next slide
Answer Key – ch. 8
4. b. …continued
nA- ⇒ (80 mL)(0.1 mol/L) = 8 mmol
nHA ⇒ (30 mL)(0.2 mol/L) = 6 mmol
pH = - log(1.9 x 10–5) + log(8/6)
pH = 4.8
Answer Key – ch. 8
5. Calculate the pH for the following:
In all 3 scenarios we’re adding a strong acid (HCl) to a salt with a
weak base (NO2–) ⇒ since we have a strong substance (HCl) we
can assume the neutralization reaction goes to completion ⇒ work
stoichiometrically in moles
a. 25 mL of 0.1 M HCl /25 mL of 0.2 M KNO2
Limiting
2.5 mmol H+
5 mmol NO2–
Reagent

H+
NO2–
I
2.5
5
0
∆
-2.5
-2.5
+2.5
F
0
2.5
2.5
HNO2
After the neutralization rxn is complete
there’s HNO2/NO2– present ⇒ buffer
using the altered HH eqn ⇒
pH = pKa + log nnA−
HA
pH = - log(4 x 10–4) + log(2.5/2.5)
pH = 3.4
note since the [HA] = [A–] ⇒ pH = pKa
Answer Key – ch. 8
5. b. 50 mL of 0.1 M HCl /25 mL of 0.2 M KNO2
Note ⇒
Equivalence
Point
5 mmol NO2–
5 mmol H+

H+
NO2–
I
5
5
0
∆
-5
-5
+5
F
0
0
5
HNO2 H2O
⇌
HNO2
H3O+
NO2–
I
0.067
N/A
0
0
∆
-x
N/A
+x
+x
E
0.067
N/A
x
x
After the neutralization rxn
is complete there’s only a
weak acid (HNO2) in
solution ⇒ change mmol to M
[HNO2] = 5 mmol = 0.067 M
75 mL
Use Ka to solve for (x)
(x)(x)
4x10–4=
0.067
x = 0.00518 = [H3O+]
pH = -log(0.00518)
pH = 2.3
Answer Key – ch. 8
c. 60 mL of 0.1 M HCl/25 mL of 0.2 M KNO2
5 mmol NO2–
6 mmol H+

H+
NO2–
I
6
5
0
∆
-5
-5
+5
F
1
0
5
Limiting
Reagent
HNO2
After the neutralization rxn is
complete there is strong acid (H+)
and weak acid (HNO2) present
A weak acid in the presence of a
strong acid becomes insignificant
[H+] = 1 mmol = 0.0118 M
85 mL
pH = - log(0.0118)
pH = 1.9
Answer Key – ch. 8
6. Calculate the pH of the following:
In all 3 scenarios we’re adding a strong base (OH–) to a weak acid
(HClO) ⇒ since we have a strong substance (OH–) we can assume
the neutralization reaction goes to completion ⇒ work
stoichiometrically in moles
a. 40 mL of 0.2 M KOH/40 mL of 0.5 M HClO
8 mmol OH–
Limiting
Reagent
OH–
HClO
I
8
∆
F

20 mmol HClO
H2O
ClO–
20
N/A
0
-8
-8
N/A
+8
0
12
N/A
8
After the neutralization rxn is complete
there’s HClO/ClO– present ⇒ buffer
using the altered HH eqn ⇒
pH = pKa + log nnA−
HA
pH = -log(3.5x10–8) + log 8
12
pH = 7.3
Answer Key – ch. 8
6.
b. 80 mL of 0.25 M KOH/40 mL of 0.5 M HClO
Note ⇒
Equivalence
Point
20 mmol OH–
OH–
20 mmol HClO
HClO  H2O
ClO–
I
20
20
N/A
0
∆
- 20
-20
N/A
20
F
0
0
N/A
20
ClO–
H2O ⇌ OH–
I
0.167
N/A
0
0
∆
-x
N/A
+x
+x
F
0.167
N/A
x
x
HClO
After the neutralization rxn
there’s only weak base (ClO–) in
solution ⇒ change mmol to M
[ClO–] = 20 mmol = 0.167M
120 mL
Need Kb to solve for (x)
1x10–14
Kb = Kw =
= 2.86x10–7
Ka 3.5x10–8
2.86x10–7= (x)(x)
0.167
–4
x = 2.18x10 = [OH–]
pOH = -log(2.18x10–4) = 3.7
pH = 14 – 3.7 = 10.3
Answer Key – ch. 8
6. c. 100 mL of 0.25 M KOH/40 mL of 0.5 M HClO
25 mmol OH–
OH– HClO 
20 mmol HClO
H2O
ClO–
I
25
20
N/A
0
∆
- 20
-20
N/A
20
F
5
0
N/A
20
Limiting
Reagent
After the neutralization rxn is
complete there’s strong base
(OH–) and weak base (ClO–)
present ⇒ A weak base
is insignificant in the presence
of a strong base ⇒
5 mmol
[OH–] =
= 0.0357 M
140 mL
pOH = - log(0.0357) = 1.45
pH = 14 – 1.45
pH = 12.5
Answer Key – ch. 8
7. Calculate the pH of the following:
a. 10 mL of 1 M HCl is added to a 100 mL buffer with 0.5 M
CH3CH2COOH (pKa = 4.87) and 0.6 M CH3CH2COONa.
A strong acid (HCl) will react with the base in the buffer
(CH3CH2COO–)

CH3CH2COO–
H+
I
60
10
50 mmol
∆
- 10
- 10
+ 10
F
50
0
60
CH3CH2COOH
After the neutralization rxn
is complete the solution is still
a buffer
using the altered HH eqn ⇒
pH = pKa + log nnA−
HA
pH = 4.87 + log 50
60
pH = 4.79
Answer Key – ch. 8
7 b. 20 mL of 0.2 M KOH is added to a 100 mL buffer with 0.5 M
CH3CH2COOH and 0.6 M CH3CH2COONa. The strong base (OH–)
will react with the acid in the buffer (CH3CH2COOH)

CH3CH2COO–
CH3CH2COOH
OH-
I
50 mmol
4 mmol
60 mmol
∆
-4
-4
+4
F
46
0
64
After the neutralization rxn
is complete the solution is still
a buffer
using the altered HH eqn ⇒
pH = pKa + log nnA−
HA
pH = 4.87 + log 64
50
pH = 4.98
Answer Key – ch. 8
8. How would you prepare 1.0 L of a buffer at pH = 9.0 from 1.0 M
HCN (Ka = 6.2 x 10-10) and 1.5 M NaCN?
The total volume of the buffer has to be 1.0 L => so if we designate X as the
volume of HCN then the volume of NaCN must be 1-X => so the moles of
HCN are (1.0 M)(X L) = X moles and the moles of NaCN are (1.5 M)(1-X) =
1.5-1.5X moles => now we can plug these into pH = pKa + log nnA−
HA
9.0 = -log(6.2 x 10-10)+log 1.5−1.5X => X=0.71
X
Therefore the buffer is made by mixing 0.71 L of 1.0M HCN with 0.29 L of
1.5M NaCN
Answer Key – ch. 8
9. How many grams of NaOH need to be added to a 50 mL of 0.3 M
HNO2 (Ka = 4.0 x 10-4) to have a solution with a pH = 4?
 H2O
HNO2
OH-
NO2-
I
15mmol
x mmol
N/A
0
Δ
-x
-x
N/A
+x
F
15-x
0
N/A
x
pH = pKa + log nnA−
HA
-4
10 )+log
x
4 = -log(4.0 x
1.5−x
x = 1.2 mmol of NaOH
Molar mass of NaOH = 40 g/mol
(1.2 mmol)(40 g/mol) = 48 mg of
NaOH
Answer Key – ch. 8
10. Determine the solubility in mol/L and g/L for the following
compounds:
a. BaCO3 (Ksp = 1.6 x 10-9) ⇒ the solubility is defined as the maximum
I
∆
Eq
amount of solute that can be dissolved in a particular amount of solvent at
any one temperature or the saturation point ⇒ the solute is at equilibrium
for saturated solutions
Use Ksp to solve for x
2+
2–
BaCO3 ⇌ Ba
CO3
1.6 x 10-9 = (x)(x)
N/A
0
0
x = 4 x 10–5
since the molar ratio is 1:1
N/A
+x
+x
the molar solubility of BaCO3
N/A
x
x
= 4 x 10–5 mol/L
or
solubility = (4 x 10–5 mol/L)(197.34 g/mol)
= 0.0079 g/L
Answer Key – ch. 8
10. b. Ag2S (Ksp = 2.8 x 10-49)
Ag2S
⇌ 2 Ag+
S2–
I
N/A
0
0
∆
N/A
+2x
+x
Eq
N/A
2x
x
Use Ksp to solve for x
2.8 x 10-49 = (2x)2(x)
x = 4.14 x 10–17
Molar solubility = 4.14 x 10–17 M
or
solubility = (4.14 x 10–17 mol/L)(247.8 g/mol)
= 1 x 10–14 g/L
Answer Key – ch. 8
11. Determine the Ksp values for the following compounds:
a. Al(OH)3 (solubility = 5 x 10-9 mol/L) ⇒ the solubility tells us how
much will dissolve in order to get to equilibrium
Al(OH)3
⇌
Al3+
3 OH–
I
N/A
0
0
∆
N/A
+ 5 x 10-9
+3(5 x 10-9 )
Eq
N/A
5 x 10-9
1.5 x 10-8
Ksp = (5 x 10-9)(1.5 x 10-8)3 = 1.7 x 10–32
Answer Key – ch. 8
11. b. MgF2 (solubility = 0.0735 g/L) ⇒ first we need the molar solubility
(0.0735 g/L)/(62.31 g/mol) = 0.00118 mol/L
MgF2
⇌
Mg2+
2 F–
I
N/A
0
0
∆
N/A
+ 0.00118
+2(0.00118 )
Eq
N/A
0.00118
0.0024
Ksp = (0.00118)(0.0024)2 = 6.6 x 10–9
Answer Key – ch. 8
12. What is the solubility of Ca3(PO4)2 (Ksp = 1 x 10-54) in 0.02M
solution of Na3PO4? The solute and the solution have the phosphate ion
in common ⇒ this will result in lower solubility than if the solute were to
be dissolved in pure water aka the common ion effect
Ca3(PO4)2
⇌
3 Ca2+
2 PO43-
I
N/A
0
0.02
∆
N/A
+ 3x
+2x
Eq
N/A
3x
0.02 + 2x
Insignificantly
small
Use Ksp to solve for x
1 x 10-54 = (3x)3(0.02)2
x = 4.5 x 10–18
molar solubility = 4.5 x 10–18 M
Answer Key – ch. 8
13. What is the solubility of Zn(OH)2 (Ksp = 4.5 x 10-17) in a solution
with pH of 11? The solute and the solution have the hydroxide ion in
common ⇒ this will result in lower solubility than if the solute were to be
dissolved in pure water aka the common ion effect ⇒ since the pH is 11
the pOH is 3 ⇒ [OH-] = 10–3 M
Zn(OH)2
⇌
Zn2+
2 OH-
I
N/A
0
10–3 M
∆
N/A
+x
+2x
Eq
N/A
x
10–3 + 2x
Insignificantly small
Use Ksp to solve for x
4.5 x 10-17= (x)(10–3)2
x = 4.5 x 10-11
molar solubility = 4.5 x 10-11M
Answer Key – ch. 8
14. Will BaCrO4 (Ksp = 8.5 x 10-11) precipitate when 200 mL of 1 x 10-5 M
Ba(NO3)2 is mixed with 350 mL of 3 x 10-5 M KCrO4?
The solution is saturated with BaCrO4 if [Ba2+][CrO42–] = 8.5 x 10-11
however if [Ba2+][CrO42–] > 8.5 x 10-11 the solution is supersaturated
and will precipitate out some BaCrO4 or if [Ba2+][CrO42–] < 8.5 x 10-11
the solution is unsaturated and there will be no noticeable change
We can use M1V1 = M2V2 to get the new concentrations
[Ba2+] = (1 x 10–5 M)(200 mL)/(550 mL) = 3.64 x 10–6 M
[CrO42–]= (3 x 10-5 M)(350 mL)/(550 mL) = 1.91 x 10 -5 M
[Ba2+][CrO42–] = (3.64 x 10–6 M )(1.91 x 10 -5 M )= 6.95 x 10 – 11
the solution is unsaturated and no precipitate will form
Answer Key – ch. 8