Transcript Acids and Bases - University of California, Santa Barbara

Acids and Bases
Chapter 7
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Acids and Bases – ch. 7
1. Arrange the following solutions in order of most acidic to most
basic.
I. [OH– ] = 0.5 M
II. [H+ ] = 0.3 M
III. pOH = 5.9
IV. pH = 1.2
V. [H+ ] = 1.0x10–4M
Acids and Bases – ch. 7
pH = -log[H3O+]
[H3O+]
pH
[H3O+] = 10-pH
Kw = [H3O+][OH-]
pKw = pH + pOH
pOH = -log[OH-]
[OH-]
pOH
[OH-] = 10-pOH
Acids and Bases – ch. 7
2. At 25°C the water ionization constant is 1.0 x 10–14 and at 98°C the
water ionization constant is 6.8 x 10–14. What is the pH of
neutral water at both temperatures?
Acids and Bases – ch. 7
3. a. Which of the following is a stronger acid?
HNO2 (Ka = 4.0 x 10–4) or HCN (pKa = 9.21)
b. Which is a stronger base?
NO2– or CN–
Acids and Bases – ch. 7
As acid strength ↑ % ionization ↑ Ka ↑ pKa ↓
As base strength ↑ % ionization ↑ Kb ↑ pKb ↓
Conjugate pairs are inversely related
As acid strength ↑ conjugate base strength ↓
Kw = (Ka)(Kb)
Acids and Bases – ch. 7
4. Calculate the pH of the following solutions:
a. 0.004 M HBr
b. 0.25 moles of Ba(OH)2 in 7.50 L of solution
Acids and Bases – ch. 7
5. Calculate the pH and the % ionization of the following solutions
a. 0.05 M HClO (Ka = 3.0 x 10–8)
b. 0.33 M CH3COOH (Ka = 1.8 x 10–5) mixed with 0.85 M HCN
(Ka = 6.2 x 10–10)
Acids and Bases – ch. 7
6. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a pH
of 2.2?
Acids and Bases – ch. 7
7. A solution with 0.5 M of weak mono-protic acid ionizes 11% . What
is the Ka for this acid?
Acids and Bases – ch. 7
8. What concentration of a weak acid that ionizes 0.1% will have a pH
of 3.8?
Acids and Bases – ch. 7
9. What concentration of HNO2 (Ka = 4.0 x 10–4) will have the same
pH as 0.03M HNO3?
Acids and Bases – ch. 7
10. If 0.1M solution of a weak acid has a pH of 3 what will be the pH
of a 0.001M solution of the weak acid?
Acids and Bases – ch. 7
11. Calculate the pH and % ionization for the following:
a. 0.01M NH3 (Kb = 1.8 x 10-5)
b. 0.5 M C6H5NH2 (Kb = 3.8 x 10-10)
Acids and Bases – ch. 7
12. If a 0.35 M solution of weak base has a pH of 11.2 what is the pH
of a 0.08 M solution of the weak base?
Acids and Bases – ch. 7
13. Rank the following 0.1 M salt solutions in order of increasing pH.
i. NaClO4
ii. LiF
iii. C5H5NHI
iv. NH4Cl
v. KCN
HF
Ka = 7.2 x 10-4
C5H5N Kb = 1.7 x 10-9
NH3
Kb = 1.8 x 10-5
HCN
Ka = 6.2 x 10-10
Acids and Bases – ch. 7
Salts ⇒ soluble ionic compounds that are the result of an acid base reaction
⇒ when a salt is dissolved in water each ion can potentially affect the pH
of the solution ⇒ you need to analyze the ions individually
Salts
Cations
Group 1 and 2
metal ions ⇒ neutral
All other cations ⇒ acids
Anions
1st 7 ⇒ neutral
All other anions ⇒ bases
Acids and Bases – ch. 7
14. Will a solution of NH4F be acidic, basic or neutral?
NH3
Kb = 1.8 x 10-5
HF
Ka = 7.2 x 10-4
Acids and Bases – ch. 7
15. How many grams of KF must be dissolved in 500 mL of water in
order to get a pH of 8.4? (Ka of HF = 7.2 x 10-4)
Acids and Bases – ch. 7
16. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L
of water, the pH of the solution is 8.07. Assume the volume of the
solid salt is negligible. What is the identity of the salt?
A) NaCN
B) NaC2H3O2
C) NaCl
D) NaF
E) NaOCl
Acids and Bases – ch. 7
You have completed ch. 7
Answer Key – ch. 7
1. Arrange the following solutions in order of most acidic to most basic.
A) [OH– ] = 0.5 M ⇒ pOH = -log(0.5) = 0.3 ⇒ pH = 14 - 0.3 = 13.7
B) [H+ ] = 0.3 M ⇒ pH = -log(0.3) = 0.52
C) pOH = 5.9 ⇒ pH = 14-5.9 = 8.1
D) pH = 1.2
E) [H+ ] = 1.0x10–4M ⇒ pH = -log(1.0x10–4) = 4
As a solution gets more acidic…
[H+ ]↑, pH↓, [OH– ]↓, pOH↑
B>D>E>C>A
Answer Key – ch. 7
2. At 25°C the water ionization constant is 1.0 x 10–14 and at 98°C the
water ionization constant is 6.8 x 10–14. What is the pH of
neutral water at these temperatures?
If a solution is neutral ⇒ [H3O+] = [OH-]
For all aqueous solutions ⇒ [H3O+]x[OH-] = Kw
at 25°C ⇒ x2 = 1.0 x 10–14 ⇒ x = 1.0 x 10-7M = [H3O+] = [OH-]
pH = -log(1.0 x 10-7M)
pH = 7
at 98°C ⇒ x2 = 6.8 x 10–14 => x = 2.61x10-7M = [H3O+] = [OH-]
pH = -log(2.61x10-7M)
pH = 6.58
Answer Key – ch. 7
3. a. Which of the following is a stronger acid?
HNO2 (Ka = 4.0 x 10–4) or HCN (pKa = 9.21)
as Ka ↑ % ionization ↑ or acid strength ↑ pKa ↓
Ka =10-pKa ⇒ Ka for HCN =10-9.21 = 6.2x 10-10
Since Ka for HNO2 > Ka for HCN ⇒ HNO2 is a stronger acid
b. Which is a stronger base?
NO2– or CN–
as acid strength ↑ conjugate base strength ↓
since HNO2 is the stronger acid ⇒ NO2– is the weaker base ⇒ CN– is
the stronger base
Answer Key – ch. 7
4. Calculate the pH of the following solutions
a. 0.004 M HBr ⇒ Strong acid
[H3O+] = 0.004 M ⇒ pH = - log(0.004) ⇒ pH = 2.4
b. 0.25 moles of Ba(OH)2 in 7.50 L of solution ⇒ Strong base
[OH-]=
(0.25 𝑚𝑜𝑙𝑒𝑠 Ba(OH)2)(2 𝑚𝑜𝑙𝑒𝑠 𝑂𝐻−/ 1 𝑚𝑜𝑙𝑒Ba(OH)2)
pOH = -log(0.067M) = 1.18
pH = 14-1.18 = 12.82
7.5 𝐿
= 0.067M
Answer Key – ch. 7
5. Calculate the pH and the % ionization of the following solutions
a. 0.05 M HClO (Ka = 3.0 x 10–8) ⇒ weak acid
⇌
H3O+
ClO2–
N/A
0
0
-x
N/A
+x
+x
0.05- x
N/A
x
x
HClO
H2O
I
0.05
∆
Eq
Insignificantly
small
Use Ka to solve for x
3 x 10–8 = (x)(x)
0.05
x = 3.87 x 10–5 M
[H3O+] = 3.87 x 10–5 M
pH = - log(3.87 x 10–5 M) = 4.41
+]
[H
O
3
% ionized =
x100
[HClO]
5
3.87
x
10–
% ionized =
x100 = 0.077%
0.05
Answer Key – ch. 7
5. …continued
b. 0.33 M CH3COOH (Ka = 1.8 x 10–5) mixed with 0.85 M HCN
(Ka = 6.2 x 10–10)
CH3COOH H2O
⇌
H3O+
CH3COO-
I
0.33
N/A
0
0
∆
-x
N/A
+x
+x
Eq
0.33 - x
N/A
x
x
Insignificantly
small
If you have 2 or more acids only the
strongest acid will contribute
significantly to the pH
Since acetic acid is stronger (higher Ka)
we can ignore the HCN entirely
2
x
1.8 x
=
0.33
x = 0.00244M = [H3O+]
pH = -log(0.00244) = 2.61
% ionized = 0.00244 = 0.74%
0.33
10–5
Answer Key – ch. 7
6. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a
pH of 2.2? Since the pH = 2.2 ⇒ [H3O+] = 10–2.2 M or 0.0063 M
⇒ since the molar ratio is 1:1 ⇒ [H3O+] = [HCOO–]
⇌
H3O+
HCOOH
N/A
0
0
-0.0063
N/A
+0.0063
+0.0063
x-0.0063
N/A
0.0063
0.0063
HCOOH
H2O
I
x
∆
Eq
1.77 x 10-4 =
(0.0063)(0.0063)
x−0.0063
x = 0.231M
Answer Key – ch. 7
7. A solution with 0.5 M of weak mono-protic acid ionizes 11% . What
is the Ka for this acid?
11%
⇌
H3O+
A–
N/A
0
0
-0.11(0.5)
N/A
+0.11(0.5)
+0.11(0.5)
0.445
N/A
0.055
0.055
HA
H2O
I
0.5
∆
Eq
Ka = (0.055)(0.055) = 6.8x10–3
(0.445)
Answer Key – ch. 7
8. What concentration of a weak acid that ionizes 0.1% will have a pH
of 3.8?
A–
N/A
0
0
-0.001x
N/A
+0.001x
+0.001x
x -0.001x
N/A
0.001x
0.001x
H2O
I
x
∆
Eq
0.1%
⇌
H3O+
HA
Since the pH is 3.8 ⇒ [H3O+] = 10–3.8 or 1.58x10–4 M
1.58x10–4 = 0.0001x
x = 1.58M
Answer Key – ch. 7
9. What concentration of HNO2 (Ka = 4.0 x 10–4) will have the same
pH as 0.03M HNO3?
Same pH ⇒ same [H3O+]
Since HNO3 is a strong acid ⇒ [H3O+] = 0.03M
⇌
H3O+
NO2–
N/A
0
0
-0.03
N/A
+0.03
+0.03
x – 0.03
N/A
0.03
0.03
HNO2
H2O
I
x
∆
Eq
4.0 x 10–4 =
(0.03)(0.03)
x−0.03
x = 2.28M
Answer Key – ch. 7
10. If 0.1M of a weak acid has a pH of 3 what will be the pH of a
0.001M solution of the weak acid?
Same acid with different concentration will have the same Ka so you
can set one Ka equation equal to another
[H3O+][A−] = [H3O+][A−]
[HA]
[HA]
If the pH = 3 ⇒ [H3O+] = 10–3
[0.001][0.001] = [x][x]
[0.1]
[0.001]
x = 10–4 M
pH = -log(10–4) = 4
Answer Key – ch. 7
11. Calculate the pH and % ionization for the following:
a. 0.01M NH3 (Ka of NH4+ = 5.6 x 10-10) NH3 is a weak base so
you need Kb ⇒ Kb = Kw = 1 x 10−14 = 1.79x10–5
K 5.6 x 10−10
𝑎
NH4+
N/A
0
0
-x
N/A
+x
+x
0.01-x
N/A
x
x
H2O
I
0.01
∆
Eq
% ionized =
⇌
OH-
NH3
4.23x10–4
𝑥100 = 4.23%
0.01
1.79x10–5 =
(x)(x)
0.01
x = 4.23x10–4 = [OH–]
pOH = -log(4.23x10–4)
pOH = 3.37
pH = 14 – 3.37
pH = 10.63
Answer Key – ch. 7
11. …continued
b. 0.5 M C6H5NH2 (Kb = 3.8 x 10-10)
3.8 x 10-10 =
⇌
OH-
C6H5NH3+
N/A
0
0
–x
N/A
+x
+x
0.5 – x
N/A
x
x
C6H5NH2
H2O
I
0.5
∆
Eq
% ionized =
0.0028%
1.38x10–5
𝑥100 =
0.5
(x)(x)
0.5
x = 1.38x10–5 = [OH–]
pOH = -log(1.38x10–5)
pOH = 4.86
pH = 14 – 4.86
pH = 9.14
Answer Key – ch. 7
12. If a 0.35 M solution of weak base has a pH of 11.2 what is the pH
of a 0.08 M solution of the weak base?
Same weak base with different concentrations will have the same Kb so
you can set one Kb equation equal to another
[OH−][HA] = [OH−][HA]
[A−]
[A−]
If the pH = 11.2 ⇒ pOH = 2.8 ⇒ [OH–] = 10–2.8 = 1.58 x 10–3
[1.58 x 10–3][1.58 x 10–3] = [x][x]
[0.35]
[0.08]
x = 7.58 x 10–4 M
pOH = -log(7.58 x 10–4 ) = 3.12
pH = 10.88
Answer Key – ch. 7
13. Rank the following 0.1 M salt solutions in order of increasing pH.
i. NaClO4 ⇒ only contains spectator or neutral ions
ii. LiF ⇒ F– is a weak base
iii. C5H5NHI ⇒ C5H5NH+ is a weak acid
iv. NH4Cl ⇒ NH4+ is a weak acid
v. KCN ⇒ CN– is a weak base
Since HF is a stronger acid than HCN ⇒ F– is a weaker base than CN–
Since NH3 is a stronger base than C5H5N ⇒ NH4+ is a weaker acid than C5H5NH+
C5H5NH+ < NH4+ < NaClO4 < F– < CN–
Answer Key – ch. 7
14. Will a solution of NH4F be acidic, basic or neutral?
NH4F is a salt that contains both a weak acid (NH4+)
and a weak base (F–)
Since HF is a stronger acid than NH3 as a base ⇒ F– is a weaker base
than NH4+ as an acid
pH of the salt will be acidic because NH4+ is the stronger component
Answer Key – ch. 7
15. How many grams of KF must be dissolved in 500 mL of water in
order to get a pH of 8.4? (Ka of HF = 7.2 x 10-4)
F– is a weak base ⇒ since the pH = 8.4 ⇒ pOH = 5.6 ⇒
[OH–] = 10–5.6 M or 2.51 x 10–6 M
F–
H2O ⇌
I
x
∆
Eq
OH–
HF
N/A
0
0
-2.51x10–6
N/A
+2.51x10–6
+2.51x10–6
x-2.51x10–6
N/A
2.51x10–6
2.51x10–6
We need Kb to solve for x ⇒
Kb = Kw = 1 x10–14
K𝑎 7.2 x 10–4
Kb = 1.4 x 10–11
1.4 x 10–11 = (2.51x10–6)(2.51x10–6)
⇒ x = 0.454
x
[F–] = 0.45 M
(0.454 mol/L)(0.5 L) = 0.227 mol KF ⇒ (0.227 mol KF)(58.1 g/mol) = 13.2 g
of KF
Answer Key – ch. 7
16. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L
of water, the pH of the solution is 8.07. Assume the volume of the
solid salt is negligible. What is the identity of the salt?
A) NaCN B) NaC2H3O2 C) NaCl D) NaF E) NaOCl
You can eliminate NaCl because it only contains spectator or neutral ions
⇒ your table has a list of Ka values ⇒ In order to get Ka for the
conjugate acid you need Kb for the weak base ⇒ since the pH = 8.07
⇒ pOH = 5.93 ⇒ [OH–] = 1.17 x 10–6
I
A–
H2O
0.1
∆
Eq
0.1
⇌
OH–
HA
N/A
0
0
N/A
+1.17 x 10–6
+1.17 x 10–6
N/A
1.17 x 10–6
1.17 x 10–6
continue to next slide…
Answer Key – ch. 7
Kb =
(1.17
x 10–6)(1.17 x 10–6)
= 1.37 x 10–7
0.1
1.0 x 10–14
Ka =
= 7.3 x 10–4 ⇒ HF ⇒ so the salt must be NaF
1.37 x 10–7