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.
A) [OH– ] = 0.5 M
B) [H+ ] = 0.3 M
C) pOH = 5.9
D) pH = 1.2
E) [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 monoprotic acid ionizes 0.62% .
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 0.004M 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. 0.05 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.
a. NaClO4
b. LiF
c. C5H5NHI
d. NH4Cl
e. KCN
HF
Ka = 7.2 x 10-4
C5H5N Ka = 1.7 x 10-9
NH3
Ka = 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. 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
15. If a 1.2 M solution of NaA has a pH of 10.4, what is the ionization
constant (Ka) for the acid HA?
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
3. Predict if the following salts are acidic, basic or neutral.
a. NaClO4 ⇒ Na+ (neutral) + ClO4– (neutral) ⇒ Neutral salt
b. LiF ⇒ Li+ (neutral) + F– (weak base) ⇒ Basic salt
c. C5H5NHI ⇒ C5H5NH+ (weak acid) + I–(neutral) ⇒ Acidic salt
d. KNO3 ⇒ K+ (neutral) + NO3– (neutral) ⇒ Neutral salt
e. NH4NO2 ⇒ NH4+ (weak acid) + F– (weak base)
since there’s an acid and a base we need to compare their
ionization constants: NH4+ ⇒ Ka = 5.6 x 10-10 verses
NO2– ⇒ Kb = Kw of HNO2 ⇒ Kb = (1(4xx10–14)
10–4)
Ka
Kb = 2.5 x 10–11 since Ka > Kb it’s an Acidic salt
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−𝑝𝑒𝑟 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
0.04- x
N/A
x
x
HClO
H2O
I
0.04
∆
Eq
Insignificantly
small
Use Ka to solve for x
3 x 10–8 = (x)(x)
0.04
x = 3.46 x 10–5 M
[H3O+] = 3.46 x 10–5 M
pH = - log(3.46 x 10–5 M) = 4.46
+]
[H
O
3
% ionized =
x100
[HClO]
5
3.46
x
10–
% ionized =
x100 = 0.086%
0.04
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 we can
ignore the HCN
2
x
–5
1.8 x 10 =
0.33
x = 0.00244M = [H3O+]
pH = -log(0.00244) = 2.61
% ionized = 0.00244 = 0.74%
0.33
Answer Key – ch. 7
5. …continued
c. 0.04 M NH4I (for NH3 Kb = 1.8 x 10–5) => if NH3 is a base then
NH4+ must be an acid
NH4+
H2O
I
0.002
∆
-x
Eq 0.002 - x
Insignificantly
small
⇌
H3O+
NH3
N/A
0
0
N/A
+x
+x
N/A
x
x
Ka is necessary to solve for x => since
NH4+ and NH3 are conjugates
Ka = Kw
K𝑏
Ka = 1x10−14 = 5.6x10-10
1.8x10−5
2
x
-10
5.6x10 =
0.002
x = 1.06x10-6
pH = -log(1.06x10-6 ) = 5.98
%ionized = 1.06x10−6x100 = 0.053%
0.002
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 ⇒ [H+] = 10–2.2 M or 0.0063 M ⇒
since the molar ratio is 1:1 ⇒ [H+] = [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
insignificantly
small
1.77 x 10-4 = (0.0063)(0.0063)
x
x = 0.224
Answer Key – ch. 7
7. If a solution with 0.5 M of unknown weak acid ionizes 0.026%
identify the weak acid. Ka is a useful identification value
⇌
H3O+
A–
N/A
0
0
-0.00026(0.5)
N/A
+0.00026(0.5)
+0.00026(0.5)
0.5
N/A
0.00013
0.00013
HA
H2O
I
0.5
∆
Eq
Ka = (0.00013)(0.00013) = 3.4x10–8 ⇒ HOCl
(0.5)
Answer Key – ch. 7
8. What concentration of a weak acid that ionizes 0.1% will have a pH
of 3.8?
⇌
H3O+
A–
N/A
0
0
-0.001x
N/A
+0.001x
+0.001x
x -0.001x
N/A
0.001x
0.001x
HA
H2O
I
x
∆
Eq
Since the pH is 3.8 ⇒ [H3O+] = 10–3.8 or 1.58x10–4
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 0.004M HNO3? Same pH ⇒ same [H3O+]
Since HNO3 is a strong acid ⇒ [H3O+] = 0.004M
⇌
H3O+
NO2–
N/A
0
0
-0.004
N/A
+0.004
+0.004
x – 0.004
N/A
0.004
0.004
HNO2
H2O
I
x
∆
Eq
4.0 x 10–4 =
(0.004)(0.004)
x = 0.04M
x
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
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
⇌
OH-
NH4+
N/A
0
0
-x
N/A
+x
+x
0.01-x
N/A
x
x
NH3
H2O
I
0.01
∆
Eq
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. 0.05 M KClO (Ka of HClO = 3 x 10-8) ⇒ ClO– is a weak
base so you need Kb ⇒ Kb = Kw = 1 x 10−14 = 3.33x10–7
K
3 x 10−8
𝑎
ClO–
H2O
I
0.05
∆
Eq
⇌
OH-
HClO
N/A
0
0
–x
N/A
+x
+x
0.05 – x
N/A
x
x
3.33x10–7 =
(x)(x)
0.05
x = 1.29x10–4 = [OH–]
pOH = -log(1.29x10–4)
pOH = 3.89
pH = 14 – 3.89
pH = 10.11
Answer Key – ch. 7
12. 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 = 0.45
x
[F–] = 0.45 M
(0.45 mol/L)(0.5 L) = 0.225 mol KF ⇒ (0.225 mol KF)(56.11 g/mol) = 12.6 g
of KF
Answer Key – ch. 7
13. If a 1.2 M solution of NaA has a pH of 10.4, what is the ionization
constant for the acid HA? In order to get Ka for HA we’ll need Kb
for A– ⇒ since the pH = 10.4 ⇒ pOH = 4.6 ⇒ [OH–] = 2.51 x 10–5
I
A–
H2O
1.2
∆
Eq
1.2
⇌
OH–
HA
N/A
0
0
N/A
+2.51 x 10–5
+2.51 x 10–5
N/A
2.51 x 10–5
2.51 x 10–5
Kb for A– = (2.51 x 10–5 )(2.51 x 10–5)
(1.2)
Kb = 5.26 x 10–10 ⇒
Ka = Kw = 1 x10–14 = 1.9 x 10–5
K 5.26 x 10–10
𝑏