Acids and Bases
Chapter 7
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Acids and Bases – ch. 7
1. Calculate the pH for the following solutions at 25 °C and determine
if the solution is acidic, basic or neutral.
a. [H+] = 5 x 10-4 M
b. [OH-] = 2 x 10-9 M
Acids and Bases – ch. 7
Useful Equations
pH = - log[H+]
pOH = - log[OH-]
[H+] = 10 –pH
[OH-] = 10 –pOH
Kw = [H+][OH-]
(Kw = 1 x 10-14 only at 25°C)
pKw = pH + pOH
(pKw = 14 only at 25°C)
Types of Solutions
Neutral Solutions:
[H+] = [OH-]
(pH=7 only at 25°C)
Acidic Solutions:
[H+] > [OH-]
(pH<7 only at 25°C)
Basic Solutions:
[H+] < [OH-]
(pH>7 only at 25°C)
Acids and Bases – ch. 7
2. Calculate the [H+] and [OH-] for the following solutions at 25 °C.
a. pH = 2
b. pOH = 4.2
Acids and Bases – ch. 7
3. At a particular temperature the water ionization constant,
Kw, is 6.8 x 10–14. What is the pH and [H3O+] in neutral water
at this temperature?
Acids and Bases – ch. 7
4. a. Which of the following is a stronger acid?
HNO2 (Ka = 4.0 x 10–4) or HCN (Ka = 6.2 x 10–10)
b. Which is a stronger base?
NO2– or CN–
Acids and Bases – ch. 7
As acid strength ↑ % ionization ↑ Ka ↑
As base strength ↑ % ionization ↑ Kb ↑
Conjugate pairs are inversely related
As acid strength ↑ conjugate base strength ↓
Kw = (Ka)(Kb)
Acids and Bases – ch. 7
5. Bicarbonate ion is an amphoteric substance. Will an aqueous
solution of HCO3- be acidic, basic or neutral?
(H2CO3 ⇒ Ka1 = 4.3x10-7 and Ka2 = 4.8x10-11)
Acids and Bases – ch. 7
6. Predict if the following salts are acidic, basic or neutral.
a. NaClO4
b. LiF
c. NH4I
d. KNO3
e. NH4F
Acids and Bases – ch. 7
Salts ⇒ the 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 metal ions ⇒ neutral
Ca2+, Sr2+ Mg2+
and Ba2+ ⇒ neutral
All other cations ⇒ acids
Anions
Cl-, Br-, I-, NO3-, ClO3-, ClO4-,
BrO3-, BrO4-, and IO4- ⇒ neutral
HSO4– ⇒ acidic
All other anions ⇒ bases
Acids and Bases – ch. 7
7. Calculate the pH of the following solutions:
a. 0.004 M HBr
b. 10 g of H2SO4 and 10 g of HCl in 4.2 L of solution
c. 2.5 g of Ba(OH)2 in 750 mL of solution
Acids and Bases – ch. 7
8. Calculate the pH and the % ionization of the following solutions
a. 0.05 M HNO2 (Ka = 4.0 x 10–4)
b. 0.33 M CH3COOH (Ka = 1.8 x 10–5) mixed with 0.85 M HCN
(Ka = 6.2 x 10–10)
c. 0.04 M NH4I (for NH3 Kb = 1.8 x 10–5)
Acids and Bases – ch. 7
9. What concentration of HCOOH (Ka = 1.77x10-4) solution will have a pH
of 2.2?
Acids and Bases – ch. 7
10. If a solution with 0.5 M of unknown weak acid ionizes 0.62%
identify the weak acid.
Acids and Bases – ch. 7
11. What concentration of a weak acid that ionizes 0.1% will have a
pH of 3.8?
Acids and Bases – ch. 7
12. What concentration of HNO2 (Ka = 4.0 x 10–4) will have the same
pH as 0.004M HNO3?
Acids and Bases – ch. 7
13. 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?
Acids and Bases – ch. 7
14. Calculate the pH and % ionization for the following:
a. 0.01M NH3 (Ka of NH4+ = 5.6 x 10-10)
b. 0.05 M KClO (Ka of HClO = 3 x 10-8)
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. 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. Calculate the pH and pOH for the following solutions and determine
if the solution is acidic, basic or neutral.
a. [H3O+] = 5 x 10-4 M
pH = -log[H3O+] ⇒ pH = -log(5 x 10-4 ) ⇒ pH = 3.3
pH + pOH = 14 ⇒ pOH = 14 – 3.3 ⇒ pOH = 10.7
acidic
b. [OH-] = 2.5 x 10-9 M
pOH = -log[OH-] ⇒ pOH = -log(2 x 10-9) ⇒ pOH = 8.6
pH = 14 – 8.6 ⇒ pH = 5.4
acidic
Answer Key – ch. 7
2. Calculate the [H3O+] and [OH-] for the following solutions.
a. pH = 2 [H3O+] = 10 -pH ⇒ [H3O+] = 10 -2 M or 0.01 M
pOH = 14 – 2 = 12 ⇒ [OH-] = 10 -pOH ⇒ [OH-] = 10 -12 M
b. pOH = 4.2 [OH-] = 10 -4.2 M or 6.3 x 10 -5 M
pH = 14 – 4.2 = 10.8 ⇒ [H3O+] = 10 -10.8 or 1.6 x 10 -11 M
Answer Key – ch. 7
3. At 20 °C, the water ionization constant, Kw, is 6.8 x 10–14. What is
the pH and H3O+ concentration in neutral water at this temperature?
If a solution is neutral => [H3O+] = [OH-]
For all aqueous solutions => [H3O+]x[OH-] = Kw
x2 = 6.8 x 10–14 => x = 2.61x10-7M
[H3O+]=2.61x10-7M
pH = -log(2.61x10-7M)
pH = 6.58
Answer Key – ch. 7
4. a. Which of the following is a stronger acid?
HNO2 (Ka = 4.0 x 10–4) or HCN (Ka = 6.2 x 10–10)
as Ka ↑ % ionization ↑ or acid strength ↑
⇒ HNO2 is a stronger acid
b. Which is a stronger base?
NO2– or CN–
as acid strength ↑ conjugate base strength ↓
since HNO2 is a stronger acid ⇒ CN– is a stronger base
Answer Key – ch. 7
5. Bicarbonate ion is an amphoteric substance. Will an aqueous
solution of HCO3- be acidic, basic or neutral?
Ka tells you a relative acid strength and Kb tells you a relative base
strength ⇒ so by comparing Ka to Kb we can determine what it prefers to
be
Ka = 4.8 x 10-11
Kb = = Kw ⇒ Kb = = (1 x 10−14) = 2.3x10-8
Ka
(4.3 x 10−7)
Since Kb > Ka bicarbonate will be a base
Answer Key – ch. 7
6. 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. NH4I ⇒ NH4+ (weak acid) + I–(neutral) ⇒ Acidic salt
d. KNO3 ⇒ K+ (neutral) + NO3– (neutral ⇒ Neutral salt
e. NH4F ⇒ 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
F– ⇒ Kb = Kw of HF ⇒ Kb = (1 x 10–14)
Ka
(7.2 x 10–4)
Kb = 1.4 x 10–11 since Ka > Kb it’s an Acidic salt
Answer Key – ch. 7
7. 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.10 g of H2SO4 and 10 g of HCl in 4.2 Lof solution ⇒ Since both
acids are strong they both contribute significantly to the [H3O+]
[H2SO4] = (10g/98.09g/mol)/(4.2L) = 0.024M => [H3O+] = 0.024M
[HCl] = (10g/36.46g/mol)/(4.2L) = 0.065 M => [H3O+] = 0.065M
[H3O+]total = 0.089M => pH = -log(0.089) = 1.05
c. 2.5 g of Ba(OH)2 in 750 mL of solution => strong base
[Ba(OH)2] = (2.5g/171.32g/mol)/(0.75L) = 0.019M
[OH-] = 2(0.019M) = 0.038M
pOH = -log(0.038) = 1.4
pH = 14-1.4 = 12.6
Answer Key – ch. 7
8. Calculate the pH and the % ionization of the following solutions
a. 0.04 M HNO2 (Ka = 4.0 x 10–4) ⇒ weak acid
H3O+
NO2–
N/A
0
0
-x
N/A
+x
+x
0.04- x
N/A
x
x
HNO2
H2O
I
0.04
∆
Eq
⇌
Use Ka to solve for x
4 x 10–4 = (x)(x)
0.04−x
use quadratic formula
x = 0.0038 M
[H3O+] = 0.0038
pH = - log(0.0038) = 2.42
+]
[H
O
3
% ionized =
x100
[HNO2]
% ionized = 0.0038x100 = 9.5%
0.04
Answer Key – ch. 7
8. …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
8. …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
9. 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
10. 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
11. 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
12. 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
13. 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
14. 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
14. …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
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 ⇌
OH–
HF
I
x
N/A
0
0
∆
-2.51x10–6
N/A
+2.51x10–6
+2.51x10–6
Eq
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
16. 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