UNIT X – ACIDS & BASES
REVIEW
I. Good things to know
 electrolyte, ionization, dissociation, salt
 6 strong acids (HCl, HBr, HI, HNO3, HClO4, and H2SO4 and 5 strong bases (Li, Na, K, Ba, Sr  OH)
 Arrhenius, Lowry-Bronsted, and Lewis definitions of acids/bases, amphoteric, conjugate pairs
 buffer system, common ion effect, hydrolysis
 acid-base indicators
II. Multiple Choice
Questions 1-3
The diagram below shows the titration of a weak monoprotic acid by a strong base:
B
D
E
1. At this point in the titration, the pH of the solution is equal to the pKa of the acid.
2. This is the equivalence point of the titration.
3. Of the points shown on the graph, this is the point when the solution is most basic.
Questions 4-6
a. 1
b
a
b
b. 3
c. 7
c
e. 13
4. The pH of a solution with a pOH of 11.
5. The pH of a 0.1-molar solution of HCl.
6. The pH of a 0.001-molar solution of HNO3
Questions 7-9
a. HNO3
b. HCN
a
a
c
e
d. 11
c. H2CO3
d. HF
e. H2O
7. This is a strong electrolyte
8. A 0.1-molar solution of this substance will have the lowest pH of the substances listed.
9. This substance is part of the buffer system that maintains the pH of blood at a constant level.
10. A 0.5-molar solution of which of the following salts will have the lowest pH?
a. KCl b. NaC2H3O2 c. NaI d. KNO3 e. NH4Cl
11. Which of the following species is amphoteric?
a. H+1 b. CO3-2 c. HCO3-1 d. H2CO3
e. H2
c
12. How many liters of distilled water must be added to 1 liter of an aqueous solution of HCl with
a pH of 1 in order to create a solution with a pH of 2?
a. 1L b. 2L c. 9L d. 10L e. 20L
b
13. The first acid dissociation constant for tartaric acid, H2C2H4O6, is 1.0 x 10-3. What is the base
dissociation constant for HC2H4O6-?
a. 1.0 x 10-13 b. 1.0 x 10-11 c. 1.0 x 10-7 d. 1.0 x 10-4 e. 1.0 x 10-1
UNIT X – ACIDS & BASES
REVIEW
III. Short Answer
1. Tell which one of the acids listed in pairs below is the stronger of the two and explain why.
a. HF and HCl
 HCl
The anion is larger, thus causing the H to be further away from the
nucleus, making it easier to remove. The easier it is to remove an H,
the stronger the acid
b. H2SO3 and HSO3-1
 H2SO3
The first hydrogen on a diprotic acid is always the easiest to remove.
After the first H is removed, the attractive force from the sulfite ion is
distributed over one less H.
c. HNO3 and HNO2
 HNO3
Oxoacids with more oxygens have easier H’s to remove. This is because
O is so electronegative. The more O’s on the molecule, the more
electron density is pulled toward those atoms, thus away from the H,
thus making it easier to remove
d. HClO4 and HBrO4
 HClO4
Since the number of O’s is the same, the central atom with the greatest
electronegativity will correspond to the strongest acid. Same reason as
#1c.
2. Why are buffer systems made with weak acids instead of strong acids?
HX  H+ + X-. In order for a buffer system to work, high concentrations of
HX and X- (through the addition of a salt) must be present. Strong acids
dissociate completely, thus making the concentration of HX effectively 0.
3. Will a solution of ammonium chloride (NH4Cl) be acidic or basic? Explain using a reaction that
supports your conclusion.
Acidic. the Cl- ion is a conjugate of a strong acid, and therefore will have no
effect on the pH. The NH4+ ion is the conjugate of a weak base (NH3). As a
result NH4+ will act as a weak acid:
NH4+
NH3 + H+
4.
H3PO4
H2PO4HPO4-2
H+1 + H2PO4-1
H+1 + HPO4-2
H+1 + PO4-3
K1 = 7.5 x 10-3
K2 = 6.2 x 10-8
K3 = 2.2 x 10-13
a. Explain why the dissociation
constant decreases with each hydrogen ion lost.
The same reason as #1b
note 3 equivalence
points; one for each
proton in the acid
pH
volume NaOH
b. Of the acids listed above, which
would be the most useful in creating a buffer
system with a pH of 7.5?
H2PO4- is the best
because its Ka is within ±1 of the
desired pH
c. Sketch the titration curve that results when H3PO4 is titrated with excess NaOH and label the
two axes.
UNIT X – ACIDS & BASES
REVIEW
IV. Problems
1. What is the pH of a solution in which you dissolve 5.00g of NaOH in 25θ mL of water? Assume no
change in volume upon addition of the salt.
5.00 g NaOH 
1mol NaOH 0.125mol

 0.500M  (OH - )  pOH  0.3  pH  13.7
0.250 L
40 g NaOH
2. What is the pH of a 0.100 M HF solution (Ka = 7.2 x 10-4)? Find the % ionization as well.
x2
0.0085
 7.2 x10 4  x  0.0085M  (H  )  pH  2.07  %i 
x100  8.5%
0.100
.100
3. What is the pH of a 0.100 M H2SO4 solution (K1 = large, K2 = 1.2 x 10-2)?
H2SO4  H+ + HSO4- This occurs to ~100% ionization (strong acid), therefore
(H+)1 = 0.20M = (HSO4-)

2
HSO 4  H +  SO 4
initial : 0.100
0.100
0
final : 0.100  x 0.100  x
x
(0.100  x)( x)
Ka 
 0.012  0.100 x  x 2  0.0012  0.012 x
(0.100  x)
x 2  0.112 x  0.0012  0  x  0.00985M  (H + ) 2
(H + ) tot  (H + )1  (H + ) 2  0.10  0.00985  0.10985  pH  0.96
4. What is the pH of a 0.100 M NH3 solution (Kb = 1.8 x 10-5)? Find the % hydrolysis as well.
NH3 will hydrolyze water to form OH- ions, so it is a weak base
x2
0.0013
 1.8 x10 5  x  0.0013M  (OH  )  pOH  2.87  pH  11.1  %i 
x100  1.3%
0.100
0.100
5. What is the pH of a mixture of 0.100 M NH4Cl and 0.200 M NH3? See the Kb above.
pOH  pK b  log
(HB)
 0.100 
  log( 1.8 x10 5 )  log 
  4.44  pH  9.56
(B)
 0.200 
6. Calculate the pH at the equivalence point for a titration in which you begin with 100.0 mL of
0.500 M acetic acid (Ka = 1.8 x 10-5) and titrate it with 0.300 M sodium hydroxide.
(0.500 M )(0.1000 L)  0.0500mol HC 2 H 3 O 2 (acid  HAc )  reacts w/ 0.0500mol NaOH(base)
V NaOH  0.0500mol 
1L
 0.167 L  Vtot  0.1000 L  0.167  0.267 L
0.300mol
HAc  OH   Ac 
 HOH
0.0500
~
0.0500
0
 0.0500  0.0500
~
 0.0500
~0
~
~0
0.0500mol
Ac  is a weak base, so find concentrat ion & K b :
0.0500mol
 0.187 M
0.267 L
1.00 x10 14
Kb 
 5.6 x10 10
5
1.8 x10
(Ac - ) =
x2
 5.6 x10 10  x  (OH  ) = 1.0 x10 5 M  pOH  5.0  pH  9.0
0.187
UNIT X – ACIDS & BASES
REVIEW
7. A 10θ mL sample of 0.100-molar NH4Cl solution was added to 8θ mL of a 0.200-molar soution of
NH3. The value of Kb for ammonia is 1.79 x 10-5.
a. What is the pH of this solution?
nNH   (0.100M )(0.100 L)  0.0100mol  nNH 3  (0.200M )(0.0800 L)  0.0160mol
4
pOH  pK b  log
(HB)
 0.0100 
  log( 1.79 x105 )  log 
  4.54  pH  9.46
(B)
 0.0160 
b. If 0.200g of NaOH were added to the solution, what would be the new pH of the solution?
Assume the volume of the solution does not change.
0.200 g NaOH 
NH 4


0.0100
1mol NaOH
 0.00500mol
40 g NaOH
OH 
0.00500
 0.00500  0.00500
~0
0.00500
pOH  pK b  log

NH3
 HOH
0.016
 0.00500
0.021
(HB)
 0.00500 
  log( 1.79 x10 5 )  log 
  4.12  pH  9.88
(B)
 0.021 
c. If equal molar quantities of NH3 and NH4+ were mixed in solution, what would be the pH of
the solution?
pOH = pKb = -log(1.79x10-5)=4.75 pH=9.25
8. The Ksp for cobalt(III) hydroxide is 2.5 x 10 -43 at a temperature of 25oC.
a. Write the dissociation equation
Co(OH)3(s)
Co+3 + 3 OH-1
b. Write the solubility expression
Ksp = (Co+3)(OH-1)3
c. Calculate the molar solubility of cobalt(III) hydroxide
2.5 x10 43  ( x)(3x)3  27 x 4
x4
2.5 x10 43
 9.8 x10 12 M
27
d. Calculate the molar solubility of cobalt(III) hydroxide in a solution with a pH of 4.50.
Is it more or less soluble than it was in distilled water?
pH  4.50  pOH  9.50  (OH  ) = 10-9.50 M  3.16 x10 10 M
2.5 x10  43  ( x)(3.16 x10 10 ) 3
2.5 x10  43
x
 7.9 x10 15 M
10
3.16 x10
7.9 x10 15  9.8 x10 12 , so salt is less soluble in the basic solution