AP Chem Test
Double Replacement Reactions and Titrations
Name and Date
Multiple Choice – Record your answers here.
1) B
2) C
3) B
4)` E
5) B
6) E
7) C
8)` D
9) D
10) E
11) B
12) D
13) Identify the Bronsted-Lowry acid-base pairs in the equilibrium of ammonia in water.
NH3 (aq)
Bronsted-Lowry
__BASE___
Accepts
Proton (H+)
Label the ACID and BASE
+
H2O (l)

NH4+ (aq)
Bronsted-Lowry
___ACID____
Donates
Proton (H+)
Bronsted-Lowry
+
OH¯ (aq)
Bronsted-Lowry
CONJUGATE
CONJUGATE
___ ACID ____
___ BASE ___
Donate
Accepts
Proton (H+)
Proton (H+)
in the reverse reaction
a) Use this equation to justify the statement, “Ammonia is a weak base.”
The presence of hydroxide ions in the product suggests base. The reversible reaction
suggests a weak base. Though technically all reaction are reversible. A better way to make this
determination is be given the equilibrium constant.
b) Write the Kb for this weak base. The Kb is 1.8x10-5. How does this show that
position of equilibrium and indicate that this is a weak base.
[NH4+][ OH¯]
[NH3]
Since the Kb is much less than one – the ratio of products (the ions including OH-) to reactants
(the molecules) is less than one – so the bottom of the fraction must be larger. So, this
is a weak base.
Kb = 1.8x10-5 =
c)
Describe equilibrium.
At equilibrium the rate of forward reaction is equal to the rate of the reverse reaction.
NH3 is reacting with water at the same rate as the ammonium ions collide with
hydroxide ions to form ammonia and water again.
Since the rates are equal the conditions – ratio of products reactants remains constant
and thus an equilibrium constant may be calculated.
14) Identify the Bronsted-Lowry acid-base pairs in the equilibrium of hydrofluoric acid in water.
HF (aq)
Bronsted-Lowry
__ACID___
Accepts
Proton (H+)
Label the ACID and BASE
+
H2O (l)

H3O+ (aq) +
Bronsted-Lowry
CONJUGATE
___BASE____
Donates
Proton (H+)
Bronsted-Lowry
F¯ (aq)
Bronsted-Lowry
CONJUGATE
___ ACID ____
___ BASE ___
Donate
Accepts
Proton (H+)
Proton (H+)
in the reverse reaction
a) Use this equation to justify the statement, “Hydrofluoric acid is a weak acid.”
The presence of hydrogen ions in the product suggests acid. The reversible reaction suggests a weak
acid, though technically all reaction are reversible. A better way to make this determination is to be
given the equilibrium constant.
b) Write the Ka for this weak acid. The Ka is 3.55 x 10¯4. How does this show that
position of equilibrium and indicate that this is a weak acid?
[H+][ F¯]
-5
Ka = 1.8x10 =
[HF]
Since the Kb is much less than one – the ratio of products (the ions including OH-) to reactants
(the molecules) is less than one – so the bottom of the fraction must be larger. So, this
is a weak base
15) Hydrofluoric acid is titrated with lithium hydroxide.
a) Write the balanced equation for this reaction.
HF (aq) + LiOH (aq)  LiF (aq) + H2O (l)
b) Write the total ionic equation.
HF (aq) + Li+(aq) + OH-(aq)  + Li+(aq) + F-(aq) + H2O (l)
c) Write the net ionic equation.
HF (aq) + OH-(aq)  + F-(aq) + H2O (l)
16) HNO3 (aq) + H2O
NO3-(aq) + H3O+(aq) Keq = a very large number
How does this show whether nitric acid is a strong or weak acid?
The Keq is much greater than one, therefore the top of the fraction (H+ and NO3 ions) must be much
larger than the bottom of the fraction (HNO3 molecules). Since almost all the molecules break up into
the ions the acid is strong acid with hydrogen ion concentration equal to or greater than the
concentration of the acid.
[H+][ NO3¯]
Ka = very large
[HNO3]
=17) If 25.0 mL of a standard 0.05 M HCl solution is required to neutralize 20.0 mL of a solution of
Sr(OH)2, what is the concentration of the base? (Start by writing a balanced equation!)
2 HCl + Ba(OH)2 
25 mL HCl soln
1L
1000 mL
0.000625 mol Ba(OH)2
20 mL Ba(OH)2 soln
0.05 mol HCl
1L
1000 mL
1L
BaCl2(aq) + 2 H2O (l)
1 mol Ba(OH)2
0.000625 mol Ba(OH)2
2 mol HCl
0.03 mol Ba(OH)2
L of sol’n
18) How many mL of a 3M NaOH solution are required to completely neutralize 20.0 mL of
1.5M H2SO4? (Start by writing a balanced equation!)
H2SO4 + 2 NaOH 
25 mL H2SO4 soln
1L
1000 mL
1.5 mol H2SO4
1L
Na2SO4(aq) + 2 H2O (l)
1 mol NaOH
2 mol H2SO4
1 L NaOH sol’n
3 mol NaOH
1000 mL
1L
6.3 mL
19) Calculate the mass of aluminum hydroxide required to completely react with 20.0 mL of
0.45M HCl. (HINT: write the balanced equation and then use stoichiometry.)
3 HCl + Al(OH)3 
20 mL HCl soln
1L
1000 mL
0.45 mol HCl
1L
AlCl3 + 3 H2O (l)
1 mol Al(OH)3
3 mol HCl
78.001 g Al(OH)3
1 mol Al(OH)3
0.234 g. Al(OH)3
AP Chem Test Double Replacement
Reactions and Titrations
DO NOT WRITE ON THIS PAGE
1) Which of the following acids is a strong acid?
(A) H3PO4 (B) HNO3 (C) H2CO3 (D) H3BO3 (E) H2SO3
2) What volume of 0.150-molar HCl is required to neutralize 25.0 milliliters of 0.120-molar Ba(OH)2?
(A) 20.0 mL
25 mL Ba(OH)2 soln
(B) 30 0 mL
1L
1000 mL
(C) 40.0 mL
(D) 60.0 mL
(E) 80.0 mL
0.12 mol Ba(OH)2
1L
2 mol HCl
1 mol Ba(OH)2
1 L HCl soln
0.15 mol HCl
1000 mL
1L
40.0 mL
3) At 25°C, aqueous solutions with a pH of 8 have a hydroxide ion concentration, [OH−], of…
(A) 1 × 10−14 M (B) 1 × 10−8 M (C) 1 × 10−6 M
(D) 1M
(E) 8M
pH = - log [H+] so 10-pH = [H+]
4) The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is…
(A) HNO2 + Na+ + OH−  NaNO2 + H2O
(B) HNO2 + NaOH  Na+ + NO2− + H2O
(C) H+ + OH−  H2O
(D) HNO2 + H2O  NO2− + H3O+
(E) HNO2 + OH−  NO2− + H2O
Nitrous acid is a weak acid – so it is present mostly as molecules in the
aqueous solution. NaOH is a strong base and completely dissociates so the ions are separated in the solution. Sodium
ions are spectator ions since they are dissociated in solution attracted to water molecules before and after the
reaction.
5) When H2SO4 and Ba(OH)2 are reacted in a double replacement reaction, one of the products of
the reaction is…
a) H2
b) H2O
d) BaH2
e) SO2
An acid/base neutralization so one of the products is water – the hydrogen ions from the acid got
together with hydroxide ions from the base
c) BaS
6) In the double replacement reaction between the weak acid, HC2H3O2 and strong base, NaOH,
which ion(s) are spectator ions?
a) Na+, C2H3O2–
d) H+, C2H3O2–
b) Na+, OH–
e) Na+ only similar to nitrous acid reaction (#4) above since acetic
acid is a weak acid as well
c) OH– only
7) What is the concentration of an NaOH solution if it takes 16.25 mL of a 0.100 M HCl solution to
titrate 25.00 mL of the NaOH solution?
a) 0.0165 M
d) 0.100 M
b) 0.151 M
e) 0.413 M
c) 0.0650 M
MV = MV 0.1 (16.25) = x (25)
Short cut MV = MV only works effective for 1:1 ratio – otherwise it can be confusing
16.2 mL HCl soln
1L
1000 mL
0.001625 mol NaOH
25 mL NaOH soln
0.10 mol HCl
1L
1000 mL
1L
8) Consider the reaction system, CoO(s) + H2(g)
1 mol NaOH
1 mol HCl
0.001625 mol NaOH
0.064 M NaOH
Co(s) + H2O(g).
The equilibrium constant expression is
[CoO][H 2 ]
[H 2 ]
[Co][H 2 O]
a)
b)
c)
[Co][H 2 O]
[H 2 O]
[CoO][H 2 ]
d)
[H 2 O]
[H 2 ]
e)
[Co][H 2 O]
[H 2 ]
leave out solids and liquids – put products on top
9)
A student pipetted five 25.00-milliliter samples of hydrochloric acid and transferred each sample to an Erlenmeyer
flask, diluted it with distilled water, and added a few drops of phenolphthalein to each. Each sample was then titrated
with a sodium hydroxide solution to the appearance of the first permanent faint pink color. The following results were
obtained:
Volumes of NaOH Solution
First Sample..................35.22 mL
Second Sample..............36.14 mL
Third Sample.................36.13 mL
Fourth Sample..............36.15 mL
Fifth Sample..................36.12 mL
Which of the following is the most probable explanation for the variation in the student's results?
(A) The burette was not rinsed with NaOH solution.
(B) The student misread a 5 for a 6 on the burette when the first sample was titrated.
(C) A different amount of water was added to the first sample.
(D) The pipette was not rinsed with the HCI solution. If the pipet were initialed rinsed with water and then not
rinsed again with the HCl solution, in the first trial the HCl was diluted by the water and thus less NaOH was
required to titrate
(E) The student added too little indicator to the first sample.
10) What is the [H+] when [OH-] = 8.1 x 10-5?
a) 8.1 x 10-5 M
b) 3.6 x 10-6 M
c) 1.0 x 10-7 M
d) 8.1 x 10-5 M
e) 1.2 x 10-10 M
Since the Kw for water = [H+][OH-] = 1.0 x 10-14
then 1.0 x 10-14 = [H+] 8.1 x 10-5
so the 1.0 x 10-14 /8.1 x 10-5 = 1.2 x 10-10 M
11) A sample of lemon juice is found to have a pH of 2.55. What is the H+ concentration of the
juice?
a) 0.0035 M
d) 0.0080 M
b) 0.0028 M
e) 355 M
c) 11.6 M
pH = - log [H+] so 10-pH = [H+] so 10-2.55 = 0.00282
12) A sample of milk is found to have a pH of 6.60. What is the OH- concentration of the milk?
a) 2.5 x 10-21 M
d) 4.0 x 10-8 M
b) 1.0 x 10-7 M
e) 2.5 x 10-7 M
c) 5.0 x 10-7 M
pH = - log [H+] so 10-pH = [H+] so 10-6.6 = 2.51 x 10-7
Since the Kw for water = [H+][OH-] = 1.0 x 10-14
then 1.0 x 10-14 = [H+] 2.51 x 10-7
so the 1.0 x 10-14 /2.51 x 10-7 = 3.98 x 10-8 M