Chapter 5 Exercise Key
Chemistry 1A - Chapter 5 Exercise Key
Exercise 5.1 – Ionic Solubility: Predict whether each of the following is soluble or
insoluble in water.
a. Hg(NO3)2
b. FeCO3
c. SnS
soluble
insoluble
insoluble
d. K3PO4 soluble
e. PbCl2
insoluble
f. Al(OH)3
insoluble
Exercise 5.2 – Precipitation Reactions: Predict whether a precipitate will form when
each of the following pairs of water solutions are mixed. If there is a precipitation
reaction, write the complete, complete ionic, and net ionic equations that describe the
reaction.
a. 3Li2CO3(aq) + 2Al(NO3)3(aq) → 6LiNO3(aq) + Al2(CO3)3(s)
6Li+(aq) + 3CO32−(aq) + 2Al3+(aq) + 6NO3−(aq)
→ 6Li+(aq) + 6NO3− (aq) + Al2(CO3)3(s)
3CO32-(aq) + 2Al3+(aq) → Al2(CO3)3(s)
b.
3KOH(aq) + Fe(NO3)3(aq) → 3KNO3(aq) + Fe(OH)3(s)
3K+(aq) + 3OH−(aq) + Fe3+(aq) + 3NO3−(aq)
→ 3K+(aq) + 3NO3−(aq) + Fe(OH)3(s)
3OH−(aq) + Fe3+(aq) → Fe(OH)3(s)
c.
NaC2H3O2(aq) + CaS(aq)
d.
K2SO4(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbSO4(s)
2K+(aq) + SO42−(aq) + Pb2+(aq) + 2NO3−(aq)
→ 2K+(aq) + 2NO3−(aq) + PbSO4(s)
SO42−(aq) + Pb2+(aq) → PbSO4(s)
Copyright 2004 Mark Bishop
→ No Reaction
Chapter 5 Exercise Key
Exercise 5.3 – Acid and Base Classification: Identify each of the following as either (1)
an Arrhenius strong acid, (2) an Arrhenius weak acid, (3) an Arrhenius strong base, (4) an
Arrhenius weak base, or (5) not acidic or basic in the Arrhenius sense (neutral).
a.
b.
c.
d.
e.
f.
g.
h.
i.
HNO2 weak acid
lithium hydroxide strong base
NaCN weak base
sodium iodide neutral
NaHSO4 weak acid
nitric acid strong acid
CH3OH neutral
hydrofluoric acid weak acid
KC2H3O2 weak base
Exercise 5.4 – Strong, Weak, and Nonelectrolyte: Each of the following substances
dissolve in water. Identify each of the following as either a (1) strong electrolyte, (2)
weak electrolyte, or (3) nonelectrolyte, and in parentheses write the type of compound
each name or formula represents.
Al(NO3)3
strong electrolyte
acetic acid
weak electrolyte
NH3
weak electrolyte
ammonium acetate
strong electrolyte
HCl(aq)
strong electrolyte
glucose
nonelectrolyte
CH3OH
nonelectrolyte
barium chloride
strong electrolyte
K2Cr2O7
strong electrolyte
nitric acid
strong electrolyte
HBrO2
weak electrolyte
2-propanol
(or isopropyl alcohol)
nonelectrolyte
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Exercise 5.5 – Writing Neutralization Equations: For each of the following pairs of
possible reactants, predict whether a neutralization reaction will take place between them.
If there is no reaction, write, “No Reaction”. If there is a reaction, write complete,
complete ionic, and net ionic equations for the reaction. (The reactions between weak
acids and weak bases given here are reversible reactions. If an acid has more than one
acidic hydrogen, assume that there is enough base to remove all of them. Assume that
there is enough acid to add as many protons to the base as possible.)
a.
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq)
→ H2O(l) + Na+(aq) + Cl−(aq)
H+(aq) + OH−(aq)
b.
e.
→
→
NH4+(aq)
NH3(aq) + HClO(aq)
NH3(aq) + HClO(aq)
NH4ClO(aq)
NH4+(aq) + ClO−(aq)
NH3(aq) + HClO(aq)
NH4+(aq) + ClO−(aq)
HC2H3O2(aq) + LiF(aq)
HF(aq) + LiC2H3O2(aq)
+
HF(aq) + Li+(aq) +
HC2H3O2(aq) + Li (aq) + F−(aq)
−
C2H3O2 (aq)
HC2H3O2(aq) + F−(aq)
f.
H2O(l) + F−(aq)
NH3(aq) + HNO3(aq) → NH4NO3(aq)
NH3(aq) + H+(aq) + NO3−(aq) → NH4+(aq) + NO3−(aq)
NH3(aq) + H+(aq)
d.
H2O(l)
HF(aq) + LiOH(aq) → H2O(l) + LiF(aq)
HF(aq) + Li+(aq) + OH−(aq) → H2O(l) + + Li+(aq) + F−(aq)
HF(aq) + OH−(aq)
c.
→
Na2CO3(aq) + 2HBr(aq) →
+
HF(aq) + C2H3O2−(aq)
2NaBr(aq) + H2O(l) + CO2(g)
2−
2Na (aq) + CO3 (aq) + 2H+(aq) + 2Br−(aq)
→ 2Na+(aq) + 2Br−(aq) + H2O(l) + CO2(g)
CO32−(aq) + 2H+(aq) → H2O(l) + CO2(g)
g.
HCl(aq) + HNO2(aq) No reaction - both acids
h.
H3PO3(aq) + 3LiOH(aq) → 3H2O(l) + Li3PO4(aq)
H3PO3(aq) + 3Li+(aq) + 3OH−(aq)
→ 3H2O(l) + 3Li+(aq) + PO43−(aq)
H3PO3(aq) + 3OH−(aq) → 3H2O(l) + PO43−(aq)
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Fe(OH)3(s) + 3HNO3(aq) → Fe(NO3)3(aq) + 3H2O(l)
Fe(OH)3(s) + 3H+(aq) + 3NO3−(aq)
→ Fe3+(aq) + 3NO3−(aq) + 3H2O(l)
i.
Fe(OH)3(s) + 3H+(aq) → Fe3+(aq) + 3H2O(l)
j.
NaI(aq) + HCl(aq)
No reaction - no base
Exercise 5.6 - Brønsted-Lowry Acids and Bases: For each of the following equations,
identify the Brønsted-Lowry acid and base.
a. HNO2(aq) + NaBrO(aq)
B/L acid
B/L base
b. H2PO4 (aq) + HNO2(aq)
B/L base
→ HBrO(aq) + NaNO2(aq)
H3PO4(aq) + NO2−(aq)
B/L acid
c. H2PO4-(aq) + 2OH−(aq) → PO43-(aq) + 2H2O(l)
B/L Acid
B/L base
d. H2SO3(aq) + 2NaOH(aq) → Na2SO3(aq) + 2H2O(l)
B/L Acid
B/L base
Exercise 5.7 – Oxidation Numbers: Determine the oxidation number for the atoms of
each element in the following formulas.
Formula
Oxidation Numbers
P4
P is 0.
PF3
P is +3, and F is −1.
PH3
P is -3, and H is +1.
P2O3
P is +3, and O is −2.
H3PO4
H is +1, P is +5, and O is −2.
N2
N is 0.
N3−
N is −3.
K3N
K is +1, and N is −3.
Co3N2
Co is +2, and N is −3.
NaH
Na is +1, and H is −1.
Na
Na is O.
HSO3
−
H is +1, S is +4, and O is −2.
Cu(NO3)2
Cu is +2, N is +5, and O is −2.
Cu3(PO4)2
Fe is +3, S is +6, and O is −2.
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Exercise 5.8 – Redox Reactions: Identify whether the following equations describe
redox reactions or not. For each of the redox reactions, identify what is oxidized, what is
reduced, what the reducing agent is, and what the oxidizing agent is.
a. Ca(s) + F2(g) → CaF2(s)
Ca goes from 0 to +2. F goes from 0 to −1. Yes, it’s redox
Ca is oxidized, and is the reducing agent.
F is reduced and is the oxidizing agent.
b. CaCO3(s)
∆
→ CaO(s) + CO2(g)
Ca goes from +2 to +2. C goes from +4 to +4. All O’s are −2. This is not
redox.
c. 2Al(s) + 3H2O(g)
→ Al2O3(s) + 3H2(g)
Al goes from 0 to +3. H goes from +1 to 0. All O’s are −2. Yes, it’s redox
Al is oxidized, and is the reducing agent.
H in H2O is reduced, so H2O is the oxidizing agent.
d. Cr2O72−(aq) + 6Cl−(aq) + 14H+(aq)
→ 2Cr3+(aq) + 3Cl2(g) + 7H2O(l)
Cr goes from +6 to +3. Cl goes from −1 to 0. All H’s are +1. All O’s are −2.
Yes, it’s redox
Cl in Cl− is oxidized, so Cl− is the reducing agent.
Cr in Cr2O72− is reduced, so Cr2O72− is the oxidizing agent.
e. 5H2C2O4(aq) + 2KMnO4(aq) + 3H2SO4(aq)
→ 10CO2(g) + 2MnSO4(aq) + 8H2O(l) + K2SO4(aq)
C goes from +3 to +4. Mn goes from +7 to +2. All H’s are +1. All O’s are −2.
All S’s are +6.
Yes, it’s redox
C in H2C2O4 is oxidized, so H2C2O4 is the reducing agent.
Mn in KMnO4 is reduced, so KMnO4 is the oxidizing agent.
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Exercise 5.9 – Stoichiometry and Molarity:
75.0 mL of 0.250 M Na2CO3?
How many milliliters of 6.00 M HNO3 are necessary to neutralize the carbonate in
⎛ 0.250 mol Na 2 CO3 ⎞ ⎛ 2 mol HNO3 ⎞ ⎛ 103 mL HNO3 soln ⎞
? mL HNO3 soln = 75.0 mL Na 2 CO3 ⎜
⎟⎜
⎟ = 6.25 mL HNO3 soln
⎟⎜
3
⎝ 10 mL Na 2 CO3 ⎠ ⎝ 1 mol Na 2 CO3 ⎠ ⎝ 6.00 mol HNO3 ⎠
Exercise 5.10 - Molarity: What is the molarity of a AgClO4 solution made by dissolving 29.993 g of silver perchlorate in water
and diluting with water to 50.00 mL total?
Molarity =
⎞ ⎛ 103 mL ⎞
2.893 mol AgClO4
? mol AgClO 4
29.993 g AgClO 4 ⎛ 1 mol AgClO 4
= 2.893 M AgClO4
=
⎜
⎟⎜
⎟ =
1 L AgClO4 soln
1 L AgClO 4 soln 50.0 mL AgClO 4 soln ⎝ 207.3185 g AgClO 4 ⎠ ⎝ 1 L ⎠
Exercise 5.11 – General Stoichiometry:
What is the maximum number of grams of silver chloride that will precipitate from a
solution made by mixing 25.00 mL of 0.050 M MgCl2 with an excess a AgNO3 solution?
⎛ 0.050 mol MgCl2 ⎞ ⎛ 2 mol AgCl ⎞ ⎛ 143.3209 g AgCl ⎞
? g AgCl = 25.0 mL MgCl2 ⎜
⎟⎜
⎟⎜
⎟ = 0.36 g AgCl
3
⎝ 10 mL MgCl2 ⎠ ⎝ 1 mol MgCl2 ⎠ ⎝ 1 mol AgCl ⎠
Exercise 5.12 – Titration Problem:
When 34.2 mL of a 1.02 M NaOH solution is added from a burette to 25.00 mL of a
phosphoric acid solution that contains phenolphthalein, the solution changes from colorless to red.
a. What is the titrant for this process? NaOH is the titrant.
b. What is the molarity of the phosphoric acid?
? mol H 3 PO 4
34.2 mL NaOH soln ⎛ 1.02 mol NaOH ⎞ ⎛ 1 mol H 3 PO 4 ⎞ ⎛ 103 mL ⎞
=
⎜
⎟ = 0.465 M H3PO4
⎜
⎟
L H 3 PO 4 soln 25.00 mL H 3 PO 4 soln ⎝ 103 mL NaOH soln ⎠ ⎜⎝ 3 mol NaOH ⎠⎟ ⎝ 1 L ⎠
Exercise 5.13 – Making solution from Solid: An experiment calls for a total of 1.50 L of 0.200 M KMnO4 for a class of
chemistry students. How would this solution be made from pure, solid potassium permanganate and water?
⎞ ⎛ 158.0339 g KMnO 4 ⎞
⎟ = 47.4 g KMnO4
⎟ ⎜ 1 mol KMnO
⎠⎝
4
⎠
Dissolve 47.4 g KMnO4 in a minimum amount of water and dilute with water to 1.50 L total.
⎛ 0.200 mol KMnO 4
? g KMnO 4 = 1.50 L soln ⎜
1 L soln
⎝
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Exercise 5.14 – Dilution Problems:
mL?
What is the molarity of a solution made by diluting 5.00 mL of a 14.8 M NH3 solution to 75.0
? mol NH 3 5.00 mL conc. soln ⎛ 14.8 mol NH 3 ⎞ ⎛ 103 mL ⎞
=
⎜
⎟
L dil. soln 75.00 mL dil. soln ⎜⎝ 103 mL conc. soln ⎟⎠ ⎝ 1 L ⎠
or
M C VC = M D VD
MD =
M C VC 14.8 M ( 5.00 mL )
=
= 0.987 M NH3
VD
75.0 mL
Exercise 5.15 – Making solution from Concentrated Acid: How would you make 250.0 milliliters of 2.00 M acetic acid from
concentrated acetic acid (called glacial acetic acid) that is 17.4 M HC2H3O2?
⎛ 2.00 mol HC 2 H 3 O 2
? mL conc. soln = 250.0 mL dil. soln ⎜
3
⎝ 10 mL dil. soln
or
M C VC = M D VD
VC =
⎞ ⎛ 103 mL conc. soln ⎞
⎟ ⎜ 17.4 mol HC H O ⎟
⎠⎝
2 3 2 ⎠
M D VD 2.00 M ( 250.0 mL )
= 28.7 mL conc. soln
=
MC
17.4 M
Carefully add 28.7 mL of 17.4 M HC2H3O2 to about 200 mL of water, mix, and dilute with water to 250.0 mL total.
Exercise 5.16 - Making solution from More Concentrated Solution:
hydrochloric acid from 2.0 M HCl?
How would you make 50.0 milliliters of 0.250 M
3
⎛ 0.250 mol HCl ⎞ ⎛ 10 mL conc. soln ⎞
? mL conc. soln = 50.0 mL dil. soln ⎜ 3
⎟
⎟⎜
⎝ 10 mL dil. soln ⎠ ⎝ 2.0 mol HCl ⎠
or
M C VC = M D VD
VC =
M D VD 0.250 M ( 50.0 mL )
= 6.3 mL conc. soln
=
MC
2.0 M
Add enough water to 6.3 mL 2.0 M HCl to yield a total volume of 50.0 mL
Copyright 2004 Mark Bishop
Chapter 5 Exercise Key
Copyright 2004 Mark Bishop