Homework Questions – Chapter 4 – Chemical Reactions
1. What allows water molecules to ‘stick’ to ions?
2. Show the partial charges on the following water molecule: HOH
3. For a soluble ionic compound, which is stronger; all the ionic bonds between ions in the solid or all the ‘bonds’
between water molecules and ions?
4. What is the name of the loose layer of water molecules that surrounds an ion in an aqueous solution?
5. Which part of the water molecule ‘bonds’ to an anion in an aqueous solution?
6. What two things are required for a substance to conduct electricity (a strong electrolyte solution, for example)?
7. What type(s) of compound; (a) dissociates in water? (b) completely ionizes in water? (c) partially ionizes in
water? (d) does not ionize or dissociate in water? (e) partially dissociates in water?
8. Predict what the light bulb conductivity apparatus would do if you initially used a barium hydroxide solution, and
then slowly added sulfuric acid solution until it exactly neutralized the base (neither reactant is in excess – this is
called the ‘stoichiometric ratio’ of reactants).
9. What is the molarity of; (a) 2.31 L of a solution containing 0.187 moles of solute? (b) 4.19 mL of a solution
containing 0.0388 moles of solute? (c) 88.6 mL of a solution containing 14 g of magnesium nitrate? (d) nitrate
ions in the solution in part c?
10. What volume of a 0.01224 mol/L solution (mL) contains 0.3867 moles of solute?
11. How many moles of sulfuric acid are in 1500. mL of a 3.921 M sulfuric acid solution? How many moles of H+
ions are in the sample?
12. Water is added to 12.0 M HCl to make 100. mL of a 0.1500 M HCl solution. How much 12.0 M HCl is used?
13. What is the molarity of the diluted solution if 86.7 mL water is added to13.3 mL of 6.0 M NaCl?
14. (a) A 3.00 M solution of compound X has a density of 1.90 g/mL. The bottle containing the solution (and the
solution inside) has an initial mass of 439.214 g. You wish to dilute a sample of this solution to make 100.0 mL
of a 0.500 M solution in a volumetric flask. If you pour the 3.00 M solution directly from the bottle into the
volumetric flask, what should be the final mass of the bottle (and remaining solution)?
(b) Follow up question of the same type as part (a), to be done for practice after working through the solution on
D2L: A 2.60 M solution of compound Y has a density of 1.45 g/mL. The bottle containing the solution (and the
solution inside) has an initial mass of 839.725 g. You wish to dilute a sample of this solution to make 140.0 mL
of a 0.370 M solution in a volumetric flask. If you pour the 2.60 M solution directly from the bottle into the
volumetric flask, what should be the final mass of the bottle (and remaining solution)?
15. For the reaction; Ba(OH)2 (aq) + H2SO4 (aq)  BaSO4 (s) + 2 H2O (l)
(a) How many moles of Ba(OH)2 react with 2.28 mol H2SO4?
(b) How many liters of 1.80 M Ba(OH)2 solution reacts with 2.28 mol H2SO4?
(c) What is the molarity of the Ba(OH)2 solution if 413 mL reacts with 0.259 mol H2SO4?
(d) How many mL of 0.54 M H2SO4 solution reacts with 0.92 L of 1.04 M Ba(OH)2 solution?
16. For the reaction; Ba(NO3)2 (aq) + K2SO4 (aq)  BaSO4 (s) + 2 KNO3 (aq)
(a) How many moles of KNO3 are produced from 0.773 mol K2SO4?
(b) How many moles of KNO3 are produced from 488 mL of 1.35 M K2SO4 solution?
(c) What volume of KNO3 solution is produced from 488 mL of 1.35 M K2SO4 solution and 300. mL of Ba(NO3)2
solution?
(d) What is the molarity of the KNO3 solution produced from 488 mL of 1.35 M K2SO4 solution and 300. mL of
Ba(NO3)2 solution?
(e) What is the molarity of the KNO3 solution produced from 0.275 L of 0.727 M K2SO4 solution and 300. mL of
0.633 M Ba(NO3)2 solution?
(f) What is the molarity of the KNO3 solution produced from 0.550 L of 0.727 M K2SO4 solution and 300. mL of
0.633 M Ba(NO3)2 solution?
(g) What is the molarity of the KNO3 solution produced from 0.275 L of 0.727 M K2SO4 solution and 800. mL of
0.633 M Ba(NO3)2 solution?
(h) Follow up question of the same type as part (g), to be done for practice after working through the solution on
D2L: What is the molarity of the KNO3 solution produced from 0.789 L of 0.848 M K2SO4 solution and 700.
mL of 0.929 M Ba(NO3)2 solution?
(i) What is the mass of KNO3 produced in part (g) – note: NOT the mass of the solution.
(j) How many moles of excess reactant remains at the end of the reaction in part (g)?
(k) Follow up question of the same type as part (j), to be done for practice after working through the solution on
D2L: How many moles of excess reactant remains at the end of the reaction in part (h)?
17.
18.
19.
20.
21.
22.
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24.
25.
26.
27.
(l) What is the molarity of NO3̶ ions in the solution at the end of the reaction in part (g)? (hint: think about ALL
the nitrate ions in the solution, not just those in KNO3)
For the following pairs of reactants; (i) write the complete, molecular equation and balance it, showing all states
(write NR if no reaction occurs), (ii) write the net ionic equation, showing all states.
(a) Ca(OH)2 (aq) + MgSO4 (aq) 
(b) Cu(OH)2 (s) + H2SO4 (aq) 
(c) Ba(C2H3O2)2 (aq) + RbOH (aq) 
(d) Pb(C2H3O2)2 (aq) + CuSO4 (aq) 
(e) Sr(OH)2 (aq) + H2SO4 (aq) 
(f) FeCl3 (aq) + Li3PO4 (aq) 
(g) BaCl2 (aq) + K2S (aq) 
(h) NiBr2 (aq) + Na2S (aq) 
(j) Cs2CrO4 (aq) + Na2SO4 (aq) 
(k) HC2H3O2 (aq) + NaOH (aq) 
What is the oxidation number of each element in the following;
(a) CaCO3
(b) CrO42
(c) I2
(d) H2O2
(e) COCl2
(f) IF3
(g) Fe3+
(h) CaH2
For the following reactants; (i) write the complete, molecular equation and balance it, showing all states (write
NR if no reaction occurs), (ii) write and label the oxidation and reduction half-reactions.
(a) Cu (s) + Sn(NO3)2 (aq) 
(b) Ag (s) + H2SO4 (aq) 
(c) Sn (s) + Cu(NO3)2 (aq) 
(d) Ca (s) + H2O (l) 
(e) Al (s) + NiSO4 (aq) 
(f) Zn (s) + HCl (aq) 
(g) Mn (s) + NiCl2 (s) 
(h) Mg (s) + H2O (l) 
For the following reactions; (i) write the complete, molecular equation and balance it, (ii) write and label the
oxidation and reduction half-reactions.
(a) Mg (s) + Br2 (l) 
(b) Al (s) + O2 (g) 
For the following reactants, write the balanced molecular equation, showing all states.
(a) C4H8 (g) + O2 (g) 
(b) C5H8O (l) + O2 (g) 
(c) C10H14O3 (s) + O2 (g) 
(d) C2H5OH (l) + O2 (g) 
Write and label the oxidation and reduction half-reactions for the following reactions.
(a) S8 (s) + 8 O2 (g)  8 SO2 (g)
(b) C2H6O (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
Give definitions for the following terms;
(a) molarity (b) ionization (c) dissociation (d) Redox reaction (e) solvation shell (f) spectator ion
Explain how an ionic compound dissolves in water. (Role of Water as a Solvent)
Explain what happens in an ionic soup during a precipitation reaction (use NaCl (aq) + AgNO3 (aq) as an example).
Draw quantitative ion-soup diagrams for before and after the following reactions;
(a) 3 molecules of HCl and 1 formula unit of Ca(OH)2 in water
(reaction is; 2 HCl (aq) + Ca(OH)2 (aq)  2 H2O (l) + CaCl2 (aq))
(b) 1 formula unit of Fe2(SO4)3 and 4 formula units of BaS in water
(reaction is; Fe2(SO4)3 (aq) + 3 BaS (aq)  Fe2S3 (s) + 3 BaSO4 (s))
(c) 1 formula unit of Na3PO4 and 3 formula units of AgNO3 in water
(reaction is; Na3PO4 (aq) + 3 AgNO3 (aq)  3 NaNO3 (aq) + Ag3PO4 (s))
Draw qualitative ion-soup diagrams for before and after the reactions described in question 26.
Answers
1. Partial charges on the water molecules are attracted to the full charges on the ions.
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6.
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19.
+ - +
HOH
All the ‘bonds’ between water molecules and ions – this is what allows the ionic compound to dissolve.
Solvation shell.
H-atoms (they have a + partial charge that ‘bonds’ to the negative charge on the anion.
(1) Charged particles (2) the charged particles must be mobile.
(a) soluble ionic compounds (b) strong acids (c) weak acids and weak bases (d) insoluble covalent compounds
and soluble covalent compounds other than strong acids or weak acids/bases (e) partially soluble (“insoluble”)
ionic compounds.
The light would initially be bright (Ba(OH)2 solution is a strong electrolyte). As the acid (another strong
electrolyte) is added, the following reaction would occur; Ba(OH)2 + H2SO4  H2O + BaSO4 Since water is a
non-electrolyte and BaSO4 is an insoluble ionic compound, both products would be non-electrolytes.
Consequently, as the reaction proceeds (acid is added), non-electrolytes replace electrolytes in the solution, so the
solution gradually becomes a weaker electrolyte and the light would grow dimmer. When the base is exactly
neutralized, the only things in the solution are non-electrolytes (no excess reactants are present), so the solution is
non-electrolytic and the light goes out.
(a) 0.0810 M (b) 9.26 M (c) 1.1 M (d) 2.1 M
31590 mL
5.882 mol sulfuric acid, 11.76 mol H+ ions.
1.25 mL
0.80 M
(a) 16.6667 mL of solution is needed, which has a mass of 31.6667 g. The final mass should be 407.5 g.
(b) 19.923 mL of solution is needed, which has a mass of 28.888 g. The final mass should be 810.8 g.
(a) 2.28 mol
(b) 1.27 L
(c) 0.627 M
(d) 1800 mL
(a) 1.55 mol
(b) 1.32 mol
(c) 788 mL
(d) 1.67 M
(e) 0.661 M
(f) 0.447 M
(g) 0.372 M
(h) 0.873 M
(i) 40.4 g
(j) 0.306 mol
(k) 0.019 mol (l) 0.9420 M
(a) (i) Ca(OH)2 (aq) + MgSO4 (aq)  CaSO4 (s) + Mg(OH)2 (s)
(ii) Ca2+(aq) + 2 OH (aq) + Mg2+(aq) + SO42 (aq)  CaSO4 (s) + Mg(OH)2 (s)
(b) (i) Cu(OH)2 (s) + H2SO4 (aq)  CuSO4 (aq) + 2 H2O (l)
(ii) Cu(OH)2 (s) + 2 H+(aq)  Cu2+(aq) + 2 H2O (l)
(c) (i) Ba(C2H3O2)2 (aq) + RbOH (aq)  NR
(ii) Ba2+(aq) + 2 C2H3O2 (aq) + Rb+(aq) + OH (aq)  NR
(d) (i) Pb(C2H3O2)2 (aq) + CuSO4 (aq)  PbSO4 (s) + Cu(C2H3O2)2 (aq)
(ii) Pb2+(aq) + SO42 (aq)  PbSO4 (s)
(e) (i) Sr(OH)2 (aq) + H2SO4 (aq)  SrSO4 (s) + 2 H2O (l)
(ii) Sr2+(aq) + 2 OH (aq) + 2 H+(aq) + SO42 (aq)  SrSO4 (s) + 2 H2O (l)
(f) (i) FeCl3 (aq) + Li3PO4 (aq)  FePO4 (s) + 3 LiCl (aq)
(ii) Fe3+(aq) + PO43 (aq)  FePO4 (s)
(g) (i) BaCl2 (aq) + K2S (aq)  NR
(ii) Ba2+(aq) + 2 Cl (aq) + 2 K+(aq) + S2 (aq)  NR
(h) (i) NiBr2 (aq) + Na2S (aq)  NiS (s) + 2 NaBr (aq)
(ii) Ni2+(aq) + S2 (aq)  NiS (s)
(j) (i) Cs2CrO4 (aq) + Na2SO4 (aq)  NR
(ii) 2 Cs+(aq) + CrO42 (aq) + 2 Na+(aq) + SO42 (aq)  NR
(k) (i) HC2H3O2 (aq) + NaOH (aq)  H2O (l) + NaC2H3O2 (aq)
(ii) HC2H3O2 (aq) + OH (aq)  H2O (l) + C2H3O2 (aq)
(a) Ca, +2; C, +4; O, -2
(b) Cr, +6; O, -2
(c) I, 0
(d) H, +1; O, -1
(e) C, +4; O, -2; Cl, -1
(f) I, +3; F, -1
(g) Fe, +3
(h) Ca, +2; H, -1
(a) (i) Cu (s) + Sn(NO3)2 (aq)  NR
(ii) NR
(b) (i) Ag (s) + H2SO4 (aq)  NR
(ii) NR
(c) (i) Sn (s) + Cu(NO3)2 (aq)  Cu (s) + Sn(NO3)2 (aq)
(ii) Oxidation: Sn0  Sn2+ + 2 e
Reduction: Cu2+ + 2 e  Cu0
(d) (i) Ca (s) + 2 H2O (l)  Ca(OH)2 (aq) + H2 (g)
(ii) Oxidation: Ca0  Ca2+ + 2 e
Reduction: 2 H+ + 2 e  2 H0
(e) (i) 2 Al (s) + 3 NiSO4 (aq)  3 Ni (s) + Al2(SO4)3 (aq)
(ii) Oxidation: 2 Al0  2 Al3+ + 6 e
Reduction: 3 Ni2+ + 6 e  3 Ni0
(f) (i) Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) 
(ii) Oxidation: Zn0  Zn2+ + 2 e
Reduction: 2 H+ + 2 e  2 H0
(g) (i) Mn (s) + NiCl2 (s)  NR (Nickel chloride is a solid, not a solution, so its ions are not mobile)
(ii) NR
(h) (i) Mg (s) + H2O (l)  NR
(ii) NR
20. (a) (i) Mg (s) + Br2 (l)  MgBr2 (s)
(ii) Oxidation: Mg0  Mg2+ + 2 e
Reduction: 2 Br0 + 2 e  2 Br
(b) (i) 4 Al (s) + 3 O2 (g)  2 Al2O3 (s)
(ii) Oxidation: 4 Al0  4 Al3+ + 12 e
Reduction: 6 O0 + 12 e  6 O2
21. (a) C4H8 (g) + 6 O2 (g)  4 CO2 (g) + 4 H2O (g) (these reactions create heat, so water is usually a gas)
(b) 2 C5H8O (l) + 13 O2 (g)  10 CO2 (g) + 8 H2O (g)
(c) C10H14O3 (s) + 12 O2 (g)  10 CO2 (g) + 7 H2O (g)
(d) C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
22. Write and label the oxidation and reduction half-reactions for the following reactions. (4.4)
(a) Oxidation: 8 S0  8 S4+ + 32 e
Reduction: 16 O0 + 32 e  16 O2(b) C2H6O (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
Oxidation: 2 C2-  2 C4+ + 12 e
Reduction: 6 O0 + 12 e  6 O223. (a) The concentration of a solution, measured in moles of solute per liter of solution (b) The breaking apart of a
molecule (which does not contain ions) to produce ions (c) The breaking apart of an ionic compound (which is
made of ions) to produce separate ions (d) A reaction in which at least one element gains one or more electrons,
and another element loses one or more electrons (e) Solvent molecules (usually water) loosely attached to an ion
in solution. The solvent molecules surround the ion in one or more layers, creating a ‘shell’. (f) An ion that is
present in solution during a chemical reaction, but which does not take part in the reaction.
24. Randomly moving water molecules that bump into an ionic solid may stick to the outer ions due to the attraction
between the partial charges on the water molecules (+ and -) and the ionic charges on the ions. At the same
time the ions within the solid are vibrating. If enough water molecules attach to an ion at just the right moment in
its vibration, the ion may break off the solid and become surrounded by water molecules (which form a solvation
shell). The loss of favorable ionic bonds between the ion and the rest of the solid is more than compensated by
the creation of many new attractive interactions between the ion and the water molecules of its solvation shell.
This process is repeated until the ionic solid is entirely broken apart and the solid is said to be dissolved.
25. The reactants in a precipitation reaction are soluble ionic compounds, so the ions are separated (the compounds
dissociate) and float around freely in the solution, surrounded by their solvation shells (we can think of this as an
‘ion-soup’). For example, in the reaction between sodium chloride solution and silver nitrate solution, Na+, Cl-,
Ag+, and NO3- ions will be present. These free-floating ions randomly bump into each other. If ions of the same
charge bump into each other, they repel. If ions from the same reactant bump into each other, they do not stick
together because they could only reform the original reactant; this doesn’t happen because the soluble reactants
prefer to dissociate into ions in water (which means their ions prefer to be separate in solution, surrounded by
their solvation shells, rather than stuck together in a solid lattice). If oppositely charged ions from different
reactants bump into each other, they must ‘decide’ if they prefer to stick together and form a solid lattice (if they
form an insoluble compound), or prefer to remain separate (if they form a soluble compound). In our example,
when Na+ bumps into NO3-, the ions prefer to remain separate, because sodium nitrate is soluble. When Ag+
bumps into Cl-, they prefer to stick together to form a solid lattice (they form insoluble silver chloride), resulting
in the appearance of a solid precipitate in the solution.
26. (a)
H+
Cl−
H+
Cl−
OH−
Ca2+
OH
Cl−
H+
After Reaction
S2−
Fe3+
2−
SO
4
Ba2+
Ba2+
Fe3+
S2−
S2− SO42−
Ba2+
Ba2+ 2− SO42−
S
Ba2+
S2−
Fe2S3 BaSO4 BaSO4 BaSO4
Before Reaction
(c)
Cl−
Cl−
H+
Cl−
−
Before Reaction
(b)
Ca2+
PO43−
Na+
−
NO
3
Ag+
Ag+
Na+
NO3− Ag+
NO3−
Na+
After Reaction
NO3−
Na
Na+
+
NO3−
Na
NO3−
+
Ag3PO4
Before Reaction
27. (a)
H+
After Reaction
Ca2+
Ca
2+
Cl−
Cl−
OH−
Before Reaction
(b)
SO42−
3+
Fe
H+
After Reaction
Ba2+
S2−
S2−
Ba2+
Fe2S3 BaSO4
Before Reaction
(c)
After Reaction
PO43−
Ag+
Na+
NO3−
Na+
NO3−
Ag3PO4
Before Reaction
After Reaction