Unit 3 Review Answers
Student Textbook pages 314–317
Answers to Knowledge/Understanding
Questions
Multiple Choice
1. (e)
2. (c)
3. (b)
4. (d)
5. (b)
6. (d)
7. (e)
8. (c)
9. (d)
10. (e)
18.
19.
20.
Answers to Short Answer Questions
11. No. Some compounds have very limited solubility in
12.
13.
14.
15.
16.
water, so a saturated solution may not be concentrated. An example is calcium carbonate.
A homogeneous mixture is uniform throughout,
while a heterogeneous mixture may have more than
one layer or phase. They can be distinguished visually.
An example of a homogeneous mixture is a salt or
sugar solution while an example of a heterogeneous
mixture would be oil and water.
mass/mass percent, mass/volume percent, volume/
volume percent, molar concentration
Some of the sodium carbonate may crystallize out of
solution. This is due to the fact that decreasing the
temperature decreases the solubility of most solid
compounds in water.
Calcium ions have larger radii than magnesium ions
and therefore the ionic attraction between ions in
Ca(OH)2 is smaller than the attraction between ions
in Mg(OH)2. Water molecules are able to separate the
ions in Ca(OH)2 more easily compared with
Mg(OH)2, and calcium hydroxide is the more soluble
of the two.
Mass of 12 L water = 1.2 × 104 g
The maximum mass of iron before staining may
occur is 0.2 ppm
4
Mass of Fe in 12 L = 1.2 × 10 g × 0.26
10
= 2 × 10−3 g
17. They have the same mass/mass or mass/volume concentration, but the molar concentrations are different.
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MHR • Unit 3 Solutions and Solubility
21.
The molar mass of salt is less than the molar mass of
sugar. Therefore, for the same mass, there will be
more moles of salt than sugar, and the molar concentration of NaCl(aq) will be larger than the molar
concentration of sugar solution.
(a) NaOH is soluble. All common sodium salts are
soluble.
(b) KCl is soluble. All common potassium salts are
soluble.
(c) Ethanol is soluble in water because the OH group
forms hydrogen bonds with water and the hydrocarbon part of the molecule is small.
(d) Benzene is insoluble in water. Benzene is a hydrocarbon with no polar groups available to form
intermolecular bonds with water.
You should correct the teacher. The volumes are
correct, but the safe dilution of a strong concentrated
acid requires that the acid must be added to the water.
(a) box II
(b) box III
(c) box II
(a) Neither is soluble in water. Non-polar substances
are not soluble in polar solvents such as water.
(b) They may dissolve in fat particles in the lungs and
bloodstream.
Answers to Inquiry Questions
22. (a) i. Add the magnesium chloride to the boiling
water. This will increase its solubility therefore
increasing the overall concentration of the
solution.
ii. Add the benzene to the boiling water. Some of
it will evaporate, but some of it may also dissolve marginally in the boiling water.
iii. Add the CO to the cold water. The solubility
of a gas in water increases with decreasing
temperature.
(b) i. Add the magnesium chloride to cold water.
The solubility of most solids decrease with a
decrease in temperature.
ii. Add the benzene to cold water. It will not
dissolve at all as it is a non-polar substance
while water is highly polar.
iii. Bubble the CO through boiling water. Its
solubility will be very low as the solubility of a
gas in a liquid decreases with an increase in
temperature.
23. (a) Hard water: the student will observe a precipitate.
24.
25.
26.
27.
Ca2+(aq) + C2O42−(aq) → CaC2O4(s)
Mg2+(aq) + C2O42−(aq) → MgC2O4(s)
Soft water: a precipitate may or may not form,
depending on the concentration of Ca2+(aq) and
Mg2+(aq). If the water is very soft the concentration
of the two oxalates may be small enough that they
will dissolve completely:
Ca2+(aq) + C2O42−(aq) → NR
Mg2+(aq) + C2O42−(aq) → NR
Distilled water: no reaction, because there are no
calcium or magnesium ions present.
(b) Add a solution of sodium carbonate. The carbonate ion will form a precipitate in a solution
containing Ca2+(aq) or Mg2+(aq).
Any of the group 1 ions, for example:
HC2H3O2(aq) + OH−(aq) + Na+(aq) →
C2H3O2−(aq) + H2O(l) + Na+(aq)
HC2H2O2(aq) + OH−(aq) + NH4+(aq) →
C2H3O2−(aq) + H2O(l) + NH4+(aq)
(a) Fe3+(aq) + 3Cl−(aq) + 3Cs+(aq) + PO43−(aq) →
FePO4(s) + 3Cs+(aq) + 3Cl−(aq)
Fe3+(aq) + PO43−(aq) → FePO4(s)
(b) 2Na+(aq) + 2OH−(aq) + Cd2+(aq) + 2NO3−(aq) →
Cd(OH)2(s) + 2Na+(aq) + 2NO3−(aq)
Cd2+(aq) + 2OH−(aq) → Cd(OH)2(s)
(c) NH4+(aq) + ClO4−(aq) + Na+(aq) + Br−(aq) → NR
Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(l)
Ca2+(aq) + 2OH−(aq) + 2H+(aq) + 2NO3−(aq) →
Ca2+(aq) + 2NO3−(aq) + 2H2O(l)
H+(aq) + OH−(aq) → H2O(l)
Pb(NO3)2(aq) + 2NaI(aq) →
PbI2(s) + 2NaNO3(aq)
−
2+
Pb (aq) + 2I (aq) → PbI2(s)
molar mass PbI2 = 461.0 g/mol
0.628 g
amount PbI2 = 461.0 g/mol = 1.36 × 10−3 mol
amount of Pb2+(aq) = 1.36 × 10−3 mol PbI2 × 1 mol Pb2+
1 mol PbI2
−3
= 1.36 × 10 mol
29. C6H8O6(aq) + Br2(aq) → 2HBr(aq) + C6H6O6(aq)
amount Br2(aq) = 0.02265 L × 0.134 mol/L
= 3.04 × 10−3 mol
amount of vitamin C
1 mol C H O
= 3.04 × 10−3 mol Br2 × 1 mol6Br8 6
−3
10 mol
concentration of vitamin C = 3.040×.125
L
= 2.43 × 10−2 mol/L
30. (a) Mg3(PO4)2(aq) + 3H2SO4(aq) →
3MgSO4(aq) + 2H3PO4(aq)
(b) amount H2SO4 = 0.0500 L × 0.100 mol/L
= 5.00 × 10−3 mol
amount H3PO4
2 mol H PO
= 5.00 × 10−3 mol H2SO 4 × 3 mol H3SO 4
2
4
= 3.33 × 10−3 mol
10−3 mol
concentration of H3PO4 = 3.330.×0450
L
= 7.41 × 10−2 mol/L
31. (a) 2NH4Cl(aq) + Pb(NO3)2(aq) →
PbCl2(s) + 2NH4NO3(aq)
(b) amount NH4Cl = 0.0210 L × 0.125 mol/L
= 2.62 × 10−3 mol
amount PbCl2
1 mol PbCl
32. (a)
(b)
(c)
(d)
(e)
10−3 mol
molar concentration Pb2+(aq) = 1.360×.035
L
= 3.9 × 10−2 mol/L
28. Amount HCl(aq) = 0.155 L × 1.25 × 10−2 mol/L
= 1.94 × 10−3 mol
2
= 3.04 × 10−3 mol
33. (a)
2
= 2.62 × 10−3 mol NH 4Cl × 2 mol NH Cl
4
= 1.31 × 10−3 mol
mass PbCl2 = 1.31 × 10−3 mol × 278.1 g/mol
= 0.364 g
119 g
You would observe undissolved solute, and a clear,
colourless liquid. This is a heterogeneous mixture.
The solution above the undissolved solute is saturated.
Heating the system to 100°C will allow all the
solute to dissolve. Only a clear, colourless solution
will be visible; this is a solution (homogeneous
mixture). The solution will be saturated
The solution will be unsaturated.
You would observe a clear, colourless liquid above
some undissolved solute. This is a heterogeneous
mixture. The solution above the undissolved
solute will be saturated.
NaCl(aq) → Na+(aq) + Cl−(aq)
amount Cl−
mol Cl −
= 0.025 L × 1.25 mol/L NaCl × 11mol
NaCl
= 0.031 mol Cl − (aq)
Unit 3 Review Answers • MHR
127
(b) ZnCl2(aq) → Zn2+(aq) + 2Cl−(aq)
amount Cl−
mol Cl −
= 1.70 L × 1.0 × 10−3 mol/L ZnCl 2 × 1 2mol
ZnCl2
−3
−
= 3.4 × 10 mol Cl (aq)
Answers to Communication Questions
34. (a) NaI(aq) will conduct a current because sodium
iodide is an electrolyte.
(b) HBr(aq) will conduct electricity because hydrobromic acid is a strong acid and dissociates 100%
into ions.
35. (a) 2NaOH(aq) + Zn(NO3)2(aq) →
Zn(OH)2(s) + 2NaNO3(aq)
+
2Na (aq) + 2OH−(aq) + Zn2+(aq) + 2NO3−(aq) →
Zn(OH)2(s) + 2Na+(aq) + 2NO3−(aq)
Zn2+(aq) + 2OH−(aq) → Zn(OH)2(s)
(b) MgBr2(aq) + KCH3COO(aq) → NR
(c) Ag2SO4(aq) + BaCl2(aq) → 2AgCl(s) + BaSO4(s)
The total ionic equation is the same as the net
ionic equation:
2Ag+(aq) + SO42−(aq) + Ba2+(aq) + 2Cl−(aq) →
2AgCl(s) + BaSO4(s)
(d) Na2SO4(aq) + Sr(NO3)2(aq) → SrSO4(s) + 2NaNO3(aq)
2Na+(aq) + SO42−(aq) + Sr2+(aq) + 2NO3−(aq) →
SrSO4(s) + 2Na+(aq) + 2NO3−(aq)
Sr2+(aq) + SO42−(aq) → SrSO4(s)
36. Molar mass C2H2O4 = 90.04 g/mol
0.5165 g
amount of oxalic acid = 90.04 g/mol
= 5.736 × 10−3 mol
concentration of 100 mL solution
= 5.736 × 10−3 mol ÷ 0.100 L
= 5.736 × 10−2 mol/L
10.00 mL × 5.736 × 10−2 mol/L = 250.0 mL × Cf
Cf = 2.295 × 10−3 mol/L
37. Wear gloves and safety goggles during the preparation
of all solutions
(a) Molar mass NiCl2 = 129.62 g/mol
mass NiCl2 required
= 1.00 L × 3.00 mol/L × 129.62 g/mol
= 389 g
Dissolve 389 g of NiCl2 in approximately 800 mL
of distilled water in a one litre volumetric flask.
Make the volume up to 1.00 L.
(b) Molar mass CuCl2 = 134.45 g/mol
mass CuCl2 required
= 0.250 L × 4.00 mol/L × 134.45 g/mol
= 134 g
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MHR • Unit 3 Solutions and Solubility
Dissolve 134 g of CuCl2 in approximately 200
mL of distilled water in a 250 mL volumetric
flask. Make the volume up to 250 mL.
(c) Molar mass MnSeO4 = 197.90 g/mol
mass MnSeO4 required
= 0.500 L × 0.133 mol/L × 197.90 g/mol
= 13.2 g
Mass 13.2 g of MnSeO4 and dissolve it in approximately 400 mL of distilled water in a 500 mL
volumetric flask. Make the volume up to 500 mL.
38. (a) Molar mass Fe(NO3)2 = 179.86 g/mol
5.23 g
concentration =
÷ 0.100 L
179.86 g/mol
= 0.291 mol/L
(b) Molar mass Pb(ClO4)2 = 406.10 g/mol
44.3 g
concentration =
÷ 0.2500 L
406.10 g/mol
= 0.436 mol/L
(c) Molar mass CoSO4 = 155.00 g/mol
9.94 g
concentration =
÷ 0.250 L
155.00 g/mol
= 0.257 mol/L
39.
Vi =
Cf × Vf
mol/L × 4.7 L = 1.4 L
= 3.511.7
Ci
mol/L
The rock carver should wear gloves and a full face
shield, and make the dilution in an operating fume
hood. About 3 L of water should be put into a large
glass or plastic container and 1.4 L of concentrated
acid added slowly, with stirring. Once the solution
has cooled, water can be added to the final volume of
4.7 L.
40. Volume of solution = 2.11 mol = 2.14 L
0.988 mol/L
The 2.5 L container must be used. The other two
containers are too small.
41. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
amount HCl = 2.11 mol/L × 2.68 L = 5.65 mol
amount NaOH = 2.28 mol/L × 3.17 L = 7.23 mol
Hydrochloric acid is limiting, therefore 5.65 mol
NaCl is formed.
Assuming the total volume is the sum of the volumes
added, the total volume = 2.68 + 3.17 = 5.85 L
.65 mol = 0.966 L
concentration NaCl(aq) = 55.85
L
mol
42. (a) Molar concentration CaCl2(aq) = 1.68
4.00 L
(b) CaCl2(aq) → Ca2+(aq) + 2Cl−(aq)
= 0.420 mol/L
concentration Ca2+(aq)
mol Ca 2+
= 0.420 mol/L CaCl 2 × 11mol
CaCl2
= 0.420 mol/L
concentration Cl−(aq)
mol Cl −
= 0.420 mol/L CaCl 2 × 1 2mol
CaCl2
= 0.840 mol/L
43. 2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l)
amount HCl = 2.22 mol/L × 3.00 L = 6.66 mol
amount Ba(OH)2 = 0.128 mol/L × 4.00 L
= 0.512 mol
The Ba2+ and Cl− ions are spectator ions that remain
in solution.
mol
concentration of Ba2+(aq) = 0.512
8.00 L = 0.0640 mol/L
mol
concentration of Cl−(aq) = 6.66
8.00 L = 0.832 mol/L
amount of Ba(OH)2 required to react with HCl
1 mol Ba(OH)
= 6.66 mol HCl × 2 mol HCl 2
= 3.33 mol Ba(OH)2
The amount of barium hydroxide is less, therefore.
Ba(OH)2 is the limiting reactant and there is excess
HCl(aq).
Amount of HCl that reacts
2 mol HCl
= 0.512 mol Ba(OH)2 × 1 mol
Ba(OH)
2
= 1.02 mol
Excess HCl = 6.66 mol − 1.02 mol = 5.64 mol
mol
concentration H+(aq) = 58.64
.00 L = 0.705 mol/L
The concentration of OH−(aq) will be negligible.
44. (a) Amount HCl = 12 mol/L × 0.150 L = 1.8 mol
(b) The amount of HCl is unchanged; the concentration will be less.
.8 mol = 0.90 mol/L
(c) Concentration HCl(aq) = 12.00
L
45. (a) Amount NaCl = 0.850 mol/L × 0.250 L
= 0.212 mol
(b) mass NaCl = 58.44 g/mol × 0.212 mol = 12.4 g
(c) mass 12.4 g of sodium chloride and dissolve it in
about 200 mL of distilled water. Make the solution up to 250 mL in a volumetric flask having
this volume.
Answers to Making Connections Questions
46. (a) Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g)
(b) Mg(OH)2(s) + 2HNO3(aq) →
Mg(NO3)2(aq) + 2H2O(l)
This is an acid-base displacement reaction.
Magnesium nitrate is soluble and can be carefully
washed off the tape.
47. (a) Cesium compounds are probably more soluble
than strontium compounds. The Cs+ ion has a
smaller charge and a larger radius compared with
the Sr2+ ion. Both of these factors tend to reduce
ionic bond strength, allowing water molecules to
separate the ions in cesium salts more easily, leading to more soluble compounds.
(b) A number of valid answers are possible, among
these are the half-life of the isotopes, the type of
radioactivity emitted and the relative amount of
each isotope.
Unit 3 Review Answers • MHR
129