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Unit B Chapter 4
Electrolytes
Precipitation (aka Double Replacement or Metathesis) Reactions
Acids & Bases
Neutralization Reactions
Oxidation Reduction
Concentration of Solutions
Solution Stoichiometry
Sucrose with Sulfuric Acid
1. put 2-3 pellets NaOH in
water with universal
indicator
1) Why this
Color?
2) What was the
temperature? the
pH?
2
Just what was that sugar reaction?
•
•
•
Sucrose or common table sugar was mixed
with concentrated sulfuric acid. Soon an
exothermic reaction takes places during which
a column of carbon rises from the beaker and a
cloud of steam is produced. Concentrated
sulfuric acid acts as a catalyst to dehydrate
sucrose to produce carbon and water. The heat
of the reaction vaporizes the water, and the
gas causing the column of carbon puff up, just
like gases during cooking cause a cake to rise.
The name carbohydrate derives from the
formula of sugars such as sucrose, C12H22O11
(notice the 2:1 H:O ratio C12(H2O)11) in which
the formula appears to be a hydrate of carbon.
C12H22O11(s) → 11H2O(g) + 12C(s)
3
Reactions in Aqueous
Solutions
Unit B Chapter 4
slide view
4
133 g of aluminum chloride in 1 L of
water would be considered (select all that apply)
1.
2.
3.
4.
5.
6.
soluble
ionized
dissociated
aqueous
concentrated
dilute
7. weak
electrolyte
8. strong
electrolyte
9. acidic
5
133 g of AlCl3 in 1 L of water would be
3+
−
considered
AlCl3 → Al + 3Cl
1.
2.
3.
4.
5.
6.
soluble
ionized
dissociated
aqueous
concentrated
dilute
generally <0.1 M is
considered dilute
7. weak
electrolyte
8. strong
electrolyte
9. acidic
(for reasons you
don’t know yet…)
slide show
6
Electrolytes
• Compounds that dissolve and dissociate into ions
Strong or weak = amount ionized
✓ Concentrated or dilute = amount dissolved
✓
• Electrolytes are
Soluble ionic compounds
✓ Acids or bases
✓
• Molecular compounds tend to be non-electrolytes
7
Dissolving & Dissociating
• When ionic compounds dissolve they
dissociate
✓
Observe the orientation of the water molecules as the
solvate around the dissociated ions
• When molecular compounds dissolve, they
do not dissociate
Slide View
8
Which solution represented below is a
nonelectrolyte?
1
2
3
9
Which solution represented below is a
nonelectrolyte?
1.
•
•
1
2
3
there are no ions in the solution
Which representation would
conduct electricity the best?
10
Rank the following in order of increasing
conductivity in water:
(equal moles of each substance put in the water)
C6H12O6, KBr, HF, Zn(OH)2
Least conductive → more conductive
1. C6H12O6 < KBr < HF < Zn(OH)2
2. Zn(OH)2 < C6H12O6 < HF < KBr
3. C6H12O6 < Zn(OH)2 < HF < KBr
4. C6H12O6 < HF < Zn(OH)2 < KBr
5. C6H12O6 < Zn(OH)2 < KBr < HF
6. KBr < Zn(OH)2 < HF < C6H12O6
11
Rank the following in order of increasing conductivity
in water:
(equal moles of each substance put in the water)
C6H12O6, KBr, HF, Zn(OH)2
Remember that conductivity is related to the amount of ions in
solution
1. C6H12O6 < KBr < HF < Zn(OH)2
2. Zn(OH)2 < C6H12O6 < HF < KBr
3. C6H12O6 < Zn(OH)2 < HF < KBr
•
Molecular sugar is least conductive, next is zinc hydroxide,
because although it is a base (and actually a strong base), it is
very insoluble, hydrofluoric acid is weak, and KBr is a soluble salt.
• Time to memorize the 7 strong acids: HCl, HBr, HI, HNO3, H2SO4,
HClO3, HClO4
4. C6H12O6 < Zn(OH)2 < KBr < HF
5. KBr < Zn(OH)2 < HF < C6H12O6
12
Rank the following in order of increasing
conductivity in water:
(equal moles of each substance put in the same amount of water)
C3H7OH, KCl, HClO, AlCl3
1. C3H7OH < HClO < KCl < AlCl3
2. AlCl3 < C3H7OH < KCl <HClO
3. HClO < C3H7OH < KCl < AlCl3
4. C3H7OH < HClO < KCl = AlCl3
5. C3H7OH < KCl < AlCl3 < HClO
6. C3H7OH < HClO < AlCl3 < KCl
13
Rank the following in order of increasing conductivity in water:
(equal moles of each substance put in the water)
C3H7OH, KCl, HClO, AlCl3
1.
C3H7OH < HClO < KCl < AlCl3
•
2.
C3H7OH is an alcohol, it is molecular and therefore least
conductive.
HClO, perchlorous acid is a weak acid.
KCl, AlCl3 are salts and you might think are equally
conductive, but because AlCl3 dissociates into 4 ions (Al+3
and 3 Cl−) it produces more ions per equal concentration and
thus is a more conductive solution.
AlCl3 < C3H7OH < KCl <HClO
3.
HClO < C3H7OH < KCl < AlCl3
4.
C3H7OH < HClO < KCl = AlCl3
5.
C3H7OH < KCl < AlCl3 < HClO
•
•
14
Calculate the concentration of chloride
ion when 5.8 g of sodium chloride and
4.8 g of magnesium chloride is
dissolved to produce 400 ml of
solution. NaCl 58.5g/mol No Calculator
MgCl2 95.21
1.
2.
3.
4.
0.05 M
0.10 M
0.15 M
0.20 M
5.
6.
7.
8.
0.50 M
2.0 M
0.40 M
0.75 M
15
Calculate the concentration of chloride ion when
5.8 g of sodium chloride and 4.8 g of magnesium
chloride is dissolved to produce 400 ml of solution.
1. 0.05 M No Calculator 5. 0.50 M
2. 0.10 M
6. 2.0 M
3. 0.15 M
7. 0.40 M
4. 0.20 M
8. 0.75 M
⎛ 1mol ⎞
−
5.8g ⎜
=
0.1molNaCl
=
0.1molCl
⎟
⎝ 58g ⎠
⎛ 2Cl ⎞
⎛ 1mol ⎞
−
4.8g ⎜
= 0.05molMgCl2 × ⎜
=
0.1molCl
⎟
⎝ 95g ⎠
⎝ 1MgCl2 ⎟⎠
−
⎛ 0.2molCl ⎞
−
=
0.5MforCl
⎜⎝ 0.4 L ⎟⎠
−
16
When 30. ml of 0.50 M K2SO4 are
added to 70. mL of 0.50 M of KMnO4,
the resulting concentration of potassium
is
1. 1.0
2. 0.85
3. 0.65
4. 0.50
5. 0.33
No Calculator
K2SO4 174 g/mol
KMnO4 158 g/mol
17
When 30. ml of 0.50 M K2SO4 are added to 70. mL of 0.50
M of KMnO4, the resulting concentration of potassium is
1.
2.
3.
4.
5.
1.0
0.85
0.65
0.50
0.33
No Calculator
⎛ 2K ⎞
+
(30ml)(0.5M ) = 15mmolK 2 SO4 × ⎜
=
30mmolK
⎟
⎝ 1K 2 SO4 ⎠
+
(70ml)(0.5M ) = 35mmolKMnO4 = 35mmolK
+
⎛ 65mmol ⎞
+
⎜⎝
⎟⎠ = 0.65molarK
100ml
18
Write the reaction that best represents
the dissolving of magnesium chloride
in water
1. MgCl → Mg(aq) + Cl(aq)
+
−
2. MgCl → Mg + Cl
2+
−
3. MgCl2 → Mg + (Cl )2
2+
−
4. MgCl2 → Mg + Cl 2
5. MgCl2 → Mg + Cl2
2+
−
6. MgCl2 → Mg + 2Cl
19
Write the reaction that best represents the
dissolving of magnesium chloride in water
1.
2.
3.
4.
5.
6.
•
•
MgCl → Mg(aq) + Cl(aq)
+
−
MgCl → Mg + Cl
MgCl2 → Mg2+ + (Cl−)2
2+
−
MgCl2 → Mg + Cl 2
MgCl2 → Mg + Cl2
MgCl2 → Mg2+ + 2Cl−
Be sure you know the correct charges, and
show the charge on the dissociated ions.
Remember that multiple amounts of the same
ion separate from each other, 2Cl− not Cl2−
20
Write the reaction that best represents
the dissolving of perchloric acid in
water
+
−
1. HClO4 → H + ClO4
+
−
2. HClO4 → H + 4ClO
+
−
2−
3. HClO4 → H + Cl + 4O
21
Write the reaction that best represents
the dissolving of perchloric acid in
water
+
−
1. HClO4 → H + ClO4
+
−
2. HClO4 → H + 4ClO
+
−
2−
3. HClO4 → H + Cl + 4O
22
Write the reaction that best represents
the dissolving of chlorous acid in water
1.
2.
3.
4.
5.
+
H
HClO2 →
+
HClO2 → H
+
HClO2 → H
+
HClO2 ! H
+
HClO2 ! H
+
+
+
+
+
ClO2
−
2ClO
−
2−
Cl + 2O
−
ClO2
−
2ClO
−
23
Write the reaction that best represents
the dissolving of chlorous acid in water
1.
2.
3.
•
•
4.
•
•
5.
HClO2 → H+ + ClO2−
HClO2 → H+ + 2ClO−
HClO2 → H+ + Cl− + 2O2−
Polyatomic ions do NOT break up in aqueous solutions.
The polyatomic ion, chlorite ion does not break apart in
solution.
HClO2 ! H+ + ClO2−
This is even better HClO2 ←
This is a weak acid so you should use the double
arrows to indicate only some of the acid ionizes.
HClO2 ! H+ + 2ClO−
→
H+ + ClO
−
2
24
In the following reaction, which ions
are spectator ions:
AgNO3 + NaCl → NaNO3 + AgCl ?
1. Ag+ and NO3−
2. Na+ and Cl−
3.
−
Cl
and NO3
4.
+
Na
and NO3
−
Can you do the
problem without
your solubility table?
−
5. Ag+ and Cl−
25
In the following reaction, which ions
are spectator ions:
AgNO3 + NaCl → NaNO3 + AgCl ?
1. Ag+ and NO3−
2. Na+ and Cl−
3.
−
Cl
4.
+
Na
•
and NO3
−
Can you do the
problem without
your solubility table?
and NO3
Since silver chloride is the precipitate, the
nitrate and alkali salts are the spectator ions.
−
5. Ag+ and Cl−
26
Identify the precipitate when sodium
hydroxide is combined with aluminum
nitrate.
1. NaNO3
2. AlOH3
Do the problem without
your solubility table.
3. NaOH
4. AlOH
5. Al(OH)3
27
Identify the precipitate when sodium
hydroxide is combined with aluminum nitrate.
1.
2.
3.
4.
5.
•
•
NaNO3
AlOH3
Do
the
problem
without
NaOH
your solubility table.
AlOH
Al(OH)3
Alkali nitrates will always be soluble.
By process of elimination, the precipitate
must be the aluminum hydroxide.
28
Which of the compounds listed below
are insoluble:
1. NiCO3
5. ZnS
2. Ba(NO3)2
6. K2CrO4
3. C6H12O6
7. Al2(SO4)3
4. Na3PO3
8. SnF2
Do the problem without
your solubility table.
29
Which of the compounds listed below are insoluble:
1. NiCO3
2. Ba(NO3)2
3. C6H12O6
4. Na3PO3
5. ZnS
Do the problem without
your solubility table.
6. K2CrO4
7. Al2(SO4)3
8. SnF2
•
•
•
•
Remember that nitrates and alkali salts are always
soluble.
Sugar is also of course soluble.
Sulfates are usually soluble
Carbonates and sulfides are usually NOT soluble
30
In the reaction done in water, shown below, which
element is oxidized?
NiCl2 + Al → AlCl3 + Ni
1.
2.
3.
4.
5.
Ni
Balance this equation
into
a
balanced
net
Cl
ionic equation.
O
Al
None of them are since there is no
oxygen in the reaction.
31
In the reaction below, which element is oxidized?
NiCl2 + Al → AlCl3 + Ni
Turn this equation into a balanced net ionic
equation.
1.
2.
3.
4.
5.
Ni
3
NiCl2 + 2Al → 2AlCl3 + 3Ni
Cl
or Ni2+ + Al → Al3+ + Ni
O
3Ni2+ + 2Al → 2Al3+ + 3Ni
Al
None of them are since there is no
oxygen in the reaction.
•
Aluminum loses three electrons per atom as it
changes from atom form to Al3+ ion form.
32
For the reaction below, select the oxidizing
agent (the substance that causes oxidation of some other substance):
Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L)
1. O
2. H
3. Fe
4. H2O
5. Fe2O3
33
For the reaction below, select the oxidizing agent:
Fe2O3(s) + 3H2(g) → Fe(s) + H2O(L)
1.
2.
3.
4.
5.
•
•
O
H
Fe
H 2O
Fe2O3
The oxidation number of iron changes from
3+ as a reactant to 0 as a product, gaining
electrons, and is therefore reduced.
Thus the substance that the iron is part of
is the oxidizing agent.
34
Which reaction below is not an oxidation
reduction reaction?
1.
2+
3Co
+ 2Al → 3Co +
3+
2Al
2. 2Na + 2H2O → 2NaOH + H2
3. 2HCl + Zn → ZnCl2 + H2
4. 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2
5. CH4 + 2O2 → CO2 + 2H2O
35
Which reaction below is not an oxidation
reduction reaction?
1. 3Co2+ + 2Al → 3Co + 2Al3+
2. 2Na + 2H2O → 2NaOH + H2
3. 2HCl + Zn → ZnCl2 + H2
4. 2HNO3 + Mg(OH)2 → 2H2O + Mg(NO3)2
5. CH4 + 2O2 → CO2 + 2H2O
• Double replacement reactions are never
oxidation reduction reactions since none of
the atoms change oxidation numbers.
36
LAD B3
Redox Titration
37
Reading the label of a cobalt supplement,
the bottle tells us that each there are 35 mg
of elemental cobalt delivered in 150 mg of
cobalt(II) acetate in a 457 mg tablet.
Is the cobalt(II) acetate in the pill a hydrate
(if so, with how many water?) or anhydrate?
38
The iron pill bottle tells us that each there are 68 mg of
elemental iron delivered in 338 mg of iron(II) sulfate.
Is the iron sulfate in the pill a hydrate (if so,with how many
water?) or is it an anhydrate?
Two ways to think about this….
✓
Yes it is a hydrate, 7 water
1. Set up a ratio to find the
mass of of the iron
compound in the tablet
✓
68g/338g = 55.85g/x
solve for x = 277.6g
then
2. Subtract the mass caused by
the anhydrate, FeSO4
✓
277.6g - 151.92g = 126g water
3. Determine the moles of
water
✓
126 / 18 = 7 H2O
✓
Yes it is a hydrate, 7 water
1. Set up a ratio to find the
mole of the iron in 68 mg
✓ 68/55.85 = 1.218 mole Fe in
FeSO4 = mole of FeSO4
2. Find the MM of the FeSO4
compound in the tablet
✓ 338 g / 1.218 mol = 277 g/mol
3. Determine the mass, then
moles of this compound that
must be water
✓ 277 g - 152 g = 125 g
✓ 125 / 18 = ~7 H2O
39
Solution Stoichiometry
If you’re not part of the solution,
you’re part of the precipitate
40
24 ml of a 0.10 M solution of
Al(NO3)3 is combined with 24 ml of
0.10 M solution of Na2CO3.
• Write out balanced “overall”
equation.
• Formulas in black, balancing in
color.
• Be sure and indicate the
precipitate by circling in color.
41
24 ml of a 0.10 M solution of Al(NO3)3 is
combined with 24 ml of 0.10 M Na2CO3.
•
Write out balanced “overall”
equation on your whiteboard.
•
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
•
Convert this to a net ionic equation
42
24 ml of a 0.10 M solution of Al(NO3)3 is
combined with 24 ml of 0.10 M Na2CO3.
•
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
2−
+
3+
2Al
•
3CO3
→ Al2(CO3)3(ppt)
•
sodium and nitrate ions are spectator ions
43
24 ml of a 0.10 M solution of Al(NO3)3 is combined with
24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the mass of the dried precipitate?
1. 7.8 g
2. 70.2 g
3. 0.19 g
4. 0.28 g
5. 0.56 g
MM (g/mol)
3Na2CO3 = 106
Al(NO3)3 = 213
Al2(CO3)3 = 234
NaNO3 = 85
6. 187 g
7. 842 g
8. 1123 g
9. 1685 g
44
24 ml of a 0.10 M solution of Al(NO3)3 is combined with
24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the mass of the dried precipitate?
#3 0.19 g
• 2.4 mmol each, and because of the 3:2 ratio
between the reactants, the 3Na2CO3 limits
1Al2 (CO3 )3
234g
2.4mmol ×
= 0.8mmol ×
= 187mg
3Na2CO3
1mol
•
which equals 0.19 g Al2(CO3)3
MM (g/mol)
3Na2CO3 = 106
Al(NO3)3 = 213
Al2(CO3)3 = 234
NaNO3 = 85
45
24 ml of a 0.10 M solution of Al(NO3)3 is
combined with 24 ml of 0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in
solution after the reaction?
46
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of
0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in solution after the
reaction?
•
•
•
•
•
Remember - the total volume of the solution has
doubled which affects the concentrations
Treat the spectator ion concentrations as completely
unreacted in twice the volume.
Don’t forget they are sometimes “buy 1 get many” ions
Assume the limiting ion in the precipitate is effectively
gone from solution.
Calculate the mmoles left over (this means a
subtraction) of the excess ion that forms the precipitate
and calculate its molarity in the total volume.
47
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of
0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in solution after the
reaction?
•
Assume the CO32− ≃ 0
+
2Na
4.8mmol
2.4mmolNa2CO3 ×
=
= 0.10MNa + Still in Solution
1Na2CO3
48ml
−
3
−
3
3NO
7.2mmolNO
−
2.4mmolAl(NO3 )3 ×
=
= 0.15M NO3 Still in Solution
Al(NO3 )3
48ml
48
24 ml of a 0.10 M solution of Al(NO3)3 is combined with 24 ml of
0.10 M Na2CO3.
3Na2CO3 + 2Al(NO3)3 → Al2(CO3)3(ppt) + 6NaNO3
What is the molar concentration of each ion in solution after the
reaction?
• Assume the CO32− ≃ 0
2Na +
4.8mmol
2.4mmolNa2CO3 ×
=
= 0.10MNa + Still in Solution
1Na2CO3
48ml
3NO3−
7.2mmolNO3−
2.4mmolAl(NO3 )3 ×
=
= 0.15M NO3− Still in Solution
Al(NO3 )3
48ml
3+
1Al
3+
2.4mmol1Al(NO3 )3 ×
= 2.4mmolAl inSolution
1Al(NO3 )3
2Al(NO3 )3
2.4mmolNa2CO3 ×
= 1.6mmolAl(NO3 )3 needed
3Na2CO3
2.4mmolAl 3+ − 1.6mmolAl 3+
0.8mmolAl 3+ Left
=
= 0.017MAl 3+ Still in Solution
48ml
49
−
NO3
Calculate the Molarity of
ions
in 40 ml of a 0.20 M solution of
Mg(NO3)2 after mixing with 60 ml of
a 0.10 M solution of Al(NO3)3
No Calculator
•
Type in a numerical answer.
50
−
NO3 ions
Calculate the Molarity of
in 40 ml of a 0.20 M solution of
Mg(NO3)2 after reacting with 60 ml
of a 0.10 M solution of Al(NO3)3
−
3
2NO
−
40ml × 0.2Μ ×
= 16mmol NO3
1Mg(NO3 )2
No Calculator
−
3
3NO
−
60ml × 0.1Μ ×
= 18mmol NO3
1Al(NO3 )3
−
3
34NO total
−
= 0.34M NO3
100ml Total
51
25 ml of 0.40-molar iron(III) bromide
solution is combined with 35 ml of
0.30-molar lead(II) nitrate solution.
Write out the balanced overall (aka
molecular) equation for this reaction
molar mass
iron(III) bromide
lead(II) nitrate
52
25 ml of 0.40-molar iron(III) bromide solution
is combined with 35 ml of 0.30-molar lead(II)
nitrate solution.
2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3
molar mass
iron(III) bromide 295.55
lead(II) nitrate 331.22
lead(II) bromide 367
iron(III) nitrate 248.88
1. What mass of precipitate forms.
2. What is the concentration of spectator
ions after the precipitation is complete?
3. What is the concentration of the other
ions after the precipitation is complete
53
25 ml of 0.40-molar iron(III) bromide solution is combined
with 35 ml of 0.30-molar lead(II) nitrate solution.
2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3
molar mass
lead(II) bromide 367
25ml × 0.4M = 10mmolFeBr2
35ml × 0.3M = 10.5mmolPb ( NO3 )2
lead(II) nitrate limits
3PbBr2
10.5mmolPb ( NO3 )2 ×
= 10.5mmolPb ( NO3 )2 × 367g / mol = 3853mg = 3.85g
3Pb ( NO3 )2
54
25 ml of 0.40-molar iron(III) bromide solution is combined
with 35 ml of 0.30-molar lead(II) nitrate solution.
2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3
molar mass
lead(II) bromide 367
25ml × 0.4M = 10mmolFeBr2
35ml × 0.3M = 10.5mmolPb ( NO3 )2
lead(II) nitrate limits
2NO3−
10.5mmolPb ( NO3 )2 ×
= 21mmolNO3−
1Pb ( NO3 )2
3+
Fe
10mmolFe
10mmolFeBr2 ×
=
FeBr2
60ml
3+
21mmolNO3−
= 0.35M NO3−
60ml
= 0.17M Fe Still in Solution
3+
55
25 ml of 0.40-molar iron(III) bromide solution is combined
with 35 ml of 0.30-molar lead(II) nitrate solution.
2FeBr3 + 3Pb(NO3)2 → 3PbBr2 + 2Fe(NO3)3
molar mass
lead(II) bromide 367
25ml × 0.4M = 10mmolFeBr2
35ml × 0.3M = 10.5mmolPb ( NO3 )2
lead(II) nitrate limits
2FeBr3
10.5mmolPb ( NO3 )2 ×
= 7mmolFeBr3used
3Pb ( NO3 )2
−
3Br
10FeBr3started − 7FeBr3used =3FeBr3leftover ×
= 9Br − leftover
1FeBr3
9mmolBr −
= 0.15M Br −
60ml
No lead ions left in solution...all in the precipitate
56
100 ml of 0.050 M lead(II) nitrate
solution is combined with 100 ml of
0.080-molar sodium chloride solution.
Here’s a series of questions....
Write out the balanced overall (aka
molecular) and net ionic equations for
this reaction
1. On your notebook paper.
57
100 ml of 0.050 M lead(II) nitrate
solution is combined with 100 ml of
0.080-molar sodium chloride solution.
Write out the balanced overall (aka
molecular) equation for this reaction
1. What mass of precipitate forms
2. What is the concentration of sodium
ions after the precipitation is complete?
3. What is the concentration of lead ions
after the precipitation is complete
58
100 ml of 0.050 M lead(II) nitrate is
combined with 100 ml of 0.080-molar
sodium chloride.
Here’s a series of questions....
Write out the overall and net ionic
equations for this reaction
1. Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(ppt)
2+
−
2. Pb + 2Cl → PbCl2
59
100.0 ml of 0.050 M lead(II) nitrate
solution is combined with 100.0 ml of
0.080-molar sodium chloride solution.
Calculate the mass of precipitate
formed (MM PbCl2 = 278 g/mol)
1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
2.
+2
Pb
+
−
2Cl
→ PbCl2
60
100.0 ml of 0.050 M lead(II) nitrate is combined with
100.0 ml of 0.080-molar sodium chloride.
Calculate the mass of precipitate formed.
(MM PbCl2 = 278 g/mol)
1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
2. (0.05 M)(0.1L) = 0.005 mol Pb(NO3)2
3. (0.08 M)(0.1L) = 0.008 mol NaCl
4. use the “limiting trick” to determine that
NaCl limits
5.
⎛ 1PbCl2 ⎞ ⎛ 278g ⎞
0.008molNaCl ⎜
= 1.1gPbCl2 precipitate
⎟
⎜
⎟
⎝ 2NaCl ⎠ ⎝ 1mol ⎠
61
100 ml of 0.050 M lead(II) nitrate is combined
with 100 ml of 0.080-molar sodium chloride.
What are the concentrations of the
sodium ions and nitrate ions in solution
after the reaction is complete?
1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
62
100 ml of 0.050 M lead(II) nitrate is combined with 100 ml
of 0.080-molar sodium chloride.
What is the concentration of the sodium and nitrate
ions in solution after the reaction is complete?
1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
2.
All of the starting sodium and nitrate ions are still
present, since they are spectator ions, swimming in the
200 ml total volume.
3.
(0.05 M)(100 ml) = 5 mmol Pb(NO3)2
4.
⎛ 2NO3− ⎞
−
5mmolPb(NO3 )2 ⎜
=
10mmolNO
3
⎟
1Pb(NO
)
⎝
3 2⎠
10mmolNO3−
= 0.050MforNO3−
200mltotalvolume
5.
(0.08 M)(100 ml) = 8 mmol NaCl
6.
⎛ 1Na + ⎞
+
8mmolNaCl ⎜
=
8mmolNa
⎝ 1NaCl ⎟⎠
8mmolNa +
= 0.040MforNa +
200mltotalvolume
63
100 ml of 0.050 M lead(II) nitrate is combined
with 100 ml of 0.080-molar sodium chloride.
What is the concentration of the lead
ions and chloride ions in solution after
the reaction is complete?
1. Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
64
100 ml of 0.050 M lead(II) nitrate is combined with 100 ml of 0.080-molar
sodium chloride.
What is the concentration of the lead and chloride ions in solution after
the reaction is complete?
1.
Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
2.
Since the NaCl limited, we can assume all the Cl− ion are
in the precipitate not the solution, thus Cl− = 0 M
3.
Some of the lead ions are in the precipitate, some are still
in solution.
4.
(0.05 M)(100 ml) = 5 mmol Pb2+ to start
5.
⎛ 1Pb(NO3 )2 ⎞
2+
8mmolNaCl ⎜
=
4mmolPb
reactedtomakeprecipitate
⎟
⎝ 2NaCl ⎠
6.
5mmol Pb2+ start − 4 mmol Pb2+ reacted = 1 mmol Pb2+ remain in soln
7.
1mmolPb 2+
= 0.0050MPb 2+ RemainInSolution
200mlTotalVolume
65
100. ml of 0.050 M (mol/L) cobalt(III) nitrate is
combined with 200. ml of 0.040-molar potassium
chromate. (molar mass of cobalt chromate = 465.86 g/mol)
Better write a balanced equation.
1.
2.
3.
4.
Is this a redox reaction?
Which ion limits the reaction?
What is the mass of the precipitate?
What is the concentration of each of the ions
left in the solution?
66
100. ml of 0.050 M (mol/L) cobalt(III) nitrate is
combined with 200. ml of 0.040-molar potassium
chromate. (molar mass of cobalt chromate = 465.86 g/mol)
Better write a balanced equation.
1. Is this a redox reaction?
• No, DR reactions are never redox
2. Which ion limits the reaction?
• Co2+
3. What is the mass of the precipitate?
• 1.2 g (rounded to 2 SF)
4. What is the concentration of each of the ions
left in the solution?
67
100. ml of 0.050 M (mol/L) cobalt(III) nitrate is
combined with 200. ml of 0.040-molar potassium
chromate. (molar mass of cobalt chromate = 465.86 g/mol)
2Co(NO3)3 + 3 K2CrO4 → Co2(CrO4)3(ppt) + 6KNO3
100mlCo(NO3 )3 × 0.05M = 5mmolCo(NO3 )3 200mlK 2CrO4 × 0.04Μ = 8mmK 2CrO4
3NO3−
15mmol
5mmolCo(NO3 )3 ×
=
= 0.050M NO3− Still in Solution
1Co(NO3 )3
300ml
2K +
16mmol
8mmolK 2CrO4 ×
=
= 0.053M K + Still in Solution
K 2CrO4
300ml
•
Assume the Co3+ ≃ 0
3K 2CrO4
5mmolCo(NO3 )3 ×
= 7.5mmolCrO42− needed
1Co(NO3 )3
2−
0.05mmolCrO
4
8mmolCrO42− − 7.5mmolCrO42− =
= 0.0017MCrO42− Still in Solution
300ml
68
A 1.0 L sample of an aqueous
solution contains 0.030 mol of KCl
and 0.020 mol of AlCl3. What is the
minimum number of moles of
AgNO3 that must be added to the
solution in order to precipitate all of
−
the Cl as AgCl(s)?
No calculator
69
A 1.0 L sample of an aqueous
solution contains 0.030 mol of KCl
and 0.020 mol of AlCl3. What is the
minimum number of moles of
AgNO3 that must be added to the
solution in order to precipitate all of
−
the Cl as AgCl(s)?
No calculator
0.09 mol AgNO3 must be added
70
A 1.0 L sample of an aqueous
solution contains 0.030 mol of KCl
and 0.020 mol of AlCl3. What is the
minimum number of moles of
Pb(NO3)2 that must be added to
the solution in order to precipitate
−
all of the Cl as PbCl2(s)?
No calculator
71
A 1.0 L sample of an aqueous
solution contains 0.030 mol of KCl
and 0.020 mol of AlCl3. What is the
minimum number of moles of
Pb(NO3)2 that must be added to
the solution in order to precipitate
−
all of the Cl as PbCl2(s)?
No calculator
0.045 mole Pb(NO3)2 must be added
72
When 10. milliliter of 0.4-molar
sodium sulfate is added to 20.
milliliters of 0.10-molar aluminum
sulfate the number of moles of
2+
Pb that must be added to
2−
precipitate out all of the SO4
would be
No calculator
1. 0.020 mol
2. 0.0060 mol
3. 0.010 mol
4. 10. mol
5. 0.0010 mol
6. 0.10 mol
7. 0.005 mol
8. 1.0 mol
73
When 10. milliliter of 0.4-molar
sodium sulfate is added to 20.
milliliters of 0.10-molar aluminum
sulfate the number of moles of
2+
Pb that must be added to
2−
precipitate out all of the SO4
No calculator
would be
1. 0.020 mol
2. 0.0060 mol
3. 0.010 mol
4. 10. mol
5. 0.0010 mol
6. 0.10 mol
7. 0.005 mol
8. 1.0 mol
74
A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to
20.0 milliliters of 0.450-molar
Ba(NO3)2 solution. Barium
phosphate precipitates. The
2+
concentration of barium ion, Ba ,
in solution after reaction is
No calculator
75
A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to
20.0 milliliters of 0.450-molar
Ba(NO3)2 solution. Barium
phosphate precipitates. The
2+
concentration of barium ion, Ba ,
in solution after reaction is
No calculator
76
A 10.0-milliliter sample of 0.200molar K3PO4 solution is added to
20.0 milliliters of 0.450-molar
Ba(NO3)2 solution. Barium
phosphate precipitates. The
2+
concentration of barium ion, Ba ,
in solution after reaction is
No calculator
3−,
2+
Ba ,
2+
Ba
2 mmol PO4 9 mmol
3 mmol
used
6 mmol Ba2+ left over/30 ml = 0.2 M Ba2+ after ppt
77
If 40.0 ml of a 0.30 M solution of barium nitrate is
combined with 25.0 ml of 0.35 M sodium
phosphate solution,
1. calculate the mass of the precipitate that should form.
2. If 1.32 g of precipitate did form in the laboratory, calculate
the % yield.
3. Determine the concentration (molarity) of any ions still
floating in the solution.
Acid Base Review
from First Year
and Preview
of What’s New This Year
weak? strong?
concentrated? dilute?
What do these words mean?
79
concentrated vs dilute
describes the amount dissolved.
• Concentration refers to the quantity of solute that
is dissolved in the solvent.
in other words, how much stuff did you put in the water….
✓ concentrated = lots
✓ dilute = little
✓
80
Reminder: When soluble ionic compounds
dissolve, they dissociate into separated ions.
•
Generic Reaction for the dissolving of any ionic compound:
+ + X−
AX
→
A
‣ (s)
The break up of the salt
and subsequent
surrounding of ions by
water molecules is called
hydrolysis.
Notice how the water
molecules arrange
differently around the +
compared to the − ions.
81
Reminder: When sugar (or other molecular
compounds) dissolves in water, the
molecules do NOT dissociate into ions.
Sugar
We represent this in an equation as:
H 2O
C6H12O6(s) → C6H12O6(aq)
Salt
We represent this in an equation as:
H 2O
NaCl(s) → Na+ + Cl−
82
Acids
• Molecular compounds made of H&(−ion)
✓
HCl, HC2H3O2, H2SO4, HClO3, HBr, HNO2, etc
• Acids are molecular compounds that “stand on
the line.” The line that differentiates ionic from
molecular.
• An acid molecule is a covalently bonded
compound, that when dissolved in water,
dissociates (all or partly) into ions.
• However, the degree to which acids ionize when
placed in water, varies from acid to acid.
✓
We call this ability to dissociate, the strength or weakness
of the acid.
83
Particle View:
• Use this as HCl
• Use this as HF
• Sketch two separate beakers (side view) depicting
no more than 5 molecules in each beaker.
Represent the two substances as they would appear
when dissolved in water
84
Particle View: strong vs weak
HCl is a strong acid. Plenty of HCl
molecules dissolve, and virtually all of
them ionize into H+ and Cl−
We represent this in an equation as:
HCl(aq) → H+ + Cl−
+
+
+
+
H+
Both strong and weak acids will be able
to neutralize a base equally well.
−
+
−
+
H+
HF is a weak acid. While plenty of HF
molecules dissolve, only a fraction of
them ionize into H+ and F−
We represent this in an equation as:
HF(aq)
H + + F−
85
concentrated vs dilute
and strong vs weak
• You can have a concentrated strong acid
• You can have a dilute strong acid
• You can have a concentrated weak acid
• You can have a dilute weak acid
86
Select the reaction below that best
represents the net ionic equation for
sulfuric acid with potassium hydroxide.
1. H2SO4 + KOH → K2SO4 + H2O
2. H2SO4 + 2KOH → K2SO4 + 2H2O
3. 2H+ + OH− → 2H2O
+
−
4. 2H + 2OH → 2H2O
+
−
5. H + OH → H2O
87
Select the reaction below that best
represents the net ionic equation for
sulfuric acid with potassium hydroxide.
1. H2SO4 + KOH → K2SO4 + H2O
2. H2SO4 + 2KOH → K2SO4 + 2H2O
3. 2H+ + OH− → 2H2O
+
−
4. 2H + 2OH → 2H2O
• this would be ok too, but it is probably best to
reduce it.
5.
+
H
+
−
OH
→ H 2O
88
Acids
• Arrhenius defined acids as substances that
produce
+
H
ions in an aqueous solution.
HBr → H+ + Br−
✓ HBr + H2O → H3O+ + Br−
✓
• Bronstead & Lowrey defined acids as proton
donors. Monoprotic acids produce 1
+
H
• Di- or Tri- protic acids produce 2 or 3 H+
H2SO4 can produce 2 H+ per acid molecule.
✓ We can think of the dissociation occurring in steps:
✓
H2SO4 → H+ + HSO4−
‣ HSO4− → H+ + SO42−
‣
✓
or all in one step: H2SO4 → H+ + SO42−
89
Strong vs Weak
• Strong electrolytes dissociate completely
✓
HNO3 → H+ + NO3−
• Weak electrolytes only partly dissociate.
Some molecules remain “whole” while some
dissociate.
✓
HF
H+
+
F−
90
Strong vs Weak
• HF
H+
+
F−
✓
The double arrow indicates that both the forward and
reverse reactions are occurring at the same rate, so
the reaction seems to “stop”.
✓
It does not mean that the reaction occurs only as far
as the half way point.
✓
This balance is called chemical equilibrium.
91
Bases
• Substances that react with H+ ions
• Arrhenius defined bases as substances that
produce OH- ions
NaOH → Na+ + OH−
✓ Ba(OH)2 → Ba2+ + 2 OH−
✓
• Bronstead & Lowrey defined bases as proton
acceptors because some common weak bases
appear not to contain OH ions.
• Ammonia is a common weak base. It reacts with
water, accepts an H+, and produces OH−
NH4+ + OH−
✓
NH3 + H2O
✓
When you see the double arrow, it tells us the reaction “comes
to equilibrium” and it is a weak base.
92
Memorize the Strong Acids & Strong Bases
• Strong acids
✓
✓
✓
✓
(All the rest we will encounter in AP are weak)
HCl, HBr, HI
HClO3, HClO4
HNO3
H2SO4
• Strong bases - metal hydroxides that
dissolve (most metal hydroxides do not dissolve very well)
✓
Alkali hydroxides
‣
✓
LiOH, NaOH, KOH, RbOH, CsOH
Heavy alkaline earth hydroxides
‣
Ca(OH)2, Sr(OH)2, Ba(OH)2
93
Acids React with Bases
• Neutralization reaction
✓ aka
Double Replacement Rx
• HNO3 + KOH → H2O + KNO3
• Acid + base
water + salt
• Net Ionic:
+
H
+
OH
→ H 2O
94
Acids
• Arrhenius defined acids as substances that produce
+
H
✓
✓
ions in an aqueous solution.
HBr → H+ + Br−
HBr + H2O → H3O+ + Br−
• Bronstead & Lowrey defined acids as proton
donors. Monoprotic acids produce 1 H+
• Di- or Tri- protic acids produce 2 or 3 H+
✓
✓
✓
✓
✓
H2SO4 can produce 2 H+ per acid molecule.
We can think of the dissociation occuring in steps:
H2SO4 → H+ + HSO4−
HSO4−
H+ + SO42−
In this case, only the first ionization is complete, thus a solution
of sulfuric acid will actualy contain a mixture of H+, HSO4−, and
SO42−
95
How do you make up 500 ml of a 2 M
solution of sodium hydroxide?
96
How do you make 500 ml of a 2 M
solution of sodium hydroxide?
1. Use the formula M = moles/Liter
to determine that you need 1
mole in the 500 ml of solution.
2. NaOH, MM = 40g/mol
3. Mass 40 g of NaOH into a volumetric
flask.
4. Add ⅔ of the water and swirl to dissolve
5. Add the remaining water to bring the
solution to 500 ml total.
97
How to make 500 ml of a 2.0-molar
solution of hydrochloric acid starting
with concentrated HCl, 12 M?
slide show
98
How to make 500 ml of a 2 M
solution of hydrochloric acid starting
with concentrated HCl, 12 M?
1. This is a dilution problem, and during a
dilution the number of moles in the flask
does not change while adding water to
dilute the more concentrated solution.
2. Use the formula M×V = M×V
3. (2M)(500 ml) = (12M)(V) V = 83.3 ml
4. Add water to the flask, ⅔ full. Measure out the
83.3ml of concentrated HCl into a graduated
cylinder and add to the water in the volumetric flask.
5. Add the remaining water to bring the solution to
500ml total.
99
On paper write an overall and a net
ionic equation to describe the reaction
between a 500 ml of 2M solution of
NaOH and 500 ml of a 2M solution of
HCl.
100
On your white boards write an overall and a net
ionic equation to describe the reaction between a
500 ml of 2M solution of NaOH and 500 ml of a
2M solution of HCl.
Calculate the moles of water that would
form during this reaction.
•
•
HCl + NaOH → H2O + NaCl
+
−
H + OH → H2O
101
On your white boards write an overall and a net ionic
equation to describe the reaction between a 500 ml of 2M
solution of NaOH and 500 ml of a 2M solution of HCl.
Calculate the moles of water that would
form during this reaction.
What volume of water is this?
•
•
•
+
H
+
−
OH
→ H 2O
+
H,
(2M)(.5L) = 1mol
combined with
−
(2M)(.5L) = 1 mol OH = 1 mole H2O
102
On your white boards write an overall and a net ionic
equation to describe the reaction between a 500 ml of 2M
solution of NaOH and 500 ml of a 2M solution of HCl.
Calculate the moles of water that would
form during this reaction.
What volume of water is this?
•
•
•
•
+
H
+
−
OH
→ H 2O
+
H,
(2M)(.5L) = 1mol
combined with
(2M)(.5L) = 1 mol OH = 1 mol H2O
1ml
1molH 2O = 18g ×
= 18mlH 2O
1g
Density of water slide show
103
In class, 500 ml of 2M solution of NaOH and 500 ml of
a 2M solution of HCl were poured together into a 1000
ml flask.
At first, you may have been surprised to see ~18 ml more
than 1000 ml of solution, however, after the calculations
shown below, you could see why this must occur.
•
•
•
•
H+ + OH− → H2O
(2M)(.5L) = 1mol H+, combined with
(2M)(.5L) = 1 mol OH- = 1 mol H2O of water will form
1mol H2O = (18g)(1ml/1g)= 18 ml
In the demonstration
done in class, the extra
water above the line was
measured and turned out
to be ~18 ml
1000 ml
mark
500 ml
mark
500 ml
+
500 ml
=
1000 ml
slide show104
Insoluble (Dilute) Strong Base
Mg(OH)2
slide view
105
Insoluble (Dilute) Strong Bases
• Mg(OH)2 is an insoluble ionic compound
in H2O
• In an acid solution, Mg(OH)2 does
dissolve
• Demonstrate
• The acid ions, H3O+ (or H+) attack the OH
−
ions that are in the solid crystal
structure creating water molecules
H+ + OH− → H2O
✓ H3O+ + OH− → 2H2O
✓
106
Adding acid to “Milk of Magnesia”
a suspension of Mg(OH)2
• Write an overall equation for the reaction
between a suspension of magnesium
hydroxide and hydrochloric acid.
✓
Mg(OH)2 + HCl → H2O + MgCl2
• Convert to a net ionic equation.
✓
✓
✓
Mg(OH)2 + H+ → H2O + Mg2+
Mg(OH)2 must be written as a compound because it is
not soluble in water (to any appreciable degree, and it
is simply a suspension in the water.)
HCl is soluble, thus the Cl− ions are spectator ions.
107
Oxidation Numbers
more practice
108
Write a balanced equation for the
reaction that represents the
combustion of magnesium in air.
1. This is a redox reaction in which magnesium
is oxidized.
2. This is a redox reaction in which oxygen is
oxidized.
3. This is a redox reaction in which magnesium
is reduced.
4. This is a redox reaction in which oxygen is
oxidized.
5. This is not a redox reaction.
109
Write a balanced equation for the reaction
that represents the combustion of
magnesium in air.
2Mg + O2 → 2MgO or Mg + ½O2 → MgO
1.
2.
3.
4.
5.
•
•
This is a redox reaction in which magnesium is
oxidized.
This is a redox reaction in which oxygen is oxidized.
This is a redox reaction in which magnesium is
reduced.
This is a redox reaction in which oxygen is reduced.
This is not a redox reaction.
Combustion reactions are always redox reactions.
In redox reactions, particles must be both oxidized
and reduced.
110
slide view
Oxidation - Reduction
• Oxidation (LEO) [OIL]
✓
When an element loses electrons
• Reduction (GER) [RIG]
✓
When an element gains electrons
• The name oxidation is used because this type of
reaction was first studied by reacting metals with
oxygen.
✓
Rusting: Fe + O2 → Fe2O3
• It was called reduction because when elements
are isolated from minerals, we say they are
“reduced” to their elemental form.
✓
AlCl3 → Al(s) + Cl2(g)
111
Select the true statement(s) about the
reaction below:
Mg(OH)2 + HCl → H2O + MgCl2
1.
2.
3.
4.
5.
6.
7.
8.
9.
This is a redox reaction in which chlorine is reduced.
This is a redox reaction in which magnesium is
reduced.
This is a redox reaction in which hydrogen is reduced.
This is a redox reaction in which oxygen is reduced.
This is a redox reaction in which chlorine is oxidized.
This is a redox reaction in which magnesium is
oxidized.
This is a redox reaction in which hydrogen is oxidized.
This is a redox reaction in which oxygen is oxidized.
This is not a redox reaction.
112
Select the true statement(s) about the
reaction below:
Mg(OH)2 + HCl → H2O + MgCl2
1.
2.
3.
4.
5.
6.
7.
8.
9.
•
This is a redox reaction in which chlorine is reduced.
This is a redox reaction in which magnesium is reduced.
This is a redox reaction in which hydrogen is reduced.
This is a redox reaction in which oxygen is reduced.
This is a redox reaction in which chlorine is oxidized.
This is a redox reaction in which magnesium is oxidized.
This is a redox reaction in which hydrogen is oxidized.
This is a redox reaction in which oxygen is oxidized.
This is not a redox reaction.
Acid base neutralization is never redox.
113
What is the oxidation number of
sulfur in SF2?
1. 0
2. +2
3. -2
4. +4
5. -4
114
What is the oxidation number of
sulfur in SF2?
1. 0
2. +2
3. -2
4. +4
5. -4
• Since fluorine is -1 × 2 = −2 the sulfur
must be +2 to balance.
115
What is the oxidation state of
chlorine in KClO4 ?
1.
2.
3.
4.
5.
−2
−4
+4
+7
+8
116
What is the oxidation state of
chlorine in KClO4 ?
1.
2.
3.
4.
5.
•
−2
−4
+4
+7
+8
O is −2 × 4 = −8, the K is +1 thus the Cl
must be the remaining +7.
117
What is the oxidation state of nitrogen
in N2O5 ?
1.
2.
3.
4.
5.
+5
+2
-10
+10
-5
118
What is the oxidation state of nitrogen
in N2O5 ?
1.
2.
3.
4.
5.
•
+5
+2
-10
+10
−5
Since oxygen is −2 × 5 = -10, the nitrogens
must total +10, but 2 nitrogens cause the
+10, so each one individually is +5.
slide show
119
Balancing Aqueous
Redox Equations
Chapter 4
and 20 (section 1 & 2)
120
Redox Reactions
• NOT double replacement
✓
✓
precipitation
acid base
• Always combustion
• Always single replacement
• Sometimes synthesis
• Sometimes decomposition
• Complex redox reactions in solution
121
Redox Reactions
−
NO3 + Al →
3+
Al
in acidic
solution
+ NO
• List the Oxidation numbers of all elements
above
• Determine which element is oxidized,
• Determine which element is reduced
• How many total number of electrons are
transferred?
122
Redox Reactions
+5 -2
−
0
NO3 + Al →
+3
3+
Al
in acidic
solution
+2 -2
+ NO
• List the Oxidation numbers of all elements
above
• Determine which element is oxidized, Al
• Determine which element is reduced N
• How many total number of electrons are
transferred? 3e
123
Redox Reactions
this Rx occurs in acid solution
SO3
2− +
−
MnO4 → SO4
2−
+
2+
Mn
• Write out each element’s oxidation number
• Determine which element is oxidized
• Determine which element is reduced
• How many electrons are transferred?
124
In an acidic solution
SO3
2− +
−
MnO4 → SO4
2−
+
2+
Mn
1. Identify which elements are oxidized and reduced
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. recheck by confirming that ion charge is balanced
125
Redox Reactions
this Rx occurs in acid solution
MnO4− + H2O2 → Mn2+ + O2 + H2O
1. Identify which elements are oxidized and reduced
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. recheck by confirming that ion charge is balanced
126
Redox Reactions
this Rx occurs in acid solution
6H+ + 2MnO4− + 5H2O2 → 2Mn2+ + 5O2 + 8H2O
• Identify the element that is oxidized.
• Identify the element that is reduced.
+1
+
6H +
+7 −2
+1 −1
−
2MnO4 + 5H2O2 →
+2
2+
2Mn
0
+1 −2
+ 5O2 + 8H2O
127
• A solution of hypochlorous acid and
hydrochloric acid is formed when chlorine
gas is dissolved in cold water.
• Write out this reaction and determine the
oxidation numbers for all elements in this
reaction.
• The balanced reaction is:
+
H
−
Cl
✓ Cl2 + H2O → HOCl +
+
• In this reaction, chlorine is both oxidized and
reduced. This is called a...
Disproportionation Reaction
128
In an acidic solution
AuCl4
−+
Cu → Au +
−
Cl
+
2+
Cu
1. Identify which elements are oxidized and reduced
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. recheck by confirming that ion charge is balanced
129
AuCl4
−+
Cu → Au +
−
Cl
+
2+
Cu
1. Identify which elements are oxidized and reduced
+ 0
→
0
+ -1
+ +2
• +3, -1
• Au reduced 3e- gained, Cu oxidized 2e- lost
2. Balance the number of redox atoms by inspection − they are
already
3. Adjust the number of redox atoms so the number of electrons lost
equals the number of electrons gained. (This is the number of
electrons transferred.)
• 2AuCl4− + 3Cu → 2Au + Cl− + 3Cu2+ (a total of 6e− transferred)
4. balance any other elements (other than H and O)
• 2AuCl4− + 3Cu → 2Au + 8Cl− + 3Cu2+
5. balance oxygen by adding water as necessary − not necessary
6. balance hydrogen by adding H+ as necessary − not necessary
7. recheck by confirming that charge is balanced − 2− = 2−
130
2AuCl4
−+
3Cu → 2Au +
−
8Cl
+
2+
3Cu
1. It will be important later, for us to identify the
number of electrons transferred.
• in this reaction, there are 6 electrons lost
• and the same 6 electrons are gained
• thus, a total of 6 (not 12) electrons are
transferred
2. In every redox reaction there are two halves.
• the oxidation
• the reduction
3. We call these the half reactions.
131
2AuCl4
−+
3Cu → 2Au +
−
8Cl
+
2+
3Cu
From time to time, AP will ask you to write out
the half reactions.
1. Pull out the two substances involved in oxidation, then
balance, first the oxidation atom, then the other atoms,
then oxygen, then hydrogen − then include the number
of electrons
2. The oxidation half reaction is
2+ + 2e−
Cu
→
Cu
•
3. reduction
− + 3e− → Au + 4Cl−
AuCl
4
•
4. this corresponds with what we had determined by the
previous method
• Au reduced 3e− gained, Cu oxidized 2e− lost
132
In an acidic solution
−
NO3 + Al →
3+
Al
+ NO
1. Identify which elements are oxidized and reduced
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. recheck by confirming that ion charge is balanced
133
In an acidic sol’n
Zn + NO3
−→
2+
Zn
+ N2
1. Identify which elements are oxidized and reduced
Alert, alert...don’t miss this step
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. recheck by confirming that ion charge is balanced
134
(in acidic solution)
Zn + NO3
−→
2+
Zn
+ N2
1. Identify which elements are oxidized and reduced
0 + +5,-2
→
+2
+ 0
•
• N reduced 5e-/atom, Zn oxidized 2e2. Balance the number of redox atoms by inspection
• Zn + 2NO3− → Zn2+ + N2
3. Adjust the number of redox atoms so the number of electrons lost
equals the number of electrons gained. (This is the number of
electrons transferred.)
• 5Zn + 2NO3− → 5Zn2+ + N2
4. balance any other elements (other than H and O) - none
5. balance oxygen by adding water as necessary
• 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O
6. balance hydrogen by adding H+ as necessary
• 12H+ + 5Zn + 2NO3− → 5Zn2+ + N2 + 6H2O
7. recheck by confirming that ion charge is balanced − +10 = +10
135
Write out the two half reactions.
+
12H
−
+ 5Zn + 2NO3 →
2+
5Zn
+ N2 + 6H2O
136
−
Zn + NO3 →
2+
Zn
+ N2 (in acidic)
1. oxidation
2+ + 2e−
Zn
→
Zn
•
• note that charge balances in half reactions as well
2. reduction
− → N
2NO
3
2
•
+ ions
this
may
involve
using
water
and
H
•
• 12H+ + 2NO3− + 10e− → N2 + 6H2O
3. this corresponds with what we had determined by the
previous method
• N reduced 5e- gained per N, but the N2 requires 10etotal, Zn oxidized 2e- lost
137
In a basic solution
Br2 + AsO2
−→
−
Br +
AsO4
3−
1.
2.
3.
4.
5.
6.
7.
Identify which elements are oxidized and reduced
Balance the redox atoms by inspection
balance the redox atoms based on total electrons transferred
balance any other elements (other than H and O)
balance O by adding water as necessary
balance H by adding H+ as necessary
if the reaction occurs in basic solution, convert to basic by
adding an equal number of OH− ions to both sides to cancel
out the H+ ions
8. recheck by confirming that charge is balanced
138
-1
Br2 + AsO2 →
-1
Br
+ AsO4
-3
“pretend” in acid, then convert to basic
1.
2.
3.
4.
5.
6.
7.
Identify which elements are oxidized and reduced
0
+ +3, -2
→ -1
+
+5, -2
•
• Br reduced 1e-/atom, As oxidized 2ebalance those atoms based on total electrons transferred
• Br2 + AsO2-1 → 2Br-1 + AsO4-3
balance any other elements (other than H and O) - none
balance O by adding water as necessary
• 2H2O + Br2 + AsO2− → 2Br− + AsO43−
balance H by adding H+1 as necessary
• 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+
if the reaction occurs in basic solution, convert to by adding an equal number
of OH-1 ions to both sides to cancel out the H+1 ions
• 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H+ + 4OH−
• 4OH− + 2H2O + Br2 + AsO2− → 2Br− + AsO43− + 4H2O
• 4OH− + Br2 + AsO2− → 2Br− + AsO43− + 2H2O
recheck by confirming that charge is balanced −5 = −5
139
In basic solution
−
Br2 → BrO3 +
−
Br
1. Identify which elements are oxidized and reduced
• This is a disproportionation reaction - the same element is both
oxidized and reduced.
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of electrons lost
equals the number of electrons gained. (This is the number of
electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. since the reaction occurs in basic solution, convert to by adding an
equal number of OH− ions to both sides to cancel out the H+ ions
8. recheck by confirming that ion charge is balanced
140
−
Br2 → BrO3 +
−
Br
“pretend” in acid, then convert to basic
1.
•
•
•
2.
3.
•
4.
5.
•
6.
•
7.
8.
•
•
•
9.
Identify which elements are oxidized and reduced
0
→
+5, -2
+
-1
This is a disproportionation reaction - the same element is both oxidized and reduced.
Br reduced 1e-/atom, and Br oxidized 5e-/atom
Balance the number of redox atoms by inspection − they are already
Adjust the number of redox atoms so the number of electrons lost equals the number of
electrons gained. (This is the number of electrons transferred.)
6Br2 → 2BrO3− + 10Br−
balance any other elements (other than H and O) − none
balance oxygen by adding water as necessary
6 Br2 + 6 H2O → 2 BrO3− + 10Br−
balance hydrogen by adding H+ as necessary
6 Br2 + 6 H2O → 2 BrO3− + 10 Br− + 12 H+
recheck by confirming that ion charge is balanced
since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions
to both sides to cancel out the H+ ions
12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H+ + 12OH−
12OH− + 6 Br2 + 6H2O → 2 BrO3− + 10Br− + 12H2O
12OH− + 6 Br2 → 2 BrO3− + 10Br− + 6H2O
recheck by confirming that charge is balanced −12 = −12
141
In basic solution
−
Br2 → BrO3 +
−
Br
Write out the half reactions which also
should be balnced as in a basic solution.
142
−
Br2 → BrO3 +
−
Br (in basic)
Write out the half reactions.
1. oxidation
− + 10e−
Br
→
2
BrO
2
3
•
− + 12 H+ + 10e− (pretending acidic)
Br
+
6
H
O
→
2
BrO
2
2
3
•
• Br2 + 12OH− → 2 BrO3− + 6 H2O + 10e−
2. reduction
−
Br
+
2e−
→
2Br
2
•
3. this corresponds with what we had determined by the
previous method
• N reduced 5e− gained per N, but the N2 requires
10e− total, Zn oxidized 2e− lost
143
−
CN +
−
MnO4 →
basic sol’n
−
CNO
+ MnO2
1. Identify which elements are oxidized and reduced
2. Balance the number of redox atoms by inspection
3. Adjust the number of redox atoms so the number of
electrons lost equals the number of electrons gained.
(This is the number of electrons transferred.)
4. balance any other elements (other than H and O)
5. balance oxygen by adding water as necessary
6. balance hydrogen by adding H+ as necessary
7. since the reaction occurs in basic solution, convert to by
adding an equal number of OH− ions to both sides to
cancel out the H+ ions
8. recheck by confirming that ion charge is balanced
144
−
CN +
1.
•
•
•
2.
3.
•
4.
5.
•
6.
•
7.
•
•
•
8.
−
basic sol’n
MnO4 →
−
CNO
+ MnO2
Identify which elements are oxidized and reduced
You may worry about what to do with the oxidation numbers - consider N to be -3 and watch the C
change
+2, -3
+
+7, -2
→
+4, -3, -2
+
+4, -2
Mn reduced 3e-, and C oxidized 2eBalance the number of redox atoms by inspection − they are already
Adjust the number of redox atoms so the number of electrons lost equals the number of electrons
gained. (This is the number of electrons transferred.)
3CN− + 2MnO4− → 3CNO− + 2MnO2
balance any other elements (other than H and O) − already done
balance oxygen by adding water as necessary
3CN− + 2MnO4− → 3CNO− + 2MnO2 + H2O
balance hydrogen by adding H+ as necessary
3CN− + 2MnO4− + 2H→ 3CNO− + 2MnO2 + H2O
since the reaction occurs in basic solution, convert to by adding an equal number of OH− ions to both
sides to cancel out the H+ ions
2OH− + 3CN− + 2MnO4− + 2H+ → 3CNO− + 2MnO2 + H2O + 2OH−
3CN− + 2MnO4− + 2H2O → 3CNO− + 2MnO2 + H2O + 2OH−
3CN− + 2MnO4− + H2O → 3CNO− + 2MnO2 + 2OH−
recheck by confirming that ion charge is balanced
−5 = − 5
145
−
CN +
−
basic sol’n
MnO4 →
−
CNO
+ MnO2
1. oxidation
− → CNO− + 2e−
CN
•
• CN− + H2O → CNO− + 2e− + 2H+ (pretending acidic)
− + 2OH− → CNO− + 2e− + H O
CN
2
•
2. reduction
− + 3e− → MnO
MnO
4
2
•
• MnO4− + 4H+ + 3e− → MnO2 + 2H2O (pretending acidic)
− + 2H O + 3e− → MnO + + 4OH−
MnO
4
2
2
•
146
Pull out the two half reactions, and
balance (basic)
2−
−
−
Pb(OH)4 + ClO → PbO2 + Cl
147
Pull out the two half reactions, and
balance (basic)
2−
−
−
Pb(OH)4 + ClO → PbO2 + Cl
oxidation
Pb(OH)4
2−
→ PbO2 + 2H2O + 2e−
reduction
H 2O +
−
ClO
+ 2e− →
−
Cl
+
−
2OH
148
Pull out the two half reactions, and
balance (acidic) each one individually.
Label which one is oxidation and
which is reduction.
I2 +
−
OCl
−
→ IO3 +
−
Cl
149
Pull out the two half reactions, and balance (acidic)
each one individually.
Label which one is oxidation and which is reduction.
I2 +
−
OCl
−
→ IO3 +
−
Cl
oxidation
−
I2 + 6H2O → 2IO3 +
+
12H +
10e−
reduction
+
2H
+
−
OCl
+ 2e− →
−
Cl
+ H 2O
150
Pull out the two half reactions, and
balance (basic)
H2O2 + Cl2O7 → O2 + ClO2
−
151
Pull out the two half reactions, and
balance (basic)
H2O2 + Cl2O7 → O2 + ClO2
−
redution
−
3H2O + Cl2O7 + 8e− → 2ClO2 +
−
6OH
oxidation
−
2OH
+ H2O2 → O2 + H2O + 2e−
152
When a basic solution of KMnO4 is added to an
SnCl2 solution, a brown precipitate of MnO2
4+
forms and Sn remains in solution. When the
same basic solution of KMnO4 is added to an
MgCl2 solution, no reaction occurs. Which of the
substances involved in these reactions serves as
the best reducing agent (i.e. the substance that is
best at causing reduction in some other substance) ?
1.
2.
3.
4.
5.
KMnO4
SnCl2
MgCl2
Sn4+
Sn2+
6.
7.
8.
9.
K+
MnO2
Mg2+
Cl−
153
When a basic solution of KMnO4 is added to an SnCl2
solution, a brown precipitate of MnO2 forms and Sn4+
remains in solution. When the same basic solution of
KMnO4 is added to an MgCl2 solution, no reaction occurs.
Which of the substances involved in these reactions
serves as the best reducing agent (e.g. the substance
that is best at causing reduction in some other
substance) ?
The Mn in the KMnO4 was reduced, and by SnCl2, not MgCl2.
1.
2.
3.
4.
KMnO4
SnCl2
MgCl2
4+
Sn
+
K
5.
6. MnO2
2+
7. Mg
−
8. Cl
154