CHEM 1020: General
Chemistry I (Fall 2022)
Chapter 6: Chemical reactions in aqueous solution
Course Instructors:
Prof. Jinqing Huang (2nd half)
Review questions:
exercise 1: In cold water, lead(II) chloride PbCl2 is
A soluble
B insoluble
Precipitation Reaction
Simple Rules for the Solubility of Salts
• Most nitrate (NO3-) salts are soluble
important
!!!
• Most salts containing the alkali metal ions (Group 1A: Li+, Na+, K+, Rb+, Cs+) and the
ammonium ion (NH4+) are soluble
• Most Cl-, Br-, and I- salts are soluble (except Ag+, Pb2+, and Hg22+)
• Most sulfate salts (XSO4) are soluble (except BaSO4, PbSO4, Hg2SO4, and CaSO4)
• Most hydroxides (OH-) are only slightly soluble (NaOH, KOH are soluble; The Group
II A compounds: Ba(OH)2, Ca(OH)2, and Sr(OH)2 are marginally soluble)
• Most sulfide (S2–), carbonate (CO32–), chromate (CrO42–), and phosphate (PO43–)
salts are only slightly soluble, except for those containing the cations in Rule 2
Review questions:
exercise 1: In cold water, lead(II) chloride PbCl2 is
A soluble
B insoluble
Limitation of solubility rule…
https://www.youtube.com/watch?v=LaWdhxlvEAE
https://pubchem.ncbi.nlm.nih.gov/compound/Lead-chloride#section=Solubility
https://en.wikipedia.org/wiki/Lead(II)_chloride
CHEM 1020: Chapter 6
CHAPTER OUTLINE
1.
2.
3.
4.
5.
6.
General properties of aqueous solutions
Types of chemical reactions
Precipitation reactions
Types of equations
Acid-base reactions
Oxidation-reduction reactions
Reference and Suggested Reading:
“Chemistry, an atoms first approach”, Zumdahl & Zumdahl, 2016: Chapter 6
inspiration
What happen if you mix potassium permanganate (KMnO4) with
vitamin C (ascorbic acid) / sodium hydroxide (NaOH) and sugar?
+
=
vitamin C (ascorbic acid)
potassium permanganate
(KMnO4)
sugar
sodium hydroxide (NaOH)
Change of color ?
Oxidation-reduction reaction?
Acid-base reactions
• Acids and bases are common chemicals
• Acids: Taste sour
Bases: Taste bitter or salty
Feel soapy, slippery
Modern Definition of Acids and Bases
Arrhenius definition
• Acid: produces hydrogen ion, H+
• Base: produces hydroxide ion, OHBrfnsted-Lowry definition
• Acid: proton donor
• Base: proton acceptor
Lewis definition (just for your information)
• Acid: electron-pair acceptor
• Base: electron-pair donor
Modern Definition of Acids and Bases
Arrhenius: A man with solutions
• Arrhenius proposed the following ionization reactions of acids in water:
• Arrhenius definition of an acid: a substance that produces H+ (hydrogen
ion, hydron, hydrogen cation, proton) when dissolved in water.
• Strong acids (i.e., HCl, HNO3, H2SO4) behave as strong electrolytes in
water due to their 100% dissociation into ions.
• Strong bases (NaOH, KOH) also behave as strong electrolytes as they
completely dissolve into OH- (hydroxide ions) and cations in water.
Arrhenius acid and base
Arrhenius acid: Substance that increases concentration of H3O+ (aq) [or H+(aq)]
in water. e,g. HCl.
HCl (aq) + H2O (l) à H3O+ (aq) + Cl- (aq)
OR
HCl (aq)
à H+ (aq) + Cl- (aq)
Chemists often use the notation H+(aq) for the H3O+(aq) ion, and call it the hydrogen
ion. Remember, that the H+ ion is actually chemically bonded to water, that is, H3O+.
H+(aq): hydrogen ion; H3O+: hydronium ion
Arrhenius base: Substance that increases
concentration of OH- (aq) in water.
NaOH(aq) à Na+ (aq) + OH-(aq)
Brønsted-Lowry acid and base
Brønsted-Lowry acid: a proton (H+) donor
Brønsted-Lowry base: a proton (H+) acceptor
It is a more general definition
HCl (aq) + H2O (l) à H3O+ (aq) + Cl- (aq)
base
acid
NH3(aq) + H2O (l) à NH4+ (aq) + OH- (aq)
acid
base
NaOH(aq) + HCl (aq) à NaCl (aq) + H2O (l)
base
acid
Lewis acid and base (just for your information)
Lewis acid: electron-pair acceptor
Lewis base: electron-pair donor
Fluoride
Boron trifluoride
Tetrafluoroborate
Reactions of acid and base (Brønsted–Lowry)
Neutralization: A reaction between an acid and a base
General reactions:
Acid + Base à Salt + Water
or Acid + Base à Salt
Examples:
NaOH(aq) + HCl (aq) à NaCl (aq) + H2O (l)
NH3(aq) + HCl (aq) à NH4Cl (aq)
Mixing of strong/weak acids with strong/weak bases:
• Strong base can be assumed to react completely with any strong acids
and weak acids. (So as strong acid)
Calculations for acid-base reactions
Suggested strategy:
• List the species present in the combined solution before any reaction
occurs, and decide what reaction will occur
• Write the balanced net ionic equation for this reaction
• Calculate moles of reactants
• Determine the limiting reactant where appropriate
• Calculate the moles of the required reactant or product
• Convert to grams or volume (of solution), as required
Calculations for acid-base reactions
example 1: What volume of a 0.100 M HCl solution is needed to neutralize 25.0
mL of 0.350 M NaOH?
• Step 1 - Identify the ions present in the combined solution
H+, Cl–, Na+, and OH–
Na + (aq) + Cl – (aq) ¾¾
® NaCl(s)
The two possible reactions are
H + (aq) + OH – (aq) ¾¾
® H 2 O(l )
• Step 2 - Determine the balanced equation for the reaction
H + (aq) + OH – (aq) ¾¾
® H 2 O (l )
• Step 3 - Determine the moles of reactant present in the solution
1 L
0.350 mol OH –
25.0 mL NaOH ´
´
= 8.75 ´10–3 mol OH –
1000 mL
L NaOH
Calculations for acid-base reactions
example 1: What volume of a 0.100 M HCl solution is needed to neutralize 25.0
mL of 0.350 M NaOH?
• Step 4 - Identify the limiting reactant
• This problem requires the addition of just enough H+ to react exactly with the
OH– ions present, so the limiting reactant is not significant here
• Step 5 - Determine the moles of H+ required
• Since H + and OH– ions react in a 1:1
0.100 mol H +
–3
+
–23
+
V
´
=
8.75
´
10
mol
H
ratio, 8.75 × 10 moles of H are
L
required to neutralize the OH– ions
8.75 ´ 10 –3 mol H +
present
–2
V =
=
8.75
´
10
L
+
• Step 6 - Determine the volume of HCl
0.100 mol H
required
L
Application of acid and base: Titrations
Volumetric analysis is used to determine the amount of a certain
substance by doing a titration
Titration is the delivery of a measured volume of solution of known
concentration (titrant) into a solution containing the substance being
analyzed (analyte)
• Equivalence point or stoichiometric point is the
stage in titration at which enough titrant has been
added to react exactly with the analyte
• An indicator is a substance added at the beginning of
the titration that changes color at the equivalence
point
• Endpoint is the stage where the indicator actually
changes color
Titration of an Acid with a Base
Application of acid and base: Titrations
Requirements for a Successful Titration
• The exact reaction between the titrant and
analyte must be known (and rapid)
• The stoichiometric (equivalence) point must be
marked accurately
• The volume of titrant required to reach the
stoichiometric point must be known accurately
Automated or Manual
Titrations?
Titration of an Acid with a Base
Application of acid and base: Titrations
exercise 1: A truck carrying sulfuric acid H2SO4 is in an accident. A laboratory analyzes
a sample of the spilled acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0
M NaOH solution. What is the concentration of the acid?
A
B
C
D
3.0 M
4.0 M
6.0 M
12.0 M
Application of acid and base: Titrations
exercise 1: A truck carrying sulfuric acid H2SO4 is in an accident. A laboratory analyzes
a sample of the spilled acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0
M NaOH solution. What is the concentration of the acid?
A
B
C
D
3.0 M
4.0 M
6.0 M
12.0 M
Application of acid and base: Titrations
exercise 1: A truck carrying sulfuric acid is in an accident. A laboratory analyzes a
sample of the spilled acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0 M
NaOH solution. What is the concentration of the acid?
2NaOH + H2SO4 ® 2H2O + Na2SO4
Balanced equation:
Mole of NaOH used:
= [NaOH] x V = 4.0 M X0.060 L) = 0.24 mol
Mole of H2SO4 reacted:
= mol NaOH/2 = 0.24 mol/2 = 0.12 mol
mol H2SO4
0.12 mol
[H2SO4] : = Volume (L) = 0.020 L
= 6.0 M
inspiration
What happen if you mix potassium permanganate (KMnO4) with
vitamin C (ascorbic acid) / sodium hydroxide (NaOH) and sugar?
+
=
vitamin C (ascorbic acid)
potassium permanganate
(KMnO4)
sugar
Change of color ?
sodium hydroxide (NaOH)
Oxidation
Oxidation-Reduction Reactions
Applications:
• Purifying metals (e.g. Al, Na, Li)
• Producing gases (e.g. Cl2, O2, H2)
• Electroplating metals
• Battery
• Protecting metals from corrosion
• Sensors and machines (e.g. pH meter)
The Nobel Prize in Chemistry 2019 is awarded to
John B. Goodenough, M. Stanley Whittingham
and Akira Yoshino for their contributions to the
development of the lithium-ion battery.
Oxidation-Reduction Reactions
Oxidation–Reduction Reactions (Redox Reactions)
• Reactions in which one or more electrons are transferred
Zn(s) + CuSO4(aq) ® ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)
2e-
Zinc is losing 2 electrons
Zn(s) ® Zn2+(aq) + 2eCopper is gaining the 2 electrons.
Cu2+(aq) + 2e- ® Cu(s)
There are changes in the oxidative states.
Oxidation-Reduction Reactions
Oxidation States (oxidation numbers)
• Defined as imaginary charges that atoms would have if the shared electrons were
divided equally between identical atoms bonded to one another
• Oxidation states on ions are represented as +n or –n
Rules for Assigning Oxidation States
•
Oxidation number of atoms in element = 0
•
Oxidation number of element in monatomic ion = charge
•
Sum of the oxidation numbers in a compound = 0
•
Sum of oxidation numbers in polyatomic ion = charge
•
F has oxidation number –1
•
•
H has oxidation number +1; except in metal hydrides where it is –1
(e.g., ZnH2, NaH)
Oxygen is usually –2. Except:
Metals usually have positive
O is –1 in peroxide, e.g. H2O2
oxidation numbers
O is –1/2 in superoxides, KO2
Non-metals usually have negative
oxidation numbers
In OF2, O is +2
-
Oxidation–Reduction Reactions
The Characteristics of Oxidation–Reduction Reactions
• At times, the transfer of electrons may not be evident
• Consider the combustion of methane
• Carbon undergoes a change in oxidation state from –4 in CH4 to
+4 in CO2
Oxidation–Reduction Reactions
The Characteristics of Oxidation–Reduction Reactions
• Oxygen undergoes a change from an oxidation state of 0 in O2 to –2 in H2O
and CO2
• Oxidation is an increase in oxidation state: Loss of electrons
• Reduction is a decrease in oxidation state: Gain of electrons
• In the combustion of methane:
• Carbon in methane is oxidized. Oxygen is reduced
• CH4 is the reducing agent. O2 is the oxidizing agent
Oxidation–Reduction Reactions
loss of electron(s).
increase in oxidation state or a ____
• Oxidation: an ________
gain of electron(s).
• Reduction: a decrease
_______ in oxidation state or a ____
reduced
acceptor (itself will be _______).
• Oxidizing Agent: an electron __________
oxidized
• Reducing Agent: the electron _______
donor (itself will be ________)
PbO(s) + CO(g) à Pb(s) + CO2(g)
+2 -2
+2 -2
0
Element being reduced:
Reducing agent: CO
+4 -2 (each)
Pb
Element being oxidized: C
Oxidizing agent: PbO
Oxidation–Reduction Reactions
example 1: Metallurgy, the process of producing a metal from its ore, always
involves oxidation–reduction reactions. In the metallurgy of galena (PbS), the
principal lead-containing ore, the first step is the conversion of lead sulfide to its
oxide (a process called roasting):
2PbS(s ) + 3O2 (g ) ¾¾
® 2PbO(s ) + 2SO 2 (g )
Lead sulfide
Oxygen
Lead oxide
Sulfur dioxide
The oxide is then treated with carbon monoxide to produce the free metal:
PbO(s) + CO(g ) ¾¾
® Pb(s ) + CO2 (g )
Lead oxide
Carbon monoxide
Lead
Carbon dioxide
For each reaction, identify the atoms that are oxidized and reduced, and specify
the oxidizing and reducing agents
Oxidation–Reduction Reactions
example 1
• Assigning oxidation states to the first reaction
• Sulfur is oxidized. Oxygen is reduced. The reducing agent is PbS.
• Assigning oxidation states to the second reaction
• Lead is reduced. Carbon is oxidized.
• PbO is the oxidizing agent. CO is the reducing agent.
Oxidation–Reduction Reactions
exercise 1: Find the oxidation states for each of the elements in each of the
following compounds:
•
•
•
•
Carbonate ion, CO32Manganese dioxide, MnO2
Phosphorus pentachloride, PCl5
Potassium dichromate, K2Cr2O7
Which element has the highest oxidation state?
A. C
B. Mn C. P
D. Cr
Oxidation–Reduction Reactions
exercise 1: Find the oxidation states for each of the elements in each of the
following compounds:
•
•
•
•
O = –2; C = +4
Carbonate ion, CO32O = –2; Mn = +4
Manganese dioxide, MnO2
Phosphorus pentachloride, PCl5 Cl = –1; P = +5
Potassium dichromate, K2Cr2O7 K = +1; O = –2; Cr = +6
Which element has the highest oxidation state?
A. C
B. Mn C. P
D. Cr
Oxidation–Reduction Reactions
exercise 2: Which of the following are oxidation-reduction reactions? Identify
the oxidizing agent and the reducing agent.
a)Zn(s) + 2HCl(aq) ® ZnCl2(aq) + H2(g)
b) Cr2O72-(aq) + 2OH-(aq) ® 2CrO42-(aq) + H2O(l)
Dichromate ion
Chromate ion
c) 2CuCl(aq) ® CuCl2(aq) + Cu(s)
Copper (I) chloride
Copper (II) chloride
Oxidation–Reduction Reactions
exercise 2: Which of the following are oxidation-reduction reactions? Identify
the oxidizing agent and the reducing agent.
a)Zn(s) + 2HCl(aq) ® ZnCl2(aq) + H2(g)
oxidizing agent: HCl
reducing agent: Zn
b) Cr2O72-(aq) + 2OH-(aq) ® 2CrO42-(aq) + H2O(l)
Dichromate ion
Chromate ion
no change in oxidation states
c) 2CuCl(aq) ® CuCl2(aq) + Cu(s)
Copper (I) chloride
Copper (II) chloride
oxidizing agent: CuCl
reducing agent: CuCl
Oxidation–Reduction Reactions
In a balanced redox equation, the number of electrons lost in oxidation
(the increase in oxidation state) must equal the number of electrons
gained in reduction (the decrease in oxidation state)
• Write the unbalanced equation
• Determine the oxidation states of all atoms in the
reactants and products
• Show electrons gained and lost using “tie lines”
• Use coefficients to equalize the electrons gained and lost
• Balance the rest of the equation by inspection
• Add appropriate states
Oxidation–Reduction Reactions
example 2: Balance the reaction between solid lead(II) oxide and ammonia gas
to produce nitrogen gas, liquid water, and solid lead.
• Step 1 - State the unbalanced equation
PbO( s) + NH 3 (g ) ¾¾
® N 2 (g ) + H 2O(l ) + Pb(s)
Lead(II) oxide
Ammonia
Nitrogen
Water
• Step 2 - Assign oxidation states to each atom
Lead
Oxidation–Reduction Reactions
example 2: Balance the reaction between solid lead(II) oxide and ammonia gas
to produce nitrogen gas, liquid water, and solid lead.
• Step 3 - Use “tie lines” to show electrons gained and lost
• The oxidation states of all other atoms are unchanged
Oxidation–Reduction Reactions
example 2: Balance the reaction between solid lead(II) oxide and ammonia gas
to produce nitrogen gas, liquid water, and solid lead.
• Step 4 - Use coefficients to equalize electrons gained and
lost
Oxidation–Reduction Reactions
example 2: Balance the reaction between solid lead(II) oxide and ammonia gas
to produce nitrogen gas, liquid water, and solid lead.
• Step 5 - Balance the rest of the elements
3PbO + 2NH 3 ¾¾
® N 2 + 3H 2 O + 3Pb
• Step 6 - Add appropriate states
3PbO(s) + 2NH 3 ( g ) ¾¾
® N 2 (g )+ 3H 2 O(l ) + 3Pb(s)
Oxidation–Reduction Reactions
Exercise 3. Balance the following equation
S +
A
B
C
D
2
3
4
4
HNO3 è SO2
2
4
4
4
2
3
3
4
2
4
4
4
1
2
2
2
+ NO +
challenging question
H2O
Oxidation–Reduction Reactions
Exercise 3. Balance the following equation
Oxidation numbers?
-4e
3
S +
0
B 3
4
3
2
H2O
4
So, coefficient 3 x 4 = 12
4 HNO3
+1 +5 -2
è
3
SO2 + 4NO +
+4 -2
+2 -2
+1 -2
+3e So, coefficient 4 x 3 = 12
Which was oxidized?
S. By how much? 0 to +4 = change of +4
Which was reduced?
N. By how much?
+5 to +2 = change of -3
2
Oxidation–Reduction Reactions
If needed, reactions that take place in acidic or basic solutions
can be balanced as follows: (higher level knowledge)
Acidic condition
Basic condition
add H2O to the side needing
oxygen
add H2O to the side needing oxygen
H+(aq)
then add
hydrogen
to balance the
add enough OH- to both sides to cancel
out each H+(aq) (making H2O) & then
cancel out water as appropriate
Oxidation–Reduction Reactions
Exercise 4. Balance the following equation,
assuming it takes place in acidic solution.
Only parts of the equation is shown below:
……
ClO4- +
Perchlorate ion
A
B
C
D
1
1
1
1
I-
Iodide ion
8
4
2
6
1
1
1
1
è ClChloride ion
4
2
1
3
+
I2
Iodine
challenging question
……
Oxidation–Reduction Reactions
Exercise 4. Balance the following equation, assuming it takes place in acidic
solution.
A 1
8
1
4
+8e
Oxidation numbers?
8 H+(aq) + ClO4- + 8 I-
è Cl-
-1
-1
Perchlorate ion Iodide ion
+7 -2
Which was oxidized?
+ 4 I 2 + 4 H2 O
Chloride ion
Iodine,
Iodine
0
-1e
-1 to 0 = change of +1
Which was reduced?
Chlorine, +7 to -1 = change of -8
add H2O to the side needing oxygen, then add H+(aq) to balance the hydrogen.
inspiration
What happen if you mix potassium permanganate (KMnO4) with
vitamin C (ascorbic acid) / sodium hydroxide (NaOH) and sugar?
+
=
vitamin C (ascorbic acid)
potassium permanganate
(KMnO4)
sugar
sodium hydroxide (NaOH)
Change of color ?
Oxidation-reduction reaction?
inspiration
What happen if you mix potassium permanganate (KMnO4) with
vitamin C (ascorbic acid) / sodium hydroxide (NaOH) and sugar?
Oxidation-reduction reaction J
+
=
vitamin C (ascorbic acid)
potassium permanganate
(KMnO4)
sugar
sodium hydroxide (NaOH)
Change of color ?
Oxidation-reduction reaction?
Summary of Chapter 6 (part 2)
Chemical reactions in aqueous solution
Arrhenius model Acids and Bases
1. General properties of aqueous solutions produce H+/OH- Bronsted-Lowry model
2. Types of chemical reactions
proton donor/acceptor
Titration
3. Precipitation reactions
neutralization reaction
4. Types of equations
stoichiometric (equivalence) point
5. Acid-base reactions
indicator
endpoint
6. Oxidation-reduction reactions
oxidation state Oxidation-Reduction Reactions
Exercise: 25, 29, 43, 55, 65, 77, 85, 89.
Challenge problems:
oxidation: increase in oxidation state (a loss of electrons)
nd
125, 131, 139 (2 Version)
reduction: decrease in oxidation state (a gain of electrons)
127, 133, 141 (3rd Version)
oxidizing agent (electron acceptor)
reducing agent (electron donor)
Balancing Oxidation–Reduction Reactions by Oxidation States