This topic covers the acid and base reactions of solutions.

1. AqueousReactionsandSolution
Stoichiometry
Acid- Base Reactions
Oxidation-Reduction
Reactions
Reporter:
1

2. Acid- Base Reactions
Learning
Goals
• Differentiate an
acid from a
base.
• Write an acid-
base reaction.
2

3. Acids
• Acids – substance that ionizes in
aqueous solution to form hydrogen
ion (H+), or a proton
• Acids, when dissolved in water
dissociate and release H+ ions.
• HCl(g) + H2O → H(aq)
+ + Cl(aq)
-
3

4. Bases
• Bases are substances that
accept (react with) H+ ions
• Bases produce hydroxide ions
(OH-) when dissolved in water
• Bases when dissolved in water
will dissociate and release OH-
ions.
• Example
• NaOH, Ca(OH)2
NaOH(s) + H2O → Na(aq)
+ + OH(aq)
-
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5. Acid-Base Reactions
In an acid-base
reaction, the acid
donates a proton (H+)
to the base.
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6. Neutralization Reactions
In an acid-base (neutralization) reaction, the acid donates a proton
(H+) to the base.
Generally, when solutions of an acid and a base are combined, the
products are a salt (ionic compound) and water.
- Water is neutral on the pH scale, unlike acids or bases, hence the
name neutralization reaction
- The formation of liquid water is the driving force for these reactions
HBr (aq) + LiOH (aq)  LiBr (aq) + H2O (l)
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) + H2O (l)
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7. Write chemical equations for the
reactions of
1. HBr (acid) + KOH (base)→
HBr (acid) + KOH (base)→ KBr (salt) + H2O
2. HCl (acid)+ NaHCO3 (base)→
Answer:
HCl (acid)+ NaHCO3 (base)→ NaCl (salt) + H2CO3
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8. Oxidation-Reduction Reactions
 Learning
Goals
• Determine the
oxidation number of an
element/compound
/ion.
• Identify the substance
oxidized and the
substance reduced in a
chemical reaction.
• Balance redox reaction.
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9. Oxidation Numbers
Oxidation numbers are for bookkeeping of
electrons.
To determine if an oxidation-reduction
reaction has occurred, we assign an oxidation
number (or oxidation state) to each element
in a neutral compound or charged entity.
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10. Rules for Assigning Oxidation Numbers
1. Elements in their elemental form have an oxidation
number of 0.
• Examples: Na, O2, H2, Cu, Ag, etc… all have oxidation numbers of 0
2. The oxidation number of a monatomic ion is the same
as its charge.
• Examples: Na+ oxidation number = +1
Cl- oxidation number = -1
O2- oxidation number = -2
Cu2+ oxdiation number = +2
Group IA metals -always have oxidation number of 1+
Group II A metals- always have oxidation numbers of 2+
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11. Rules for Assigning Oxidation Numbers
3. Oxygen has an oxidation number of −2, except in
the peroxide ion O2
2- in which it has an oxidation
number of −1.
4. Hydrogen is −1 when bonded to a metal, +1 when
bonded to a nonmetal.
5. Fluorine always has an oxidation number of −1.
• The other halogens have an oxidation number of −1
when they are monatomic ions; they can have positive
oxidation numbers, however, most notably in
oxyanions.
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12. Rules for Assigning Oxidation Numbers
6. The sum of the oxidation states of all
atoms present in the formula of a
compound is zero
- in a neutral compound is 0.
• Ex: MgCl2 Mg2+ oxidation number = +2
Cl- oxidation number = -1
(+2) + 2(-1) = 0
• Aluminum Oxide, Al2O3
• 2(+3) + 3(-2) = 0
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13. Rules for Assigning Oxidation Numbers
7. In a polyatomic ion( an ion that contains a few
atoms), the sum of the oxidation numbers of all
atoms equals the charge on the ion.
Example: sulphate ion, SO4
-2
SO4
-2 sum of the oxidation numbers
is the same as the charge on
(+6) + 4(-2) = -2 charge on suphate ion
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14. Try this!
• Assign oxidation numbers to each element
in the following examples:
a) H2
b) MoS2
c) ClO2
-
d) Ag
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15. Try this!
• Assign oxidation numbers to each element
in the following examples:
a) H2 H = 0
b) MoS2 S = -2 Mo = +4
c) ClO2
- O = -2 Cl = +3
d) Ag Ag = 0
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16. EXAMPLE : In a polyatomic ion
1. What is the oxidation number of Cr in
dichromate ion Cr2O7
2-?
- The sum of the oxidation number must be
equal the charge of the ion, 2-. Therefore,
2 (ox.no.Cr) + 7 (ox.no.O) = 2-
Let X equal the ox.no. of Cr.
O is assigned an oxidation number 2-
2X + 7(2-) = 2-
2X + 14- = 2-
2X = 2- + 14+
X = 6 + ( oxidation number of Cr)
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17. Try this!
Determine the Oxidation State of the bold
element in each of the following:
1. Na3PO3
2. H2PO4
-
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18. Try this!
Determine the Oxidation State of the bold element in each of the
following:
1. Na3PO3
SOLUTIONS
• The oxidation numbers of Na and O are +1 and -2. Because sodium
phosphite is neutral, the sum of the oxidation numbers must be
zero. Letting x be the oxidation number of phosphorus, 0= 3(+1) + x
+ 3(-2). x=oxidation number of P= +3.
2. H2PO4
-
• Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion
has a charge of -1, so the sum of the oxidation numbers must be -1.
Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-
2), y= oxidation number of P= +5.
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19. Oxidation-Reduction (Redox) Reactions
19
• An oxidation occurs when
an atom or ion loses
electrons.
• A reduction occurs when an
atom or ion gains electrons.
• One cannot occur without
the other.

20. Oxidation-Reduction (Redox) Reactions
20
• OIL RIG, meaning
"oxidation is loss" and
"reduction is gain,".
mnemonic devices
• LEO says GER, meaning
"loss of e- = oxidation"
"gain of e- = reduced "

21. Displacement Reactions
A + BX  AX + B
• In displacement reactions, ion B oxidizes an element A.
• The ion B, then, is reduced to elemental form.
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22. Displacement Reactions
A + BX  AX + B
• In displacement reactions, ion B oxidizes an element A.
• The ion B, then, is reduced to elemental form.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
O +1 -1 +2 -1 O
• Oxidation number of Mg increases from 0 to +2
Mg is losing electrons so Mg is being oxidized!
• Oxidation number of H decreases from +1 to 0
H is gaining electrons so H is being reduced!
• Oxidation number of Cl remains the same
Cl is a spectator ion 22

23. 2Na + Cl2  2 NaCl
2Na + Cl2  2 NaCl
0 0 +1 -1
2 Na0  2Na +1 oxidized
Cl2
0  2Cl- reduced
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24. Try this!
Determine which element is oxidized and which
element is reduced in the following reactions (be sure
to include the oxidation state of each):
• Zn + 2H+ → Zn2+ + H2
• 2Al + 3Cu2+→2Al3+ +3Cu
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25. Try this!
Determine which element is oxidized and which element is reduced in the following
reactions (be sure to include the oxidation state of each):
1. Zn + 2H+ → Zn2+ + H2
answer
Zn is oxidized (Oxidation number: 0 → +2);
H+ is reduced (Oxidation number: +1 → 0)
2. 2Al + 3Cu2+→2Al3+ +3Cu
answer
Al is oxidized (Oxidation number: 0 → +3);
Cu2+ is reduced (+2 → 0)
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26. Balancing Redox Equations
1.Divide the equation into two
skeleton partial equations. Balance
the atoms that change their
oxidation numbers in each partial
equation.
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27. 2. Balance the O and H atoms in each partial equation.
A. Acid solution
> For each O atom that is needed, add one H2OTo
the side of the partial equation that is deficient in
oxygen.
> Add H+ where needed to bring the hydrogen
into balance.
B . For reactions in alkaline solution:
> For each O atom that is needed, add one H2O
to the side of the partial equation that is
deficient in Oxygen.
> For each H atom that is needed, add one H2O
to the side of the partial equation that is
deficient in H, and add one OH- to the opposite side
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28. 3. To each partial equation, add electrons in
such a way that the net charge on the left side
of the equation equals the net charge on the
right side.
4. If necessary, multiply one or both partial
equations by numbers that will make the
number of electrons lost in one partial equation
equal the number of electrons gained in the
other partial equation.
5.Add the partial equations. In the addition,
cancel terms common to both sides of the final
equation.
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29. Acid solution: Cr2O7
2- + Cl-  Cr+ 3 + Cl2
Step 1: Cr2O7
2-  2 Cr+ 3
2 Cl-  + Cl2
Step 2: 14 H+ + Cr2O7
2-  2Cr+ 3 + 7 H2O
2 Cl-  + Cl2
Step 3 : 6 e- + 14 H+ + Cr2O7
2-  2Cr+ 3 + 7 H2O
2 Cl-  Cl2 + 2 e-
Step 4: 6 e- + 14 H+ + Cr2O7
2-  2Cr+ 3 + 7 H2O
6 Cl-  3Cl2 + 6 e-
14 H+ + Cr2O7
2- + 6 Cl-  2Cr+ 3 + 3Cl2+ 7 H2O

30. Alkaline solution MnO4
- + N2H4  MnO4 + N2
Step 1: MnO4
-  MnO4
N2H4  N2
Step 2: MnO4
-  MnO4 + 2 H2O
4 H2O + MnO4
-  MnO4 + 2 H2O + 4 OH-
2H2O + MnO4
-  MnO4 + 4 OH-
4 OH- + N2H4  N2 + 4 H2O
step 3: 3 e- + 2 H2O + MnO4
-  MnO2 + 4 OH-
4 OH- + N2H4  N2 + 4 H2O + 4 e-
Step 4: 12 e- + 8 H2O + 4MnO4
-  4MnO4 + 16 OH-
12 OH- + 3N2H4  3N2 + 12 H2O + 12 e-
Final answer : 4MnO4
- + 3N2H4  4MnO4 +3 N2 + 4H2O + 4 OH-

31. Try This!
MnO4
- + As4O6 Mn+2 + H3AsO4

32. Try This!
MnO4
- + As4O6 Mn+2 + H3AsO4
Final answer :
24 H+ + 18 H2O + 5 As4O6 + 8 MnO4
-  20H3AsO4 + 8 Mn+2

33. Thank you
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