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Lecture March 11
1
No HW and Quiz due this week
Chap. 4 - Part IIAcid-Base Reactions
Oxidation-Reduction Reactions
All regrades must be submittedby THIS Friday, March 14
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Acid-Base Reactions
3
Ionic compoundin water
Ions in solution
are surrounded bywater molecules
HCl
Note the H+hasattached to the water to
give H3O+thehydronium ion
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Strong Acids and Bases
4
The HCl dissociates into H
+
and Cl
-
ionsH+then attaches to H2O to produce H3O+ ion
Strong base is NaOHNaOH(aq) !Na+(aq) + OH-(aq)
Produce OH-ions in solution
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Acid-Base Reactions
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Acid + Base !water + salt
HCl(aq) + NaOH(aq) !H2O(l) + NaCl(aq)
Net ionic equation for strongacid + strong base
H+(aq) + OH-(aq) !H2O(l)
H
+
(aq) +Cl-(aq) Na
+
(aq) +OH-(aq) Na+
(aq) + Cl-
(aq)
Neutralization Reaction
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Strong Bases
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Soluble Hydroxides - give OH-inaqueous solution
LiOH, NaOH, KOH very soluble
Ba(OH)2and Ca(OH)2slightly solublebut will still react with
strong acid even
as a solid.
2 HCl (aq) + Ca(OH)2(s) !CaCl2(aq) + 2 H2O(l)
OR Ca(OH)2(aq)
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Solubility and Molarity
7
Solubility of Ca(OH)2is 0.173 g/100mL.What is the concentration
of a saturated
solution of Ca(OH)2?
Molar Mass = 57.10 g/mol
M = n/Vn = (0.173 g) (1 mol/57.10 g) = 0.003030 mol
M = .003030 mol/.100 L = 0.0303 M
Note: concentration of OH-= 2 x (0.0303) = 0.0606 M
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Calculations - Titrations
8
Example of M1V1= M2V2
Moles of Acid = Moles of base
If it takes 23.7 mL of a 0.105 M solution ofNaOH to neutralize
(equivalence point)50.0
mL of an unknown strong acid, what wasthe initial concentration
of that acid?
Ans: 0.0498 M
See 4.19 and 4.20 for a titration example
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More Acid-Base Reaction
Using Concentrations
9
What volume of 0.135 M HNO3isrequired to fully neutralize 100.0
mL of a
0.030 M Ca(OH)2solution?
n base = (0.030)(.100)(2) = .0060 mol base
nbase= nacid = MV
nacid= .0060 = (0.135) Vacid
Vacid= 44 mL of HNO3
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Another Example
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What volume of a 0.75 M HCl solution isrequired to neutralize
100.0 mL of a 0.23
M strong base
Ans: 31 mL
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Oxidation-Reduction
Reactions
11
Determine the OXIDATION state(number) of an element in a
reaction
Will use the rules on p. 164 to determineoxidation number - when
possible
2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
assign the
S -2, thenPb is +2
O = 0 O = -2
Pb = +2
O = -2
S = +4
S goes from -2 to +4 O goes from 0 to -2
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Determining the Oxidizingand Reducing Agents
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2 Mg(s) + O2(g) 2MgO(s)
oxidation - increase in ox #, loss of electronsreduction -
decrease in ox #, gain of electrons
One compound is oxidizedand one is reduced
O2is reduced (0!-2)
Mgis oxidized (0!+2)
0 0 Mg=+2 O=-2
OIL RIG
oxidizing agent
reducing agent
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More on Ox-Red
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2N2H4(g) + N2O4(g) ! 3N2(g) + 4H2O(g)
Note: H and O are unchanged so willconcentrate on the N in each
compound
N2H4:H=(+1)4=+4
N = -4/2 = -2
N2O4:O=(-2)4=-8
N = +8/2 = +4
N2:N=0
N -2!0oxidation
N2H4reducing agent
N +4!0reduction
N2O4
oxidizing agent
Note: checkingthe ox # change
-2 x 4 = -8
+4 x 2 = +8
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Quiz of the Day
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If the solubility of Ba(OH)2is 3.89 g/100 mL,what would be the
concentration of OH-(inmolarity) in a saturated solution of
Ba(OH)2?
Molar mass of Ba(OH)2= 171.35 g/mol
Ans: 0.454 M
