exp. 16 exp. 16: volumetric analysis: redox titration...
TRANSCRIPT
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Exp. 16: Volumetric Analysis: Redox Titration
Normality = eq wt of solute
L solution
Acid/bases: #eq = # H+ or OH- ionized
Redox reactions – transfer of e-
reduction – oxidation reactions
Exp. 16 – video (time: 1 hr and 23:08 minutes)
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Redox reaction
Equivalent wt - one equivalent of any oxidizing agent
reacts with one equivalent of any reducing agent.
This means #eq/mol is equal to the number of e-
transferred.
MnO4-(aq) + 8H+
(aq) + 5e- Mn2+(aq) + 4H2O(l) (net)
MnO4- : 5eq same for KMnO4
mol MnO4-
Fe2+(aq) Fe3+
(aq) + 1e- 1eq
mol Fe2+
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N M or M N
N (eq) = M (mol) x #eq
L L mol
Note: N equal to or greater than M
0.1 M KMnO4 N? Goal: eq KMnO4
L soln
MnO4-(aq) + 8H+
(aq) + 5e- Mn2+(aq) + 4H2O(l)
Calc:
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Solubility Rules for Ionic Compounds (Dissociates 100%)
1.) All compounds containing alkali metal cations and the ammonium ion are soluble.
2.) All compounds containing NO3-, ClO4
-, ClO3
-, and C2H3O2
- anions are
soluble.
3.) All chlorides, bromides, and iodides are soluble except those containing Ag+, Pb2+, or Hg2
2+.
4.) All sulfates are soluble except those containing Hg22+, Pb2+, Ba2+, Sr2+,
or Ca2+. Ag2SO4 is slightly soluble.
5.) All hydroxides are insoluble except compounds of the alkali metals and Ca2+, Sr2+, and Ba2+ are slightly soluble.
6.) All other compounds containing PO43-, S2-, CO3
2-, CrO42-, SO3
2- and most other anions are insoluble except those that also contain alkali metals or NH4
+.
Generally, compound dissolves > 0.10 M - soluble (aq)
< 0.01 M - insoluble (s)
in between - slightly soluble
(this class we will assume slightly soluble as soluble)
Hg2Cl2 (s) insoluble
KI (aq) soluble
Pb(NO3)2 (aq) soluble
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Strong Acids (Ionizes 100%)
HCl, HBr, HI, HClO4, HNO3, H2SO4
Strong Bases (Dissociates 100%)
NaOH, KOH, LiOH, Ba(OH)2, Ca(OH)2,
Sr(OH)2
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• A molecular/formula unit equation is one in which the reactants and products are written as if they were molecules/formula units, even though they may actually exist in solution as ions.
Calcium hydroxide + sodium carbonate
M.E.
Ca(OH)2
Ions in Aqueous Solution Molecular and Ionic Equations
+ Na2CO3 CaCO3 + NaOH 2 (aq)
strong base strong base soluble salt insoluble salt
(aq) (s) (aq)
s solid
l liquid
aq aqueous (acid/bases and soluble salts dissolve in water)
g gases
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• An total ionic equation, however, represents strong electrolytes as
separate independent ions. This is a more accurate representation of the
way electrolytes behave in solution.
– A complete ionic equation is a chemical equation in which strong
electrolytes (such as soluble ionic compounds, strong acids/bases) are
written as separate ions in solution. (note: g, l, insoluble salts (s), weak
acid/bases do not break up into ions)
M.E.
Ca(OH)2 (aq) + Na2CO3 (aq) CaCO3 (s) + 2 NaOH (aq)
Total ionic
Ions in Aqueous Solution
Molecular and Ionic Equations
Ca2+ (aq) + 2OH- (aq)
strong base soluble salt insoluble salt strong base
+ 2Na+ (aq) + CO32-
(aq) CaCO3 (s) + 2Na+ (aq) + 2OH- (aq)
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Net ionic equations.
– A net ionic equation is a chemical equation from
which the spectator ions have been removed.
– A spectator ion is an ion in an ionic equation that
does not take part in the reaction. M.E.
Ca(OH)2 (aq) + Na2CO3 (aq) CaCO3 (s) + 2 NaOH (aq)
Total Ionic Ca2+ (aq) + 2OH- (aq) + 2Na+ (aq) + CO3
2- (aq) CaCO3 (s) + 2Na+ (aq) + 2OH- (aq)
Net
Ca2+ (aq) + CO32-
(aq) CaCO3 (s)
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Types of Chemical Reactions
• Oxidation-Reduction Reactions (Redox rxn)
– Oxidation-reduction reactions involve the
transfer of electrons from one species to another.
– Oxidation is defined as the loss of electrons.
– Reduction is defined as the gain of electrons.
– Oxidation and reduction always occur
simultaneously.
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27.1 Reduction and Oxidation
Redox reactions – transfer of e-
reduction – oxidation reactions
Reduction – gain of e- / gain of H / lost of O
Fe3+ + 1e- Fe2+ (lower ox state)
note: must balance atoms and charges
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Oxidation - loss of e- / loss of H / gain of O
Fe2+ Fe3+ + 1e- (higher ox state)
H2O + BrO3- BrO4
- + 2H+ + 2e-
(Br oxidized: charge 5+ 7+)
2H+ + 2e- H2 (H reduced: charge 1+ 0)
Oxidizing agent is species that undergoes reduction.
Reducing agent is species that undergoes oxidation.
Note: need both for reaction to happen; can’t have
something being reduced unless something else is being
oxidized.
Br + 3(-2) = -1
Br = -1 +6 = +5
Br + 4(-2) = -1
Br = -1 +8 = +7
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27.3 Balancing Redox Reactions
- Must know charges (oxidation numbers) of species
including polyatomic ions
- Must know strong/weak acids and bases
- Must know the solubility rules
Oxidation Numbers – hypothetical charge assigned to the
atom in order to track electrons; determined by rules.
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Rules to balance redox
1.) Convert to net ionic form if equation is originally in molecular form
(eliminate spectator ions).
2.) Write half reactions.
3.) Balance atoms using H+ / OH- / H2O as needed:
– acidic: H+ / H2O put water on side that needs O or H (comes from
solvent)
– basic: OH- / H2O put water on side that needs H but if there is no H
involved then put OH- on the side that needs the O in a 2:1 ratio
2OH- / H2O balance O with OH, double OH, add 1/2 water to
other side.
4.) Balance charges for half rxn using e-.
5.) Balance transfer/accept number of electron in whole reaction.
6.) Convert equation back to molecular form if necessary (re-apply
spectator ions).
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Zn(s) + AgNO3(aq) Zn(NO3)2(aq) + Ag(s)
Total ionic:
Net ionic:
Zn(s) + Ag+(aq) + NO3-(aq) Zn2+(aq) + 2NO3
-(aq) + Ag(s)
Zn(s) + Ag+(aq) Zn2+(aq) + Ag(s)
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Net: Zn(s) + Ag+(aq) Zn2+
(aq) + Ag(s)
Oxidation:
Reduction:
Balanced net:
Balanced eq:
Zn(s) Zn2+(aq) + 2e-
Ag+(aq) Ag(s) 1e- +
Zn(s) + 2 Ag+(aq) Zn2+(aq) + 2 Ag(s)
[ ] 2
Zn(s)
15
+ 2 AgNO3(aq) Zn(NO3)2(aq) + 2 Ag(s)
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H+
Net: MnO4-(aq) + Fe2+
(aq) Mn2+(aq) + Fe3+
(aq)
Ox:
Red:
Balanced net:
Fe2+(aq) Fe3+(aq) + 1e- [ ] 5
MnO4-(aq) Mn2+(aq) + H2O(l) 4 8 H+(aq) + 5e- +
8 H+(aq) + MnO4-(aq) + 5 Fe2+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
16
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KMnO4(aq) + NaNO2(aq) + HCl(aq) NaNO3(aq) + MnCl2(aq) + KCl(aq) + H2O(l)
Net:
Ox:
Red:
Balanced net:
Balanced eq:
MnO4-(aq) Mn2+(aq) + NO2
-(aq) NO3
-(aq) + + H+(aq) + H2O(l)
NO2-(aq) NO3
-(aq)
MnO4-(aq) Mn2+(aq) + 4 H2O(l) 8 H+(aq) +
H2O(l) + + 2 H+(aq)
5 e- +
+ 2 e- [ ] 5
[ ] 2
2 MnO4-(aq) + 5 NO2
-(aq) + 16 H+(aq) + 5 H2O(l) 2Mn2+(aq) + 8 H2O(l) + 5 NO3-(aq) +10 H+(aq)
2 MnO4-(aq) + 5 NO2
-(aq) + 6 H+(aq) 2Mn2+(aq) + 3 H2O(l) + 5 NO3-(aq)
2 KMnO4(aq) + KCl 2 17
+ 5 NaNO2(aq) + 6 HCl(aq) 2MnCl2(aq) + 3 H2O(l) + 5 NaNO3(aq)
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Net: OH-
CrI3 (s) + Cl2 (g) CrO42-
(aq) + IO4-(aq) + Cl-(aq)
Ox:
Red:
Balanced net:
CrI3(s) CrO42-(aq) + IO4
-(aq)
Cl2(g) Cl-(aq) 2
3 32 OH-(aq) + + 16 H2O(l)
2 e- +
+ 27 e- [ ] 2
[ ] 27
64 OH-(aq) + 2 CrI3(s) + 27 Cl2(g) 2 CrO42-(aq) + 6 IO4
-(aq) + 54 Cl-(aq) + 32 H2O(l)
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Exp 16:
S2O32-
(aq) + I2 S4O62-
(aq) + I-(aq)
thiosulfate ion iodine
Ox:
Red:
Balanced net:
Outside exercise II page 199 – posted on my website
S2O32-(aq) S4O6
2-(aq)
I2(aq) I-(aq) 2
2 + 2 e-
2 e- +
2 S2O32-(aq) + I2(aq) S4O6
2-(aq) + 2 I-(aq)
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S2O32-
2eq = 1 eq
2mol S2O32- mol S2O3
2-
I2 2eq
mol I2
Exp today
First: Standardize thiosulfate against 0.100 N I2 standard
solution.
Changes in sample preparation:
10 mL I2, 30 mL deionized H2O, 1 mL starch (20 drops)
Starch – indicator (add from beginning)
Starch + I2 gives blue color
At end pt (all I2 consumed), solution will be colorless
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Since using normality can use
NiodineViodine = NthiosulfateV thiosulfate
minimum 3 runs ± 0.005 N (around ± 0.5 mL)
report
Avg N ± s N thiosulfate ion (S2O32-)
Convert average N to M
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Second: Same exact procedure as standardization
except using unknown conc. of I2.
minimum 3 runs ± 0.005 N (around ± 0.5 mL)
report
Avg N ± s N iodine (I2) unknown
Convert average N to M
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Amount of chemicals to obtain in small beaker per
group:
Na2S2O3.5H2O – 150 mL (source of thiosulfate ions)
0.100 N I2 standard solution – 50 mL
Unknown I2 solution – 45 mL