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9: Charge transfer rxns: acids and bases and oxidation-reduction
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What are acids?
• Acids:
• sharp,sour taste, feel prickly on skin,
• react with metals to release H2(g),
• react with carbonates to release CO2(g),
• turn litmus red,
• neutralize bases.
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What are bases?
• Bases:
• bitter taste
• slippery to touch
• turn litmus blue
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Arrhenius definitions
• Acid increases __ conc (hydrogen ion, proton) when dissolved in water.
• HCl (aqCl-(aq)
• Acid has to have __ to be acid.
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Arrhenius definitions
• Base increases ____ (hydroxide ion) conc when dissolved in water
• NaOH Na+(aq) + OH-(aq)
• Base has to have __
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Arrhenius acid-base rxn
• Arrhenius acid-base neutralization rxn yields salt + water.
• Arrhenius limited to aqueous solution.
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But NH3 is a base!!
• Arrhenius says bases have to have OH-’s and acids H+’s.
• However NH3 is a base but has no OH- in its formula.
• Now have enlarged defn of acid and base that’s not restricted to aqueous soln
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• A Bronsted-Lowry (B-L) acid
• A B-L base
• The acid-base neutralization rxn is the transfer of a H+ from an acid to a base.
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• Concept of base in enlarged (H+ acceptor)
• Acid still has to have H+.
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Identify acids and bases in
• SO32- + HCl HSO3- + Cl-
• Note that the products of the rxn are acids and bases themselves.
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• H3PO4 + H2O H3O+ + H2PO4-
• H3O+: hydronium ion
• HCO3- + H2O H2CO3 + OH-
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Water
• Note that water can act both as a base and as an acid.
• H3PO4 + H2O H3O+ + H2PO4-
• HCO3- + H2O H2CO3 + OH-
• Amphiprotic (amphoteric)
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Conjugate acid-base pairs
• Conjugate acid-base pairs are species on opposite sides of the rxn that differ by one proton only.
• Are these conjugate acid/base pairs?
• H 2O/OH-
• H2SO4/SO42-
• H2SO4/ HSO3-
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• Given base, to get conjugate acid: ___
• What is the conjugate acid of:HS- SO4
2-
NH3 S2-
• Given acid to get conjugate base: _________
• What is the conjugate base of:
• HS- NH3 HSO4- S2-
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• 9.36: Identify conjugate acid-base pairs in:
• NH4+(aq) + CN--(aq) NH3(aq) +HCN(aq)
• CO32-(aq) + HCl(aq) HCO3
-(aq) + Cl-(aq)
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Strength of acids and bases
• Some species are better proton donors than others; some species are better proton acceptors than others.
• The better the species is at donating a proton to the base, the stronger the acidic properties of the species.
• Strength is a measure of the degree of dissociation of the acid or base in solution. Stronger the acid(base) the larger the value of K for that dissociation.
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• The better the species is at accepting the proton from the acid, the stronger the basic properties of the species.
• We are talking about strong and weak here in relative terms.
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• Generally talk about aqueous sol’ns.• Some acids and bases are strong acids and
bases meaning that they are
• HCl + H2O Cl- orHCl(aq) H+(aq) + Cl-(aq)
• NaOH(aq) Na+(aq) + OH-(aq)• Ca(OH)2(aq)Ca2+(aq) + 2OH-(aq)
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• Some acids and bases are not 100% ionized. These are weak acids (and bases). (weak electrolytes)This means that K < 1 for these rxns. Instead an equilbrium is set up:
• HF + H2O H3O+ + F- Ka= 7.1 x 10-4
• HCOOH + H2O HCOO- + H3O+
Ka = 1.7 x 10-4
• Larger value of Ka stronger the acid
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How do you know what acids and bases are strong?
• See fig 9.2 p239Strong acids Strong bases
• HCl OH- (metal )
• H2SO4 (1st ioniz. only) S2-
• HNO3 O2-
• H3O+
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• The stronger the acid, the weaker the conj. base of the acid. The weaker the acid, the stronger the conj. base of the acid, etc.
• What does this imply?
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• Acid (dec ) Base (inc • HCl Cl-
• H3O+ H2O
• HSO4- SO4
2-
• HF F-
• CH3COOH CH3COO-
• NH4+ NH3
• Which is the stronger acid--HF or HSO4-?
• Which is the stronger base--H2O or F-?
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Dissociation of water• HOH(l) + HOH(l) H3O+ + OH-• Acid1 base2 acid2 base1 • Or H2O(l) H+(aq) + OH-(aq)• Find from experiment that for pure water at 25oC
that [H3O+] = [OH-] = • Can write the ion product for the dissociation of
water as ion product = [H3O+][OH-] = ____________
at 25oC• H3O+ and H+ stand for the same thing.
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• Can write the ion product for the dissociation of water as
ion product = [H3O+][OH-] = 1 x 10-14 at 25oC
• This is also called Kw.
• This holds for all solutions not just pure HOH
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• Kw = [H3O+][OH-] = 1 x10-14
• as [H3O+]
• If [H3O+] = 4.6 10-3M, [OH-] =?
• as [OH-]
• If [OH-] = 1.22 x 10-4M, [H3O+]=?
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pH or how to avoid the exponents
• define pH = -log10[H3O+]
• Neutral sol’n: [H3O+] =[OH-] = 1 x10-7MpH=_
• Acidic sol’n; [H3O+] >OH-] ; [H3O+] > 1x10-7M pH<_
• Basic sol’n: [OH-]> [H3O+]; [OH-]>1x10-7MpH >_
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• [H+]M pH • 10 -1• 1 0• 0. 1 1 • 0.001 3• 1x10-7 7 • 1x10-10 10• 1x10-14 14 • 1x10-15 15 • As pH [H+]
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• What’s pH for a sol’n whose [H3O+] = 4.46 x10-3M?
• If pH = 9.72, what’s [H3O+]?• If pH = 4.45, what’s [H3O+] ?• What is [OH-] in the above solutions?
• Calc pH of 0.0158M HCl
• Calc pH of 1 x 10-4M KOH and of 0.036M KOH
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• 9.53 and 54. How many times more basic is a soln at pH 6 relative to pH 4?
• How many times more acidic is a soln at pH 7 relative to pH 11?
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9.3 Rxns btn acids and bases• Neutralization rxn
HCl(aq) + NaOH(aq) NaCl(aq) + HOH(l)• acid + base salt + water• In the dissociated form
H+(aq) +Cl-(aq) +Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + HOH(l)
• H+(aq) + OH-(aq) H2O(l)• This is the same for any strong acid-strong
base neutralization
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Titration
• Used to determine the conc of unknown acid or base solution.
• 6.00mL of a 0.500M HCl soln is needed to neutralize 10.00mL of a NaOH soln. What is the concentration (molarity) of the NaOH soln?
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• 15.00mL of a 0.200M NaOH soln is needed to neutralize 25.00mL of a HCl soln. What is the concentration (molarity) of the HCl soln?
• Equivalence pt and end point, indicators
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Polyprotic acids
• Some acids have more than one acidic hydrogen as: H2SO4, H2CO3, H3PO4
• H2CO3 +2NaOH Na2CO3 + 2H2O
• Undergo stepwise dissociation as:
• H2CO3 + H2O HCO3- + H3O+
• HCO3- + H2O CO3
2- + H3O+
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9.4 Acid-base buffers
• Solutions that contain a
• Buffers have the ability of withstanding small additions of added acid or base without large fluctuations in pH.
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How do buffers work?
• How do buffers work?
• HA H+ + A-
• add acid A- + H+ HA
• add base HA + OH- HOH + A-
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• Have 100mL of 3.6x10-5M HCl; pH = 4.44
• Add 1mL of 0.10M HCl to the above soln, the pH changes to 2.99; add 1mL of 0.10M NaOH to the above soln, pH changes to 11.99.
• Have 100mL of a buffer that is 0.050M Na acetate and0.10M acetic acid, pH = 4.44
• Add 1mL of 0.10M HCl to the buffer, pH changes to 4.43; add 1mL of 0.10M NaOH to the buffer, pH changes to 4.46.
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• Are these buffers:
• NaOH/NaCl
• NaOH/HCl
• NH3/NH4Cl
• NaF/HF
• CH3COOH and CH3COONa
• HNO3 and KNO3
• HBr and MgCl2
• H2CO3 and NaHCO3
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Calculating the pH of a buffer
• HA(aq) + H2O(l) H3O+(aq) + A-(aq)
• Ka = [H3O+][A-] [HA]
• Or [H3O+] = Ka[HA][A-]
• [H3O+] = Ka[acid] [salt]
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• 9.65 and 66 What is [H3O+] for a buffer soln that is 0.200M in acid and 0.500M in the corresponding salt if the weak acid has Ka= 5.80 x 10-7? What is the pH of the solution?
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• What is the pH of a solution that is 0.1M in CH3COOH and 0.1M in CH3COONa? Ka for acetic acid is 1.8 x 10-5.
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• 9.49 and 50 ForCH3COOH(aq)+H2O(l) CH3COO-(aq)
+H3O+(aq)
• Write the K expression• Explain what happens if • Add strong acid
• Dilute with water
• Add strong base
• Add more acetic acid
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Henderson-Hasselbalch equation
• pH = pKa + log
•
• Solve: What is [H3O+] for a buffer soln that is 0.200M in acid and 0.500M in the corresponding salt if the weak acid has Ka= 5.80 x 10-7? What is the pH of the solution? Usong Henderson-Hasselbalch
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9.5 Oxidation-reduction processes
• Oxidation (oxd’n) ; originally rxn in which O2 combined with another substance as
C(s) + O2 CO2(g)
• Also defined as
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• Reduction (red’n) : originally rxn in which O2 was removed from a substance as
Fe2O3(s) + 3C(s) 3CO(g) + 2 Fe(s)
• Also defined as
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Let’s look at gain and loss of electrons more closely
• Look at the reaction2NaCl(s) 2Na(s) + Cl2(g)
• We can divide this rxn into half rxns as
•
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And we identify oxidation and reduction as
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Na+ + e- Na Cl- Cl2 + 2e-
• The species (reactant, Cl-) that is oxidized is the _________ agent (reducing agent donates e-’s to the species that is reduced)
• The species (reactant, Na+) that is reduced is the ___________ agent (oxidizing agent causes oxd’n of another species by accepting e-’s from that species)
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Identify species oxidized, reduced, oxidizing agent,
reducing agent
• Cu2+ + Mg Mg2+ + Cu
• Br2 + 2KI 2KBr + I2
• 4Fe + 3O2 2Fe2O3
• Zn + Cu 2+ Zn2+ + Cu
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Corrosion
• Corrosion: eating away of metals by oxidation-reduction processes.
• Rusting of iron4Fe(s) + 3O2(g) 2Fe2O3(s)
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Voltaic cells
• Voltaic cells: spontaneous chemical rxn is used to generate electricity
• Cu2+(aq) +Zn(s) Zn2+(aq) + Cu(s)
• Oxidation occurs at the______ and reduction at the _________.
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• Half cell rxns: Cu2+ + 2e- Cu Zn Zn2+ + 2e-
• Electrochemical cells: generate electricity (e-’s moving through a wire) by spont. chem. rxn --voltaic cells
• Right now we have a direct transfer of e-’s from the anode to the cathode: short circuit
• Cu2+ + 2e- Cu GER; cathode Zn Zn2+ + 2e- LEO; anode
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• Need to design a system such that keeps Cu2+ and Zn from being in direct contact with each other.
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Eocell = 1.10V
Zn --> Zn2+ + 2e- Cu2+ + 2e- -->Cu
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Batteries: portable source of electricity
• Flashlight battery• anode: Zn(s) Zn2+(aq) + 2e-
• cathode: 2NH4+(aq) + 2MnO2(s) + 2e-
Mn2O3(s) + 2NH3(aq) + H2O(l)
• overall: Zn(s) + 2NH4+ (aq) +MnO2(s) Zn2+(aq)
+ Mn2O3(s) + 2NH3(aq) + H2O(l)
• supplies 1.5V new; not rechargeable; voltage decreases on use (why?)
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(inert
electrolyte
Why?
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Alkaline version of flashlight battery
• anode: Zn(s) + 2OH-(aq) ZnO(s) + H2O(l) + 2e-
• cathode: 2MnO2(s) + H2O(l) + 2e- Mn2O3(s) + 2OH-
• overall: Zn(s) + 2MnO2(s) Mn2O3(s) + ZnO(s)
• advantage: supplies constant voltage over time period
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Mercury battery
• anode: Zn(Hg) + 2OH-(aq) ZnO(s) + H2O(l) + 2e-
• cathode: HgO(s) + H2O(l) +2e- Hg(l) + 2OH-(aq)
• overall: Zn(Hg)+ HgO(s) ZnO(s) + Hg(l)
• Eo constant(1.35V)--why?; long life: good for pacemakers, watches, hearing aids
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•Zn(Hg)+ HgO(s) ZnO(s) + Hg(l)
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Electrolysis
• Electricity used to cause a nonspontaneous chemical reaction to occur as
• Definitions for oxd’n, etc still hold
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Electrolysis of water
• Electrolysis of pure water slow. Why?
• Instead use 0.1M H2SO4 soln
• cathode: 2H+ + 2e- H2
• anode: 2H2O O2(g) + 4H+ + 4e-
• Overall 2H2O 2H2(g) + O2(g)
• cathode anode