1 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Names of Acids Perchloric...

1

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Names of Acids• Perchloric

• Sulfuric

• Hydrochloric

• Nitric

• Sulfurous (H2SO3)

• Phosphoric

• Acetic

• Nitrous (HNO2)

• Dihydrogen phosphate ion H2PO4

-

• Hydrogen phosphate ion,

HPO4 2-

• Ammonium,NH4+

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Weak Acids & Bases

• Work examples writing reactions. MUST INCLUDE CHARGES FOR IONS!!

• H3PO4

• HOCl

• Co(H2O)6 2+

• HCO3-

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FORMULAS

• Ka (Kb) = Kw

pH = - log [H3O+]

pOH = -log [OH-]

pH + pOH = 14

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Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hconcs. of HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Step 1.Step 1. Define equilibrium concs. Define equilibrium concs.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial

changechange

equilibequilib

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Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

Step 2.Step 2. Write K Write Kaa expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

This is a quadratic. Solve This is a quadratic. Solve using quadratic formula or using quadratic formula or method of approximations method of approximations (see Appendix A).(see Appendix A).

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Useful Rule of Thumb:Useful Rule of Thumb: If 100 • KIf 100 • Kaa < C < Coo, then okay , then okay

to assume not much to assume not much dissociateddissociated

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Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.

HCOOH + HHCOOH + H22O O HCOO HCOO-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid

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Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1.Step 1. Define equilibrium concs. Define equilibrium concs.

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

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NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(a)(a) Reaction of ClReaction of Cl-- with H with H22OO

ClCl-- + + HH22O ---->O ----> HCl + HCl + OHOH--

basebase acidacid acidacid basebase

ClCl-- ion is a VERY weak base because its ion is a VERY weak base because its conjugate acid is strong. conjugate acid is strong.

Kb ClKb Cl-- (very small) (very small)

Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts

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NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(b)(b) Reaction of NHReaction of NH44++ with H with H22OO

NHNH44++ + H + H22O ---->O ----> NHNH33 + + HH33OO++

acidacid basebase basebase acidacid

NHNH44++ ion is a moderate acid because its ion is a moderate acid because its

conjugate base is weak. conjugate base is weak.

Ka NHKa NH44++ = 5.6 x 10 = 5.6 x 10 -10-10


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Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .

NaNa++ + H + H22O ---> neutralO ---> neutral

COCO332-2- ++ HH22OO HCOHCO33

-- ++ OHOH--

basebase acidacid acidacid basebase

KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 1.Step 1. Set up concentration tableSet up concentration table

[CO[CO332-2-]] [HCO[HCO33

--]] [OH[OH--]]

initialinitial

changechange

equilibequilib


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Exercise 17.15

• NaCl

• FeCl3

• NH4NO3

• Na2HPO4

Why?Why?Why?Why?• Why do acids and bases vary Why do acids and bases vary

in strength?in strength?

• Can we predict variations in Can we predict variations in acidity or basicity?acidity or basicity?

• Why do acids and bases vary Why do acids and bases vary in strength?in strength?

• Can we predict variations in Can we predict variations in acidity or basicity?acidity or basicity?

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

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1. Electronegativity 1. Electronegativity of Oof O

2.2. Polarity of O—H Polarity of O—H bond bond

3.3. The H atom of OThe H atom of O—H is readily —H is readily attracted to polar attracted to polar HH22O. O.

Acetic acidAcetic acid

Partial Partial chargescharges

(-)(-)(+)(+)

(+)(+)

(-)(-)

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High electronegativity of Cl withdraws High electronegativity of Cl withdraws electrons from the rest of the molecule. electrons from the rest of the molecule.

This makes the O—H bond highly polar. The H This makes the O—H bond highly polar. The H of O—H is very positive.of O—H is very positive.

Acetic acidAcetic acid Trichloroacetic acidTrichloroacetic acidKKaa = 1.8 x 10 = 1.8 x 10-5-5 KKaa = 0.3 = 0.3

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These ions are These ions are BASES.BASES.As the charge goes up, they interact more As the charge goes up, they interact more

strongly with polar water molecules & strongly with polar water molecules & become more and more basic .become more and more basic .

NONO33--

COCO332-2-

Basicity of OxoanionsBasicity of OxoanionsBasicity of OxoanionsBasicity of Oxoanions

POPO443-3-

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• Lewis acid = electron Lewis acid = electron pair acceptor (BFpair acceptor (BF33))

• Lewis base = Lewis base = electron pair donor electron pair donor (NH(NH33))

Lewis Acids & BasesLewis Acids & BasesLewis Acids & BasesLewis Acids & Bases

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A Lewis acid and base can interact by A Lewis acid and base can interact by sharing an electron pair.sharing an electron pair.

Lewis Acids & BasesLewis Acids & BasesLewis Acids & BasesLewis Acids & Bases

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A Lewis acid and base can interact by A Lewis acid and base can interact by sharing an electron pair.sharing an electron pair.

Formation of Formation of hydronium ion hydronium ion is an excellent is an excellent example.example.

Lewis Acids & BasesLewis Acids & Bases

HH

H

BASEACID

O—H••••

••O—H

H+H

H

H

BASEACID

O—H••••

••O—H

H+++

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Other good examples involve metal ions.Other good examples involve metal ions.

HH

Co2+ ••

BASEACID

O—H••••

••O—H

Co2+


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Other good examples involve metal ions.Other good examples involve metal ions.

COORDINATE COVALENT BONDSCOORDINATE COVALENT BONDS because both electrons are supplied by one because both electrons are supplied by one of the atoms of the bond.of the atoms of the bond.

HH

Co2+ ••

BASEACID

O—H••••

••O—H

Co2+

HH

Co2+ ••

BASEACID

O—H••••

••O—H

Co2+


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The combination of metal The combination of metal ions (Lewis acids) with ions (Lewis acids) with Lewis bases such as Lewis bases such as HH22O and NHO and NH33 ------> ------>

COMPLEX IONSCOMPLEX IONS

All metal ions form All metal ions form complex ions with water complex ions with water —and are of the type —and are of the type [M(H[M(H22O)O)xx]]n+n+ where x = 4 where x = 4

and 6.and 6.


[Cu(NH[Cu(NH33))44]]2+2+

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[Ni(H[Ni(H22O)O)66]]2+2+ + 6 NH + 6 NH33 ---> [Ni(NH ---> [Ni(NH33))66]]2+2+

+ DMG+ DMG


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The Fe2+ in heme can interact with O2 or CO in a Lewis acid-base reaction.


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Many complex ions containing water Many complex ions containing water undergo undergo HYDROLYSISHYDROLYSIS to give acidic to give acidic solutions. solutions.


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Many complex ions containing water undergo Many complex ions containing water undergo HYDROLYSISHYDROLYSIS to give acidic solutions. to give acidic solutions.

[Cu(H[Cu(H22O)O)44]]2+2+ + H + H22O ---> [Cu(HO ---> [Cu(H22O)O)33(OH)](OH)]++ + H + H33OO++


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Many complex ions containing water Many complex ions containing water undergo undergo HYDROLYSISHYDROLYSIS to give acidic to give acidic solutions. solutions.

This explains why water solutions of FeThis explains why water solutions of Fe3+3+, , AlAl3+3+, Cu, Cu2+2+, Pb, Pb2+2+, etc. are acidic., etc. are acidic.


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AMPHOTERICAMPHOTERIC ( (some metal hydroxides)some metal hydroxides)

Al(OH)Al(OH)33(s) + 3 H(s) + 3 H++ --> Al --> Al3+3+ + 3 H + 3 H22OO

Here Al(OH)Here Al(OH)33 is a Brønsted base. is a Brønsted base.

Al(OH)Al(OH)33(s) + OH(s) + OH-- --> Al(OH) --> Al(OH)44--

Here Al(OH)Here Al(OH)33 is a Lewis acid. is a Lewis acid.


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Neutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidCarbon dioxide is a neutral Lewis acid.Carbon dioxide is a neutral Lewis acid.

+1.5+1.5 -0.75-0.75-0.75-0.75

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Many complex ions are very Many complex ions are very stable.stable.

CuCu2+2+ + 4 NH + 4 NH33 [Cu(NH[Cu(NH33))44]]2+2+

K for the reaction is called K for the reaction is called

KKformationformation

or a “formation constant”or a “formation constant”

Here K = 6.8 x 10Here K = 6.8 x 101212. Reaction is . Reaction is strongly product-favored.strongly product-favored.


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Formation of complex ions explains why you Formation of complex ions explains why you can dissolve a ppt. by forming a complex can dissolve a ppt. by forming a complex ion. ion.

AgCl(s) + 2 NHAgCl(s) + 2 NH33

Ag(NHAg(NH33))22+ + + Cl+ Cl--


AgCl(s)

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Formation of complex ions explains why you Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion. can dissolve a ppt. by forming a complex ion.

AgCl(s) AgCl(s) AgAg++ + Cl + Cl-- K Kspsp = 1.8 x 10 = 1.8 x 10-10-10

AgAg++ + 2 NH + 2 NH33 --> Ag(NH --> Ag(NH33))22++ K Kformform = 1.6 x 10 = 1.6 x 1077

--------------------------------------------------------------------------

AgCl(s) + 2 NHAgCl(s) + 2 NH33 Ag(NHAg(NH33))22+ + + Cl+ Cl--

KKnetnet = K = Kspsp • K • Kformform = 2.9 x 10 = 2.9 x 10-3-3


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Exercise 17.18

• Lewis acid or base?

• PH3

• BCl3

• H2S

• HS-

1 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Names of Acids Perchloric...

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