Acid-Base Equilibria: Acids and Bases What makes an

Acid-Base Equilibria: Acids and Bases What makes an

Acid-Base Equilibria: Acids and Bases What makes an Acid an Acid? An acid possess a sour taste An acid dissolves active metals magnesium An acid causes certain vegetable dyes to turn characteristic colors

What makes a Base a Base? A bases possess a bitter taste A base feels slippery to the touch 7 strong acids and 8 strong bases Acids - HI, HBr, HCl, HClO3, HClO4, H2SO4, HNO3 Bases LiOH, NaOH, KOH,

RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 The Arrhenius Definition of an Acid and a Base An acid is a substance that produces H+ ions in water solutions HCl HH+ H+ HClA base is a substance that produces

OH- ions in a water solution NaOH HNa+ H+ HOH- Hydrogen atom acid dissociation equations HC6H5O3 C6H5O31- + H+ Fe(H2O)63+ Fe(H2O)5(OH)2+ + H+

CH3CH2NH31+ CH3CH2NH2 + H+ The Proton in Water When HCl dissolves in water we write: HCl(g) H+(aq) + Cl-(aq) Reality for the Hydronium ion + H5O2

H 9 O4 + Acidic solutions are formed by a chemical reaction in which + and acid transfers a proton (H ) to water, so we can write them either way. HCl(aq) + H2O(aq) H3O+(aq) + Cl-(aq)

or HCl(aq) H+(aq) + Cl-(aq) Nitrogen compounds are Bronsted acids when they are protonated. protonated NH4Cl

NH4+ NH3 + H+ CH3)2NH2+ (CH3)2NH + H+ The Bronsted-Lowry definition for Acids and Bases Acids may be defined as a substance that is capable of donating protons Bases may be defined as substance that accepts protons.

HCl + NH3 acid base H H HNH4+ + Cl-

conjugate acid conjugate base Is Water an Acid? NH3(aq) +H2O(aq) NH4+(aq) + OH-(aq)

Is Water a Base? HC2H3O2(aq) + H2O(aq) H3O+(aq) +C2H3O21-(aq) The auto ionization of water The reaction occurs to a very small extent; about 1 in 108 molecules is ionized at any given moment

H H H H H O

+ : + :O

: : :O H +

H .. :O .. H

Dissociation of Water, pH Scale H2O(l) H+(aq) + OH-(aq) K H=[H+] H[OH-] H H H H H[H2O] since water is a liquid and its concentration is therefore constant, this expression may be written as:

Kw H= H[H+] H[OH-] H= H1.0 Hx H10-14 [H+] H= H[OH-] H= H1.0 Hx H10-7 HM Sample Hexercise: H HIndicate Hwhether H each Hof Hthe Hfollowing Hsolutions His H neutral, Hacidic, Hor Hbasic: H a) [H+] H= H2 Hx H10-5 HM H H b) [OH-] H= H0.010 HM H H c) [OH-] H= H1.0 Hx H10-7 HM

Calculate Hthe Hconcentration Hof HH+(aq) Hin H (a)a Hsolution Hin Hwhich Hthe H[OH-] His H 0.020M H (b)a Hsolution Hin Hwhich Hthe H [OH-] H= H2.5 Hx H10-6 HM. H H Indicate Hwhether Hthe Hsolution His Hacidic Hor H basic

The HpH HScale pH H= H-log H[H+] If H[H+] H= H2. H5 Hx H10-5 H Hthe HpH His? pH H= H-log H[2. H5 Hx H10-5 H] H= H H4.6 If HpH His H3.8 Hthe HH+ Hconcentration His Antilog H-3.8= H1.58 Hx H10-4 HM In Ha Hsample Hof Hlemon Hjuice, H[H+] H= H3.8 Hx H 10-4 HM. HWhat His Hthe HpH? H

A Hcommonly Havailable Hwindow Hcleaner H has Ha H[H+] H= H5.3 H Hx H10-9 HM. HWhat His Hthe H pH? H In Ha Hsample Hof Hfreshly Hpressed Happle Hjuice H has Ha HpH Hof H3.76. H HCalculate Hthe H[H+] What Hif Hwe Htook Hthe Hlog Hof Hthe HKw H expression Kw H= H[H+] H[OH-] H= H1.0 Hx H10-14 H

H H H HpKw H= HpH H+ HpOH H= H14 H What His Hthe HpH, H[H+], H[OH-], Hof H a Hsolution Hwith Ha HpOH Hof H2.5? H H H Is Hthe Hsolution Hacidic Hor Hbasic? Major species

HCl(aq) + H2O(aq) H3O+(aq) + Cl-(aq) or HCl(aq) H+(aq) + Cl-(aq) HC2H3O2(aq) + H2O(aq) H3O+(aq) +C2H3O21-(aq) Pb(NO3)2 + NaCl NaNO3 + PbCl2 Indicators

What His Hthe HpH Hof H0.010 HM H solution Hof HHCl? If Hit Hionizes Hcompletely Hwhich His Hwhat H strong Hmeans Hthen Htake Hthe Hnegative H strong log Hof Hthe Hconcentration. HCl(aq) H H HH+(aq) H+ HCl-(aq) .

01M H H H H H H H H H.01M H H H H H.01M H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H HpH H= H2 What His Hthe HpH Hof Ha Hsolution H made Hfrom H20mL Hof H2.0M HHCl H and H35mL Hof H3.2M HHNO3? What about H2SO4

H2SO4 HH+ H+ HHSO41- H(Strong) HSO41- H HH+ H+ HSO42- H(Weak) What about weak acids HX(aq) H H HH+(aq) H+ HX-(aq), Hthen H H H Ka H H= H[H+][X-] [HX] The Hsmaller Hthe Hvalue Hof Hthe Hacid H

dissociation Hconstant HKa, Hthe H weaker Hthe Hacid What His Hthe HKa Hof Ha H0.10 HM Hsolution Hof H formic Hacid H(HCHO2) Hwhich Hhas Ha H pH H H= H2.38? H I C

E HCHO2 H+ + 0.10 0 - .00417 + .00417 .0958

.00417 CHO210 + .00417 .00417 Ka = (.00417)2 = 1.8 x 10-4 .0958

What His Hthe Hconcentration Hof HH+ Hions H in Ha H0.10 HM Hsolution Hof HHC2H3O2 H(Ka H = H1.8 Hx H10-5)? HpH? H% Hionization? HC2H3O2 I .10 C -X E

.10 X 1.8 x 10-5 = X2 .10 X H+ + C2H3O210 0 +X +X

X X X = 1.3 x 10-3 pH = 2.87 percent dissociation 1.3 X 10-3 x 100 = 1.3% .10

What His Hthe HpH Hand Hpercent H ionization Hof Ha H0.20 HM Hsolution Hof H HCN? H HKa H= H4.9 Hx H10-10 H Acid-Base Equilibria: Strong Bases The Hmost Hcommon Hsoluble Hstrong HBases Hare Hthe Hhydroxides H of Hgroup HIA Hand HCa, HBa Hand HSr

What His Hthe HpH Hof Ha H H0.010 HM H solution Hof HBa(OH)2? Anions Hof HWeak HAcids Amines Dealing with Weak Bases The Hbase Hdissociation Hconstant HKb Hrefers H Hto Hthe H

equilibrium Hin Hwhich Ha Hbase Hreacts Hwith HH2O Hto form Hthe Hconjugate Hacid Hand HOHWeak H Hbase H+ HH2O H H Hconjugate Hacid H+ HOHNH3 H(aq) H+ HH2O H(l) H H HNH4 H+(aq) H+ HOH-(aq) + H H [NH ] [OH ] 4 Kb H=

[NH3] Calculate Hthe H[OH-] Hin Ha H0.15 HM Hsolution Hof HNH3. HNH3 H H H H H+ HH2O H H H H H H H H HNH4+ H H H H H+ H H HOHI C E Polyprotic HAcids H2SO3(aq) H HH+(aq) H+ HHSO3-(aq) H H H H H H H H H H H Ka1 H= H1.7 Hx H10-2

HSO3-(aq) H H HH+(aq) H+ HSO32-(aq) H H H H H H H H H H H H H Ka2 H= H6.4 Hx H10-8 H H HCalculate Hthe HpH Hof Ha H.1M Hsolution What His Ha HSalt HNaOH(aq) +HC2H3O2(aqH2O+NaC2H3O2 NaC2H3O2 H HNa+(aq) H+ H HC2H3O2-(aq) Na+(aq) H+ HH2O H HNaOH(aq) H+ HH+(aq)

C2H3O2-(aq)+H2O HHC2H3O2 H(aq)+OH- H(aq) NH4Cl H HNH4+ H+ HClCl- H+ HHOH HHCl H+ HOH- NH4+ H HNH3 H+ HH+ Acid HBase Hor HNeutral NH4Cl H NaC2H3O2

NH4C2H3O2 Salt Hderived Hfrom Ha Hstrong Hbase Hand Ha Hstrong Hacid H will Hhave Ha HpH Hof H7 Salt Hderived Hfrom Ha Hstrong Hbase Hand Ha Hweak Hacid H will Hhave Ha HpH Habove H7 Salt Hderived Hfrom Ha Hweak Hacid Hand Ha Hweak Hbase H depends Hupon Hwhether Hthe Hdissolved Hion Hacts Has H an Hacid Hor Ha Hbase Has Hdetermined Hby Hthe Hsize Hof H

the HKa Hor HKb Ka and Kb NH4 + H HNH3(aq)+ HH

(aq) [H ][NH3] + (aq) NH3(aq)+ HH2O HNH4+ H(aq)+ HOH-(aq)

+ Ka H= Kb H= [NH4+]

[NH4][OH- H] [NH3] NH4+(aq) H H HNH3(aq) H+ HH+ H(aq) NH3(aq) H H+ HH2O(l) HNH4+(aq) H H+ HOH- H(aq) H2O H HH+(aq) H+ HOH-(aq) When Htwo Hreactions Hare Hadded Hto Hgive Ha Hthird Hreaction, Hthe H equilibrium Hconstant Hfor Hthe Hthird Hreaction Hreaction His Hgiven Hby Hthe Hproduct Hof Hthe Hequilibrium Hconstants Hfor Hthe Htwo H

added Hreactions pKa H H+ H HpKb H= HpKw K H Hx H HK H= HK a b w

pH of Salt Solutions Calculate Hthe HpH Hof Ha H0.01 HM Hsolution Hof Hsodium H hypochlorite H H(NaClO) ClO- H H H H+ H H H HH2O H H H H H HHClO H H H H H+ H H H HOHI C E Now Hits Hyou Hturn: H Hthe HKb Hfor HBrO- His H5.0 Hx H10-6. H Calculate Hthe HpH Hof Ha H0.050 HM Hsolution Hof HNaBrO

Calculate Hthe H(a) Hbase-dissociation Hconstant, HKb, Hfor Hthe H fluoride Hion, Hif Hthe HpKa Hof HHF H= H3.17 pKa H= H-log HKa 3.17 H= H-log HKa Antilog H-3.17 H= H6.76 Hx H10-4 Since H H Ka H Hx H HKb H= HKw (6.76 Hx H10-4)x HKb H= H1.0 Hx H10-14 Kb H= H1.0 Hx H10-14/ H6.76 Hx H10-4 H= H1.5 Hx H10-11

or pKa H+ HpKb H= H14 Hso HpKb H= H10.83 H H H H H H H H H H H H H H H H H10-10.83 H= H1.5 Hx H10-11 Calculate Hthe HpKb Hfor Hcarbonic Hacid H(Ka H= H4.3 Hx H10-7) Now Hits Hyour Hturn Anions Hof Hpolyprotic Hacids, Hsuch Has HHCO3-, H Hthat Hstill Hhave H ionizable Hprotons are Hcapable Hof Hacting Has Heither Hproton H

donors Hor Hacceptors Hdepending Hupon Hthe Hmagnitudes Hof Hthe H Ka Hor HKb Predict Hwhether Hthe Hsalt HNa2HPO4 Hwill Hform Han Hacidic Hor Hbasic Hsolution Hon Hdissolvingin Hwater. Na2HPO4 H H2Na+ H(aq) H+ HHPO4HPO4- Hacting Hlike Han Hacid K3 H= H4.2 Hx H10-13 HPO4 (aq) H+ HH2O H HH3O H+ HPO4 (aq) H H H H H H H

HPO4- Hacting Hlike Han Hbase HPO4- H(aq) H+ HH2O H HH2PO42-(aq) H+ HOH-(aq) - + 3- 1.0 Hx H10-14

Kw = Kb H= 6.2 Hx H10-8 Ka So HHPO- His Hthe Hconjugate Hbase Hof HH2PO4-. H H Since Hthe HK2 Hof HH2PO4- H= H6.2 Hx H10-8 H Hthen: = 1.6 Hx H10-7 H

Since HKb His Hlarger Hthan HKa, HHPO4- Hwill Hact Hlike Ha Hbase Acid-Base HCharacter Hand HChemical HStructure two Hthings Hto Hconsider Hpolarity Hdifference Hand H strength Hof Hthe Hbond HF H> HHCl H> HHBr H> HHI (most Hpolar H H H H H H H H H H Hleast) H H H HBased Hon Helectronegativity H

difference HHF His Hthe Hmost Hpolar Hbut Ha Hweak Hacid H because Hthe Hbond His Hso Hstrong Acid Hstrength Hof Hoxyacids The Hmore Hoxygen's Hthe Hstronger H the Hacid Hbecause Hof Hthe Hoxygen H pulling Hthe Helectrons Htowards H themselves. H HClO4> HHClO3> HHClO2> HHOCl

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