Chapt. 18 – Acids and Bases 18.1 Acids and Bases: An Introduction

Chapt. 18 – Acids and Bases 18.1 Acids and Bases: An Introduction
18.2 Strengths of Acids and Bases 18.3 Hydrogen Ions and pH 18.4 Neutralization

Section 18.1 Introduction to Acids and Bases
Different models help describe the behavior of acids and bases. Identify the physical and chemical properties of acids and bases. Classify solutions as acidic, basic, or neutral. Compare the Arrhenius, Brønsted-Lowry, and Lewis models of acids and bases.

Section 18.1 Introduction to Acids and Bases
Key Concepts The concentrations of hydrogen ions and hydroxide ions determine whether an aqueous solution is acidic, basic, or neutral. An Arrhenius acid must contain an ionizable hydrogen atom. An Arrhennius base must contain an ionizable hydroxide group. A Brønsted-Lowry acid is a hydrogen ion donor. A Brønsted-Lowry base is a hydrogen ion acceptor. A Lewis acid accepts an electron pair. A Lewis base donates an electron pair.

Acids have sharp taste or are sour
Acid/Base Properties Acids have sharp taste or are sour Carbonic acid (CO2 in water) - soda Phosphoric acid (colas) Citric acid (lemons, citrus fruits) Acetic acid (vinegar) Bases taste bitter, feel slippery Taste and feel are extremely dangerous ways to classify chemicals!

Industrial Chemicals - US
Acids & bases occupy 7 of top 20 spots in list of most widely produced industrial chemicals These are commodity chemicals – relatively cheap Sulfuric has many industrial uses

Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Typical Reactions Single replacement reaction: reaction of metals & acids to form metal salt and hydrogen Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)    Double replacement reaction: generation of gas & water NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) Base reacting with acid: neutralization

Practice Reactions of acids Problems 1(a-b), 2 page 635
Problems 1 – 2 page 989

Properties - Litmus Acid Blue litmus  Pink Base Red litmus  Blue

Properties – Conductivity
Pure water essentially nonconductor Any electrolyte (material that produces ions when dissolved in water) raises conductivity Acids and bases are electrolytes – raise conductivity

Relation to Ions in Solution
All aqueous solutions contain hydrogen ions (H+) and hydroxide ions (OH-) Acid – more H+ than OH- Base – more OH- than H+ Neutral – equal concentrations + -

Water – Self Ionization
Water produces small but equal numbers of H+ and OH- (it is neutral) H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) H3O+ = hydronium ion (water molecule covalently bonded to H+) Mostly OK to use H+ instead of H3O+ H2O(l)  H+(aq) + OH-(aq)

Arrhenius Acid/Base Model
Acid: Contains hydrogen & ionizes to produce H+ in aqueous solution HCl(g)  H+(aq) + Cl-(aq) Base: Contains OH- group & ionizes to produce OH- in aqueous solution NaOH(s)  Na+(aq) + OH-(aq) Model limitation: NH3(g) when dissolved in water produces OH- (base)

Brønsted-Lowry Acid/Base Model
Broadens Arrhenius model Acid: Hydrogen ion donor Base: Hydrogen ion acceptor Uses concept of conjugate acid-base pairs

Conjugate Acid-Base Pairs
Generalized acid dissociation (n=1) HX(aq) + H2O(l)  H3O+(aq) + X-(aq) Forward reaction: HX donates H+ to H2O acid H2O is base in this process Acid Base Conjugate Acid Conjugate Base Reverse reaction: H3O+ donates H+ to X conjugate acid X- is conjugate base in this process

HF(aq) + H2O(l)  H3O+(aq) + F-(aq) +  - Acid? HF Base? H2O Conjugate Acid? H3O+ Conjugate Base? F-

HF(aq) + H2O(l)  H3O+(aq) + F-(aq) All acids and bases that fit Arrhenius definition also fit Brønsted-Lowry definition

Brønsted-Lowry Base NH3 not Arrenhius base + + base conjugate acid
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) base conjugate acid acid conjugate base

Practice Brønsted-Lowry Model Problems 3(a-c), 4 page 640
Problem 9 page 643 Problems 19 (a-b) page 649 Problem 69, page 672 Problems 3 – 4 page 989

Amphoteric Substances
NH3(aq) + H2O(l)  NH4+(aq) OH-(aq) acid HF(aq) + H2O(l)  H3O+(aq) + F-(aq) base H2O(l) acting as either an acid or a base Compounds with this capability called amphoteric

Amphoteric Substances
Can act as either acid or base Water most common example Also common is bicarbonate ion, HCO3- HCO3-(aq) + OH-(aq)  CO32-(aq) + H2O(l) acid base conjugate conjugate base acid HCO3-(aq) + H3O+(aq)  H2CO3(aq) + H2O(l) base acid conjugate conjugate acid base

Lewis Acid/Base Model Most general model of acids & bases
Same G.N. Lewis that gave us Lewis electron dot structures and shared electron pair model for covalent bond Acid: electron pair acceptor ( / share) Base: electron pair donor ( / share) Includes all substances classified as Brønsted-Lowry acids & bases & many more; also reactions that do not necessarily occur in aqueous solution

Other Lewis Bases .. .. - :Cl: H - N - H .. H (HCl) (NH4+)

Lewis Acid/Base Model Formation of HF molecule
Fluoride ion donates electron pair; empty 1s orbital of H accepts pair (covalent bond formed)

Lewis Acid/Base Model Reaction of gaseous boron trifluoride and gaseous ammonia Electron deficient B accepts lone pair from ammonia (formation of coordinate covalent bond)

Lewis Acid/Base Model Gaseous sulfur trioxide reacts with solid magnesium oxide SO3(g) + MgO(s)  MgSO4(s) Lewis acid/base part of reaction: Will return to this reaction shortly

3 Acid-Base Models (Table 18.2)

Mono- and Polyprotic Acids
Monoprotic – capable of donating only one hydrogen ion HCl Polyprotic – can donate more than one H2SO4 Above examples simple - inorganic acids, all hydrogens ionize More complicated for organic acids

Mono- and Polyprotic Acids - HnX
HnX acids that dissociate into nH+(aq) ions and Xn-(aq) ions X=Cl hydrochloric acid n=1 X=NO3 nitric acid n=1 X=C2H3O2 acetic acid n=1 X=SO4 sulfuric acid n=2 X=CO3 carbonic acid n=2 X=PO4 phosphoric acid n=3 monoprotic polyprotic

Step-Wise Ionization Sulfuric acid Conjugate Base Acid
H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq) HSO4-(aq) + H2O(l)  H3O+(aq) + SO42-(aq) Conjugate Base Acid

Step-Wise Ionization Phosphoric acid Conjugate Base Acid Amphoteric H1
H3PO4(aq) + H2O(l)  H3O+(aq) + H2PO4-(aq) H2PO4-(aq) + H2O(l)  H3O+(aq) + HPO42-(aq) HPO42-(aq) + H2O(l)  H3O+(aq) + PO43-(aq) H1 Conjugate Base Acid H2 H3 Amphoteric

Monoprotic Acids Acetic acid: organic acid; HF: inorganic acid; each has one ionizable hydrogen atom (one that is involved in a polar bond)

Organic Acids Organic acids contain carboxylic acid group, COOH O = O
R - C - O - H = O Ionizable hydrogen R - C – O H+ = O –  Formula often written as RCOOH R = (almost) anything For acetic acid, R = CH3 (methyl) CH3 hydrogens not ionizable – monoprotic even though has 4 H atoms

Organic Acids Malonic acid (1,3-Propanedioic acid)
How many ionizable hydrogens? ? 2

Anhydrides Oxides that become acids or bases when added to water
Nonmetal oxides (C, S, N) – acids CO2(g) + H2O(l)  H2CO3 (Carbonic Ac) N2O3(g) + H2O(l)  2HNO2 (Nitrous Ac) SO3(g) + H2O(l)  H2SO4 (Sulfuric Ac) Metal oxides – bases CaO(s) + H2O(l)  Ca2+(aq) + 2OH-(aq)

Anhydrides and Acid Rain
Natural rain water: pH 5.6 Naturally slightly acidic due mainly to CO2 Acid rain: Rain water with pH < 5.6 Due to pollutant gases dissolved in air (nonmetal oxides = acid anhydrides) SO2, SO3, NO2 Acid rain linked to damage in ecosystems and structures

Damage from Acid Rain Reacts with metals and materials that contain carbonates Damages bridges, cars, and other metallic structures Damages buildings and other structures made of limestone or cement Acidifies lakes affecting aquatic life Dissolves and leaches minerals from soil Makes it difficult for trees to grow

Practice Acids and Bases Intro – other topics
(neutrality, Arrenhius model, polyprotic acids, ionizable hydrogens, anhydrides) Problems 5 – 8, 10 – 11 page 643 Problems , 64 (c-d) page 672 Problems 5 – 6 page 989

Section 18.2 Strengths of Acids and Bases
In solution, strong acids and bases ionize completely, but weak acids and bases ionize only partially. Relate the strength of an acid or base to its degree of ionization. Compare the strength of a weak acid with the strength of its conjugate base. Explain the relationship between the strengths of acids and bases and the values of their ionization constants.

Section 18.2 Strengths of Acids and Bases
Key Concepts Strong acids and strong bases are completely ionized in a dilute aqueous solution. Weak acids and weak bases are partially ionized in a dilute aqueous solution. For weak acids and weak bases, the value of the acid or base ionization constant is a measure of the strength of the acid or base.

Acid Strength and Concentration
...are not the same thing e.g. 1.0 M HCl and 1.0 M H2CO3 stronger acid same concentrations! Can have a dilute solution of a strong acid and a concentrated solution of a weak acid

Acid Strength 0.10 M HCl conducts electricity better than 0.10 M HC2H3O2 (acetic acid) Acetic Hydrochloric

3 binary halogen acids except HF
Acid Strength Strong acids dissociate completely Limited number of them (see Table 18.3) HCl hydrochloric HBr hydrobromic HI hydroiodic 3 binary halogen acids except HF HClO4 perchloric HNO3 nitric H2SO4 sulfuric

Relationship Between Strengths of Acids and Their Conjugate Bases
The stronger the acid, the weaker is the attraction of the ionizable H for the rest of the molecule The better the acid is at donating H+, the worse its conjugate base will be at accepting an H+ Strong acid: HCl + H2O → Cl– + H3O+ Weak conj base Weak acid: HF + H2O  F– + H3O+ Strong conj base

Acid Strength & Brønsted-Lowry Model
HA + H2O(l) acid base  H3O+(aq) conjugate acid of B + A- conjugate base of HA Strong acid HA has weak conj. base A- H2O(l) stronger base than A- Reaction equilibrium lies far to right

Strong/Weak vs Concentrated/Dilute
Number of acid/base molecules per volume 6 M HCl concentrated, 0.01 M HCl dilute Strong/Weak Measure of degree of ionization HCl strong acid CH3COOH (acetic) weak acid

Acid Strength & Brønsted-Lowry Model
HA + H2O(l) acid base  H3O+(aq) conjugate acid of B + A- conjugate base of HA Weak acid HA has strong conj. base A- A- stronger base than H2O(l) Reaction equilibrium lies far to left

Acid Ionization Constants
Focus on equilibrium for weak acids HCN(aq) + H2O(l)  H3O+(aq) + CN-(aq) Write K following standard rules Ka = [H3O+(aq)] [CN-(aq)] = 6.2x [HCN(aq)] Ka acid ionization constant Small number for weak acids – major species is undissociated form of acid Table 18.4, page 647 lists values at 25C

Acid Ionization Constants - pKa
HCN(aq) + H2O(l)  H3O+(aq) + CN-(aq) Ka = [H3O+(aq)] [CN-(aq)] = 6.2x [HCN(aq)] Ka acid ionization constant Many texts will utilize quantity pKa pKa = log(Ka) For HCN, pKa = 9.21 [Note: SF after DP for log based on SF of number in front of power of ten]

Weak Acids & Ka Values Strongest Acids Weakest Acids Weakest Bases
Conjugate Base Weakest Bases Strongest Bases HF 6.3 x 10-4 F- HNO2 5.6 x 10-4 NO2- HCO2H 1.8 x 10-4 HCO2- CH3CO2H 1.8 x 10-5 CH3CO2- HOCl 4.0 x 10-8 OCl- NH4+ 5.6 x 10-10 NH3 HCN 6.2 x 10-10 CN-

Practice Strengths of Acids Problems 12 – 14 page 647
Dilute/concentrated vs weak/strong Problem 74 page 672

Strong Bases Alkali & some alkaline earth metal hydroxides are strong Arrhenius bases M(OH)n(s)  Mn+(aq) + n OH-(aq) Dissociate completely Limited number (see table 18.5) NaOH KOH RbOH CsOH (alkali metals) Ca(OH)2, Sr(OH)2, Ba(OH)2 (alkaline earths)

Strong Bases Ca(OH)2(s)  Ca2+(aq) + 2OH-(aq)
Picture complicated by low solubility of some Ca(OH)2(s)  Ca2+(aq) + 2OH-(aq) Calcium hydroxide Ksp = 6.5x10-6 Base strength based on dissolved form Ca(OH)2 strong base – dissolved form [Ca(OH)2(aq)] completely dissociates

Weak Bases Weak bases mostly Brønsted-Lowry bases
B(aq) + H2O(l)  OH-(aq) + BH+(aq) Most common weak base either ammonia or amine (organic compound with –NH2) CH3NH2(aq) + H2O(l)  OH-(aq) + CH3NH3+(aq) Kb = [CH3NH3+(aq)] [OH-(aq)] = 4.3x [CH3NH2(aq)] Kb= base ionization constant (see table 18.6)

Weak Base Equilibria Ammonia NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
Methyl amine CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH−(aq) Ethyl amine C2H5NH2(aq) + H2O(l)  C2H5NH3+(aq) + OH−(aq) Bicarbonate HCO3−(aq) + H2O(l)  H2CO3 (aq) + OH−(aq)

Practice Strengths of Bases Problems 15 (a-d), 16, 20 page 649
Problems 70, 72, 73 page 672 Problems 7 – 8 page 989

Section 18.3 Hydrogen Ions and pH
pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions. Explain pH and pOH. Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of aqueous solutions.

Section 18.3 Hydrogen Ions and pH
Key Concepts The ion product constant for water, Kw, equals the product of the H+ ion concentration and the OH– ion concentration. Kw = [H+][OH–] The pH of a solution is the negative log of the hydrogen ion concentration. The pOH is the negative log of the hydroxide ion concentration. pH plus pOH equals 14. pH = –log [H+] pOH = –log [OH–] pH + pOH = 14.00 A neutral solution has a pH of 7.0 and a pOH of 7.0 because the concentrations of hydrogen ions and hydroxide ions are equal.

Ion Product Constant for Water (Kw)
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) or H2O(l)  H+(aq) + OH-(aq) Keq = [H3O+(aq)] [OH-(aq)] = [H+(aq)] [OH-(aq)] Keq given special symbol, Kw Kw = 1.0 x at 25oC Ion product=Kw in any aqueous solution

Ion Product Constant for Water (Kw)
 H2O(l) H3O+(aq) OH-(aq) + - Kw = [H3O+(aq)] [OH-(aq)] = 1.0 x 10-14 In pure water only source of ions is water dissociation, so at 25oC (298 K) [H3O+] = [OH-] = 1.0 x 10-7 M

[H+] and [OH-] for General Case
H2O(l)  H+(aq) + OH-(aq) Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14 H+ or OH- can come from source other than water (i.e., an acid or a base) Le Châteliers principle  equilibrium shifts to left, depressing either H+ or OH- concentrations Kw must remain constant

Calculation of [H+] and [OH-]
Example problem 18.1, page 651 Kw = [H+(aq)] [OH-(aq)] = 1.0 x 10-14 Given [H+] = 1.0x10-5 M, [OH-] = ? [OH-]= Kw / [H+] [OH-]= 1.0 x /1.0x10-5 = 1.0x10-9 M

Practice Kw and Water Ionization Equilibrium
Problems 22(a-d), 23 page 651 Problems 37, 38 page 658 Problems 77 – 78, 80 page 673 Problems 11 – 12 page 989

pH Defined as pH = – log[H+] Book says scale 0 to 14
Not strictly true! 10 M HCl pH = -1 10 M NaOH pH = 15 Remember log properties pH 3 solution has 100x [H+] of pH 5 pH acid pH basic

pH – Common Materials Similar to Fig 18.14, page 652

Calculating pH Example problem 18.2, page 653
pH of neutral solution at 298 K? Kw = 1.00x10-14 = [H+] [OH-] Neutral solution, [H+] = [OH-]  [H+] = 1.00x10-7 pH = – log[H+] = 7.00

Practice Calculate pH from [H+] Problems 24(a-b), 25 page 653
Problems 82, 83 page 673 Problem 13 page 989

pOH pOH = – log[OH-] pOH basic pH acidic pH + pOH = 14.00

Calculating pOH & pH Sample problem 18.3, page 654
Ammonia cleaner [OH-] = 4.0x10-3 M pOH, pH? pOH = – log[OH-] = 2.40 pH + pOH =  pH = 11.60

Practice Calculate pH or pOH from [H+] or [OH-]
Problems 27(a-d), 28(a-b), 29 page 654 Problems 41, 42(c-d), 43 page 658 Problems 80, 81 page 673 Problems page 989

Calculating Ion Concs from pH
[H+] = antilog(-pH) If pH=5, [H+] = antilog(-5) = 10-5 Problem 18.4, page 655 Blood pH = 298 K; [H+], [OH-] ? [H+] = antilog(-7.40) = 4.0x10-8 M [OH-] = Kw / [H+] = 1.0x10-14 / 4.0x10-8 [OH-] = 2.5 x10-7 M (book uses pH + pOH = and antilog(-pOH)

Practice Calculate [H+] or [OH-] from pH Problems 30(a-d), 31 page 655
Problem 40 page 658 Problem 79 page 673 (also other topics) Problems page 989

pH of Strong Acids & Bases
Strong acids and bases ~100% ionized HCl(aq)  H+(aq) + Cl-(aq) (Also HBr, HI, HClO4, HNO3, H2SO4)  [H+] = Acid concentration NaOH(aq)  Na+(aq) + OH-(aq)  [OH-] = Base concentration Ca(OH)2(aq)  Ca2+(aq) + 2OH-(aq)  [OH-] = 2 x Base concentration

Using pH to Calculate Ka
HF(aq)  H+(aq) + F-(aq) (weak acid) pH of M solution = 2.12; Ka? Ka = [H+(aq)] [F-(aq)] / [HF(aq)] [H+(aq)] = antilog(-2.12) = 7.6x10-3 M (2 SF) If ignore contribution of H2O to H+(aq) [F-(aq)] = [H+(aq)] [HF(aq)] = M – 7.6x10-3 M Ka = 7.6x x10-3 / = 6.3x10-4

Using pH to Calculate Ka
Sample problem 18.5, page 657 pH M formic acid = 2.38; Ka? HCOOH(aq)  H+(aq) + HCOO-(aq) [H+] = antilog(-2.38) = 4.2x10-3 (2 SF) [HCOO-(aq)] = [H+(aq)] [HCOOH(aq)] = M - 4.2x10-3 M Ka = [H+(aq)] [HCOO-(aq)] / [HCOOH(aq)] Ka = 4.2x x10-3 / = 1.8x10-4

Practice Use pH or pOH to calcuate Ka
Problems 32(a-b), 33(a-c), 34 page 657 Problem 39 page 658 Problem 84* (avoid) page 673 Problems page 989 * Relies on assumption that 1st ionization of chromic acid acts like strong acid (no hydrolysis of HCrO4-)

Measuring pH Indicators – quick, cheap but not precise
pH meter – can measure to pH units under ideal conditions with proper calibration

Methods for Measuring pH
pH (indicator) paper pH meter

Section 18.4 Neutralization
In a neutralization reaction, an acid reacts with a base to produce a salt and water. Write chemical equations for neutralization reactions. Explain how neutralization reactions are used in acid-base titrations. Compare the properties of buffered and unbuffered solutions.

Section 18.4 Neutralization
Key Concepts In a neutralization reaction, an acid and a base react to form a salt and water. The net ionic equation for the neutralization of a strong acid by a strong base is H+(aq) + OH–(aq) → H2O(l). Titration is the process in which an acid-base neutralization reaction is used to determine the concentration of a solution. Buffered solutions contain mixtures of molecules and ions that resist changes in pH.

Neutralization Reaction
Double replacement reaction of acid and base in aqueous solution to produce a salt and water Salt – ionic compound, cation from base and anion from acid Mg(OH)2(aq) + 2HCl(aq)  MgCl2(aq) + 2H2O(l) Base Acid Salt Water

Neutralization & Net Ionic Equation
Salt formed from neutralization of strong acid and strong base will be a strong electrolyte (fully ionized) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Complete ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l) Net ionic: H+(aq) + OH-(aq)  H2O(l)

Practice Neutralization reactions Problem 49 page 668
Problems 85, 90 page 673 Problems 22, 23 page 989

Acid-Base Titration Method for determining concentration of an unknown acid (base) solution by reacting it with a base (acid) of known concentration HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Stoichiometry applies - if know concentration & volume of NaOH needed to neutralize & initial volume of HCl, can calculate initial concentration of HCl

Titration Technique that uses reaction stoichiometry to determine concentration of unknown solution Titrant (known solution) added from buret Indicators – color-changing chemicals added to help determine when reaction complete (not required if pH measured) Endpoint of titration occurs when reaction complete

Acid-Base Titration Titrant (solution in buret) is a base
As base added to acid, H+ reacts with OH– to form H2O; still have excess acid present, so no color change At titration’s endpoint, just enough base has been added to neutralize all acid. At this point, indicator changes color

Acid-Base Titration Steps
Measure volume of unkown. Determine initial pH Fill buret with titrating solution of known concentration (standard solution) Slowly add standard solution while recording pH until neutralization reaction reaches stoichiometric point (equivalence point)

At equivalence point, pH changes rapidly with small addition of NaOH

Acid-Base Titration - Indicators
Often convenient to use pH indicator to determine equivalence point (point at which number of moles of base added equals number of moles of acid originally present) [OHadd-(aq) = H+start(aq) for 1:1 type system) End point: pH at which indicator changes color Must choose indicator so pH at end point and pH at equivalence point are ~ same

Acid-Base Titration Equivalence Point
Equivalence point for strong acid-strong base titration: pH 7 For weak acid-strong base, pH > 7 For strong acid-weak base, pH < 7 Different from 7 due to salt hydrolysis (topic discussed later) Must choose pH indicator end point accordingly

Matching of equivalence point and pH indicator end point – titration of strong acid with strong base

Matching of equivalence point and pH indicator end point – titration of weak acid with strong base

Acid Base Indicators are weak acids or bases themselves
Hin(aq) + H2O(l)  H3O+(aq) + In-(aq) Color A Color B

Hin(aq) + H2O(l)  H3O+(aq) + In-(aq)
Acid Base Indicators Hin(aq) + H2O(l)  H3O+(aq) + In-(aq) Ka = [H3O+] [In-] [Hin] [In-] [Hin] = Ka [H3O+] i.e. ratio of colors changes with pH

Indicators – See Fig 18.24, p 662

3 Very Common Indicators

Known volume of unknown in flask with indicator
Step 1 Known volume of unknown in flask with indicator Known concentration of titrant in buret

Pink color appears momentarily but then disappears upon mixing
Step 2 Titrant added Pink color appears momentarily but then disappears upon mixing

Just enough titrant added so that pink color permanently remains
Step 3 Just enough titrant added so that pink color permanently remains

Titration Calculation - Prob. 18.6 p 664
18.28 mL of M NaOH for mL of formic acid (HCOOH) Write balanced neutralization eqn. HCOOH(aq)+ NaOH(aq)  HCOONa(aq)+H2O(l) 2. Calculate moles titrant added 18.3 mL NaOH x 1 L NaOH = L NaOH mL NaOH L NaOH x mol NaOH = mol L NaOH NaOH

Acid-Base Titration Calculation
HCOOH(aq)+ NaOH(aq)  HCOONa(aq)+H2O(l) 3. Use mole ratio: calculate moles unknown 1.83x10-3 mol NaOH x 1 mol HCOOH mol NaOH = 1.83x10-3 mol HCOOH 4. Calculate concentration of unknown using moles and initial volume MHCOOH = 1.83x10-3 mol HCOOH L HCOOH MHCOOH = 7.31x10-2 M

Practice Acid-Base Titration Problems 44 – 46 page 664

Salt Hydrolysis NaNO3 KF NH4Cl Neutral Basic Acidic
Put salt in pure water Unless salt product of strong acid/strong base reaction, pH won’t be neutral Neutral Basic Acidic NaNO3 KF NH4Cl

F-(aq) + H2O(l)  HF(aq) + OH-(aq)
Salt Hydrolysis KF(s)  K+(aq) + F-(aq) KOH strong base, HF weak acid (see table 18.4, page 647) F-(aq) + H2O(l)  HF(aq) + OH-(aq) F-(aq): Brønsted-Lowry base Solution basic (OH- generated)

NH4Cl(s)  NH4+(aq) + Cl-(aq)
Salt Hydrolysis NH4Cl(s)  NH4+(aq) + Cl-(aq) NH3 weak base, HCl strong acid NH4+(aq) + H2O(l)  NH3(aq) + H3O+ (aq) NH4+(aq): Brønsted-Lowry acid Solution acidic (H3O+ generated)

Practice Salt Hydrolysis Problems 47(a-d), 48 page 665
Problems 52, 54 page 668 Problems 91(a-b) page 673 Problems 26 – 27 page 990

Buffered Solutions Buffer: Solution that resists change in pH when limited amounts of acid or base added Add 0.01 mol HCl to solution 1 L water – pH change: 7.0 to 2.0 Buffer – typical pH change 0.1 unit

HF (weak acid) + KF (F- conj base)
Buffer Composition Mixture of either: (a) Weak acid & salt of its conjugate base HF (weak acid) + KF (F- conj base) (b) Weak base & salt of its conjugate acid NH3 (weak base) + NH4Cl (NH4+ conj acid) Standard buffers mostly made using (a) Having both acid & base present makes buffer resist pH changes

How Buffers Work Weak acid in buffer mixture can neutralize added base
Conjugate base in buffer mixture can neutralize added acid Net result: little to no change in solution pH in either case

How Buffers Work H2O New HA HA HA A− A−  H3O+ + Added H3O+

How Buffers Work H2O New A− A− HA A− HA  H3O+ + Added HO−

Equimolar Buffer Systems Table 18.7

HF(aq)  H+(aq) + F-(aq)
HF/KF Buffer 0.1 M HF (weak ac) M KF (F- conj base) HF(aq)  H+(aq) + F-(aq) F-(aq) from 2 sources: Dissociation of HF(aq) Dissociation of KF(aq) Add acid – equilibrium shifts to left so H+ consumed Add base – H+ neutralized to H2O, equil. shifts to right to generate more H+ Either way, pH change minimized

Acetic Acid / Sodium Acetate Buffer
Mix equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2: buffer pH = 4.75 Add 10 mL of 0.1 M HCl to: 1 L of buffer, pH = 4.75 1 L of DI water, pH = 3.0 Add 10 mL of 0.1 M NaOH to: 1 L of DI water, pH = 11.0

HF(aq)  H+(aq) + F-(aq)
Buffer Capacity Amount of acid or base buffer can absorb without significant pH change 0.1 M HF (weak ac) M KF (F- conj base) HF(aq)  H+(aq) + F-(aq) Add more H+ than available F-, pH must drop Remove more H+ than can be replaced by remaining HF, pH must rise

pH of Buffer pH = -log[H+] = -log(Ka) = -log(6.2x10-8)
Easy to determine for equimolar buffers Equimolar mix NaH2PO4 & Na2HPO4 H2PO4-(aq)  H+(aq) + HPO42-(aq) Acid conj. Base Ka = [H+(aq)][HPO42-(aq)] / [H2PO4-(aq)] Since [HPO42-(aq)] = [H2PO4-(aq)] pH = -log[H+] = -log(Ka) = -log(6.2x10-8) pH = (see table 18.7, p 667 for others)

Practice Buffers Problems 51, 53 page 668
Problems 88, 102, 103 pages 673-4

Chapt. 18 – Acids and Bases 18.1 Acids and Bases: An Introduction

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Chapt. 18 – Acids and Bases 18.1 Acids and Bases: An Introduction

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