Acid-Base Chemistry

Assignment #7Assignment #7

Acids and Bases - 3 Definitions

ArrheniusArrhenius

Bronsted-LowryBronsted-Lowry

LewisLewis

Arrhenius Definition

Acid = proton donorAcid = proton donorHA = HHA = H++ + A + A--

Base = hydroxide donorBase = hydroxide donorBOH = BBOH = B++ + OH + OH--

Dilemma: NHDilemma: NH33

Bronsted-Lowry Definition

Acid = proton donorAcid = proton donor

Base = proton acceptorBase = proton acceptorNHNH33 + H + H++ = NH = NH44

++

Dilemna: ferrocene (organometallics)Dilemna: ferrocene (organometallics)

Lewis Definition

Acid = electron pair acceptor (electrophile)Acid = electron pair acceptor (electrophile)

Base = electron pair donor (nucleophile)Base = electron pair donor (nucleophile)

Acid and Base Strength Strong acids and Bases Strong acids and Bases completelycompletely

dissociate (ionize) in aqueous solutiondissociate (ionize) in aqueous solution EX:EX:

HClHClaqaq ->-> H H++aqaq + Cl + Cl--

aqaq

Weak acids and Bases Weak acids and Bases incompletelyincompletely dissociate in aqueous solutiondissociate in aqueous solution

EX:EX:HCHC22HH33OO2aq2aq == H H++

aqaq + C + C22HH33OO22--aqaq

Not All Mineral Acids are Strong! HCNHCN

HCN = HHCN = H++ + CN + CN-- KKaa = 2.1 x 10 = 2.1 x 10-9-9

HFHFHF = HHF = H++ + F + F-- KKaa = 6 x 10 = 6 x 10-4-4

Polyprotic Acids

Protons are always lost one at a time!Protons are always lost one at a time! acids produced by proton loss from acids produced by proton loss from

polyprotic acids are weak acids, polyprotic acids are weak acids, characterized by a unique Kcharacterized by a unique Kaa value value

HH22SOSO44 = HSO = HSO44-- + H + H++ KKa1a1 not measurable not measurable

HSOHSO44-- = SO = SO44

2-2- + H + H++ KKa2a2 = 1.1 x 10 = 1.1 x 10-2-2

Strong Acids

Most mineral acidsMost mineral acids HClHCl HH22SOSO44

HNOHNO33

HClOHClO44

Not: HF, HNot: HF, H33POPO44

Weak Acids

Organic acids (need C and usually have Organic acids (need C and usually have COOH)COOH)EXAMPLES:EXAMPLES:formic acidformic acidacetic acidacetic acidpropionic acidpropionic acid

Strong Bases

Inorganic hydroxides containing metals Inorganic hydroxides containing metals from families IA or IIAfrom families IA or IIA

Note: names of these families: Alkali Note: names of these families: Alkali metals, alkaline earth metalsmetals, alkaline earth metals

Weak Bases

NHNH33, organic amines, and hydroxides other , organic amines, and hydroxides other

than group Ia or IIa hydroxidesthan group Ia or IIa hydroxides

organic amines contain amino group:organic amines contain amino group:R-NHR-NH22 + H + H++ = R-NH = R-NH33

++

Scheme for Identification of Acids and Bases

Proton donor Proton donor or proton acceptor?or proton acceptor?

If proton donor,If proton donor, then acidthen acid

If proton acceptor,If proton acceptor,Then base Then base

InorganicInorganic OrganicOrganic InorganicInorganic OrganicOrganic

Strong exceptStrong exceptHF, HCN, HHF, HCN, H22S, HS, H33POPO44

Weak acidWeak acid

Weak acidWeak acid Weak baseWeak baseIf IA or IIA,If IA or IIA,then strongthen strong

If NHIf NH44OH or not IA and IIAOH or not IA and IIA

Then weakThen weak

Practice

Identify the acid/base nature of the following Identify the acid/base nature of the following compounds:compounds:

Hydroxyl amineHydroxyl amine Calcium hydroxideCalcium hydroxide Carbon dioxideCarbon dioxide Pthalic acidPthalic acid Hydrogen sulfideHydrogen sulfide pyridinepyridine

Conjugates

Acids and Bases exist in a conjugate Acids and Bases exist in a conjugate relationship:relationship:

HAHA = H = H++ + + AA--

acid baseacid base

BOHBOH = = BB++ + OH + OH--

base acidbase acid

Example:

NHNH44OHOH = = NHNH44++ + OH + OH--

basebase acid acid

HCHC22HH33OO22 = H = H++ + + CC22HH33OO22--

acidacid base base

Problem

Identify the conjugate acid-base pairs for Identify the conjugate acid-base pairs for each of the following compounds:each of the following compounds:

Ammonium hydroxideAmmonium hydroxide DiethylamineDiethylamine Iodic acidIodic acid Formic acidFormic acid HPOHPO44

--

Amphoterism Some compounds can function as both Some compounds can function as both

acids or bases depending on the situationacids or bases depending on the situatione.g., He.g., H22OO

HCHC22HH33OO22 + + HH22OO = H = H 3 3OO++ + C + C22HH33OO22--

acidacid base acid base base acid base

NHNH33 + + HH22OO = NH = NH44++ + OH + OH--

base acid acid basebase acid acid base

Dissociation Constants for Weak Acids and Bases Recall for HA:Recall for HA: KKaa = = [H[H++][A][A--]]

[HA] [HA]

The bigger KThe bigger Kaa, the _____ the [H, the _____ the [H++] and the ] and the

_____ the [HA]_____ the [HA]

Question

What is the comparatively strongest weak What is the comparatively strongest weak acid on Table A?acid on Table A?

ANS: iodic acid, KANS: iodic acid, Kaa = 0.18 = 0.18

What is the comparatively strongest weak What is the comparatively strongest weak base on Table B?base on Table B?

ANS: diethylamine and piperidine are ANS: diethylamine and piperidine are equally strong, Kequally strong, Kbb = 0.0013 = 0.0013

Conjugate Acid-Base Strength

For HA = HFor HA = H++ + A + A--

Recall,Recall,KKaa = = [H[H++][A][A--]] [HA] [HA]

HA is the conjugate _____ and AHA is the conjugate _____ and A-- is its conjugate is its conjugate ________

HA is a ____ ____HA is a ____ ____ AA-- is a ____ ____ is a ____ ____ If HA is a relatively strong weak acid, then AIf HA is a relatively strong weak acid, then A-- is a is a

comparatively ____ ____ basecomparatively ____ ____ base

Problem

HSOHSO44-- is is

a) the conjugate acid of SOa) the conjugate acid of SO44-2-2

b) a strong acidb) a strong acidc) the conjugate base of Hc) the conjugate base of H22SOSO44

d) a strong based) a strong basee) the conjugate acid of He) the conjugate acid of H22SOSO44

Problem:

Which of the following is/are amphoteric:Which of the following is/are amphoteric:a) Ha) H22POPO44

--

b) Cb) C22HH33OO22--

c) CHc) CH33CHCH22NHNH22

d) HCOd) HCO33--

e) CHe) CH33CH(NHCH(NH22)COOH)COOH

Conjugate Acid-Base Strength

The stronger the The stronger the conjugate acid is, the conjugate acid is, the weaker the conjugate weaker the conjugate base is and vice versabase is and vice versa

HAHA

AA--

Salts

There are four kinds of salts:There are four kinds of salts: salts of weak acidssalts of weak acids

example: sodium acetateexample: sodium acetate salts of weak basessalts of weak bases

example: ammonium chlorideexample: ammonium chloride salts of strong acids and strong basessalts of strong acids and strong bases

example: sodium chlorideexample: sodium chloride salts of weak acids and weak basessalts of weak acids and weak bases

example: ammonium acetateexample: ammonium acetate

Identification of Salts

Salts hydrolyze in water:Salts hydrolyze in water:

Salt + water Salt + water acid + base acid + base

EX: EX:

NaCl + HOH NaCl + HOH Na Na++ + OH + OH-- + H + H++ + Cl + Cl--

Identification of Salts

Salts are obtained by reacting acids and Salts are obtained by reacting acids and bases:bases:

Acid + Base = Salt + waterAcid + Base = Salt + water

Note: this is just the reverse of the Note: this is just the reverse of the hydolysis reaction:hydolysis reaction:

Salt + water = Acid + BaseSalt + water = Acid + Base

Identification of Salts

So, salts are classified based on their So, salts are classified based on their parentparent acids acids Their parents are the acids and bases used to form Their parents are the acids and bases used to form

themthem

EX: NaClEX: NaCl

NaNaOHOH + + HHCl Cl Na Na++ + Cl + Cl-- + H + H22OO

So, parents are strong acid and strong base andSo, parents are strong acid and strong base and

NaCl is the salt of a strong acid and strong baseNaCl is the salt of a strong acid and strong base

Examples:

NaCNaC22HH33OO22 – salt of ________________ – salt of ________________

NHNH44Cl – salt of ___________________Cl – salt of ___________________

NHNH44COOH – salt of _______________COOH – salt of _______________

LiF – salt of _____________________LiF – salt of _____________________

Homework Problem #1:

Classify each of the following as a strong or Classify each of the following as a strong or weak acid, base, or salt (identify parents):weak acid, base, or salt (identify parents):CsOHCsOH AgOHAgOH

sodium salicylatesodium salicylate HH22POPO44--

HClOHClO44 HH22COCO33

ferric hydroxideferric hydroxide oxalic acidoxalic acidNHNH44CC22HH33OO22

trimethylaminetrimethylamine

The Autoionization of Water

HOHHOH(l)(l) + HOH + HOH(l)(l) = H = H33OO++ + OH + OH--

hydronium ionhydronium ion This is an This is an equilibrium processequilibrium process and is and is

characterized by an equilibrium constant, characterized by an equilibrium constant, KKww::

KKww = [H = [H33OO++][OH][OH--] = 10] = 10-14-14 at 25 at 2500CC

Kw DOES vary with Temperature

Temperature, 0C Kw

0 1.13 x 10-15

25 1.008 x 10-14

60 9.614 x 10-14

higherhigher

lowerlower

The Relationship between [H+] and [OH-]

KKww = [H = [H++][OH][OH--] = 10] = 10-14-14

Useful Equivalent forms:Useful Equivalent forms: [H[H++] = 10] = 10-14-14/[OH/[OH--]]

[OH[OH--] = 10] = 10-14-14/[H/[H++]]

The pH Scale

pH = -log [HpH = -log [H++] ] ; no units; no units

[H[H++] = antilog[-pH]] = antilog[-pH]

pH of pH of purepure water = 7 water = 7

A Brief Review of LOG Math

Taking a logarithm corresponds to Taking a logarithm corresponds to answering the question:answering the question:To what power do I raise 10 in order to To what power do I raise 10 in order to represent the number of interest?represent the number of interest?

log 100 => 10log 100 => 10?? = 100 = 100

NOTE: LOG is not same thing as LNNOTE: LOG is not same thing as LNnatural log is based on enatural log is based on e??= number= number

A Brief Review of LOG Math

log (ab) = log a + log blog (ab) = log a + log b

log (a/b) = log a - log blog (a/b) = log a - log b

log alog abb = b log a = b log a

The pOH scale

pOH = -log [OHpOH = -log [OH--]] ; no units ; no units

[OH[OH--] = antilog[-pOH]] = antilog[-pOH]

What is the pOH in pure water?What is the pOH in pure water? ANS: pOH = -log (10ANS: pOH = -log (10-7-7) = 7) = 7

How is pH related to pOH? recall: recall:

KKww = [H = [H33OO++][OH][OH--] = 10] = 10-14-14 at 25 at 2500CC

DerivationDerivation call call pKpKww = -logK = -logKww = 14 at 25 = 14 at 2500CC

then:then:

pKpKww = pH + pOH = 14 = pH + pOH = 14

The pH of Some Common SubstancesCommon Substance pHlemons 2.3seawater 8.5blood 7.4urine 5.5 - 7.0saliva 6.5 - 7.5gastric juices 1.0 - 3.0

Aspirin and IbuprofenCOOH

OH

O O

O

CH3

COOH

(H3C)2HCH2C CH2COOH

salicylic acid

acetyl salicylic acid (aspirin)

ibuprofen

COOH

OH

O O

O

CH3

COOH

(H3C)2HCH2C CH2COOH

salicylic acid

acetyl salicylic acid (aspirin)

ibuprofen

Clean and Clear Clean and Clear Sensitive Skin Deep Sensitive Skin Deep Cleaning AstringentCleaning Astringent active ingredient 0.5% active ingredient 0.5%

salicylic acidsalicylic acid

AdvilAdvil active ingredient active ingredient

ibuprofenibuprofen

Orange Juice

Tropicana Pure Tropicana Pure Premium with Premium with CalciumCalcium active ingredients: active ingredients:

calcium hydroxide, calcium hydroxide, malic acid and citric malic acid and citric acidacid

Toothpaste

Aquafresh Whitening Aquafresh Whitening ToothpasteToothpaste active ingredient: active ingredient:

sodium fluoridesodium fluoride

Colgate Maximum Colgate Maximum Cavity Protection Cavity Protection Fluoride ToothpasteFluoride Toothpaste active ingredient: active ingredient:

0.76% sodium 0.76% sodium monofluorophosphatemonofluorophosphate

The Relationship between pH and pOH

[H+], M pH pOH [OH-], M

10-1 1 ?? 10-13

10-4 ?? 10 10-10

10-7 7 7 ??

?? 11 3 10-3

10-14 14 ?? 1

acidicacidic

neutralneutral

basicbasic

Acidic Solution

high [Hhigh [H++] concentration] concentration low pH valuelow pH value

value below 7value below 7

low [OHlow [OH--] concentration] concentration high pOH valuehigh pOH value

value greater than 7value greater than 7

Basic Solution

high [OHhigh [OH--] concentration] concentration low pOH valuelow pOH value

value less than 7value less than 7

low [Hlow [H++] concentration] concentration high pH valuehigh pH value

value greater than 7value greater than 7

pKa and pKb

pKpKaa ≡ - log K≡ - log Kaa

pKpKbb ≡ - log K ≡ - log Kbb

pKpKaa * pK * pKbb = pK = pKww

Conjugate Acid-Base Strength

The stronger the conjugate The stronger the conjugate acid is, the weaker the acid is, the weaker the conjugate base is and vice conjugate base is and vice versaversa

KKaa * K * Kbb = 10 = 10-14-14

pKpKaa + pK + pKbb = 14 = 14

HAHA

AA--

pKa , pKb, and Weak Acid/ Base Strength

The lower the pKThe lower the pKaa the the

______ the weak acid______ the weak acid The higher the pKThe higher the pKaa, the , the

______ the weak acid______ the weak acid The lower the pKThe lower the pKaa of a of a

weak acid, the ______ the weak acid, the ______ the pKpKbb of its conjugate weak of its conjugate weak

base and the _____ its base and the _____ its conjugate baseconjugate base

HAHA

AA--

pKpKaa * pK * pKbb = 14 = 14

Question

Which is the comparatively stronger weak Which is the comparatively stronger weak acid ammonium or pyridinium?acid ammonium or pyridinium?

Strong Acids and Strong Bases

Photographs from Atkins, P.W. Photographs from Atkins, P.W. MoleculesMolecules; W.H. Freeman: New York, 1987.; W.H. Freeman: New York, 1987.

sea squirts squirt HNOsea squirts squirt HNO33

sea slugs secrete Hsea slugs secrete H22SOSO44

HA HA → H→ H++ + A + A--

BOH → BBOH → B++ + OH + OH--

Strong acids and strong Strong acids and strong bases completely bases completely dissociate in water so dissociate in water so their concentration their concentration gives us the [Hgives us the [H++] in ] in solution directlysolution directly

Problem:

Determine the pH of the following Determine the pH of the following solutions. Are the solutions acidic or basic?solutions. Are the solutions acidic or basic?a) 0.001 M HCl solutiona) 0.001 M HCl solutionb) a solution whose [OHb) a solution whose [OH--] is 10] is 10-3-3 M Mc) 0.0001 M NaOH solutionc) 0.0001 M NaOH solution

ANS:ANS:a) pH 3, acidic; b) pH 11, basic; c) pH 10, a) pH 3, acidic; b) pH 11, basic; c) pH 10, basicbasic

Homework Problem #1:

Calculate the hydrogen ion concentration Calculate the hydrogen ion concentration and pH of a solution prepared by placing and pH of a solution prepared by placing 11.5 g of HClO11.5 g of HClO44 (perchloric acid; FW 100) (perchloric acid; FW 100)

in a 500 mL volumetric flask subsequently in a 500 mL volumetric flask subsequently filled to the mark with water.filled to the mark with water.(Hint: What kind of acid is HClO(Hint: What kind of acid is HClO44???)???)

Problem:

What are the pH and pOH of a solution What are the pH and pOH of a solution prepared by mixing 25 mL of 0.20 M prepared by mixing 25 mL of 0.20 M NaOH with 60 mL of 0.10 M HCl?NaOH with 60 mL of 0.10 M HCl?(Suggestion: Draw a picture)(Suggestion: Draw a picture)

ANS: pH = 1.93; pOH = 12.07ANS: pH = 1.93; pOH = 12.07

Weak Acids and Bases

HA = HHA = H++ + A + A--

BOH = BBOH = B++ + OH + OH--

Weak acids and weak bases incompletely Weak acids and weak bases incompletely dissociate so their concentration does not dissociate so their concentration does not provide meaningful insight into the aqueous provide meaningful insight into the aqueous pH of their solutionspH of their solutions

Fire ants venom contains formic acidFire ants venom contains formic acid

Photograph from Atkins, P.W. Photograph from Atkins, P.W. MoleculesMolecules; W.H. Freeman: New York, 1987.; W.H. Freeman: New York, 1987.

Weak Acids (Table A)

Weak Acid Ka

HF 7.2 x 10-4

HCOOH (formic) 1.8 x 10-4

HC2H3O2 1.8 x 10-5

HCN 6.2 x 10-10

H3BO3 (boric) 5.8 x 10-10

Weak Acids

The stronger a weak acid the greater the The stronger a weak acid the greater the [H[H++] in solution] in solution

The weaker a weak acid, the greater the The weaker a weak acid, the greater the [HA] in solution[HA] in solution

Recall: pKRecall: pKaa = -log K = -log Kaa

Acid-Base Strength

AA--

HAHA

Reminder: Weak Acids

Weak Acid Ka pKa

Acetic 1.75 x 10-5 4.76

Hydrocyanic 2.1 x 10-9 8.68

The stronger the weak acid, the larger the Ka and the lower the pKa

Calculating the pH of Weak Acids the dissociation of weak acids determines the dissociation of weak acids determines

the equilibrium concentration of Hthe equilibrium concentration of H++ and and therefore the pHtherefore the pH

the dissociation of weak acids is the dissociation of weak acids is characterized by the equilibrium constant characterized by the equilibrium constant KKaa

DerivationDerivation

Problem:

What is the pH and pOH of a solution of What is the pH and pOH of a solution of 0.05 M butyric acid (CH0.05 M butyric acid (CH33CHCH22CHCH22COOH) COOH)

given the pKgiven the pKaa for butyric acid 4.81? for butyric acid 4.81?

Q: Where found in nature?Q: Where found in nature?

Weak Bases

The stronger a weak base the greater the The stronger a weak base the greater the [OH[OH--] in solution] in solution

The weaker a weak base, the greater the The weaker a weak base, the greater the [BOH] in solution[BOH] in solution

Define: pKDefine: pKbb = -log K = -log Kbb

DerivationDerivation

Problem

What is the pH of an 0.026 M solution of What is the pH of an 0.026 M solution of hexamethylenetetramine (Khexamethylenetetramine (Kbb 10 10-9-9)?)?

Salts

There are four kinds of salts:There are four kinds of salts: salts of weak acidssalts of weak acids

example: sodium acetateexample: sodium acetate salts of weak basessalts of weak bases

example: ammonium chlorideexample: ammonium chloride salts of strong acids and strong basessalts of strong acids and strong bases

example: sodium chlorideexample: sodium chloride salts of weak acids and weak basessalts of weak acids and weak bases

example: ammonium acetateexample: ammonium acetate

Salts of Strong Acids and Bases

Example: NaClExample: NaCl

NaCl + HNaCl + H22O -> NaO -> Na++aqaq + Cl + Cl--

aqaq + H + H22OO

pH pH 7 7

These salts do not directly perturb the water These salts do not directly perturb the water equilibriumequilibrium

Salts of Weak Acids

Example: NaCExample: NaC22HH33OO22

NaCNaC22HH33OO22 + H + H22O = NaO = Na++ + HC + HC22HH33OO22 + + OHOH--

pH = 0.5(14 + pKpH = 0.5(14 + pKaa + log[salt]) + log[salt])

solutions are weakly basic solutions are weakly basic

Salts of Weak Acids

DerivationDerivation pH = 0.5(14 + pKpH = 0.5(14 + pKaa + log[salt]) + log[salt])

the the weakerweaker the weak acid, the the weak acid, the more basicmore basic the solution (the more tightly the weak acid the solution (the more tightly the weak acid holds onto the Hholds onto the H++ and the higher the [OH and the higher the [OH--] ] in solutionin solution

Question:

Solutions of which salt would be more basic Solutions of which salt would be more basic - sodium acetate or sodium cyanide?- sodium acetate or sodium cyanide?

Salts of Weak Bases

Example: NHExample: NH44ClCl

NHNH44Cl + HCl + H22O = NHO = NH44OH + OH + HH++ + Cl + Cl--

DerivationDerivation pH = 0.5(14 - pKpH = 0.5(14 - pKbb - log[salt]) - log[salt])

solutions are weakly acidicsolutions are weakly acidic

Salts of Weak Bases

pH = 0.5(14 - pKpH = 0.5(14 - pKbb - log[salt]) - log[salt])

The weaker the weak base, the lower the The weaker the weak base, the lower the solution pH (the more tightly the weak base solution pH (the more tightly the weak base holds onto the OHholds onto the OH-- and the higher the [H and the higher the [H++] ] in solution)in solution)

Salts of Weak Acids and Weak Bases Example: NHExample: NH44CC22HH33OO22

NHNH44CC22HH33OO22 + H + H22O = NHO = NH44OH + HCOH + HC22HH33OO22

DerivationDerivation pH = 0.5(14 + pKpH = 0.5(14 + pKaa - pK - pKbb))

solution pH depends on the relative strength of solution pH depends on the relative strength of the acid vs. the basethe acid vs. the base

Homework Problem #2

Calculate the pH of an 0.1 M solution of Calculate the pH of an 0.1 M solution of each of the following compounds:each of the following compounds:

A) sodium acetateA) sodium acetate B) sodium nitrateB) sodium nitrate C) sodium hydroxideC) sodium hydroxide D) hydrogen fluorideD) hydrogen fluoride E) pyridineE) pyridine

Problem:

Identify the following compounds in terms Identify the following compounds in terms of their acid/base properties and predict of their acid/base properties and predict whether aqueous solutions of these whether aqueous solutions of these compounds will be acidic, basic, or neutral:compounds will be acidic, basic, or neutral:a) sodium cyanidea) sodium cyanideb) ammonium nitrateb) ammonium nitratec) potassium nitratec) potassium nitrate

Buffers

DEFINITION:DEFINITION:a solution containing both a weak acid/base a solution containing both a weak acid/base and its salt which resists change in pH due to:and its salt which resists change in pH due to: temperaturetemperature dilutiondilution and and

addition of SMALL amounts of strong acid or addition of SMALL amounts of strong acid or basebase

Examples:

HCHC22HH33OO22 and NaC and NaC22HH33OO22

HCOOH and HCOONaHCOOH and HCOONa NaHNaH22POPO44 and Na and Na22HPOHPO44

pyridine and pyridinium chloridepyridine and pyridinium chloride ammonia and ammonium chlorideammonia and ammonium chloride

Problem:

Solutions are made by combining equal Solutions are made by combining equal volumes of the following. Which is/are a volumes of the following. Which is/are a buffer(s)?buffer(s)?a) 0.1 M NHa) 0.1 M NH44Cl + 0.1 M NHCl + 0.1 M NH44

++

b) 0.1 M HF + 0.05 M NaOHb) 0.1 M HF + 0.05 M NaOHc) 0.05 M HF + 0.1 M NaOHc) 0.05 M HF + 0.1 M NaOHd) 0.1 M NaF + 0.05 M HCld) 0.1 M NaF + 0.05 M HCle) 0.1 M NaF + 0.05 M Nae) 0.1 M NaF + 0.05 M Na++

Henderson Hasselbalch Equation

pH = pKpH = pKaa + log [conj. base/conj. acid] + log [conj. base/conj. acid]

ororpH = pKpH = pKaa + log [salt/acid] + log [salt/acid]

for a weak acid and its saltfor a weak acid and its salt

Recall: Recall: pKpKaa = -log K = -log Kaa

derivationderivation

Problem:

Calculate the pH of a solution that is 0.25 Calculate the pH of a solution that is 0.25 M sodium acetate and 0.30 M acetic acid M sodium acetate and 0.30 M acetic acid given Kgiven Kaa = 1.8 x 10 = 1.8 x 10-5-5 for acetic acid. for acetic acid.

(hint: what is the pK(hint: what is the pKaa?)?)

ANS:ANS:pH = pKpH = pKaa + log [salt/acid] + log [salt/acid]

pH = 4.74 + log(0.25/0.30) = 4.74 - 0.08pH = 4.74 + log(0.25/0.30) = 4.74 - 0.08pH = 4.66pH = 4.66

Question:

If we have a base buffer containing If we have a base buffer containing ammonia and ammonium chloride, what is ammonia and ammonium chloride, what is the correct form of the Henderson-the correct form of the Henderson-Hasselbach equation based on these Hasselbach equation based on these species?species?

pH = pKpH = pKaa + log [ammonia]/[ammonium + log [ammonia]/[ammonium chloride]chloride]

Problem: Calculate the pH of the solution that results Calculate the pH of the solution that results

when 200 mL of 0.300 M ammonium when 200 mL of 0.300 M ammonium hydroxide are mixed with 250 mL of 0.150 hydroxide are mixed with 250 mL of 0.150 M ammonium chloride.M ammonium chloride.

ANS:ANS:pH = pKpH = pKaa + log (conj. base/conj. acid) + log (conj. base/conj. acid)

= 9.25 + log [ammonia/ammonium] = 9.25 + log [ammonia/ammonium] = 9.25 + log (0.06/0.0375) = = 9.25 + log (0.06/0.0375) = = 9.25 + 0.20 = 9.45 = 9.25 + 0.20 = 9.45

Buffers - Effect of Dilution

Consider the pH of the solution that results Consider the pH of the solution that results from mixing 100 mL 0.1 M NaHfrom mixing 100 mL 0.1 M NaH22POPO44 and and

100 mL 0.1 M Na100 mL 0.1 M Na22HPOHPO44? What is the pH if ? What is the pH if

the solution is diluted by a factor of 2? 5? the solution is diluted by a factor of 2? 5? Generalize your findings.Generalize your findings.

Buffers - Effect of Temperature

Name pKa @ 200C pKa/0C

MES 6.15 -0.011

HEPES 7.55 -0.014

Tris 8.30 -0.031

Phosphoric acid (K2)

7.21 -0.0028

Buffers Calbiochem Corp., Doc. No. CB0052-591; Perrin & Dempsey Buffers Calbiochem Corp., Doc. No. CB0052-591; Perrin & Dempsey Buffers for pH and Metal Ion Control Chapman & Hall: London, 1979.Buffers for pH and Metal Ion Control Chapman & Hall: London, 1979.

Comparison - Effect of Addition of SMALL Amount of Strong Acid or Base

Buffer itself (100 mL 0.1 M NaHBuffer itself (100 mL 0.1 M NaH22POPO44 and 100 and 100

mL 0.1 M NamL 0.1 M Na22HPOHPO44) pH 7.20) pH 7.20

vs. 200 mL water pH 7.00vs. 200 mL water pH 7.00 Upon addition 0.005 moles strong acid, buffer pH Upon addition 0.005 moles strong acid, buffer pH

6.726.72water pH 1.60water pH 1.60

Upon addition 0.005 moles strong base, buffer pH Upon addition 0.005 moles strong base, buffer pH 7.687.68water pH 10.40water pH 10.40

Preparing Buffers - Useful References Perrin & Dempsey “Buffers for pH and Perrin & Dempsey “Buffers for pH and

Metal Ion Control” New York: Wiley, Metal Ion Control” New York: Wiley, 1974.1974.

Chemical Company booklets.Chemical Company booklets.Example: Calbiochem Doc. No. CB0052-Example: Calbiochem Doc. No. CB0052-591591

Practical Preparation

Practical:Practical: Identify reagent based on pKIdentify reagent based on pKaa

Prepare appropriate molarity and add Prepare appropriate molarity and add NaOH/HCl to adjust pHNaOH/HCl to adjust pH

Special Types of Buffers

GOOD Buffers - temperature resistantGOOD Buffers - temperature resistant

Volatile Buffers - can be removed by freeze Volatile Buffers - can be removed by freeze dryingdrying

Universal Buffers - wide effective pH rangeUniversal Buffers - wide effective pH range

GOOD Buffers

zwitterionic - have zwitterionic - have both amino and both amino and sulfonyl groupssulfonyl groups

EXAMPLES:EXAMPLES:MES pK 6.15MES pK 6.15HEPES pK 7.55HEPES pK 7.55

HOCH2CH2N NCH2CH2SO3H

O NCH2CH2SO3H

HEPES:

MES:

GOOD Buffers

pK typically 6-8 (physiological pH)pK typically 6-8 (physiological pH) No complexation with metal ions (no No complexation with metal ions (no

inhibition of enzymes)inhibition of enzymes) High aqueous solubilityHigh aqueous solubility Minimal salt effectsMinimal salt effects No UV-vis absorption (240-280 nm)No UV-vis absorption (240-280 nm)

Desirable Characteristics of GOOD Buffers pH independent of temperaturepH independent of temperature Compare with TRIS:Compare with TRIS:

Temperature, C pH

4 (cold room) 8.8

20 (room temp) 7.0

37 (incubation) 5.95

Volatile Buffers

Can be removed by simple evaporation or Can be removed by simple evaporation or lyophilizationlyophilization

good for electrophoresis or preparative ion good for electrophoresis or preparative ion exchange chromatographyexchange chromatography

Volatile Buffers

EXAMPLES:EXAMPLES: ammonium acetate pH 4-6ammonium acetate pH 4-6

pyridinium formate pH 3-6pyridinium formate pH 3-6

ammonium carbonate pH 8-10ammonium carbonate pH 8-10

Universal Buffers

Mixture of two or more buffersMixture of two or more buffers Effects of buffers are additiveEffects of buffers are additive

greater buffering capacitygreater buffering capacity wider effective pH rangewider effective pH range

EXAMPLE:EXAMPLE:citric acid 3.13, 4.76, 6.40citric acid 3.13, 4.76, 6.40phosphoric acid 2.15, 7.20, 12.15phosphoric acid 2.15, 7.20, 12.15boric acid 9.24, 12.74, 13.80boric acid 9.24, 12.74, 13.80

Amino Acids

Given:Given: pKpKaa (COOH) 2.3 (COOH) 2.3

pKpKaa (NH (NH33++) 9.6) 9.6

What form do amino What form do amino acids assume at pH 7?acids assume at pH 7?

Titrimetry

Purpose: Purpose: Determine concentration of an acid or base of unknown Determine concentration of an acid or base of unknown

concentration (Mconcentration (MAAVVAA = M = MBBVVBB)) Identification of unknown acid or base based on pKIdentification of unknown acid or base based on pKa a

(pH = pK(pH = pKaa at ½ volume at equivalence point) at ½ volume at equivalence point)

Method: volumetrically using biuretMethod: volumetrically using biuret

At endpoint: moles acid = moles baseAt endpoint: moles acid = moles baseMMaa V Vaa = M = Mbb V Vbb

Terminology

Titrant = standardized strong acid or base Titrant = standardized strong acid or base delivered from a biuretdelivered from a biuret

Standardized = concentration made known Standardized = concentration made known both in terms of accuracy and precisionboth in terms of accuracy and precision

Endpoint=pH at which visual indicator Endpoint=pH at which visual indicator changes colorchanges color

Equivalence point=pH at which moles of Equivalence point=pH at which moles of acid equal moles of baseacid equal moles of base

Typical Experimental Methodology - Weak Acid Standardize titrant (NaOH)Standardize titrant (NaOH)

titrate NaOH with KHP of known titrate NaOH with KHP of known concentrationconcentration

Titrate unknown (weak acid)Titrate unknown (weak acid)titrate unknown with standardized titranttitrate unknown with standardized titrant

Indicators

Organic weak acids that have different Organic weak acids that have different colors in their acid and conjugate base colors in their acid and conjugate base formsforms

EX:EX:phenolphthaleinphenolphthaleinHA = HHA = H++ + A + A--

colorlesscolorless pinkpink

Indicators

pH = pKpH = pKaa + log[A + log[A--/HA]/HA]

Your eye can detect color for 10-fold excess of Your eye can detect color for 10-fold excess of AA--/HA/HA

At equivalence point pH changes rapidlyAt equivalence point pH changes rapidly

Bottom line: endpoint may not be equivalence Bottom line: endpoint may not be equivalence point if indicator pKpoint if indicator pKaa not near equivalence point not near equivalence point

Titration of a Weak Acid

Let’s calculate the pH of the solution Let’s calculate the pH of the solution produced by adding 0, 10, 20, 25, 50, and produced by adding 0, 10, 20, 25, 50, and 70 mL of 0.1 M sodium hydroxide to 50 70 mL of 0.1 M sodium hydroxide to 50 mL of 0.1 M formic acidmL of 0.1 M formic acid

Titration of Weak Acid

1 – weak acid1 – weak acidpH = 0.5 (pKpH = 0.5 (pKaa – log[acid] – log[acid]

2 – buffer2 – bufferpH = pKpH = pKaa + log[base/acid] + log[base/acid]

3 – equivalence point; salt of 3 – equivalence point; salt of weak acidweak acidpH = 0.5 (14 + pKpH = 0.5 (14 + pKaa + log[salt]) + log[salt])

4 – strong base4 – strong basepH = 14+log[OHpH = 14+log[OH--]]

Vol. Titrant, mLVol. Titrant, mLpHpH

11 22

33

44

XX

XX

Identification of Weak Acid

At equivalence point:At equivalence point:MMacidacid V Vacidacid = M = Mbasebase V Vbasebase

At ½ volume corresponding to equivalence At ½ volume corresponding to equivalence point:point:pH = pKpH = pKaa

Titration of a Weak Base

Let’s calculate the pH of the solution Let’s calculate the pH of the solution produced by adding 0, 10, 20, 25, 50, and produced by adding 0, 10, 20, 25, 50, and 70 mL of 0.1 M hydrochloric acid to 50 mL 70 mL of 0.1 M hydrochloric acid to 50 mL of 0.1 M ammonium hydroxideof 0.1 M ammonium hydroxide

Titration of Weak Base

1 – weak base1 – weak basepH = 14 - 0.5 (pKpH = 14 - 0.5 (pKbb – log[base] – log[base]

2 – buffer2 – bufferpH = pKpH = pKaa + log[base/acid] + log[base/acid]

3 – equivalence point; salt of 3 – equivalence point; salt of weak baseweak basepH = 0.5 (14 - pKpH = 0.5 (14 - pKbb - log[salt]) - log[salt])

4 – strong acid4 – strong acidpH = -log[HpH = -log[H++]]

Vol. Titrant, mLVol. Titrant, mLpHpH

11

22

33

44

XX

XX