152 lecture 7 Acids Bases II slides = 78

Lecture 7Professor Hicks

Inorganic Chemistry (CHE152)

Strong acids HCl HNO3 H2SO4

H+ Cl- H+ NO3- 2H+ SO4

2-

hydrochloric acid nitric acid sulfuric acid

Weak acidsHC2H3O2 HF

H+ C2H3O2- H+ F-

acetic acid hydrofluoric acid

metalsform cations

non-metalsform anions

AcidsH+ + anion

H+ -anion

substances that release H+ ions when dissolved

molecularcompounds

all other molecular

compoundsdissolve

Dissolved molecules

ioniccompounds

dissolve

cations(+ ions)

anions(- ions)

separated ionsH+ + F- ~5%(separated ions)

HCl H+ + Cl-

~ 100%

HF ~ 95%(molecules)

acids

dissolve

Acids aremolecular compounds

strong acid

AcidsH+ -anion

• Acids are molecular compound because they candissolve without dissociating into ions

• Ionic compounds must separate into ions to dissolve

• Weak acids have a small percentage of moleculesseparated into H+ and an anion, the rest stay together as one particle

HF ~ 95% H+ and F- 5%

Strong acids separate 100% into H+ and anion in waterHCl ~ 0 % H+ and Cl- ~ 100%

Hydrogen ion • has no electrons!

• to bond other atom must provide both electrons (lone pair)

H+

H+

N

H

H

H

NH3 (aq) + H+ NH4+ (aq)

H N

H

H

H +curved arrows used to showelectron pair movement

1) start at lone pair 2) end at electron acceptor

carboxylic acids

• weak acids

• most common acids in nature C=O

O

R

H

C=O

O

RC=O

O

C=O

O

R

H

R = benzenebenzoic acid

R = C,H

C=O

O

C=O

O

CH3

H

R = CH3acetic acid

food preservative

5% solution = vinegar

curved arrows1) start at lone pair 2) end at electron acceptor

C=O

O

C=O

O

CH3

-

+ H+

hydronium ion H3O+

• H3O+ = H2O + H+

• form H+ takes in water

• reactions of acids in water can be written with H+ or H3O

+

+

0 Q 1010

Gib

bs

Fre

e E

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rgy

0 Q 2 x 10-4

Gib

bs

Fre

e E

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rgy

acid ionization constant Ka

• equilibrium constant for acid to hydrolyze

• acid + H2O H3O+ + anion

• larger Ka = stronger acid

strong acidsalmost 100%dissociated

weak acidsmuch less

100% dissociated

Q = Ka

Benzoic Acid = C6H5CO2H

C6H5CO2H (aq) + H2O (l) C6H5CO2- (aq) + H3O

+ (aq)

Q =[C6H5CO2

-][H3O+]

[C6H5CO2H]

Ka = 6.5 x 10-5

HClO4 Ka= 1010 !

reactants products reactants products

(weak acid)

Bases

- molecular compounds with lone pairs

+ cation

OH-

any cation

hydroxideion

• Bronsted-Lowry base = proton acceptor

• proton acceptors must have lone pair

- ionic compounds with OH- ion

N

H

H

H

lone pair

both proton acceptors = both bases

3 lone pairs O-H

Examples of baseshydroxide ion

ammonia

C

H

H

H

methane

H

no lone pairsnot a base

conjugate acid/base pairs

Whenever an acid-base reaction occurs:

1) the product that is “ acid minus H+ ” is called the conjugate base of the acid

2) The product that is “base plus H+ “ is called the conjugate acid of the base

HC2H3O2 (aq) + NaOH (aq) NaC2H3O2 (aq) + H2O (l)

acid conjugate acidbase conjugate base

strength conjugate acids/bases

• the stronger a base the weaker itsconjugate acid

0 Q 1010

Gib

bs

Fre

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rgy

0 Q 2 x 10-4

Gib

bs

Fre

e E

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rgy

Q closer to pure conjugate base

Q closer to pure acid

Benzoic acidKa = 6.5 x 10-5

perchloric acid Ka= 1010 !

acidconjugate

base acidconjugate

base

• the weaker an acid the stronger its conjugate base

Water autoionizes

• Keq for this reaction called Kw

• Kw = 10-14 at room temperature

H2O (l) H+ (aq) + OH- (aq) Q = [H+][OH-]

2H2O (l) H3O+ (aq) + OH- (aq)

or written with hydronium ion

Q = [H3O+][OH-]

• Neutral solution has [H+] =10-7 and [OH-] =10-7

Water is amphoteric(an acid and a base)

water acting as a base

H2O (l) + HCl (aq) H3O+ (aq) + Cl- (aq)

water acting as an acid

H2O (l) + NH3 (aq) OH- (aq) + NH4+ (aq)

Conjugate acid of water H3O + = hydronium ion

Conjugate base of water OH- = hydroxide ion

Le Chateliers principle and Kw

2H2O (l) H3O+ (aq) + OH- (aq)

disturbance = add HNO3increase H3O

+ to 0.10 Msystems response= to decrease H3O

+

what if neutral water is perturbed by adding an acid or base?

10-7 M10-7 M

[H3O+][OH-] = K w = 10-14

Q = 10-7 x 10-7 = 10-14

[H3O+][OH-] = Kw

0.10 * [OH-] = 10-14

[OH-] = 10-14 / 0.10=10-13 M

is system is at equilibrium?

system finds a new equilibrium

+ 0.10 M

Why is [H3O+] ~ 0.10 M?

(about)

If rxn used no OH-

[H3O+] = 0.10 + 10-7

= 0.1000001If rxn used all OH-

[H3O+] = 0.10 + 10-7 -10-7 =0.10

initial M

yes b/c Q = Ka

? ?equilibrium M

Approximately = 0.10Acidic solution has [H3O

+] > 10-7

Basic solution has [OH-] > 10-7

[H3O+]

pH

• scale of [H+] concentration

• more convenient than scientific notation

• pH = -log [H3O+]

still not sure? take the log of 10 it should be 1

pOH• same idea as pH

• scale of OH- concentration

• pOH = -log [OH-]

[H3O+]

pH + pOH = 14

Why?

[H3O+][OH-] = 10-14

pH + pOH = -log(10-14) = 14

pH, pOH, and pKw

1) take –log both sides2) log(A*B) = log(A) + log(B)

pH is what most people think in terms of some problems we get a result [OH-] or pOH

use this equation to express it as a pH

pH = 14 - pOH

pH fun facts!

• more bacteria that are not harmful grow in acidic conditions (acidophilus strains)

• more bacteria that are harmful grow in basic conditions

• blood pH about 7.4• stomach pH 1.5 !• H+ and OH- are catalysts for the

reactions that hold together, fats, carbohydrates, and proteins!!!!

control of pH important for life

The Strong Acids

• molarity of a monoprotic strong acid = molarity of [H3O+]

• b/c strong acids completely, 100% dissociate

For example a 1.0 M solution of HCl has a [H3O+] = 1.0 M

HCl, HBr, HI, HNO3, HClO4, H2SO4

Dissociation of weak acids

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)0.55 - 0 0-x - +x +x0.55-x - +x +x

initial (M)

Ka =[H3O+][F-]

[HF]

change (M)

equilibrium (M)

x2

0.55-x3.5 x 10-5 = x = ???

a quicker alternative is to look for an approximationif x is very small compared to 0.55

x2

0.553.5 x 10-5 =

x = sqrt{ 0.55* 3.5 x 10-5 } = 0.004387 Mx = change in [HF] to reach equilibrium,

and final equilibrium molarities [H3O+], [F-]

Calculate [H3O+] at equilibrium for a 0.55 M solution of HF in water.

look up Ka for HF

write hydrolysis reactionAcid + H2OH3O

+ + conj base

to solve for x you could use the quadratic equation

HF beforedissociation

HF (aq)H+ (aq)F- (aq)

dissolve

% dissociated =

- x

initial molarity

initial molarity x

“x”

initial molarity

x 100%PercentDissociated

at equilibrium

Dissociation of weak acidsHF (aq) + H2O (l) H3O+ (aq) + F- (aq)0.55 - 0 0-x +x +x0.55-x +x +x

initial (M)

Ka =[H3O+][F-]

[HF]

change (M)

equilibrium (M)

x2

0.55-x3.5 x 10-5 = x = ???

to solve for x you could use the quadratic equation

a quicker alternative is to look for an approximation

if x is very small compared to 0.55x2

0.553.5 x 10-5 =

x = sqrt{ 0.55* 3.5 x 10-5 } = 0.004387 M

is 0.004387 is small compared to 0.55 M? 5% error is widely accepted in science

0.004387

0.55 x 100% = 0.79%

approximation is valid b/c %dissociated is less than 5% so

error could not be larger than 5%

% dissociated

The method of successive

approximations

CO32- (aq) + H2O (l) HCO3

- (aq) + OH- (aq)0.010 - 0 0-x +x +x0.01 - x +x +x

initial (M)change (M)

equilibrium (M)

x = sqrt{ 0.01* 1.8 x 10-4 } = 0.00134 M

% dissociated = 0.00134

0.010x100% = 13.4%

more than 5%!approximation not good!

x2

0.010 -x1.8 x 10-4 =

1) substitute this approximationof x back into equation

2) calculate new approximate value of x

x2

0.010 - 0.00134

x = sqrt { 1.8x10-4 (0.01 - 0.00134 } = 0.00128

repeat steps 1, 2x = sqrt { 1.8x10-4 (0.01 - 0.00128 }

=0.00125

repeat steps 1, 2 again

x = sqrt { 1.8x10-4 (0.01 - 0.00125 } = 0.00125

the approximations have stopped changing in the 5th decimal place so the

approximation good to this precision

an improved approximation of x

use the method of successive approximations if % dissociation

is more than 5%

a more improved approximation of x

1) Break into groups of 2-3 - each group will be assigned an acid2) Determine the pH and % dissociated.

acid Ka

assigned M

1) HClO2 1.10 x 10 -02 5.0

2) HCHO2 1.80 x10 -04 9.0E-02

3) C7H6O2 6.50 x10 -05 3.0E-02

4) HC2H3O2 1.80 x10 -05 1.0E-02

5) HClO 2.90 x10 -08 2.0E-05

6) HCN 4.90 x10 -10 2.5E-07

7) HC6H5O 1.30 x10 -10 6.0E-08

8) HF 3.50 x10 -04 1.8E-01

9) HNO2 4.60 x10 -04 2.0E-01

Determine the pH and % dissociated for a 1.5 x 10-4 M solution of acetic acid using the method of successive approximations

Le Chateliers Principle and % dissociated

HF + H2O F + H3O+ Ka = 3.5 x 10-4

disturbance = increase H2O (dilute acid)

response = decrease [H2O]

1%10%30%40%50%60%70%80%90%~100%

percent dissociated

at “infinite” dilution

1.0 M 0.10 M

0.00010 M

0.001 M

etc

molarity declines with dilution

Polyprotic Acids• often acid molecules have more than one ionizable H –

these are called polyprotic acids

– 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic• HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

• ionizable H ’s have different Ka’s

• polyprotic acids ionize in steps– each ionizable H removed sequentially

• removing of the first H+ makes removal of the second H+

harder– H2SO4 is a stronger acid than HSO4

– H3PO4 is a stronger acid than H2PO4-

– H2PO4- is a stronger acid than HPO4

2-

Sulfuric acid is a diprotic acid

H2SO4 (aq) H2O (l) H3O+ (aq) + HSO4

-

~ 100%

HSO4- (aq) H2O (l) H3O

+ (aq) + SO42-

weak acidmuch less 100%

a 1.0 M solution of H2SO4 has [H3O+] = 1.0 M (first step)

+ a little more H3O+ from second dissociation

strong acid

• Estimate the concentration of sulfate ion in a 5.0 M solution of sulfuric acid.

Hints:

• In the first hydrolyses of sulfuric acid it acts as a strong acid

• For the second hydrolyses Ka = 1.0 x10-2

Strong bases

• calculate pOH from molarity [OH-]

• calculate pH

pH = 14 - pOH

Example: A 0.0010 M solution of NaOH has a [OH-] = 0.0010 M

pOH = -log(0.0010) = 3.0

pH = 14 - 3 = 11

ionic compounds with OH- ionKOH, NaOH, LiOH.

Weak bases

• react with water to produce OH-

Base + H2O Base-H+ + OH-

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

• do not react completely

• equilibrium constant Kb

hydrolysis reaction

Dissociation of weak bases


- (aq) + OH- (aq)0.10 - 0 0-x +x +x

0.10 - x x x

initial (M)

Kb =[OH-][HCO3

-]

[CO32-]

change (M)

equilibrium (M)

x2

0.10-x1.8 x 10-4 =

x2

0.101.8 x 10-4 =

x = sqrt{ 0.10* 1.8 x 10-4 } = 0.00424 M

Calculate pH at equilibrium for a 0.10 M solution of Na2CO3 in water.

look up Kb for CO32-

write hydrolysis reactionbase + H2O conj acid + OH-


0.10x100% = 4.2 %

less than 5%!approximation is OKpOH = -log[OH-]=-log(0.00424)=2.37

pH =14-pOH =14-2.37=11.62

dissociation of weak bases


- (aq) + OH- (aq)0.010 - 0 0-x +x +x

0.010 - x x x

initial (M)

Kb =[OH-][HCO3

-]

[CO32-]

change (M)

equilibrium (M)

x2

0.010-x1.8 x 10-4 =

x2

0.0101.8 x 10-4 =

x = sqrt{ 0.010* 1.8 x 10-4 } = 0.00134 M

Calculate pH at equilibrium for a 0.010 M solution of Na2CO3 in water.

look up Kb for CO32-

write hydrolysis reactionbase + H2O conj acid + OH-


0.010x100% =13.4%

more than 5%!approximation not good enough!

Need to use the quadratic equationor successive approximations method

% dissociated increases asthe acid/base is more dilute

acidsH+ + anion

H+ -anion

+ anion

Na+ -anion

salts of acids

K+ -anion

Strong acids HCl HNO3 H2SO4

H+ Cl- H+ NO3- 2H+ SO4

2-

hydrochloric acid nitric acid sulfuric acid

Weak acidsHC2H3O2 HF

H+ C2H3O2- H+ F-

acetic acid hydrofluoric acid

Salts of Strong acids LiCl NaNO3 K2SO4

Li+ Cl- Na+ NO3- 2K+ SO4

2-

lithium chloride sodium nitrate potassium sulfate

Salts of Weak acidsMg(C2H3O2)2 CsF

Mg2+ 2C2H3O2- Cs+ F-

magnesium acetate cesium fluoride

often from group Isince all group I salts are soluble

replace H+

any cation

if an acid is unchargedits conjugate base is

negatively charged

conjugate bases of acids exist as ionic compoundsaka salts

0 % dissociated 100

Gib

bs

Fre

e E

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rgy

0 % dissociated 100

Gib

bs

Fre

e E

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rgya weak acid

conjugate base acetate ion

is a weak base

acetate ion acetic acidacetic acid acetate ion

conjugate bases of weak acids• most conjugate acid/base pairs are both weak• exception: conjugates of the strong acids and

strong bases (OH-) are weak bases/acids

HCl, HBr, HI, HNO3, HClO4, H2SO4

Weak acids

Increasing Acid Strength

“Strong Acids”weaker

weak acidsstronger

weak acids

Increasing Base Strength

Cl-, Br-, I-, NO3-, ClO4

-,

Conjugate bases of “Strong Acids”

Conjugate bases of weak acids

weakerweak bases

strongerweak bases

OH-

alcoholsC2H5OH

“Strong bases”

H2O

C2H5O-

HClO2HCN

ClO2-CN-

(so weak they do not affect pH)

(so weak they donot affect pH)

weak base salt of itsconjugate acid

compound with lone pairsoften a N containing compound

N

H

H

H

ammonia amines

N

H

H

H

R

N

H

H

H

H + N

H

H

H

R

H +

-anion

-anion

and

when they act as bases gaining H+ they become positively charged

examples NH4Cl CH3NH3(ClO4)

weak bases and the salts of their conjugate acidsif a base is unchargedits conjugate acid is

positively charged

conjugate acids of bases exist as ionic compoundsaka salts

Ka Kb Kw

HC2H3O2(aq) + H2O(l) C2H3O2-(aq) + H3O+(aq) Ka=1.76 x10-5

for acetic acid the hydrolysis reaction is

for acetate ion the hydrolysis reaction is

C2H3O2- (aq) + H2O HC2H3O2 (aq) + OH- (aq) Kb=5.68 x10-10

notice if you add them the conjugate acid and base cancel

2H2O(l) H3O+(aq) + OH- (aq) K = ?Kw = 10-14

when reactions are added the overall Keq is the product of the Keq’s

KaKb = 1.76 x10-5 x 5.68 x10-10 = 10-14 !!!!!!!!!

overall reaction becomes

=

0 %ionized 100

Gib

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0 %ionized 100

Gib

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Fre

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NH3 a weak baseNH4

+ is a weak acid

NH4+

NH3NH3 NH4

+

conjugate acids of weak bases

• most weak bases conjugate acids are weak

• the conjugate acid of the strongbase OH- = H2O is a weak acid

Kb = 1.76 x 10-5

Ka = 5.68 x 10-10

same free energy bowl looked at from other side

not on table Ka’s

Ka = 10-14

Kb

use

salts of weak acids/bases

• if soluble fully dissociate into ions

• initial molarity calculated from chemical formula

Example

0.33 M NaC2H3O2 a solution 0.33 M in C2H3O2-

0.24 M Ca(C2H3O2)2 a solution 0.48 M in C2H3O2-

2 C2H3O2- per 1 Ca(C2H3O2)2

ICE tables for salts of weak acidsWhat is the pH of a 0.66 M solution of sodium acetate?

C2H3O2- (aq) + H2O HC2H3O2 (aq) + OH- (aq)

Kb=

initial

change

equil

0.66 0 0

-x +x +x

0.66-x +x +x

by the usual approximation x = square root (0.66*5.68 x 10 -10) = 1.936 x 10-5

= 1.936 x 10-5[OH-]

pOH = 4.71

pH = 14 - 4.71 = 9.29a basic solution b/c we added

the conjugate base of acetic acid

acetate ion

this problem is setup like other weak base problems but you will not find acetate in table of bases you must recognize

it as the conjugate base of a weak acid and calculate its Kb

= 5.68 x10-10

10-14

Ka (acetic acid) 10-14

1.76 x10-5=

trends in strengths of acids

2 factors

• electronegativity of conjugate base

• bond strength

weaker bonds break more easily stronger acidA H

A H - +

higher electronegativity of Amakes H closer to H+1 stronger acid

bond more polarized

Strengths of Binary Acids

• binary acid strength increases to the right across a period

example H-C < H-N < H-O < H-F

• binary acid strength increases down the column

example H-F < H-Cl < H-Br < H-I

1) the more + H-X - polarized the bond, the more acidic the bond

2) the weaker the H-X bond, the stronger the acid

Strengths of Oxoacids

• more oxygen atoms stronger acid– helps polarize the H-O bond

• more oxygen atoms in the chemical formula, like adding a single atom of greater electronegativity– used to compare similar acids

Example: H2SO4 is a strong acid but H2SO3 is a weak acid

Lewis Acids• Any substance that can accept an electron pair

to form a new bond is called a Lewis acid• H+ qualifies as a Lewis acid because it has no

electrons – it must accept electron pairs to form bonds

• Many metal ions accept electron pairs to form Coordinate Covalent Bonds

• Electron deficient species in general - such as boron compounds that violate the octet rule are Lewis acids

H

B

H

H

H

N

H

H

H

B

H

H

H

N

H

H

complex ions

• metal ion + base new complex• structure of base stays intact• metal ion is acting as a Lewis Acid• bases are called ligands when they bind to

metals

Al3+H-O-H repeat 6x

Al(H2O)63+

ligand metal ion complex ion

complex ions

• metal ion + base new complex

• water becomes more acidic when bound to metal

(H2O)5Al O H

H

3+

+ H2O

Al(H2O)5(OH)2+

+ H3O+

Ka = 1.4 x 10-5Al(H2O)63+

acidity of complex ions• increases as metal ion becomes smaller

and/or more highly charged

size decreases

charge increases

both increase acidity

not acidic

metal ion + water complex ion

Classifying Salt Solutions asAcidic, Basic, or Neutral

• cations of group 1 (Li+, Na+, K+, etc) will not change the pH

• anions that are conjugate bases of strong acids are such weak bases they will not change the pHNaCl LiNO3 KBr

group 1 ions conjugate bases of strong acids

neutral solutions

Cl– NO3– Br –


• if the anion is the conjugate base of a weak acid, it will form a basic solution

NaF KNO2

group I ions

Na+ K+

conjugate bases of weak acids

F– NO2-

basicneutral

solution will be basic


• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution

NH4Cl

weak acid conjugate base of a strong acid

acidic

neutral


• if the salt cation is a small / highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution

Al(NO3)3

weak acid conjugate base of a strong acid

acidic

neutral


• if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base

NH4F

is larger than Kb of the F−;

solution will be acidic

5.68 x 10-102.86 x 10-11

Ka of NH4+

Example: Determine whether a solution of the following salts is acidic, basic, or neutral

a) SrCl2

Sr2+ slightly acidic

Cl− is the conjugate base of a strong acid

pH neutral

solution will be slightly acidic

b) AlBr3

Al3+ is a small, highly charged metal ion

weak acid

Br− is the conjugate base of a strong acid,

pH neutral



c) CH3NH3NO3

CH3NH3+ conjugate acid of a weak base

acidic

NO3− is the conjugate base of a strong acid, pH

neutral


Example: Determine whether a solution of thefollowing salts is acidic, basic, or neutral


d) NaCHO2

Na+ is in group I,

neutral

CHO2− base of a weak acid

basic

solution will be basic


e) NH4(HCO2)

NH4+ conjugate acid of a weak base

acidic

HCO2− conjugate base of a weak acid

basic

Ka(NH4+) > Kb(F

−); solution will be acidic(5.68 x 10 -10) (2.8 x 10 -11)


NH4OH

weak acid strong base

Basic

Estimate the pH of a 0.10 M NH4OH solution(when bad approximations go good)

Forms NH4

+ and OH-

conjugate acid of NH3

base

will the solution be basic or acidic?

Write down an equilibrium reaction that includes OH- and NH4+

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Kb = 1.76 x 10-5

initial (M) 0 0.10 0.10

change (M) x -x -x

equilibrium (M) x 0.10 - x 0.10 - x

(0.10 – x)2

x1.76 x 10-5 =

(0.10 – x)(0.10 – x)

x1.76 x 10-5 = (0.10 )2

1.76 x 10-5x =

= 568 !!!!!!!!! Very bad approximation !!!!!!!!

0 100%

Gib

bs

Fre

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rgy

Estimate the pH of a 0.10 M NH4OH solution

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Kb = 1.76 x 10-5

initial (M) 0.10 0 0

change (M) -x +x +x

equil (M) 0.10 -x x x

x2

0.10 - x1.76 x 10-5 =

x = sqrt {0.10 x 1.76 x10-5}

x = 0.001326

If approximation is very bad then x is large

x small0 100%

Gib

bs

Fre

e E

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rgy

x large

placing the ball in a different initial position does not change where it will end

up at equilibrium

reactants and products can beimagined to react into any set of

concentrations and then “allowed”to move to equilibriumlike the ball in the bowl

% dissociated = (0.001326/0.010) x100% = 1.3%

152 lecture 7 Acids Bases II slides = 78

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Transcript of 152 lecture 7 Acids Bases II slides = 78