Lecture 7 Professor Hicks Inorganic Chemistry (CHE152) substances - - PDF document

Lecture 7

Professor Hicks Inorganic Chemistry (CHE152)

Strong acids HCl HNO3 H2SO4 H+ Cl- H+ NO3

• 2H+

SO4

2- hydrochloric acid nitric acid sulfuric acid

Weak acids HC2H3O2 HF H+ C2H3O2

• H+

F-

acetic acid hydrofluoric acid

metals form cations non-metals form anions

Acids

H+ + anion

H+

• anion

substances that release H+ ions when dissolved

molecular compounds all other molecular compounds

dissolve

Dissolved molecules ionic compounds

dissolve

cations (+ ions) anions (- ions)

separated ions H+ + F- ~5% (separated ions) HCl  H+ + Cl- ~ 100% HF ~ 95% (molecules)

acids

dissolve

Acids are molecular compounds

strong acid

Acids

H+

• anion
• Acids are molecular compound because they can

dissolve without dissociating into ions

• Ionic compounds must separate into ions to dissolve
• Weak acids have a small percentage of molecules

separated 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 water

HCl ~ 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 show electron 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

R

C=O O C=O O

R

H

R = benzene benzoic acid R = C,H

C=O O C=O O

CH3

H

R = CH3 acetic acid food preservative 5% solution = vinegar curved arrows 1) 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

Gibbs Free Energy

0 Q 2 x 10-4

Gibbs Free Energy

acid ionization constant Ka

• equilibrium constant for acid to hydrolyze
• acid + H2O H3O+ + anion
• larger Ka = stronger acid

strong acids almost 100% dissociated weak acids much 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 hydroxide ion

• 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 bases hydroxide ion ammonia

C

H H H

methane

H

no lone pairs not 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 acid base conjugate base

strength conjugate acids/bases

• the stronger a base

the weaker its conjugate acid

0 Q 1010

Gibbs Free Energy

0 Q 2 x 10-4

Gibbs Free Energy

Q closer to pure conjugate base Q closer to pure acid Benzoic acid Ka = 6.5 x 10-5 perchloric acid Ka= 1010 ! acid conjugate base acid conjugate 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)

• r 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 HNO3 increase H3O+ to 0.10 M systems response = to decrease H3O+

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

10-7 M 10-7 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 = 0.1000001 If rxn used all OH- [H3O+] = 0.10 + 10-7 -10-7 =0.10 initial M yes b/c Q = Ka ? ? equilibrium M Approximately = 0.10 Acidic 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 sides 2) 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 +x 0.55-x - +x +x

initial (M)

Ka = [H3O+][F-] [HF]

change (M) equilibrium (M)

x2 0.55-x 3.5 x 10-5 = x = ???

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

x2 0.55 3.5 x 10-5 = x = sqrt{ 0.55* 3.5 x 10-5 } = 0.004387 M

x = 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 reaction Acid + H2OH3O+ + conj base

to solve for x you could use the quadratic equation

HF before dissociation

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

dissolve % dissociated =

• x

initial molarity initial molarity

x

“x” initial molarity

x 100%

Percent Dissociated

at equilibrium

Dissociation of weak acids

HF (aq) + H2O (l)  H3O+ (aq) + F- (aq) 0.55 - 0 0

• x +x +x

0.55-x +x +x

initial (M)

Ka = [H3O+][F-] [HF]

change (M) equilibrium (M)

x2 0.55-x 3.5 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 0.55

x2 0.55 3.5 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

CO3

• (aq) + OH- (aq)

0.010 - 0 0

• x +x +x

0.01 - x +x +x initial (M) change (M) equilibrium (M)

x = sqrt{ 0.01* 1.8 x 10-4 } = 0.00134 M % dissociated = 0.00134 0.010 x100% = 13.4% more than 5%! approximation not good!

x2 0.010 -x 1.8 x 10-4 =

1) substitute this approximation

• f 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, 2 x = 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 acid 2) 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) + SO4

2-

weak acid much 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- ion KOH, 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

CO3

2- (aq) + H2O (l)  HCO3

• (aq) + OH- (aq)

0.10 - 0 0

• x +x +x

0.10 - x x x

initial (M)

Kb = [OH-][HCO3

• ]

[CO3

2-]

change (M) equilibrium (M)

x2 0.10-x 1.8 x 10-4 = x2 0.10 1.8 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 CO3

2-

write hydrolysis reaction base + H2O conj acid + OH-

% dissociated = 0.00424 0.10 x100% = 4.2 % less than 5%! approximation is OK

pOH = -log[OH-]=-log(0.00424)=2.37 pH =14-pOH =14-2.37=11.62

dissociation of weak bases

CO3

2- (aq) + H2O (l)  HCO3

• (aq) + OH- (aq)

0.010 - 0 0

• x +x +x

0.010 - x x x

initial (M)

Kb = [OH-][HCO3

• ]

[CO3

2-]

change (M) equilibrium (M)

x2 0.010-x 1.8 x 10-4 = x2 0.010 1.8 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 CO3

2-

write hydrolysis reaction base + H2O conj acid + OH-

% dissociated = 0.00134 0.010 x100% =13.4% more than 5%! approximation not good enough! Need to use the quadratic equation

• r successive approximations method

% dissociated increases as the acid/base is more dilute

acids

H+ + 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 acids HC2H3O2 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 acids Mg(C2H3O2)2 CsF Mg2+ 2C2H3O2

• Cs+

F-

magnesium acetate cesium fluoride

• ften from group I

since all group I salts are soluble replace H+ any cation if an acid is uncharged its conjugate base is negatively charged conjugate bases of acids exist as ionic compounds aka salts 0 % dissociated 100

Gibbs Free Energy

0 % dissociated 100

Gibbs Free Energy

a weak acid conjugate base acetate ion is a weak base acetate ion acetic acid acetic 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 acids stronger weak acids

Increasing Base Strength

Cl-, Br-, I-, NO3

• , ClO4
• ,

Conjugate bases of “Strong Acids”

Conjugate bases of weak acids

weaker weak bases stronger weak bases OH- alcohols C2H5OH “Strong bases” H2O C2H5O- HClO2 HCN ClO2

• CN-

(so weak they do not affect pH) (so weak they do not affect pH)

weak base salt of its conjugate acid

compound with lone pairs

• ften 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 acids

if a base is uncharged its conjugate acid is positively charged conjugate acids of bases exist as ionic compounds aka 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 !!!!!!!!!

• verall reaction becomes

= 

0 %ionized 100

Gibbs Free Energy

0 %ionized 100

Gibbs Free Energy

NH3 a weak base NH4

+ is a weak acid

NH4

+

NH3 NH3 NH4

+

conjugate acids

• f weak bases
• most weak bases conjugate

acids are weak

• the conjugate acid of the strong

base 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 acids

What 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

• 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 = 9.29 a 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 acid

A H A H

 - +

higher electronegativity of A makes 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)6

3+ 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 10-5

Al(H2O)6

3+

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 as Acidic, 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 pH NaCl LiNO3 KBr 

group 1 ions conjugate bases of strong acids

neutral solutions Cl– NO3

– Br –

Classifying Salt Solutions as Acidic, Basic, or Neutral

• 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

• basic

neutral

solution will be basic

Classifying Salt Solutions as Acidic, Basic, or Neutral

• 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

Classifying Salt Solutions as Acidic, Basic, or 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

Classifying Salt Solutions as Acidic, Basic, or 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 10-10 2.86 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 solution will be acidic Example: Determine whether a solution of the following salts is acidic, basic, or neutral

c) CH3NH3NO3 CH3NH3+ conjugate acid of a weak base acidic NO3− is the conjugate base of a strong acid, pH neutral solution will be acidic

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

Example: Determine whether a solution of the following 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

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

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)

Classifying Salt Solutions as Acidic, Basic, or Neutral

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 x 1.76 x 10-5 = (0.10 – x)(0.10 – x) x 1.76 x 10-5 = (0.10 )2 1.76 x 10-5 x = = 568 !!!!!!!!! Very bad approximation !!!!!!!! 0 100%

Gibbs Free Energy

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 - x 1.76 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 small 0 100%

Gibbs Free Energy

x large

placing the ball in a different initial position does not change where it will end up at equilibrium reactants and products can be imagined to react into any set of concentrations and then “allowed” to move to equilibrium like the ball in the bowl

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