Ch. 16: Equilibrium in Ch. 16: Equilibrium in Acid-Base SystemsAcid-Base Systems

16.3a: Acid-Base strength and 16.3a: Acid-Base strength and equilibrium lawequilibrium law

DefinitionsDefinitions

ArrheniusArrhenius A: produce H+ in aqueous solutionA: produce H+ in aqueous solution B: produces OH- in aqueous solutionB: produces OH- in aqueous solution very limitedvery limited

Bronsted-LowryBronsted-Lowry A: H+ donorA: H+ donor B: H+ acceptorB: H+ acceptor more generalmore general

Acid ionization constantAcid ionization constant

equilibrium expression where Hequilibrium expression where H++ is is removed to form conjugate baseremoved to form conjugate base

so for: HA + Hso for: HA + H22O O <--><--> H H33OO++ + A + A--

][

]][[

][

]][[ 3

HA

AH

HA

AOHKa

StrengthStrength

determined by equilibrium position of determined by equilibrium position of dissociation reactiondissociation reaction

strong acid: strong acid: lies far to right, almost all HA is dissociatedlies far to right, almost all HA is dissociated large Klarge Kaa values values creates weak conjugate basecreates weak conjugate base

weak acid: weak acid: lies far to left, almost all HA stays as HAlies far to left, almost all HA stays as HA small Ksmall Kaa values values creates strong conjugate basecreates strong conjugate base

Water is a stronger base than the CB of a strong acid but a weaker base than the CB of a weak acid

Water is a stronger acid than the CA of a strong base but a weaker acid than the CA of a weak base

[H[H22O], pH and KO], pH and Kww

conc. of liquid water is omitted from the Kconc. of liquid water is omitted from the Kaa expressionexpression we assume that this conc. will remain constant in we assume that this conc. will remain constant in

aqueous sol’n that are not highly concentratedaqueous sol’n that are not highly concentrated pH= -log[HpH= -log[H++]] pOH = -log[OHpOH = -log[OH--]] 14.00= pH + pOH14.00= pH + pOH

143 100.1]][[ OHOHKw

Example 1Example 1

The [OHThe [OH--] of a solution at 25] of a solution at 25ooC is 1.0x10C is 1.0x10-5-5 M. M. Determine the [HDetermine the [H++], pH and pOH.], pH and pOH. KKww = 1.0x10 = 1.0x10-14-14 = [OH = [OH--] x [H] x [H++]] [H[H++] = 1.0x10] = 1.0x10-9-9

pH= -log(1.0x10pH= -log(1.0x10-9-9) = 9.00) = 9.00 pOH = -log(1.0x10pOH = -log(1.0x10-5-5) = 5.00) = 5.00 acidic or basic?acidic or basic? basicbasic

ApproximationsApproximations

If K is very small, we can assume that the If K is very small, we can assume that the change (x) is going to be negligiblechange (x) is going to be negligible

““rule of thumb” is if initial conc. of the acid is rule of thumb” is if initial conc. of the acid is >1000 times its K>1000 times its Kaa value then cancel x value then cancel x

this makes the answer true to +/- 5% and why Kthis makes the answer true to +/- 5% and why Kaa values are given to 2 sig. digsvalues are given to 2 sig. digs

32

2

2

2

4)0.1(

)2)((

)20.1(

)2)((K x

xx

x

xx

0

Calculating Weak AcidsCalculating Weak Acids

1.1. Write major speciesWrite major species2.2. Decide on which can provide HDecide on which can provide H++ ions ions3.3. Make ICE tableMake ICE table

4.4. Put equilibrium values in KPut equilibrium values in Kaa expressionexpression

5.5. Check validity of assumption (x must Check validity of assumption (x must be less than 5% of initial conc)be less than 5% of initial conc)

6.6. Find pHFind pH

Example 2Example 2

Calculate the pH of 1.00 M solution of Calculate the pH of 1.00 M solution of HF (KHF (Kaa = 7.2 x 10 = 7.2 x 10-4-4)) HF, HHF, H22OO HF HF H H++ + F + F-- KKaa = 7.2x10 = 7.2x10-4-4

HH22O O H H++ + OH + OH-- KKww = 1.0 x 10 = 1.0 x 10-14-14

HF will provide much more HHF will provide much more H++ than than HH22O – ignore HO – ignore H22OO

Example 2Example 2 HF HF H H++ + F + F--

II 1.00 M1.00 M 00 00

CC -x-x +x+x +x+x

EE 1.00 -x1.00 -x xx xx

Mxx

x

x

xx

HF

HFKa

027.0102.7

102.700.100.1

))((

][

]][[

42

42

Example 2Example 2

Check assumption:Check assumption:

%5%7.210000.1

027.0

%5100][

HA

x

pH = -log(0.027) = 1.57pH = -log(0.027) = 1.57

Example 3Example 3

Find pH of 0.100 M solution of HOCl (KFind pH of 0.100 M solution of HOCl (Kaa = 3.5x10 = 3.5x10-8-8))

HOCl, HHOCl, H22OO

HOCl will provide much more HHOCl will provide much more H++ than H than H22O, so we ignore O, so we ignore

HH22OO

HOCl HOCl H H++ + OCl + OCl--II 0.100 M0.100 M 00 00

CC -x-x +x+x +x+x

EE 0.100 -x0.100 -x xx xx

Example 3Example 3

Mxx

x

x

xx

HOCl

HOClKa

592

82

109.5105.3

105.3100.0100.0

))((

][

]][[

Check assumption:Check assumption:

%5%059.0100100.0

109.5 5

pH = -log(5.9x10pH = -log(5.9x10-5-5) = 4.23) = 4.23

Homework Textbook p743 #2a,c,e 5,7,9 LSM 16.3A and 16.3D

Ch. 16: Equilibrium in Ch. 16: Equilibrium in Acid-Base SystemsAcid-Base Systems

16.3b: Base strength and 16.3b: Base strength and equilibrium lawequilibrium law

Base Strength and Kb

follows same standard rules as for calculating Ka for acids

Kb is used with weak bases that react only partially with water (<50%)

BasesBases

KKbb base ionization constantbase ionization constant refers to reaction of base with water to make refers to reaction of base with water to make

conjugate acid and OHconjugate acid and OH--

liquid water is again ignored like in Kliquid water is again ignored like in Kaa

BB(aq)(aq) + H + H22OO (l)(l) BH BH+ +

(aq)(aq) + OH + OH- - (aq)(aq)

][

]][[

B

OHBHKb

Example 4Example 4

Find the pH for 15.0 M solution of NHFind the pH for 15.0 M solution of NH33

(K(Kbb = 1.8x10 = 1.8x10-5-5)) NHNH33 will create more OH will create more OH-- than water so than water so

self- ionization (Hself- ionization (H22O) can be ignoredO) can be ignored

NHNH33 + H + H22O O NH NH44++ + OH + OH--

II 15.015.0 00 00

CC -x-x +x+x +x+x

EE 15.0-x15.0-x xx xx

NHNH33 + H + H22O <--> NHO <--> NH44++ + OH + OH--

Sol’n

Example 4 con’tExample 4 con’t

%5%11.01000.15

016.0

016.0

0.15108.1

]0.15[

]][[ 25

x

x

x

xxKb

1413

13

[ ] 0.016

1.0 10[ ] 6.2 10

0.016

log(6.2 10 ) 12.20

x OH M

H

pH

% ion.

Example 5Example 5

Codeine (CCodeine (C1818HH2121NONO33) is a weak organic base. A ) is a weak organic base. A 5.0x105.0x10-3-3 M solution of codeine has a pH of 9.95. M solution of codeine has a pH of 9.95.

Calculate the KCalculate the Kbb for this substance. for this substance.

Sol’nSol’n What is chemical reaction?What is chemical reaction? CC1818HH2121NONO3 3 + H+ H22O O <--> <--> HCHC1818HH2121NONO33

++ + OH + OH--

Find [OHFind [OH--] using given pH] using given pH pOH= 14.00-9.95 = 4.05pOH= 14.00-9.95 = 4.05 [OH[OH--] = 10] = 10-4.05-4.05 = 8.9x10 = 8.9x10-5-5

Example 5 con’tExample 5 con’t

CC1818HH2121NONO3 3 + + HH22OO<--> <--> HCHC1818HH2121NONO33++ + OH + OH--

II 5.0x105.0x10-3-3 00 00

CC -x-x +x+x +x+x

EE 5.0x105.0x10-3-3 - x - x xx xx

x = [OHx = [OH--] = ] = 8.9x108.9x10-5-5

6

53-

25

3-

106.1

]109.85.0x10[

]109.8[

]5.0x10[

]][[

b

b

K

x

xxK

Ka - Kb relationship for conjugate pairs

Acid-Base strength tables do not give Kb values we use the Ka - Kb relationship to solve this

problem this can be used with any conjugate acid-base pair

Kw = Ka x Kb

or

Kb = Kw / Ka

Kw = 1.0 x 10-14

Example 6 What is the Kb value for the weak base present when

sodium cyanide dissociates into an aqueous solution?

Sol’n

NaCN --> Na+ + CN- (complete dissociation)

the CN- is the weak base so we write the equilibrium equation with water to determine its conjugate acid

CN- + H2O <--> HCN + OH-

the conjugate acid is found to be HCN and its Ka is 6.2 x 10-10

Using Kb = Kw / Ka find Kb for CN-

Kb = 1 x 10-14 / 6.2 x 10-10 = 1.61 x 10-5

Homework Textbook p746 #11,12 Textbook p750 #2,6,9 LSM 16.3 A,B & D