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 HA: produce H++ in aqueous solution in aqueous solution B: produces OHB: produces OH-- in aqueous solution in aqueous solution very limitedvery limited

Bronsted-LowryBronsted-Lowry A: HA: H++ donor donor B: HB: H++ acceptor 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 is stays as HAlies far to left, almost all HA is 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

ExampleExample 44 Find KFind Kaa for propanoic acid given the following information for propanoic acid given the following information

[C[C22HH55COOH] = 0.10M and pH = 2.96COOH] = 0.10M and pH = 2.96

[H[H33OO++] = 1.1 x 10] = 1.1 x 10-3-3 M M

CC22HH55COOH + HCOOH + H22O O C C22HH55COOCOO- - + H+ H33OO++

II 0.10 M0.10 M 00 00

CC -x-x-1.1 x 10-1.1 x 10-3-3 M M

+x+x+1.1 x 10+1.1 x 10-3-3 M M

+x+x+1.1 x 10+1.1 x 10-3-3 M M

EE 0.10 – x(1.1 x 100.10 – x(1.1 x 10-3-3 M) M) 1.1 x 101.1 x 10-3-3 M M 1.1 x 101.1 x 10-3-3 M M

1) C1) C22HH55COOH + HCOOH + H22O O C C22HH55COOCOO-- + H + H33OO++ Sol’n

2) Calculate [H3O+] using pH [H3O+] = 10-pH [H3O+] = 10-2.96 [H3O+] = 1.1 x 101.1 x 10-3-3 M M

Pure water is not included as it does not change

3)

Example 4 con’tExample 4 con’t

3 22 5 3 5

2 5

[ ][ ] ( )( ) (1.1 10 )1.2 10

[ ] 0.10 0.10a

C H COO H O x xK M

C H COOH x

% ionization:% ionization:

31.1 10100 1.1%

0.10

4) Solve for Ka

< 5% indicates a weak acid

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