The first six acids in the table from the data booklet are strong acids because they all react quantitatively with water to form hydronium ions.

The first six acids in the table from the data booklet are strong acids

because they all react quantitatively with water to form hydronium ions.

Most other acids are weak, in that they react with water to a much lesser extent.

For example, hydrofluoric acid:

The acid ionization constant (K

) indicates the extent to which an acid will react with water. It is a ratio of the dissociated form of the acid to the undissociated form.

undissociated molecules

dissociated molecules

 

+ 3 a

H O (aq) F (aq) HF(aq)

K

  



   



+

2 3

HF(aq) + H O(l)  H O (aq) + F (aq)

Eg 1:

Hydrofluoric acid normally has an ionization constant of 6.4 x 10

.

In an experiment 0.05 mol/L of HF (aq) is introduced in a 1.0 L reaction beaker. At equilibrium the pH was found to be 2.3. Calculate the K

under these conditions.

R I C E

HF H

F

1 1 1

0.05 0 0

HF(aq)  H

(aq) + F

(aq)

R I C E

HF H

F

1 1 1

0.05 0 0

-0.005 +0.005 +0.005 0.045 0.005 0.005

K

= [H [HF]

][F

]

= [.045]

[0.005]

= 5.6 x 10

[H

] = 10

= 10

= 0.005

Eg 2)

Nicotinic acid is a weak acid related to the B class of vitamins. It ionizes according to the following reaction:

HC

H

NO

(aq)  H

(aq) + C

H

NO

(aq)

if 0.010 mol/L of the acid is placed in a beaker

and the acid has a K

of 1.4 x 10

, Calculate the pH.

[HC

H

NO

]

Ka =

[H

][C

H

NO

]

=

HC₂H₄NO₂ (aq)  H⁺ (aq) + C₂H₄NO₂^– (aq)

R I C E

HC₂H₄NO₂ H⁺ C₂H₄NO₂^–

1 1 1

0.010 0 0

-x +x +x

0.010 - x x x

[HC₂H₄NO₂] Ka = [H⁺][C₂H₄NO₂^–]

= [0.010 - x]

[x]²

= 1.4 x 10 ^{– 5}

Since x is small 0.010 – x = 0.010 X = 3.74 x 10^–4 M [0.010]

[x]²

=1.4 x 10 ^{– 5}

pH = 3.43

K

summary

• K

follows the pattern of other “K” equations

• I.e. for HA(aq) + H

O(l)  H

O

(aq) + A

(aq)

• K

= [H

O

][A

] [HA]

• Notice that H

O is ignored because it is liquid

• HA cannot be ignored because it is aqueous

• Acids can be in solution whether ionized or not

• The solubility of acids makes sense if you think

back to the partial charges in HCl for ex.

K

summary

• Generally K

tells you about acid strength

• Strong acids have high Ka values

• A “strong” acid is an acid that completely ionizes.

E.g. HCl + H

O  H

O

+ Cl

• A “weak” acid is an acid that doesn’t ionize completely. E.g. HF + H

O  H

O

+ F

• Note: don’t get confused between strength and

concentration. 1 M HCN has a smaller [H

], thus a higher pH, than 0.001 M HCl

• In general: K

< 10

Weak acid 10

< K

< 1 Moderate acid

K

> 1 Strong acid

Dissociation vs. Ionization

• Ionization and dissociation indicate ions form

• Dissociation: ions form when a chemical comes apart.

E.g. NaCl melts to form Na

, Cl

• Ionization: ions form when two chemicals react. E.g.

HCl(aq) + H

O  H

O

(aq) + Cl

(aq)

• Even though we write HCl  H

+ Cl

, this is just an abbreviation. In reality HCl reacts with H

O, thus it is an ionization not a dissociation

• Note that NaCl can also dissociate in water. This is not

an ionization, since water is only required to stabilize

ions (it is not needed as a reactant involved in forming

ions)

 Read pgs. 737 – 742

 pg. 743 Practice #’s 1, 6 – 9

Homework:

Base Strength and the Ionization Constant, K

All ionic hydroxides completely dissociate upon dissolving,

so they are considered to be strong bases.

Most other bases are weak, in that they react with water to a much lesser extent.

For example, citrate ions:

The base ionization constant (K

) indicates the extent to which a base will react with water.

6 5 7 2 6 5 7

C H O (aq) + H O(l)

 HC H O (aq) + OH (aq)

6 5 7

b

6 5 7

HC H O (aq) OH (aq)

= C H O (aq)

K

 



   

   

 

 

Calculating K

from Amount Concentrations

The K

–K

Relationship for Conjugate Acid–Base Pairs

Consider the case of a general weak acid of the form HX(aq) and its reaction with water:

+

2 3

HX(aq) + H O(l)  H O (aq) + X (aq)

 

+ 3 a

H O (aq) X (aq)

K = HX(aq)

  



   

For the conjugate base X^–(aq):

Notice what happens when we multiply these two equilibrium expressions:

K

× K

= K

X (aq) + H O(l)

 HX(aq) + OH (aq)

 

b

HX(aq) OH (aq)

K = X (aq)





 

 

 

 

   

+ 3

a b

H O (aq) X (aq) HX(aq) OH (aq)

HX(aq) X (aq)

K K

 



     

     

  

 

 

+

H O (aq)

OH (aq)

   

     

K _b

• K

is similar to K

except b stands for base

• The general reaction involving a base can be written as B(aq) + H

O  BH

(aq) + OH

(aq)

• Thus K

= [BH

] [OH

] [B]

Eg. 1)

An unknown base (M) has an initial concentration of

0.010 mol/L, after a period of time it is found that the

equilibrium mixture has a pH of 10.10. Calculate K

.

M + H

O  MH

+ OH

R I C E

M MH

OH

1 1 1

0.010 0 0

-0.00013 +0.00013 +0.00013 0.00013 0.00013

0.00987

Kb = [MH [M]

] [OH

]

= =1.6 x 10

pOH = 14 - pH = 14 - 10.10 = 3.90 [OH-] = 10

= 10

= 1.26 x 10

[0.00987]

[0.00013] [0.00013]

Eg.2)

Ammonia is a weak base. If ammonia with an initial concentration of 0.020 mol/L is present in a beaker, what is the pH at equilibrium?

NH

(aq) + H

O (l)  NH

(aq) + OH

(aq)

R I C

E

NH

NH

OH

1 1 1

0.020 0 0

NH

(aq) + H

O (l)  NH

(aq) + OH

(aq)

R I C E

NH

NH

OH

1 1 1

0.020 0 0

-x +x +x

x x

0.020 - x

[0.020]

Kb = [x] [x]

= = 1.8 x 10

pOH = -log[OH

] = 3.22

[0.020]

x

x = 6.0 x 10

pH = 14 - pOH = 10.78

Strength of Conjugates

Consider:

HCl(aq) + H

O(l)  Cl

(aq) + H

O

(aq) The K

for HCl is: [Cl

(aq)] [H

O

(aq)]

[HCl(aq)]

The K

for Cl

is: [HCl(aq)]

[Cl

(aq)] [H

O

(aq)]

Reciprocal!

Relative values of K

Recall for HX  H

+ X

, Ka = [H+][X

] / [HX]

Q - what does a large K

indicate?

A - equilibrium is far to the right (all dissociates) Thus a large K

= strong acid

Look at Table on page 829 or blue book The text uses this definition:

Ka < 10

is a weak acid

10

< Ka < 1 is a moderate acid 1 < Ka is a strong acid

These definitions are somewhat arbitrary, we will

not focus on this. Just remember a high K

means

the acid is strong.

The Effect of Amphoteric Entities

Remember, amphoteric entities can act as an acid or as a base.

To decide which one, compare the values of the K

and K

: If K

> K

, then it acts as an acid.

If K

< K

, then it acts as a base.

The Rule Of 1000

The value of x in the denominator can be omitted whenever the original concentration of the base is at least 1000 times the numerical value of the K_b or K_a.

For any weak base: For any weak acid:

 

2 b

original K  x

 

2 a

original

K  x

Other Formulas

That could be used for weak acids K

= [ H

]

.

[ HA] - [ H

]

Can be rearranged when solving for [H

] [H

]

= K

[HA] - K

[H

]

Put in “ax

+ bx + c = 0 “ format

[H

]

+ K

[H

] - K

[HA] = 0

Then

Put in “ax

+ bx + c = 0“ format [H

]

+ K

[H

] - K

[HA] = 0

Quadratic Eq’n Form x = -b

√b

- 4ac 2a

[H

] = -K

+ √K

– 4(1) (-K

x [HA])

2 (1)

Eg. Given 0.10 mol/L HCl

and 0.10 CH

COOH

find the pH of each.

HCl (aq) has 100 % : pH = -log(0.10) = 1.00 0.10 mol/L CH

COOH

calculate 3 ways:

1. Ka= [H

]

[HA]

1.8 x 10

x 0.10 mol/L = [H

]

[H

] = 0.0013416408 mol/L

pH = -log(0.0013416408) = 2.87

K

= [ H

]

. [ HA] - [ H

]

[H

]

= 1.8 x 10

[0.10] - 1.8 x 10

[H

]

Put in “ax

+ bx + c = 0 “ format

[H

]

+ 1.8 x 10

[H

] - 1.8 x 10

[0.10] = 0

x = -b

√b

- 4ac 2a

[H

] = -K

+ √K

– 4(1) (-K

x [HA]) 2 (1)

ax

+ bx + c = 0 [H

]

+ 1.8 x 10

[H

] - 1.8 x 10

[0.10] = 0

[H

] = -1.8 x 10

+ √(1.8 x 10

)

-(4)(0.10)(-1.8 x 10

) 2(1)

[H

] = 1.3326106 x 10

mol/L

pH = -log(0.0013326106) = 2.875296737 = 2.88

Not much difference!

General Rule is if…

[acid] / Ka > 1000 USE Regular Formula

[acid] / Ka < 1000 USE Quadratic Formula

HOWEVER YOU DO NOT HAVE TO USE QUADRATIC ON

DIPLOMA EXAM. ALL QUESTIONS WILL WORK WITHOUT IT!

 Read pgs. 744 – 750

 pg. 746 Practice #’s 10 – 13

 pg. 750 Section 16.3 Questions #’s 1 – 10

Homework: