DISSOCIATION OF WATER

H2O + H2O H3O+ + OH-

Acid Base

Kw = [H3O+] [OH-]

Where, Kw: water dissociation

constant for pure water at 25 ˚C

• [H3O+] = [OH-] = 1.0 mol dm-3

• Thus, Kw = [H3O+] [OH-] = (1.0 ) (1.0 ) mol2 dm-6

= 1.0 mol2 dm-6

• log [H3O+] + log[OH-] = log (1.0 )log [H3O+] + log[OH-] = -14.0-log [H3O+] + log[OH-]= 14.0

• Thus, pH + pOH = 14.0or pH + pOH = pKw (-log = p)

THE pH of AQUEOUS SOLUTION

• Calculate the pH of an aqueous solution of Ba(OH)2 0.25 mol dm-3

Strong bases

• Ba(OH)2 (aq) Ba2+ (aq) + 2OH- (aq)

• pOH = -log [OH-] = -log (0.50) = 0.3

Initial molarity (mol

dm-3)

0.25 0 0

Final molarity (mol dm-3)

0 0.25 0.50

pH = 14.0 – 0.3 = 13.7

What is the molarity of OH- in an aqueous solution of NaOH if its pH is 13.5?

• pOH = 14.0 – pH = 14.0 – 13.5 = 0.5

• pOH = -log [OH-]0.5 = -log [OH-]log [OH-] = -0.5[OH-] = 0.3 mol dm-3

SOLUTION :

• Calculate the pH value of 1.0 mol dm-3 of HNO3 acid. Explain your answer.

EXERCISE : (STRONG ACID)

• HNO3 H+ + NO2-

1.0 1.0 1.0

• pH = -log [H+] = -log (1.0 ) = 8

• For acid, pH 7, [H3O+] 1.0 consider [H3O+] from dissociation of water.

• [H3O+] = [H3O+]w + [H3O+]acid = 1.0

ANSWER :

• Anions derived from strong acids (such as Cl- from HCl) and cations from strong bases (such as Na+ from NaOH) do not react with water to affect the pH.

• Weak acid :CH3COOH (aq) H+ (aq) + CH3COO- (aq)

The relation between Ka, Kb and Kw

Acid Base

Ka =

* CH3COO- (aq) + H2O CH3COOH (aq) + OH-

Kb =

• Ka . Kb =

Ka . Kb = [H+] [ OH-]

Ka . Kb = Kw

To obtain the value of Kb for the anion derived from a weak acid

•

• ,

• For the hydrolysis of acetate ion (*),

• Kb = = 5.6

Kb = Ka =

Weak acid and weak base

• The degree of dissociation of weak acids and bases, less than 1 (or less 100%).

• The molarity of H3O+ ions is less than the molarity of the acid (HA).

•

Weak acid

[H3O+] [Acids]

The molarity of H3O+ for weak acids (HA) can be calculated using equilibrium law.

• HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Initial molarity (mol dm-3 ) :

C 0 0

Reacted concentration :

-X +X +X

Equilibrium molarity (mol dm-3)

( C – X ) X X

For a weak acid, X is small.

Thus, ( C-X ) C

Acid dissociation constant, Ka

• Each weak acid has an acid dissociation constant (Ka).

• The relative strength of weak acids are deduced from the Ka values.

• Strong acids have higher Ka (lower pKa) values than weaker acids.

Example ; Methanoic acid (HCOOH); Ka = 1.6 mol pKa = 3.8Is a stronger acid than ethanoic acid (CCOOH); Ka = 1.7 mol pKa = 4.8

Ka =

Example;Calculate (a) pKa, (b) pH, and (c) degree of dissociation , of 0.01 mol dm-3 nitrous acid, HNO2 [Ka = 5.1 10-4 mol dm-3]

pKa = - log Ka

Answers ;(a) pKa = - log Ka

= - log (5.1 10-4) = 3.3(b) HNO2 (aq) + H2O H3O+ (aq) + NO2

- (aq)Initial : 0.01 mol dm-3

Reacted : - X X X

At equilibrium : (0.01 – X) mol dm-3 X X

Ka = 5.1 10-4 M = , X is small (0.01 – X) 0.01 = X2 = 5.1 M2

X = 2.26 M [H3O+] = 2.26 MpH = -log [H+] = -log (2.26 = 2.6

X = = = = 2.26 M

Or :

(c) degree of dissociation, = 100 = 100 = 23 %

• The degree of dissociation of weak bases • The molarity of OH- ion is less than the

molarity of undissociated base.

• The molarity of OH- from weak bases (B) can be calculated using equilibrium law.

WEAK BASE

[OH-] [Base]

• B (aq) + H2O BH+ (aq) + OH- (aq)

Initial molarity , moldm-3

C O O

Reacted -X X X

Equilibrium molarity , moldm-3

C – X X X

For a weak base, X is small.

Thus, (C-X) C

• Weak bases has a base dissociation constant, Kb.

• The relative strength of weak bases can be deduced by comparing the Kb values.

• Strong bases have higher Kb (lower pKb) values than weaker bases.

Base dissociation constant, Kb

Example ;Ammonia, (NH3 ; Kb = 1.7 pKb = 4.8 ) Is a weaker base than methanamine ( CH3NH2; Kb = 4.2 moldm-3

pKb = 3.4)

pKb = -log Kb Kb =

Example ;Calculate (a) pKb, (b) pH and (c) degree of dissociation of 0.1 moldm-3 NH2OH. [Kb = 9.1 mold]

(a) pKb = -log Kb

= -log ( 9.1 = 8.0(b) NH2OH + H2O NH3OH+ (aq) + OH- (aq)

Solution

Initial : 0.1 0 0

Reacted : -X +X +X

At equilibrium : (0.1 – X) +X +X

Kb = 9.1 M = , X is small, (0.1 – X) 0.1

= = 9.1 M2 X = X = 3.02 M [OH-] = 3.02 M

POH = -log [OH-] = -log (3.02 ) = 4.5pH = 14.0 – 4.5 = 9.5

(c) degree of dissociation, = = = = 0.03 %

NH2OH is a very weak base.

!

NOTE :

Both in (a) and (b)- neglected the contribution of the autoionization of water to [H+]and[OH-] because 1.0 M is so small compared with 1.0 M and 0.040M.

Worked example :

Question 1a) Calculate the pH of a 1.8M Ba(OH)2 solution.b) Calculate the pH of HNO3 0.001 moldm-3.

Answer a) Ba(OH)2 is a strong base. It ionise completely to form

1.8 x 10-2 moldm-3 OH- (aq).

pOH = -log[OH-] = -log(1.8x10-2) = 1.74pH + pOH = 14 pH = 14 – pOH = 14 – 1.74 = 12.3

(b)HNO3 is a strong acid.It ionise completely to form 0.001 moldm-3 H+ (aq).pH = -log[H+] = -log(0.001) = 3

Question 2The pH of rainwater collected in a certain region of north Malaysia on a particular day was 4.82. calculate the H+ ion concentration of the rainwater.

Answer pH = 4.82pH = -log[H+]log[H+] = -pHlog10[H+] = -pH[H+] = 10-pH

[H+] = 10-4.82

= 0.000015 = 1.5 x 10-5 mol dm-3

Anti log

logab = cb= ac

Question 3The pH of a certain fruit juice is 3.33. calculate the H+ ion concentration.

Answer pH = 3.33[H+] = ?pH = -log[H+]3.33 = -log[H+]log[H+] = -3.33log10[H+] = -3.33[H+] = 10-3.33

= 0.000468 = 4.7 x 10-4 M

anti log

log a b = cb = ac

Question 4In a NaOH solution, [OH-] is 2.9M. calculate the pH of the solution.

Answer pOH = -log[OH-] = -log(2.9 x 10-4) = 3.5pOH + pH = 14 pH = 14 – pOH = 14 – 3.5 = 10.5

Question 5The OH- ion concentration of a certain ammonia solution is 0.88 M. What is the pH of the solution?

Answer

pOH = -log [OH-] = -log(0.88) = 0.055517 = 0.06pH + pOH = 14 pH = 14 – pOH = 14 – 0.06 = 13.94

[OH-] = 0.88 MpH = ?

CALCULATION OF PH

A. Strong acid and strong bases

Example ; Calculate the pH of a) 1.0 M of HClb) 0.020 M Ba(OH)2 solution

Answer a) HCl is a strong acid, completely ionized in

solution : HCl(aq) H+

(aq) + Cl-(aq)

HCl(aq) H+(aq) + Cl-

(aq) Initial (M) : 1.0 M 0.0 0.0

Change (M) : -1.0 M +1.0 +1.0

Final (M) : 0.0 1.0 1.0

[ H+] = 1.0 M pH = -log (1.0 ) = 3.00

pH = -log [H+]

(b) Ba(OH)2 is a strong base. Each Ba(OH)2 unit produces two OH- ions: Ba(OH)2 (aq) Ba2+

(aq) + 2OH- (aq)

[OH-] = 0.040 M pOH = -log (0.040) = 1.40 pH = 14.00 – 1.40 = 12.60

Initial (M) : 0.020 0.00 0.00

Change (M) : - 0.020 + 0.020 + 2(0.020)

Final (M) : 0.00 0.020 0.040

Exercise!!!

1. What are the concentration of all species present in 1.00 M acetic acid at 25? For HC2H3O2, Ka is 1.8 .

X = 4.2

2. What are the concentration of all species present in a 0.10 M solution of HNO2 at 25. For HNO2 Ka is 4.5 .

X = 6.5

3. Give the definition of pH.

pH is a measure of hydrogenion concentration; a measure of the acidity or alkalinity of asolution

4. a) What are [H+] and [OH-] in a 0.020 M solution

of HCl.b) What are [H+] and [OH-] in a 0.0500 M

solution of NaOH.

5.a) What is the pH of a solution that is 0.050 M

in H+?b) What is the pH of a solution for which [OH-] =

0.030 M ?

6. What is [H+] of a solution with a pH of 10.60 ?

7. The pH of a 0.10 M solution of a weak acid HX is 3.30. what is the ionization constant of HX?(Ka = 2.5 )

8.a) Propanoic acid, HC3H5O2 a weak monoprotic

is 0.72% ionized in 0.25 M solution. What is the ionization constant for this acid?

b) In 0.25 M solution of benzylamine, C7H7NH2, the concentration of OH-

(aq) is 2.4 M.

C7H7NH2 + H2O C7H7NH3+ + OH-

what is the value of Kb for the aqueous ionization of benzylamine?

• Answer :a) Ka = 1.3 b) Kb = 2.3