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ACID AND BASE EQUILIBRIUM PREPARED BY: NURUL HIDAYU BT MOHD YUSOF D20051021999

THEORY Ka =

[A – ] [H3O + ] [HA]

K b = [BH+ ] [OH-] [B]

Ka AND Kb VALUES - Equation - Calculation

DEFINITION THEORY

SELF IONIZATION OF WATER -Animation - Equation

pH = pKa ± 1

ACID

LEWIS Electron pair LEWIS Equation - Equation reactionacceptor reaction

ACID AND BASE EQUILIBRIUM

BUFFER SOLUTION -Types - Calculation

BASE

A substance thatA substance that ARRE HENIUS ARREHENIUS contains hydrogencontain hydroxide -Analogy Analogy - Equation reaction (H) and produces(OH) and produces - Equation H+ in aqueoushyroxide ion OHsolution. in aqueous reaction solution.

BRONSTEDLOWRY Proton donor BR ONSTED- Equation LOWRY - Equation reaction pH AND

w = [H3O+] [OH-] = 1.0 x 10-14 ACID BASE INDICATOR - Steps

DEFINITION

pOH = pKb

Is a solution which resist a change in pH when a small amount of acid or base is added

Electron pair donor

Proton acceptor

pOH SCALE -Equation Calculation + log [salt] [base] pH = pKa + log [salt] [acid]

pH = -log[H+] pOH = -log[OH-] pH + pOH = 14

ARREHENIUS THEORY

ACID

BASE

DEFINITION: A substance that contains hydrogen (H) and produces H+ in aqueous

solution.

HCl

H+

+

Cl-

EXAMPLE OF ARREHENIUS ACIDS HCl, HNO3 , H2SO4

DEFINITION: A substance that contain hydroxide (OH) and produces hyroxide ion OHin aqueous solution.

NaOH

Na+

+ OH-

EXAMPLE OF ARREHENIUS BASE NaOH, Ca(OH)2

BRONSTED-LOWRY THEORY

ACID

BASE

 Proton donor

 Proton acceptor

EQUATION REACTION: proton acceptor NH3 + H2O BASE

ACID

CONJUGATE ACID

NH4+

+

OH-

CONJUGATE BASE

proton donor

LEWIS THEORY

ACID Electron pair acceptor

:O: -O

H

BASE Electron pair donor

+

H

..

.. :O=C=O:

:O: .. II .. O–C .. :

H

H

Ka AND Kb VALUES EQUATION FOR WEAK ACID (HA) HA + H2O

H3O+ + A –

Ka =

[A – ] [H3O + ] [HA]

K a = acid dissociation constant @ acid ionization constant K a smaller, acid become

EQUATION FOR WEAK BASE(B) B + H2O

BH+ + OH-

K b = [BH+ ] [OH-] [B]

K b = base dissociation constant @ Base ionization constant K b bigger, base becomes stronger

Example 1 Calculate the concentration of OH- form and percent of ionization for 0.10M NH3 .(K b = 1.8 x 10 -5 ) Solution

NH3 + H2O

NH4+ +

OHInitial [ ] , M

0.10

0

Equilibrium [ ], M 0.10 – x x K b = [x] [x] = 1.8 x 10 -5 [0.10 – x] Using approximation method: 0.10 – x ≈ 0.10 So, x 2 /0.10 = 1.8 x 10 -5 x = [OH- ]= 1.3 x 10 -3 M

0 x

% ionization = 1.3 x 10 -3 M 100% 0.10M = 1.30 %

X

Dissociation, self-ionization of water, Kw

The animation shows one of the effective collisions between two water molecules to form hydroxide and hydronium ions. The process is reversible. Water normally exists as a mixture of molecules, hydroxide ions and hydronium ions.

 The equation is H2O + H2O <---> H3O+ + OH The equilibrium expression is the normal products over reactants. K= [H3O+] [OH-] / [H2O] [H2O]  The molarity for the water is a constant at any specific temperature. This means the equation can be rewritten as K[H2O] [H2O] = [H3O+] [OH-]  The numerical value for Kw is different at different temperatures. At 25oC Kw = 1.0 x 10-14 Kw = K[H2O] [H2O] Kw = [H3O+] [OH-] = 1.0 x 10-14

pH AND pOH SCALE pOH = -log[OH-]

pH = -log[H+]

pH + pOH = 14

The relationship In a water solution the ion product for water is: [H+] [OH-] = Kw = 1 X 10-14 Take the -log of both sides of the equation

- log [H+] +(- log [OH- ]) = - log [1 X 10-14 ] pH + pOH = 14

Example 2 Determine the pH and the percent of ionization 0.010M NH 3 solution. K b = 1.8 X 10 -5 NH3 + H2O Initial [ ], M Equilibrium [ ], M

0.010 (0.010 – x)

K b = [x] [x]

NH4+ + OH0 x

0 x

= 1.8 X 10 -5

[0.010 – x] Since K is small, use approximation method: x = [OH-] = 4.2 X 10 -4 M % ionization = (4.2 X 10 -4 M / 0.010 M ) X 100% = 4.20% From the equation, pOH = - log [OH-] = -log (4.2 X 10 - 4 ) = 3.38 pH = 14.00 – 3.38 = 10.62

BUFFER SOLUTION Is a solution which resist a change in pH when a small amount of acid or base is added to the solution. 2 types Acidic type – to maintain pH < 7 Basic type – to maintain pH > 7

ACIDIC TYPE OF BUFFER SOLUTION • Buffer solution contains a solution of weak acid and the salt of the weak acid. CH3COOH + H2O CH3COO- + H3O+ CH3COONa

CH3COO- + Na+

• When a small amount of OH- is added to the buffer solution, it will be neutralized by CH3COOH. CH3COOH + OH-

CH3COO-

• If a small amount of H3O+ neutralized by CH3COO- . CH3COO- + H3O+

+

H2O

is added, it will be CH3COOH + H2O

• Consequently, there is no change in pH.

BASIC TYPE OF BUFFER SOLUTION • Buffer solution contains a solution of weak base and the salt of the weak base. NH3 + H2O NH4+ + OHNH4Cl

NH4+ + Cl-

• When a small amount of OH- is added to the buffer solution, it will be neutralized by NH4+ . NH4+ + OH-

NH3 + H2O

• If a small amount of H3O+ is added, it will be neutralized by NH3 . NH3 + H3O+

NH4+

+ H2O

EQUATION FOR BUFFER SOLUTION Henderson- Haselbalch Equation

pH = pKa + log [salt] [acid]

pOH = pKb + log [salt] [base]

Example 3 1.00 L buffer was prepared by mixing 0.100 mole HC2H3O2 and 0.100 mole NaC2H3O2 (K a = 1.8 x 10 -5 ) 1)Determine the pH of the buffer solution.

pH = pKa + log [salt] [acid] = - log 1.8 x 10 -5 + log (0.100/0.100) = 4.74

2) Determine the pH of the buffer solution when 0.020 mole NaOH is added to it.(assume there is no change in volume when acid is added) BASE ADDED WILL REACT HC2H3O2 HC2H3O2 + Initial [ ] ,M Equilibrium[ ], M

OH-

0.10 0.020 (0.10 - 0.020) 0

C2H3O2- + H2O 0.100 (0.100 + 0.02)

pH = - log 1.8 x 10 -5 + log (0.120/0.080) = 4.92

3) Determine the pH of the buffer solution when 0.020 mole HCl is added to it.(assume there is no change in volume when acid is added) ACID ADDED WILL REACT WITH C2H3O2C2H3O2- + H3O+ Initial [ ] ,M 0.10 0.020 Equilibrium [ ], M (0.10 - 0.020) 0

HC2H3O3 + H2O 0.100 (0.100 + 0.02)

pH = - log 1.8 x 10 -5 + log (0.080/0.120) = 4.57

ACID-BASE INDICATOR • Used to determine the equivalence point of titration. • Because the indicator has a different colour in acid to that of base solution. • Indicator changes colour is called the end point of titration. • Normally, the equivalence is very closed to the end point.

1) Consider the phenophthalein indicator (Hln) which has a pink colour in base solution and colourless in acidic. Hln + H2O In- + H3O+ colourless

pink

Ka = [In-] [H3O+] / [Hln] pH = pKa + log [In-] / [Hln] 2) The point where the colour changes: [In-] = [Hln], so [In-] / [Hln] = 1 pH = pKa + log 1 pH = pKa the colour of the end point occurs at the pKa indicator.

value of

3) For phenolphthalein Ka = 3 X 10 -10 It change colour at pH = pKa = 9.5 However our eyes can only detect the colour change when, [In-] / [Hln] = 10 The range of colour change observe by our eyes will be pH = pKa ± log10 pH = pKa ± 1

4) In the case of phenolphthalein the colour can only detect by the eyes in the pH of 9.5 ± 1 That is from pH 8.50 to pH 10.50 5) Therefore the equivalence point of titration should be in the range if phenolphthalein is choosen to be the indicator. The exact colour change at pH = 9.5 cannot be detected by the eyes.

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