chapter 13 chemistry foundation 2014

8/12/2019 Chapter 13 chemistry foundation 2014

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CHAPTER 13 16)ACID BASE

EQUILIBRIA


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Contents

Definition of Acid and Base : Arrhenius

Brnsted-Lowry , conjugate acid-base pairs

Lewis

Strength of Acid and Base : Strong acid weak acid

Strong base weak base

Ionization of Acid and Base

Concepts of pH, pOH, pKa, pKb

Dissociation Constant : Ka, Kb, Kw


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Learning Outcomes

Able to differentiate and calculate acid

dissociation constants for weak and strongacids (applicable to bases)

Calculate pH and pOH


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Whenever an acid dissociates (ionizes) in water,

solvent molecules participate in the reaction

HA (g or aq) + H2O(l) A-(aq) + H3O

+(aq)

The H3O+is

called

hydronium ion

The terms hydrogen

ion = proton = H+ are

used interchangeably

Arrhenius Definition


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Strong acids: dissociate/ionize completely intoions in water.

Eg:HNO3(aq)+ H2O(l) H3O+(aq) + NO3

-(aq)

Weak acids: dissociate/ionize very slightly intoions in water.

Eg: HCN(aq) + H2O (l) H3O+(aq) + CN-(aq)

This classification correlates with classification ofelectrolytes: strong electrolytes dissociate/ionize completely, and

weak electrolytes dissociate/ionize partially.

Arrhenius Definition


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Brnsted-Lowry Acid-Base Definition

Acid

pro ton donor.Any species that donates an H+

such as HCl, HNO3, H3PO4

All Arrhenius acids areBrnsted-

Lowry acids

Base

proto n acceptor.

Any species that accepts an H+,

must contain lone pair of electrons to

bind the H+, such as NH3, CO32-, F- andOH-

Brnsted-Lowry bases are not

Arrhenius bases, but all Arrhenius bases

contain Brnsted-Lowry base OH-. -


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HCl (g) + H2O (l) H3O+(aq) + Cl-(aq)

HCl acts as a Brnsted-Lowry acid ( it donates a

proton to H2O)

H2O acts as a Brnsted-Lowry base ( it accepts a

proton from HCl)



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NH3(aq) + H2O (l) NH4+ (aq) +OH-

(aq) H2O acts as an acid, it donates the H

+

NH3

acts as a base, it accepts H+.

Thus, H2O is ampotheric: it acts as a base in

one case and as an acid in the other.



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NH3(aq) + H2O (l) NH4+ (aq) +OH-

(aq)



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Example

Determine acid and base in the following reactions:

1. HCO3-(aq) + HF (aq) H2CO3(aq) + F

-(aq)

2. HCO3-(aq) + OH-(aq) CO3

2-(aq) + H2O (aq)

3. SO32-(aq) + NH4

+(aq) HSO3-(aq) + NH3(aq)

4. HSO3-

(aq) + NH3(aq)

SO32-

(aq) + NH4+

(aq)


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Answer

Acid Base

1. HF (aq) HCO3-

(aq)2. HCO3

-(aq) OH-(aq)

3. NH4+(aq) SO3

2-(aq)

4. HSO3-(aq) NH3(aq)


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Conjugate Acid-Base Pairs

Consider the reaction of acid HA:

HA (aq) + H2O(l)

A

-

(aq) + H3O

+

(aq) Forward reaction: HA is the acid, H2O is the base

Reverse reaction: H3O+is the acid, A-is the base

HA and A-: differ in the presence or absence of aproton (base)

a conjugate acid-base pair.


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Add H+

In any acid-base reaction:

HNO2(aq)+ H2O (l) NO2- (aq) + H3O

+ (aq)

Acid Base Conjugate

base

Conjugate

acid

Remove H+



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Acids such as HCl , HNO3 , or H2SO4ionize in water to form a

hydrated proton and a species called the con jugate baseof

the acid;HCl (aq) H+(aq) + Cl-(aq)

AcidConjugate base

of HCl

Conjugate base of HCl results from the loss of one proton.



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Bases such as NH3react with water to form a hydrated

hydroxide ion and a species called the conjugate acidof

base.NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Base Conjugate acid

of NH3

Conjugate acid of NH3results from the gain of one proton.



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In NH3and H2O

NH3

(aq) + H2

O (l) NH4

+(aq) + OH-(aq)Base Acid Conjugate

acid

Conjugate

Base

Add H+

Remove H+

NH3 NH4+

base Conjugate acid

Add H+H2O OH

-

Acid Conjugate Base

Remove H+Note:



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a) What is the conjugate base of the following acids:HClO4; HCO3

-

- remove one proton ( H+) from the formula.

ClO4-; CO32-

b) What is the conjugate acid of the following bases:

CN-; H2O ; HCO3-

- add one proton to the formulaHCN ; H3O

+; H2CO3

Note: Hydrogen Carbonate ion, HCO3-is amphoteric: a

substance that can act as an acid or as a base.

Example


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Example

Determine the conjugate acid-base pairs in the

following:

1. HCO3-(aq) + HF (aq)H2CO3(aq) + F

-(aq)

2. HCO3-(aq) + OH-(aq)CO3

2-(aq) + H2O (aq)

3. SO32-(aq) + NH4+(aq)HSO3-(aq) + NH3(aq)

4. HSO3-(aq) + NH3(aq)SO3

2-(aq) + NH4+(aq)


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Answer

Acid Base Conj. acid Conj. Base

HF (aq) HCO3-(aq) H2CO3(aq) F-(aq)

HCO3-(aq) OH-(aq) H2O (l) CO3

2-(aq)

NH4+(aq) SO3

2-(aq) HSO3-(aq) NH3(aq)

HSO3-

(aq) NH3(aq) NH4+

(aq) SO32-

(aq)


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Other example: HSO3-

Acid

HSO3-(aq) + H2O(l) SO3

2-(aq) + H3O+

Remove H+

Add H+

Base

HSO3-(aq) + H2O(l) H2SO3(aq) + OH

- (aq)

Add H+

Remove H+


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Conjugate pairs in acid-base reactions


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Lewis Acid-Base Definition

OH-, H2O , an amine - are electron-pair donors

A base in the Brnsted-Lowry ( a proton acceptor)also base in the Lewis (an electron pair donor )

In the Lewis theory: a base can donate its electronpair to species other than H+

Acid Electron pair acceptor

Base Electron pair donor


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Lewis Acids with Electron-Deficient Atoms:

Eg: Reaction between NH3and BF3.

BF3has a vacant orbital in its valence shell- acts as

an electron pair acceptor (a Lewis acid) toward

NH3.NH3donates the electron pair.


N

H

H

H

B

F

F

F N

H

H

H B

F

F

F

Lewis base Lewis acid


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Lewis acid - molecules have an incomplete octet ofelectrons.

Eg. Cations Fe3+interacts with cyanide ions;

Fe3+ion has vacant orbitals - accept the electronpairs donated by the CN-ions.


Fe3+

+ C N6 C NFe 63-


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Note: Lewis acids and bases do not need to containprotons.

Therefore, the Lewis definition is the most generaldefinition of acids and bases.

Lewis acids generally have an incomplete octet (eg.BF3).

Transition metal ions are generally Lewis acids.

Lewis acids must have a vacant orbital (into which theelectron pairs can be donated).



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Metal Cation as Lewis Acid


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Definitions Comparison

Theory

Arrhenius

Brnsted-Lowry

Lewis

Acid Base

Forms H3O+ions in

water

Forms OH-ions in

water

Proton donorHCl + H2O H3O

++ Cl-Proton acceptor

NH3(aq) + H2O

NH4 + OH-

Electron pair acceptor

B

F

F

F

Electron pair donor

N

H

H

H


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Strength of Acid and Base:Objectives

Strength of acid-base

Strong acid - strong base / weak acid - weakbase

Strength in conjugate acid base determinethe direction of acid-base reaction

Water auto-ionization

Concept of pH


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Variation in Acid Strength

Strong acids: Ionize completely into ions in water.

Eg: HNO3(aq)+ H2O(l) H3O+(aq) + NO3

-(aq)

Weak acids: ionize very slightly into ions in water.

Eg: HCN(aq) + H2O (l) H3O+(aq) + CN-(aq)

This classification correlates with classification ofelectrolytes:

strong electrolytes dissociate/ionize completely

weak electrolytes dissociate/ionize partially.


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Strong Acids

Hydrohalic acidsHCl,

HBr,

HI

Oxoacids

(number of O exceeds no. of

ionisable H atoms by two or more)

HNO3H2SO4

HClO4HClO3

HCl (aq) + H2O (l) H3O+(aq) + Cl-(aq)

HNO3(aq) + H2O (l) H3O+(aq) + NO3

-(aq)

Strong acid will fully dissociate/ionize in water


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Weak acids

Hydrohalic acids HF

Acids in which H is not bonded

to O or to halogen

HCN

H2S

Oxoacids

(in which no. of O equals or exceed

by one the no. of ionizable H)

HClO,

HNO2

H3PO4

Organic acids

(general formula RCOOH)

CH3COOH

C6H5COOH


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Example

HF (aq)+ H2O (l)H3O+(aq)+ F

-(aq)

CH3COOH (aq)+ H2O (l)H3O+

(aq)+ CH3COO-

(aq) HNO2 (aq)+ H2O (l)H3O

+(aq)+ NO2-(aq)

Weak acid will partially ionize in water


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Strong Base

Soluble

compoundscontaining O2-

or OH-

The cations are

usually those ofthe most active

metalsM2O or MOH;

Where M= Group IA

metal

(Li, Na, K, Rb, Cs)

MO or M(OH)2;

Where M= Group IIA

metal

(Be, Mg, Ca, Sr, Ba)

Ionic hydrides

and nitrides

H-(aq) + H2O (l) H2(g) + OH-(aq)

N3-(aq) + 3 H2O (l) NH3(aq) + 3OH-(aq)


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Weak Bases

Compounds with

an electron-rich

nitrogen are weak

bases(none are

Arrhenius bases).

The common

structural

feature is an

N atom thathas a lone

pair

Ammonia: NH3

Amines

such asCH3CH2: NH2C5H5N:


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Relative Strengths of Acid-Base

Some acids are better proton donors than others.

The stronger the acid, the weaker its conjugate base.

Some bases are better proton acceptors.

The stronger the base, the weaker its conjugate acid

The strength of an acid tells the strength of its conjugatebase.

Strong acids: completely transfer their protons 100%dissociation Their conjugate bases have a negligibletendency to be protonated.

(E.g: HCl + H2O H3O++ Cl-)


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Consider the proton transfer in this:

HA(aq) + H2O (l) H

3O+(aq) + A-(aq)

If HA is a stronger acid than H3O+;

HA will transfer its proton to H2O ; equilibrium lies to

the right

If H3O+is a stronger acid than HA (if HA is a weak

acid);

equilibrium lies to the left

In every acid-base reaction, the position of theequilibrium favors transfer of the proton from thestronger acid to the stronger base to form the

weaker acid and the weaker base.


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Example: CN-reacts with water as follows:

CN- (aq) + H2O(l) HCN (aq) + OH-(aq)

A Cl-ion does not react with water to form HCl.

Which acid is stronger, HCl (aq) or HCN (aq)? Why?

HCl and HCN are the conjugated acids of the bases Cl-and

CN-.The information provided tells us that Cl- is a weaker

base than CN-.CN-accepts a proton from H2O to form HCN.

Principle: The weaker a base, the stronger its conjugated

acid, can conclude that HCl is a stronger acid than HCN


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Example

Predict the product of the following acid-base reactionsand also predict whether the equilibrium lies to the left

or to the right of the equation:1. Cl-(aq) + H3O

+(aq)

2. H2PO4-(aq) + H2O (l)

3. HS-

(aq) + HC2H3O2(aq)

4. HCO3- (aq) + OH (aq)

5. NH4+ (aq) + H2PO4

-(aq)

HCl (aq) + H2O(aq) Lies to the left

HPO42-(aq) +H3O

+(aq)Lies to the

leftH

2S (aq) + C

2H

3O

2

-

Lies to the

right

CO32-(aq) + H2O (l) Lies to the

rightNH3(aq) + H3PO4(aq)

Lies to the left


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Autoionization of water

Water ionises into H+and OH-(v. small extent).This process is called the autoionizationof water.


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The Ion-Product Constant for Water, Kw

We can write an equilibrium constant expression for the

autoionization of water: 2H2O(l) H3O+(aq) + OH-

(aq)

Because H2O(l) is a pure liquid, the expression can be

simplified:

22

eq

OH

OHOHK

3

weq KOHOHKOH 32

2


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The Ion-Product Constant for Water, Kw

Kw is called the ion-product constant.

At 25C the ion-product of water is always:

In neutral solution: [H3O+] = [OH-]

In acidic solution : [H3O

+

] exceeds [OH-

] In basic solution : [OH-] exceeds [H3O

+]

Note: The ion -product constant Kw is unaffected bywhether a solution is acidic or basic.

OHOHKw 3

14101


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Calculate the concentration of H3O+(aq) in a

solution in which [OH-] is 2.0 x 10-9M at25C.

M10x5.010x2.010x1.0

OH

10x1.0H

10x1.0OHH

69

1414

3

14

3

O

O

Example


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The pH scale

Express H+in terms of pH.

For [H+] = 1.0 x 10-3M.

pH = -log (1.0 x 10-3) = -(- 3.00 ) = 3.00

pH = -log [H+] = -log [H3O+]

Basic

[H+

] (M)

1.0x10-7[OH-] (M)

>1.0x10-7

=1.0x10-7


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Acidic Neutral Basic

[H3O+]

1 10-7

[OH-]

1 10-7

Neutral

[H3O+]1

[OH-]1 10-14

Acidic

1 10-1

1 10-2

1 10-31 10-4

1 10-5

1 10-6

1 10-13

1 10-12

1 10-111 10-10

1 10-9

1 10-8

[OH-][H3O+]1 10-6

1 10-5

1 10-4

1 10-3

1 10-13

1 10-8

1 10-9

1 10-10

1 10-111 10-12

1 10-14

Basic

1 10-2

1 10-1

1


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Example

Given [OH-] = 2.0 x 10-3, calculate [H+] (or [H3O+])

and pH at 25C.

pH = -log ( 5.0 X 10-12 )

= 11.3

MOH

KH W

12

3

14

100.5100.2

100.1


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The concentration of OH-can be expressed as pOH

Example:

A solution has a pH of 5.6 . Calculate the hydrogenion concentration.

pH = -log [H+], log [H+] = -5.6

[H+

] = 2.5 10-6

M

Other p scale

pOH = -log [OH-]

pH + pOH = pKw= -log Kw= 14.00


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Calculate [H+] for a solution with pOH of 4.75

pOH is defined aslog [OH-

] The pH and the pOH are related: pH + pOH = 14.00

pH = 14.00 - pOH = 14 - 4.75 = 9.25

pH = 9.25

log [H+] = -pH = -9.25

[H+] = 5.6 x 10-10 M.

Example


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Acid-base indicator

pH paper

pH meter

Methods for measuring pH


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Strong Acids

Strong acids are strong electrolytes and they ionizecompletely in solution.

HNO3(aq) + H2O (l) H3O+(aq) + NO3-(aq)

Example

What is the pH of a 0.04 M solution HClO4?

HClO4 is completely ionized: [H+] = 0.04 M

Hence, pH= -log (0.040) = 1.40


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Strong Bases

Most ionic hydroxides are strong bases (eg. NaOH,KOH and Ca(OH)2.)

Strong bases are strong electrolytes and dissociatecompletely in solution:

NaOH (aq)

Na

+

(aq) + OH

-

(aq)

The pOH of a strong base is given by the initialmolarity of the base.


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Example

What is the pH of a 0.011 M solution of Ca(OH)2?

Ca(OH)2 is a strong base.

Ca(OH)2 Ca2+ + 2OH-

0.011M 0.011M 2 0.011M

pOH = -log (0.022) = 1.66pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.66 = 12.34


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Acid Dissociation Constant, Ka

HA (g or aq) + H2O(l) A-(aq) + H3O

+(aq)

Stronger acidlarger Ka Weaker acidSmaller Kalower % HA ionizes

HAorHA

AH

KAOH

K eqeq3

HA

AOH

K 3

a


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Ka values for weak acids

Acid Ka % HA dissociated

1 M HClO2 1.12 x10-2 10 %

1 M CH3COOH 1.8 x10-5 0.42 %

1 M HCN 6.2 x10-10 0.0025 %


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Solving Problems Involving Weak-AcidEquilibria

1. Given equilibrium concentrations, find Ka.

2. Given Ka and other info, find other equilibriumconcentration.

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

Ionization is incomplete and some significantamount of undissociated acid remains at equilibrium.

[HA]

][A][HKa


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Problem-Solving Approach : ProblemsInvolving Weak-Acid Equilibria

1. Write the balanced equation and Kaexpression

2. Define x as unknown concentration that changesduring the reaction.

3. Construct a reaction table that incorporates theunknown.

4. Make assumptions that simplify the calculations.

5. Substitute the values into the Ka expression and solveforx

6. Check that the assumptions are justified.


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

Iffis the formal concentration of acid ,xbe theequilibrium concentration of H3O

+. Then,

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

f - x x x

xf

x

[HA]

][A]O[HK

2

3a



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The equation can be solved using quadratic eqn.

If f>>x, then

x2

Kafx=

Ifxis 5% of the initial concentration, we can use

the assumption.

Ifxis 5% of the initial concentration, it maybe

best to solve the quadratic equation.

fKa



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Percent ionisation is another method to assess acid

strength.

% ionisation

The higher the percent ionisation, the stronger theacid.

x100[HA]

]O[H

initial

eqm3



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A student prepared 0.10 M solution of formic acid HCHO2.

The pH= 2.38 at 25C.

1. Calculate Ka

2. Percentage of ionisation.

Example

MH

H

HpH

HCHO

CHOHK

a

3

2

2

102.4

38.2log

38.2log


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Calculate Ka

HCHO2(aq) H+(aq) + CHO2

-(aq)

Example

MHHHpH

HCHO

CHOHKa

3

2

2

102.4,38.2log,38.2log


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AnswerHCHO2(aq) H

+

(aq) + CHO2-

(aq)

Initial

Change

Eqm

(0.10 4.2 10-3)M 0.10 M

-4.2 10-3M

0.10 M 0 0

+4.2 10-3M +4.2 10-3M

4.2 10-3M 4.2 10-3M(0.10 - 4.2 10-3)M

4

33

108.110.0

102.4102.4

a

K


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The percentionisation of a weakacid decreases as its

concentrationincreases. 1.0

2.0

3.0

4.0

5.0

0.05 0.10 0.15

Per

centionised

Acid concentration (M)

%2.4%10010.0

102.4

%100%

3

2

initial

mequilibriu

HCHO

Hionisation


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Calculate the percentage of HF (Ka= 6.8 x 10-4)

molecules ionised in:

a. 0.10M HF solution

b. 0.010 M HF solution

Example


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Acids have more than one ionizable H atom

Example: Sulfurous acid, H2SO3.

H2SO3 (aq) H+(aq) + HSO3

-(aq) Ka1= 1.7 x 10-2

HSO3-

(aq) H+

(aq) + SO32-

(aq) Ka2= 6.4 x 10-8

Ka2 is much smaller thanKa1; easier to remove the firstproton from polyprotic acid than the second

Ka1is much larger thanKa2; can estimate the pH byconsidering onlyKa1.

Polyprotic Acid


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Example

CO2dissolved in water at 25C and 0.1 atm to form H2CO3 withconc. 0.0037 M.

What is the pH ?

CO2(aq) + H2O(l) H2CO3(aq)

[H2CO3] = 0.0037 M

H2CO3is a polyprotic acid:

H2CO3(aq) H+(aq) + HCO3

-(aq) ka1= 4.3 10-7

HCO3-(aq) H+ (aq) + CO3

2-(aq) ka2= 4.710-11

ANSWER = 4.40


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Weak Bases

There is an equilibrium between the base and theresulting ions

Weak base + H2O(l)conjugate acid + OH-(aq)Eg: NH3(aq) + H2O NH4

+(aq) + OH-(aq)

The base-dissociation constant , Kb, is defined as

The larger the Kb the stronger the base

][NH]][OH[NHK

3

4b


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Example

Calculate the concentration of OH- in a 0.15 Msolution of NH3. (Kb= 1.8 x 10

-5)

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

0.15 -x x x

x= [NH4+] =[OH-] = 1.6 x10-3 M

5

3

4b 10x1.8x0.15

(x)(x)

][NH

]][OH[NHK


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Example

A solution is made by adding Sodium Hypochlorite,NaClO, to water (2.0 L soln). The pH of solution is10.50. How many moles of NaClO were added?

Note:

NaClO is an ionic compound, strong electrolyte.

NaClO

Na+

+ ClO-

ClO-(aq) + H2OHClO(aq) + OH-(aq)

Kb= 3.3 x 10-7.


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Unit in molarity

pOH + pH = 14

pOH = - log [OH-]

pH = - log [ H+]

Given: pH = 10.50 ; base. Whats the [OH-]

Knowing the [OH-] then we know the Molarity of ClO-.

Molarity of NaClO mole ( in the same volume: 2.0 L )

NaClO Na+ + ClO-

1 mole 1 mole 1 mole

Answer


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ClO-(aq) + H2O(l) HClO(aq) + OH-(aq)

Let say we start with 1.0 M NaClO : conc. of ClO-is 1.0 M. Then,


1 - x x x

x; concentration of OH-

* If we dont know the ClO-

. Let say ClO-

: zBut we know the [OH-] x . Then,

z - x x x

Answer


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Given pH = 10.50

from : pH + pOH = 14.0pOH = 14.0 - 10.50 = 3.50

pOH = - log [OH-]

[OH-] = 10 -3.50

= 3.2 x 10-4M

x = 3.2 x 10-4M

Answer


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ClO-(aq) H2O(l) HClO(aq) OH-(aq)

Initial z - 0 0

Change -3.2 10-4 - 3.2 10-4 3.2 10-4

final z - 3.2 10-4 - 3.2 10-4 3.2 10-4

Tabulate in equilibrium table:


M0.31z

10x3.310x3.2z

)10x(3.2

][ClO][HClO][OHK

7

4

24

b


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Z = [ClO-] = 0.31 M

NaClO = 0.31 M

Mol of NaClO = 0.31 (mol/ L) x (2 L)= 0.62 mol

2L

Mol0.31

Volume(L)

MolMMolarity,

Answer


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Relationship between Kaand Kb

We need to quantify the relationship between

strength of acid and conjugate base.

NH4+(aq)NH3(aq) + H

+(aq)

NH3(aq) +H2O(l)NH4+(aq) + OH-(aq)

][NH

]][OH[NHK

][NH]][H[NHK

3

4b

4

3a


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NH4+(aq) NH3(aq) + H

+(aq)

NH3(aq) +H2O(l) NH4+(aq) + OH-(aq)

H2O(l) H+(aq) + OH-(aq)

When two reactions are added to give a third, the equilibrium

constant for the third reaction is the product of the equilibrium

constant for the two added reactions:

reaction 1+ reaction 2 = reaction 3

K1x K2 = K3

Relationship between Kaand Kb


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For a conjugate acid -base pair

wba

w

3

4

4

3ba

KKKK]][OH[H

][NH

]][OH[NH

][NH

]][H[NHKK

Therefore , the larger the Kathe smaller the Kb. That is, the

stronger the acid, the weaker the conjugate base.

Taking negative logarithms:

pKa = -log (Ka) and pKb= -log(Kb)

then pKa+ pKb= pKw


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END of CHAPTER 3

chapter 13 chemistry foundation 2014

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