IN THE NAME OF ALLAH THE MOST GRACIOUS, THE MOST MERCIFUL
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IN THE NAME OF ALLAH THE MOST GRACIOUS, THE MOST MERCIFUL
CHEM 122. LEVEL-2 LECTURE# 2CHAPTER 9 – ACIDS & BASES
Chemistry by Timberlake p.266
RCDP Presented by:Presented by:
Department OfDepartment OfChemistryChemistry
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Ionization of Water P-277
Occasionally, in water, a H+ is transferred between H2O molecules
. . . . . . . .H:O: + :O:H H:O:H + + :O:H-
. . . . . . . . H H H
water molecules hydronium hydroxide ion (+) ion (-)
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Pure Water is Neutral
Pure water contains small, but equal amounts of ions: H3O+ and OH-, these are
called ion product of water.
H2O + H2O H3O+ + OH-
hydronium hydroxide ion ion
1 x 10-7 M 1 x 10-7 MH3O+ OH-
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Water itself is a very weak electrolyte.
H2O + H2O ⇌ H3O+aq + OH-
aq
( auto ionization )
Kc = [H3O+] [OH-] [H2O][H2O]
This is called equilibrium expression
Dissociation of water and Kw:
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Ion Product of Water Kw
[ ] means Molar concentration
Kw = [ H3O+ ] [ OH- ]
= [ 1 x 10-7 ][ 1 x 10-7 ]
= 1 x 10-14
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H2O ⇌ H+ + OH- ( here H+ is called H3O+)
Kw = [H3O+] [OH-] [H3O+] [OH-] = Mass action expression, is called ion product for water Kw is called Ion product constant for water
orIonization constant orDissociation constant of water
Kw = [H+] [OH-] (This equation is used to calculate molar concentrations of H+ and OH-
ions in pure water)
At equilibrium [H+] = [OH-] If; [H2O][H2O] =1 mol x 1 mol
Kw = 1.0 x 10 -14 at 250C
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Acids
Increase H+
HCl (l) + H2O (l) H3O+ (aq) + Cl-
(aq)
More [H3O+] than water: [H3O]> 1 x 10-
7M
As H3O+ increases, OH- decreases
[H3O+] > [OH-]H3O+
OH-
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Bases
Increase the hydroxide ions (OH-)
H2O
NaOH (s) Na+(aq) + OH- (aq)
More [OH-] than water, [OH-] > 1 x 10-7M
When OH- increases, H3O+ decreases
[OH] > [H3O+]H3O+OH-
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Using Kw
Q:The [OH- ] of a solution is 1.0 x 10- 3 M. What is the [H3O+]?
Kw = [H3O+ ] [OH- ] = 1.0 x 10-14
[H3O+] = 1.0 x 10-14
[OH-]
[H3O+] = 1.0 x 10-14 = 1.0 x 10-11 M
1.0 x 10- 3
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PH [P-279]
Indicates the acidity [H3O+] of the solution
pH = - log [H3O+]
From the French pouvoir hydrogene (“hydrogen power” or power of hydrogen)
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In the expression for [H3O+]
1 x 10-exponent
the exponent = pH
[H3O+] = 1 x 10-pH M
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-To specify the hydrogen ion concentration in a solution, we speak about pH. This is defined as, pH = log 1/ [H+] = - log [H+]
-In a solution, if the hydrogen ion concentration is 10-3M . The pH of this solution is, pH = - log ( 10-3 ) = - ( -3 )
pH = 3 -If the hydrogen ion concentration is 10-8M, the pH will be “ 8“
-We already know that Kw = [H+] [OH-]
thus, pKw = pH + pOH
Since Kw = 1 x 10-14 , so pKw = 14.0
pH + pOH = 14.0-In the neutral solution [H+] = [OH-] = 1.0 x 10-7
therefore pH = pOH = 7.0 pH for neutral solution = 7.0
The Concept of pH p279
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In acidic solution hydrogen ion concentration is greater than 1.0 x 10-7 M ( 1.0 x 10-1 to 1.0 x 10-6 M ) , it means pH is less than 7.0 i.e. from 1.0 to 6.0 In basic solution the hydrogen ion concentration is less than 1.0 x 10-7 M ( 1.0 x 10-8 to 1.0 x 10-10 M ), it means pH is more than 7.0 i.e. from 8.0 to 10.0
-The relationship of H+, OH-, pH, pOH at 250C, Kw=1.0x10-14
[H+] [OH-] pH pOH
Acidic solution > 1 x 10-7 < 1 x 10-7 < 7.0 > 7.0Neutral solution = 1 x 10-7 = 1 x 10-7 = 7.0 = 7.0Basic solution < 1 x 10-7 > 1 x 10-7 > 7.0 < 7.0
Substances having pH less than 7.0 are acidic , Lemon juice( citric acid ) Vinegar ( acetic acid ). Acids are sour in taste
Substances having pH more than 7.0 are basic , Milk of magnesia ( a suspension of Mg (OH)2 ). Bases are bitter in taste.
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pH Range or scale P-283
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Neutral
[H+]>[OH-] [H+] = [OH-] [OH-]>[H+]
Acidic Basic
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Some [H3O+] and pH
[H3O+] pH
1 x 10-5 M 5
1 x 10-9 M 9
1 x 10-11 M 11
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pH of Some Common Acids
o gastric juice 2.0
o lemon juice 2.3
o vinegar 2.8
o orange juice 3.5
o coffee 5.0
o milk 6.6
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pH of Some Common Bases
blood 7.4
tears 7.4
seawater 8.4
milk of magnesia 10.6
household ammonia 11.0
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Q: The pH of a soap is 10. What is the [H3O+] of the soap solution?
Answer:
[H3O+] = 1 x 10-pH M
= 1 x 10-10 M
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An indicator is a weak organic acid or base, which on dissociation undergoes a change in color and this change in color depends on the pH of the solution.
Acid form ⇌ H+ + Base form1st. Color 2nd. Color
Litmus paper are of two types, (1).Pink (2). Blue(1). To test whether a solution is acidic, blue litmus paper is used.
In acidic solution blue litmus paper turns pink.(2). To test whether a solution is basic, pink litmus paper is used.
In basic solution pink litmus paper turns blue. For any accurate measurement of pH, a pH meter is used
Measuring pH
and acid –base indicators :
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Acid – Base indicators Indicators Color in
acidic solution
Color in basic solution
pK pH range
Thymol blue acid red yellow 1.7 1.2 to 2.8
Methyl orange red orange 3.5 3.1 to 4.5
Bromophenol blue yellow purple 4.1 3.0 to 4.6
P – Nitro phenol Colorless yellow 7.1 5.0 to 7.0
Phenolphthalin Colorless pink 9.3 8.0 to 9.8
Trinitrobenzene colorless brown 13.5 13.5 to 14.0
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Acid – Base indicators
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HA H+ + A- , Ka= [H+][A-] , pKa = - log Ka
pKa for acetic acid = pKa = - log Ka
= - log ( 1.8 x 10-5 ) = 4.74
BOH B+ + OH-, Kb=[B+][OH-], pKb = - log Kb
pKb for pyridine = pKb = - log kb = - log ( 1.7 x 10-9)= 8.77
Equilibria Involving Weak Molecular Acids and Bases
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The smaller the value of pka or pKb ,the stronger is the acid or base.
For example
Acetic acid , CH3-COOH , pka = 4.7
Chloroacetic acid, ClCH2-COOH, pKa = 2.85
Dichloroacetic acid, Cl2CH-COOH ,pKa = 1.30
The order of increasing acidity Acetic acid < Chloro acetic acid < Dichloro acetic acid
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Equilibria Involving Weak Molecular Acids and Bases Percentage dissociation:
For weak acid: HA H+ + A-
For weak base: BOH OH- + B+
For 0.1 M acetic acid, [H+] = [A-]=1.3 x10 -3:
Percentage dissociation = [A-] x 100
[HA]
So, percentage dissociation of acetic acid =
1.3 x10 -3 x 100 = 1.3 %
0.1
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Equilibria Involving Weak Molecular Acids and Bases
Dissociation of weak acids or bases is known by its percentage dissociation which is define as;
eableinterchang are ionization andon dissociati termsThe
.ionizationpercent called also ison dissociatiPercent
% 1.3 100 M 0.10
M 10 1.3 on dissociatiPercent Therefore;
M 0.1ion concentrat original theis available L
mol The M. 10 1.3 toequal iswhich
formed, H of L
mol the toequals ddissociate
L
mol acid aceticFor
100
available baseor acid of L
mol
ddissociate baseor acid of L
mol
on dissociatiPercent
3-
3-
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Practicing questions
1) The pH of a 0.009300 molar solution of unknown monoprotic acid was measured and found to be 2.990. Calculate the percentage dissociation of this acid.
2) A weak acid, HA, has a Ka of 1 x 10-5. If 0.100 mole of this acid is dissolved in one liter of water, the percentage of acid dissociated at equilibrium is closest to:
a) 0.100%b) 1.00%c) 99.0%d) 99.9%e) 100%
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What is the pH of a solution that contains 1.00 x 10-4 M hydronium ion?
Calculate the [H3O+] of a solution that has a pH of 3.70. Find the pH and pOH of a solution that contains 0.00350 M H3O+
ion. Calculate the pH of a solution of NaOH if [OH]= 5x10-5 Calculate the pH for a solution of HCl contains 0.2 M H3O+?
What is the pH of 0.2 M NaoH? IF Kb = 1x10-11
pH= 0.699
pH= 8.15
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Calculate [OH-] of a solution whose [H+] = 0.00375 M Calculate the [H+] of a solution whose pH = 11.93 Calculate the pOH of a solution whose [OH-] = 7.24 x 10-3 M Calculate the pH of a solution whose pOH = 12.00 Calculate the [H+] of a solution whose pOH = 8.00 The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is
1.3x10-5. Calculate the hydrogen ion concentration, [H+], in a 0.20 molar solution
of propanoic acid. Calculate the percentage of propanoic acid molecules that are ionized in
the solution.