assignment 4 power point
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Power Point presentations~Chapter 7TRANSCRIPT
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GROUP MEMBERS
MAZLIYANA BINTI MUHSINONM20112001387
NUR HARLYANA BINTI HARUNM20112001372
NURAINI BINTI NORUDDINM2O112001368
MAGENTIRAN A/L NAWAMANIM20102001019
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ACID AND BASE
Learning Outcome:At the end of the lesson, student should be able to:
State the meaning of acid, based and alkali.
State uses of acid, based and alkali in daily life.
Explain the role of water in the formation of hydrogen ions to show the properties of acids
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http://www.youtube.com/watch?v=RF40cI2O16U
VIDEO
CLICK BUTTON TO VIEW VIDEOINTRODUCTION TO ACID AND BASES
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ALKALI Chemical substance that produce hydroxide ion, OH-, when it dissolve in water. Alkali can dissolve in water. H
20
NaOH Na+ + OH-
BASE Chemical substance that can react with acid to produce salt and water. Base can’t dissolve in water. CuO (p) + H2SO4 CuSO4 (ak) + H2O
ACIDChemical substance that produce hydrogen ion, H+ (hydroxonium ion, H3O+) when it dissolve in
water. H
20
HCl H+ + Cl-
CONCEPT OF ACID, BASE
AND ALKALI
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Inorganic Acid Organic AcidHCl (Hydrochloric acid) CH3COOH
(Acetic acid)
HNO3 (Nitric acid) HCOOH (Methanoic acid)
H2SO4 (Sulphuric acid) HOOCCOOH (Ethanedioic acid)
H3PO4 (Phosphate acid)
H2CO3 (Carbonic acid)
TYPES OF ACID
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BASE
UNDISSOLVE BASE
CALLED: BASE CALLED: ALKALI
DISSOLVE BASE
TYPES OF BASE
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CuO (p) + H2SO4 CuSO4 (ak) + H2O
Fe2O3 (p) + 6HNO3 2Fe(NO3)3 (ak) + 3H2O
Mg(OH)2 (p) + 2HCl MgCl2 (ak) + 2H2O
CHEMICAL REACTION OF BASE
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CHEMICAL REACTION OF ALKALI
H2O NaOH Na+ + OH-
NH3 + H2O NH4+ + OH-
H2O Ca(OH)2 Ca+ + 2OH-
K2O + H2O 2K+ + 2OH-
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Do you know that you are using acid,
base and alkali everyday?
Yes, in our daily life,
things such as vinegar,
soap, vitamin C, pineapple,
orange, toothpaste
and shampoo
that based on acid, base
and alkali
A ?
USES OF ACID, BASE AND ALKALI IN DAILY LIFE
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Uses of acid in daily life
Uses of acid in daily life
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Uses of alkali and base in daily life
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Every liquid you see will probably be either an acid or a base. Most water you drink has
ions in it. Those ions in solution make something acidic or basic.
ACIDS AND BASES ARE EVERYWHEREACIDS AND BASES ARE EVERYWHERE
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ROLE OF WATER IN THE FORMATION OF HYDROGEN IONS TO SHOW THE PROPERTIES OF ACID
Arrhenius said that acid will ionized in water to produce hydrogen ion, H+. Example hydrochloric acid, HCl
HCl H+ + Cl-
Hydrogen ion, H+ that produced is hydrated to form hydroxonium ion, H3O+
H+ + H2O H3O+
So, ionization of hydrochloric acid, HCl in water canshow by this equation:
HCl + H2O H3O+ + Cl-
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ClH
OH H
+ OH H
H
Cl+
+-
When the hydrogen chloride, HCl is dissolve inwater, one of the hydrogen ion H+ or proton is donated to water molecule to form hydroxonium ion.
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The role of water
If you have an ionic compound and you put it in water, it will break apart into two ions. If one of those ions is H+,
the solution is acidic. If one of the ions is OH-, the solution is basic.
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PROPERTIES OF ACIDS AND ALKALIS
At the end of the lesson, students should be able to:
describe chemical properties of acids and alkalis
LEARNING OUTCOME
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VIDEO
CLICK BUTTON TO VIEW VIDEOACID AND BASE PROPERTIES
Source: http://www.youtube.com/watch?v=tjewLktzy9k
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SOLID AQUEUS
NaOH (p) Na+ + OH- NaOH (p) Na+ + OH-
KOH (p) K+ + OH- KOH(p) K+ + OH-
DISSOCIATION OF BASE
NaOH and KOH dissolve NaOH and KOH dissolve in water to form ion in water to form ion OHOH--
Without the water, ion Without the water, ion OH-OH-not formnot form
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DISSOCIATION OF ACID
SOLID AQUEUS
HCI (p) ) H+ + Cl-
HCl (p ) H+ + Cl-
CH3COOH(ce) CH3COO- + H+
CH3COOH CH3COO- + H+
CONCLUSIONCONCLUSIONHCl and HCl and CH3COOH
dissolve in water and form dissolve in water and form ion ion HH++
CONCLUSIONCONCLUSIONWithout the water, ion Without the water, ion
HH++ not formnot form
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MATERIAL LITMUS PAPER
DISSOCIATION TO FORM
OH-
ALKALINE CONCLUSION
Toothpaste
Toothpaste + water
No changes
Red to blue
-
Soap
Soap + water
No change
Red to blue
-
-
NaOH
NaOH + water
No change
Red to blue NaOH
Na+ + OH-
- Hydroxyl ion caused the litmus paper paper changes from blue to red
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MATERIAL LITMUSPAPER
DISSOCIATIONEQUATION
ACID CONCLUSION
Tamarind
Asam gelugor + water
No change
Blue to red-
Lemon
Lemon+ water
No change
Blue to red
-
Etanoik acid
Etanoik acid + water
No change
Blue to red CH3COOH CH3COO- + H+
- Hydrogen ion caused the litmus paper changes from blue to red
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Explanation
Acid react with water by donating an H+ ion to a neutral water molecule to form the H3O+ ion.
Without the presence of water, acid not show the acid properties because the dry acid does not have hydrogen ion.
Base react with water by donating an OH- ion.
Without the presences of water, bases does not show the base properties because the dry base does not show hydroxide ion.
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Experiment Litmus paper
Metal Carbonate
HCl ( p) No changes No changes No changes
HCl ( ak ) Blue to red The bubble gas yield and produce “ pop”sound when is test Mg ( p) + 2HCl MgCl2( ak)
+ H2
CaCO3( p) + 2HCl
CaCl2 + CO2 +
H2O
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Conclusion
HCl solution contain the H+ and Cl-
Hydrogen chloride was dissociated completely
HCl (g) of solution shows the chemical properties of acid because it has a hydrogen ion.
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Experiment Litmus paper
Ca ( OH )2 ( p) No changes
Ca ( OH )2 + H2O Ca+2 +
2 OH-
Red to blue
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Conclusion
Ca(OH)2 solid does not show the characteristic of alkaline.
In water,
Ca (OH)2 (ak) Ca2+ + 2OH-
Calcium hydroxide solution dissolve in water will dissociated and show the characteristic of alkaline
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Chemical properties of bases
Has a pH more than 7
Litmus paper was change from blue to red colour
Reacts with acid to form salt and water
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Chemical properties of acid
Litmus paper changes from red to blue colour
React with bases to form salt and water as the only products
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WEB BASED SIMULATION
CLICK LINK TO VIEW SIMULATIONACID AND BASE PROPERTIES
Source: ONLINE LABS
http://amrita.olabs.co.in/?sub=73&brch=3&sim=6&cnt=72
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LEARNING OUTCOME
At the end of the lesson, students should be able to:
state the use of a pH scale
relate pH value with acidic or alkaline properties of a substance
relate concentration of hydrogen ions with pH value
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pH SCALE
Bases solution blue Acidic solution red
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pH METERS
There are two different apparatus that use to measure the pH scale.
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Universal Indicators Colors
Universal Indicator is a mixture of different indicators which covers the full range of pH values
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pH VALUE
[ H + ] = 7 neutral [ H + ] > 7 alkali
[ H + ] < 7 acid
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How to measure the pH value?
take measurements and record the results
[ H + ] = 10For bases
[ H + ] = 3For acids
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WHEN THE CONCENTRATION OF HYDROGEN IONS [H+] IN SOLUTION
INCREASE
SO THE pH VALUES ARE
LOWER
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Concentration of acids and bases with pH values
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ACID AND ALKALIACID AND ALKALI
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Synthesizing the concept of strong Synthesizing the concept of strong acid, weak acid, strong alkalis and acid, weak acid, strong alkalis and weak alkalis.weak alkalis.
LEARNING OUTCOMES
Relate strong or weak acid with degree of dissociation.
Relate strong or weak alkali with degree of dissociation.
Conceptualize qualitatively strong and weak acid.
Conceptualize qualitatively strong and weak acid.
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ACID
ALKALI
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For Strong Acid
For Weak Acid
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Back
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HNO3
HClH2SO4
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Next
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Diagram of strong acid dissociation
Before dissociation After dissociation
H A
HA (ak) H+ (ak) + A-(ak)
H2O
Back
HA
A
A
H
H
H+
H+
H+
H+A-
A-A-
A-
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Back
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HCOOHH3PO3
Back
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Next
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Diagram of weak acid dissociation
Before dissociation
HA (ak) H+ (ak) + A-(ak)
H2O
Back
After dissociation
AH
H
H
HA
A
A
A-A-
A-A-
H+
H+
H+
H+
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STRONG ALKALI AND WEAK STRONG ALKALI AND WEAK ALKALIALKALI
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Back to Menu
For Strong Alkali
For Weak Alkali
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Diagram of strong alkali dissociation
Before dissociation
BOH (ak) B+ (ak) + OH-(ak)
H2O
After dissociation
Back
B
B
B
B B+
OH
OH
OH
OH OH-
B+ B+
B+
OH-
OH-
OH-
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Back
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NaOH KOH
Back
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Diagram of weak alkali dissociation
Back
Before dissociation
BOH (ak) B+ (ak) + OH-(ak)
H2O
After dissociation
B
B
B
B B+
OH
OH
OH
OH OH-
B+
B+
B+
OH-
OH-
OH-
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Back
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NH4OH
CH3OH
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LEARNING OUTCOME
At the end of the lesson, students, should be able to:
state the meaning of concentration state the meaning of molarity state the relationship between the number of moles with molarity and volume of solution
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quantity of solute in gram or mole in 1 dm3 solution.
Concentration
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Concentration = mass ( g )
volume ( dm 3 )
Unit : g / dm 3
Concentration = number of mole
volume ( dm 3 )
Unit : mole / dm 3
Formula :
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1) 2.0 g NaCl is dissolve in 10 dm 3 water. Calculate the
( Na = 23, Cl = 35.5 )
i . Concentration of the solution in g / dm 3. Answer
ii . Molarity of the solution.Answer
Example :
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quantity of solute in mole in 1dm 3
solution.
Unit : mole / dm 3
Formula :
Molarity, M = number of mole
volume of solution ( dm 3 )
Molarity, M
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Number of mole, n
Molarity, M Volume, V
Number of mole,n = MV
1000
M = molarity ( mole / dm 3 )
V = volume ( cm 3 )
Relationship between number of mole with molarity and volume of solution
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Answer :Number of mole, n = MV
1000
= ( 0.2 mole / dm 3 )( 50 cm 3 )
1000
= 0.01 mole
Example:
Calculate the number of mole of HCl in 50 cm 3 HCl
aqueus 0.2 mole / dm 3 .
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Number of mole = mass ( g )
relative molecular mass ( g / mole )
1 dm 3 = 1000 cm 3
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Answer :( 1 ) i. Concentration = mass ( g )
volume ( dm 3 )
= 2.0 g
10.0 dm 3
= 0.2 g / dm 3
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ii . Molarity, M = number of mole
volume of solution ( dm 3 )
Relative molecular mass = 23 + 35.5
= 58.5
Number of mole,n = 2.0 g
58.5 g / mole
= 0.03 mole
Molarity, M = 0.03 mole
10.0 dm 3
= 0.3 mole / dm 3
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1) 5.0 g NaOH is dissolve in 10 dm 3 water. Calculate the
( Na = 23, O = 16, H = 1 )
i . Concentration of the solution in g / dm 3. Answer
ii . Molarity of the solution.Answer
Questions :
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2 ) Calculate the number of mole of H2SO4 in 25 cm 3
H2SO4 aqueus 0.5 mole / dm 3 .
Answer
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( 1 ) i. Concentration = mass ( g )
volume ( dm 3 )
= 5.0 g
10.0 dm 3
= 0.5 g / dm 3
Answer :
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ii . Molarity, M = number of mole
volume of solution ( dm 3 )
Relative molecular mass = 23 + 16 + 1
= 40
Number of mole,n = 5.0 g
40 g / mole
= 0.125 mole
Molarity, M = 0.125 mole
10.0 dm 3
= 1.25 mole / dm 3
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2. Number of mole, n = MV
1000
= ( 0.5 mole / dm 3 )( 25 cm 3 )
1000
= 0.0125 mole
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ANALYSINGANALYSINGCONCENTRATIONCONCENTRATION
OF ACIDS & OF ACIDS & ALKALISALKALIS
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At the end of the lesson, students should be able to:
describe the methods of preparing standard solution
describe the preparation of a solution with a specified concentration using dilution method
relate pH value with molarity of acid and alkali
solve numerical problems involving molarity of acids and alkalis
LEARNING OUTCOME
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Standard solution is a solution that we
had already knew its concentration.
Volumetric flask is a apparatus with certain volume to use preparation standard solution
STANDARD SOLUTION
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STEPS HOW TO PREPARE THE STANDARD SOLUTION
a) Calculate mass (m g) of the chemical substance was needed to preparation solution V cm3. V is a volume volumetric flusk.
b) Weight the m g chemical substance accurately.
c) M g chemical substance was soluble into the desolve in volumetric flusk.
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PREPARATION STANDARD SOLUTION NATRIUM HIDROXIDE 0.1 MOLE DM-3
(A) Calculate mass NaOH was needed to preparing 100cm3 NaOH solution 0.1 mole dm-3 like below:
= number of mole NaOH x mass molecule relative
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Number of mole = concentration x volume
1000
= MV / 1000 1
= mass(g)
J.M.R (g/mole) 2
STANDARD SOLUTION
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1 = 2
MV / 1000 = mass(g)
J.M.R (g/mole)
mass (g) = MV x J.M.R
1000
MASS (g) NaOH = MV x (23+16+1)
1000
= 0.1 x 100 x 40 =0.4g
1000
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(B) PREPARATION SOLUTION NaOH
1)Weight 0.4g NaOH solid accurately.
2) Move a solid into the small bikar and soluble
in 20cm3 disolve water.
3) With used the funner, NaOH solution moved
into the volumetric flusk.
4) Add the disolve water until arrive the desire
level.’
5) Closed the volumetric flusk and shake.
6) Now, the standard is 100 cm3 NaOH 0.1 mol
dm-3
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DILUTION
preparation
formula
meaning
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Concentration (in gdm-3 unit) refers to an amount of substance (in gram) per unit volume (1000cm3). The unit is gdm-3.
Concentration (gdm-3) = Amount of substance (g) Volume (dm3)Volume (dm3)
REVISION
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DILUTION
Dilution is a process by adding more solvent
( eg:water ) into a hard solution.
To dilute, the solution means to increase its
volume, by adding more solvent but no more
solute.
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DILUTION
Dilution must be prepared using distilled water.
Clean water is required to prepare dilutions so
that the concentration of the diluted standard
can be known exactly.
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FORMULA?The total amount of = The total amount of solute
solute before dilution after dilution
M1V1 = M2V2
1000 1000
M1V1 M2V2
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EXAMPLE
Given that the concentration of diluted standard is 100mg/L fluoride is diluted by dispensing 10mL of standard into a 1000mL volumetric flask and filling it to the line with distilled water. Calculate the concentration of the diluted standard.
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CALCULATION
Use this equation,
M1V1 = M2V2
(10mg/L) (10mL) = M2 (1000mL)
M2 = (100mg/L) (10mL)
(1000mL)
M2 = 1.00mg/L
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TRY THIS
Given that the volume of a standard required for dilution 10mg/L nitrate standard is required for testing. The lab has a 50mg/L nitrate, some pipettes, and a 100mL volumetric flask. Calculate the volume of 50mg/L standard to prepare 100mL of a 10mg/L standard.
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ANSWER
Use this equation,
M1V1 = M2V2
(10mg/L) (100mL) = (50mg/L) V2V2 = (10mg/L) (100mL)
(50mg/L)V2 = 20.0mL
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PREPARING DILUTION
Steps to follow :
1)Use a volumetric or automatic pipette to dispense the chosen volume of concentrate standard into a clean volumetric flask.
2) Fill the flask to volume with distilled water until the bottom of the meniscus rests on the top of the volumetric mark that is etched on the flask.
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PREPARING DILUTION
3) Invert the flask several times to
thoroughly mix the solution. Once
the dilution is prepared, the test
can be run on the diluted
standard.
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ACID & ALKALI
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Related pH value with molarity of acid and alkali.
Solve numerical problems involving molarity of acids and alkalis.
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Molarity is one unit of concentration which show the mole number of solute that contain in 1 dm3 solution
pH is a measure of the concentration of hydrogen ions.
More hydrogen ions become more acidity and less pH.
Acid have lower pH than alkali which showing in the pH scale.
We measure the concentration using meter pH.
Introduction
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pH scale
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Example of food and substance using daily that have difference pH value.
Substance pH Substance pH
Lime juice 2.3 Coffee 5.0
Vinegar 2.8 Milk 6.6
Soft Drink 3.0 Eggs 7.8
Orange juice 3.5 Toothpaste 8.0
Banana 4.6 Soap 8.2
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pH meter
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The pH value of an acid or alkali depends on 2 factors:
a) degree of ionisation / dissociation
b) molarity of the solution
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Litmus paper: blue
Observation: blue colour red colour
pH scale:
1 7 14acid alkali
Litmus paper: red
Observation: red colour blue colour
pH scale:
Litmus paper
ACID ALKALI
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Example: Acid (H2SO4)
0.1 M 0.01 M 0.001 M
pH
Acidity
[H+]
H+H+
H+H+
H+
H+H+
H+H+
H+H+
H+ H+
H+
H+
H+H+
H+
H+H+
H+
H+H+
H+
Highest Higher Low
Highest Higher Low
Low Higher Highest
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Example: Alkali (NaOH)
0.1 M 0.01 M 0.001 M
pH
Alkalinity
[OH-]
H+
H+H+
H+
H+
H+H+
H+H+
H+
H+
H+
H+
H+
H+
H+H+H+
H+H+
H+
Highest Higher Low
Highest Higher Low
Highest Higher Low
[H+] Low Higher Highest
OH-
OH-
OH-
OH-
OH-
OH-
OH-
OH-
OH-
OH-
OH- OH-
OH-
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH scale
acid alkali
pH
M/[H+] M/[H+]
pH Increasing pH
Decreasing concentration of [H+]
Increasing alkalinityIncreasing acidity
neutral
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CALCULATIONS ON MOLARITY
The molarity of a solution changes when :
Water is added to it An acid or alkali is added to it
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CALCULATIONS ON MOLARITY
M1 = Initial molarityM2 = Final molarityV1 = Initial volumeV2 = Final volume
Thus the formula
M1V1 = M2V2
can be used to find the new molarity
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EXAMPLE 1
Find the volume of distilled water that is added to 100cm3 of hydrochloric acid, 0.5 mole dm-3, to
obtain an acid solution of strength 0.2 mole dm-3.
M1V1 = M2V2
Find final volumeV2 = M1V1
M2
= 0.5 x 100 0.2= 250 cm3
the water added to obtain 250 cm3 acid = 250 – 100= 150 cm3
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Find the resulting molarity of sulphuric acid if 200 cm3 of HCl 2 mole dm-3, is added to 600 cm3 of HCl, 0.5
mole dm-3.
Total number of moles of HCl
= (2+0.5) = 2.5 mole dm-3
Total volume of HCl
= (200+600) = 800 cm3
Resulting molarity
= number of moles
volume
= 2.5
800
= 0.003125 mole dm-3
EXAMPLE 2
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Conclusion Relationship;
acid; the higher molarity the lower pH value because of the higher
concentration of H+
(pH , molarity)
bes; the higher molarity the higher pH value because of the higher concentration of OH+ (pH , molarity)
Solve numerical problems involving molarity of acids and alkalis with using the formula .
M1V1 = M2V2
M1 = Initial molarityM2 = Final molarityV1 = Initial volumeV2 = Final volume
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LEARNING OUTCOME
At the end of the lesson, students should be able to:
explain the meaning of neutralisation
explain the application of neutralisation in daily life
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Is a reaction between acid and alkali and it’s produce salt and water.- So, the acid and alkali will lost it’s properties
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Acid + Base Salt + Water
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Example :
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(aq)
CuO(s) + 2HCl(aq) CuCl2(aq) + 2H2O(aq)
Ca(OH)2(aq) + H2SO4(aq) CaSO4(aq) +
2H2O(aq)
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The Ion Equation
NaOH(aq) + HCl(aq) NaCl(aq)+H2O(aq)
# H+ (aq) + OH-
(aq) H2O(l)
So, in neutralisation ion hidrogen (H+) from acid will combine with hidroxide ion (OH-) to produce water molecule.
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Soil treatment – farming If the soil is too acidic, it is treated
with base (chemical opposite to an acid) in order to neutralise it.
Common treatment use is quicklime (calcium hydroxide) or chalk (calcium carbonate)
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A bee sting contains acid. To relieve the painful symptoms of the sting we need to neutralise the acid, by rubbing on calamine lotion (zinc carbonate) or baking soda, so the acid can be neutralised.
Bee sting + calamine lotion neutral( acid ) (alkaline )
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Wasp stings are alkaline, hence acid is needed to
neutralise and remove the painful sting. Vinegar ( ethanoic acid)
is needed
Wasp sting + vinegar neutral( alkaline ) (ethanoic acid )
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LEARNING OUTCOME
At the end of the lesson, students should be able to:
give the right definition of acid bases titration.
can determine the end point of titration during neutralization.
can solve numerical problems involving neutralization reaction to calculate either unknown concentration or unknown volume.
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Definition of titration:
Titration is a neutralization of an acid with a base. It
is used to find concentrations of unknown solutions.
Definition of end point:
The end point is found by when the indicators change
color. It happen when the mol hydrogen ion from acid
is equal to the mol of hydroxide ion from the base
solution. It shows when the indicator change the
color.
MaVa = MbVb
(acid) (base)
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Indicator
Buret fill with acid or base
Erlenmeyer flask fill with acid or base
Apparatus position in Titration
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Medium Orange methyl color
Phenolphthalein color
Litmus paper
Acid Red None Blue – red
Neutral Orange None None
Base Yellow Pink Red –blue
Acid base indicator:
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The end point
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Na+
Cl-
Na+
Na+
Cl-
Cl-
H+
OH-
OH-
OH-
H+
H+
Sodium hidroxide solution
Hidroclorix acid solution
Contain 3 drops of Phenolphthalein
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acidbase
INSTRUCTION ON HOW TO USE A RED CABBAGE AS INDICATOR
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EXAMPLE 1
In an experiment, 25 cm3 natrium Hydroxide with unknown concentration needs 26.5 cm3 Sulfuric acid 1.0 mol dm-3 for complete reaction in titration. Calculate the molarity of natrium Hydroxide.
Answer :
2 NaoH (ak) + H2SO4 (ak) Na2SO4 (ak) + 2 H2OFrom the equation, 2 mol NaOH react with 1 mol H2SO4.
The number of NaOH = 2The number of Sulfuric Acid 1
MBVB = 2MAVA 1
where MB= molarity of NaoH
MB(25.0) =2 , MB = 2 x 26.5 x 1.01.0(26.5) 1 1 25.0 =2.12 mol dm-3.
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BROMIN GROUPThe equation shows the reaction between sulphuric acid and sodium hydroxide.
H2SO4 + 2NaOH Na2SO4 + H2O
What is the volume of 1.0mol dm-3 sodium hydroxide solution which can neutralize 25 cm3 of 1.0 mol dm-3 sulphuric acid?
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FLUORIN GROUP
Figure above shows the set-up of apparatus for the titration of potassium hydroxide solution with sulphuic acids. What is the total volume of the mixture in the conical flask at the end point of the titration?
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CHLORINE GROUPDilute sulphuric acid added exessly to 100 cm3 potassium hydroxide solution 0.1mol. calculate the mol dm-3 number of potassium sulphate that produce.
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IODINE GROUP
Which of the acid base pair that produce orange colour when using orange methyl indicator in titration?
Acid Base
I 25cm3 hydrocloric acid 1.0 mol dm-3
25 cm3 sodium hidroxide solution 1.0mol dm-3
II 25cm3 hydrocloric acid 1.0 mol dm-3
25 cm3 ammonia aques 1.0mol dm-3
III 25cm3 sulphuric acid 1.0 mol dm-3
25 cm3 sodium hidroxide solution 1.0mol dm-3
IV 25cm3 nitric acid 1.0 mol dm-3 25 cm3 ammonia aques 1.0mol dm-3
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