Assignment 4 power point

GROUP MEMBERS

MAZLIYANA BINTI MUHSINON
M20112001387

NUR HARLYANA BINTI HARUN
M20112001372

NURAINI BINTI NORUDDIN
M2O112001368

MAGENTIRAN A/L NAWAMANI
M20102001019

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

VIDEO

CLICK BUTTON TO VIEW VIDEO
INTRODUCTION TO ACID AND BASES

http://www.youtube.com/watch?v=RF40cI2O16U

ACID ALKALI
Chemical substance that produce Chemical substance that
hydrogen ion, H+ (hydroxonium produce hydroxide ion, OH-,
ion, H3O+) when it dissolve in when it dissolve in water.
water. Alkali can dissolve in water.
H 0 H 0
2 2

HCl H+ + Cl- NaOH Na+ + OH-

CONCEPT OF ACID, BASE
AND ALKALI

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

TYPES OF ACID

Inorganic Acid Organic Acid
HCl (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 BASE

BASE

UNDISSOLVE BASE DISSOLVE BASE

CALLED: BASE CALLED: ALKALI

CHEMICAL REACTION OF BASE

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 ALKALI

H2O
NaOH Na+ + OH-

NH3 + H2O NH4+ + OH-

H2O
Ca(OH)2 Ca+ + 2OH-

K2O + H2O 2K+ + 2OH-

USES OF ACID, BASE AND ALKALI IN DAILY LIFE

Yes, in our A ? Do you know
daily life, that you are
things such using acid,
as vinegar, base and alkali
soap, vitamin everyday?
C, pineapple,
orange,
toothpaste
and shampoo
that based on
acid, base
and alkali

Uses of acid in daily life
Uses of acid in daily life

Uses of alkali and base in daily life

ACIDS AND BASES ARE EVERYWHERE
ACIDS AND BASES ARE EVERYWHERE

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.

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 can
show by this equation:

HCl + H2O H3O+ + Cl-

+
-
H H H H H
Cl + O O + Cl
H

When the hydrogen chloride, HCl is dissolve in
water, one of the hydrogen ion H+ or proton is donated
to water molecule to form hydroxonium ion.

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.

PROPERTIES OF
ACIDS AND ALKALIS

LEARNING OUTCOME

At the end of the lesson, students should be able
to:

 describe chemical properties of acids
and alkalis

VIDEO

CLICK BUTTON TO VIEW VIDEO
ACID AND BASE PROPERTIES

Source: http://www.youtube.com/watch?v=tjewLktzy9k

DISSOCIATION OF BASE

SOLID AQUEUS

NaOH (p) Na+ + OH- NaOH (p) Na+ + OH-

KOH (p) K+ + OH- KOH(p) K+ + OH-

Without the water, ion OH- NaOH and KOH dissolve in
not form water to form ion OH-

DISSOCIATION OF ACID

SOLID AQUEUS

HCI (p) ) H+ + Cl- HCl (p ) H+ + Cl-

CH3COOH(ce) CH3COO- + H+ CH3COOH CH3COO- + H+

CONCLUSION CONCLUSION
Without the water, ion H+ HCl and CH3COOH dissolve
not form in water and form ion H+

MATERIAL LITMUS DISSOCIATION ALKALINE CONCLUSION
PAPER TO FORM OH-

Toothpaste No changes  -

Toothpaste + Red to blue
water 
Soap No change -  -

Soap + water Red to blue 
 - Hydroxyl ion
NaOH No change caused the
litmus paper
NaOH + water Red to blue NaOH paper changes

from blue to red
Na+ + OH-

MATERIAL LITMUS DISSOCIATION ACID CONCLUSION
PAPER EQUATION
Tamarind No change 
-
Blue to red
Asam gelugor 
+ water
Lemon No change - 
Lemon+ water Blue to red 
 - Hydrogen ion
Etanoik acid No change caused the
litmus paper
Etanoik acid + Blue to red CH3COOH changes from

water blue to red
CH3COO- + H+

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.

Experiment Litmus Metal Carbonate
paper
HCl ( p) No changes No changes No changes

HCl ( ak ) Blue to red The bubble gas yield and CaCO3( p) + 2HCl
produce “ pop”sound CaCl2 + CO2 +
when is test H2O
Mg ( p) + 2HCl MgCl2( ak)
+ H2

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.

Experiment Litmus paper
Ca ( OH )2 ( p) No changes

Ca ( OH )2 + H2O Ca+2 + Red to blue
2 OH-

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

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

Chemical properties of acid

 Litmus paper changes from red to blue colour

 React with bases to form salt and water as the
only products

WEB BASED SIMULATION

CLICK LINK TO VIEW SIMULATION
ACID AND BASE PROPERTIES

Source:
ONLINE LABS
http://amrita.olabs.co.in/?sub=73&brch=3&sim=6&cnt=72

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

pH SCALE

Bases solution blue Acidic solution
red

pH METERS

There are two different apparatus that use to
measure the pH scale.

Universal Indicators Colors
Universal Indicator is a mixture of
different indicators which covers the full
range of pH values

pH VALUE
[ H + ] = 7 neutral
[ H +] > 7 alkali
[ H + ] < 7 acid

How to measure the pH
value?
take measurements and
record the results

[ H + ] = 10
For bases

[ H +] = 3
For acids

WHEN THE CONCENTRATION OF HYDROGEN
IONS [H+] IN SOLUTION

INCREASE

SO THE pH VALUES ARE

LOWER

Concentration of acids and bases with pH
values

Synthesizing the concept of strong acid,
weak acid, strong alkalis and 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.

For Strong Acid
For Weak Acid

Back to Menu

HNO3
HCl
H2SO4

Back

Diagram of strong acid
dissociation

Before dissociation After dissociation

H A H+
H A-
A H2 O H+
H+
A-
H H A-
A A A- H +

HA (ak) H+ (ak) + A-(ak)

Back

Diagram of weak acid
dissociation


H H2 O H+ H+
A H A A- A-
A H+
H A H A- A- H+

HA (ak) H+ (ak) + A-(ak)

Back

For Strong Alkali
For Weak Alkali

Back to Menu

Diagram of strong alkali
dissociation


OH B+
B H2 O OH-
B
OH B+ B+
OH-
B OH B OH- OH-
OH B+

BOH (ak) B+ (ak) + OH-(ak)

Back

 NaOH
 KOH

Back

Diagram of weak alkali
dissociation


OH B+ - B+
B H2 O OH
B
OH OH-

B OH OH- B OH B+
+ -
B
OH

BOH (ak) B+ (ak) + OH-(ak)

Back

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

Concentration
quantity of solute in gram or mole in 1
dm3 solution.

Formula :
Concentration = mass ( g )
volume ( dm 3 )
Unit : g / dm 3

Concentration = number of mole
volume ( dm 3 )
Unit : mole / dm 3

Example :
• 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

Molarity, M
• quantity of solute in mole in 1dm 3
solution.

Formula :
Molarity, M = number of mole
volume of solution ( dm 3 )
• Unit : mole / dm 3

Relationship between number of mole
with molarity and volume of solution
Number of mole, n

Molarity, M Volume, V

Number of mole,n = MV
1000
M = molarity ( mole / dm 3 )
V = volume ( cm 3 )

Example:
Calculate the number of mole of HCl in 50 cm 3 HCl
aqueus 0.2 mole / dm 3 .

Answer :
Number of mole, n = MV
1000
= ( 0.2 mole / dm 3 )( 50 cm 3 )
1000
= 0.01 mole

Number of mole = mass ( g )
relative molecular mass ( g / mole )

1 dm 3 = 1000 cm 3

Answer :
( 1 ) i. Concentration = mass ( g )
volume ( dm 3 )
= 2.0 g
10.0 dm 3
= 0.2 g / dm 3

ii . Molarity, M = number of mole
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

Questions :
• 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

2 ) Calculate the number of mole of H2SO4 in 25 cm 3
H2SO4 aqueus 0.5 mole / dm 3 .

Answer

Answer :
( 1 ) i. Concentration = mass ( g )
volume ( dm 3 )
= 5.0 g
10.0 dm 3
= 0.5 g / dm 3

ii . Molarity, M = number of mole
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

2. Number of mole, n = MV
1000
= ( 0.5 mole / dm 3 )( 25 cm 3 )
1000
= 0.0125 mole

ANALYSING
CONCENTRATION
OF ACIDS &
ALKALIS

LEARNING OUTCOME

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

STANDARD SOLUTION

 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

STEPS HOW TO PREPARE THE
STANDARD SOLUTION

c) 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.

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

STANDARD SOLUTION

Number of mole = concentration x volume
1000

= MV / 1000 1

= mass(g)
J.M.R (g/mole) 2

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

(B) PREPARATION SOLUTION NaOH
2) 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

DILUTION

meaning

formula

preparation

REVISION
 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)

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.

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.

? FORMULA

The total amount of = The total amount of solute
solute before dilution after dilution

M1V1 = M2V2
1000 1000
M1V1 M2V2

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.

CALCULATION

Use this equation,
M1V1 = M2V2
(10mg/L) (10mL) = M2 (1000mL)
M2 = (100mg/L) (10mL)
(1000mL)
M2 = 1.00mg/L

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.

ANSWER

Use this equation,

M1V1 = M2V2
(10mg/L) (100mL) = (50mg/L) V2
V2 = (10mg/L) (100mL)
(50mg/L)
V2 = 20.0mL

PREPARING DILUTION

Steps to follow :
2) 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.

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.

Related pH value with molarity of acid and
alkali.

Solve numerical problems involving molarity of
acids and alkalis.

Introduction
 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.

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

 The pH value of an acid or alkali depends on
2 factors:
a) degree of ionisation / dissociation
b) molarity of the solution

ACID ALKALI

Litmus paper: blue Litmus paper: red

Observation: Observation:
blue colour → red colour red colour → blue colour
pH scale: pH scale:

Litmus
paper
1 7 14
acid alkali

Example: Acid (H2SO4)

H+ + H+ H+ H+
H+ H H+
+
H
H+ + H+H H+
+

H H+ H+ H+ H+
H H+ H+
+ H+ H+
H+ H+
0.1 M 0.01 M 0.001 M

[H+] Highest Higher Low

pH Low Higher Highest

Acidity Highest Higher Low

Example: Alkali (NaOH)

H+ -OH- +
OH H+ H+ OH- + H+ H+
H OH- + H H+
OH- OH- H OH- H+ + OH-
H OH- H+
+
H
OH+OH H
- +
- H+
H OH- OHH+ H+
-
H H+ H+
+

0.1 M 0.01 M 0.001 M

[H+] Low Higher Highest

[OH-] Highest Higher Low

pH Highest Higher Low

Alkalinity Highest Higher Low

M/[H+] Decreasing concentration of [H+]
M/[H+]
Increasing pH
pH pH

pH
scale
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

acid neutral alkali

Increasing acidity Increasing alkalinity

CALCULATIONS ON MOLARITY

The molarity of a solution
changes when :

Water is added to it An acid or alkali is added
to it

CALCULATIONS ON
MOLARITY

M1 = Initial molarity
M2 = Final molarity
V1 = Initial volume
V2 = Final volume

Thus the formula
M1V1 = M2V2
can be used to find the
new molarity

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 the water added to obtain
Find final volume 250 cm3 acid
V2 = M1V1 = 250 – 100
M2 = 150 cm3
= 0.5 x 100
0.2
= 250 cm3

EXAMPLE 2
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 Resulting molarity
HCl = number of moles
= (2+0.5) = 2.5 mole dm-3 volume
= 2.5
Total volume of HCl 800
= (200+600) = 800 cm3 = 0.003125 mole dm-3

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 .

M1 = Initial molarity
M1V1 = M2V2 M2 = Final molarity
V1 = Initial volume
V2 = Final volume

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

Is a reaction between acid and alkali and
it’s produce salt and water.
- So, the acid and alkali will lost it’s
properties

Acid + Base  Salt +
Water

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)

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.

 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)

 ABee sting +acid. To relieve the painful symptoms of the sting
bee sting contains calamine lotion neutral
weacid ) to neutralise the acid, by rubbing on calamine lotion (zinc
( need (alkaline )
carbonate) or baking soda, so the acid can be neutralised.

Wasp sting + vinegar neutral
( alkaline ) (ethanoic acid )

Wasp stings are alkaline, hence
acid is needed to
neutralise and remove the painful
sting. Vinegar ( ethanoic acid)
is needed

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.

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)

Apparatus position in Titration

Buret fill with
acid or base

Erlenmeyer
Indicator
flask fill with
acid or base

Acid base indicator:

Medium Orange methyl Phenolphthalein Litmus
color color paper

Acid Red None Blue – red

Neutral Orange None None

Base Yellow Pink Red –blue

Cl-
H+

Cl-
H+
Hidroclorix acid
solution
Cl-
H+

Na+
OH -
Sodium hidroxide
OH- solution
Na
+

Contain 3 drops of OH- Na+
Phenolphthalein

INSTRUCTION ON HOW TO USE A RED CABBAGE
AS INDICATOR

base acid

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 H2O
From the equation, 2 mol NaOH react with 1 mol H2SO4.

The number of NaOH =2
The number of Sulfuric Acid 1

MBVB =2
MAVA 1

where MB= molarity of NaoH

MB(25.0) =2 , MB = 2 x 26.5 x 1.0
1.0(26.5) 1 1 25.0

BROMIN GROUP
The 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?

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?

CHLORINE GROUP
Dilute sulphuric acid added exessly to 100 cm3
potassium hydroxide solution 0.1mol. calculate
the mol dm-3 number of potassium sulphate that
produce.

Which of the acid base pair that produce orange colour when using
orange methyl indicator in titration?

Acid Base
IODINE GROUP
I 25cm3 hydrocloric acid 1.0 25 cm3 sodium hidroxide
mol dm-3 solution 1.0mol dm-3
II 25cm3 hydrocloric acid 1.0 25 cm3 ammonia aques
mol dm-3 1.0mol dm-3
III 25cm3 sulphuric acid 1.0 mol 25 cm3 sodium hidroxide
dm-3 solution 1.0mol dm-3
IV 25cm3 nitric acid 1.0 mol dm-3 25 cm3 ammonia aques
1.0mol dm-3

Assignment 4 power point

More Related Content

What's hot (18)

Viewers also liked (9)

Similar to Assignment 4 power point (20)

Recently uploaded (20)

Assignment 4 power point