Assignment 4 power point

GROUP MEMBERS

MAZLIYANA BINTI MUHSINONM20112001387

NUR HARLYANA BINTI HARUNM20112001372

NURAINI BINTI NORUDDINM2O112001368

MAGENTIRAN A/L NAWAMANIM20102001019

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

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

VIDEO

CLICK BUTTON TO VIEW VIDEOINTRODUCTION TO ACID AND BASES

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

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

BASE

UNDISSOLVE BASE

CALLED: BASE CALLED: ALKALI

DISSOLVE BASE

TYPES 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 BASE

CHEMICAL REACTION OF ALKALI

H2O NaOH Na+ + OH-

NH3 + H2O NH4+ + OH-

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

K2O + H2O 2K+ + 2OH-

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

Uses of acid in daily life

Uses of acid in daily life

Uses of alkali and base in daily life

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

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-

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.

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

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

describe chemical properties of acids and alkalis

LEARNING OUTCOME

VIDEO

CLICK BUTTON TO VIEW VIDEOACID AND BASE PROPERTIES

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

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

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

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

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

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 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

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 +

2 OH-

Red to blue

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 SIMULATIONACID AND BASE PROPERTIES

Source: ONLINE LABS

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

LEARNING OUTCOME


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 + ] = 10For bases

[ H + ] = 3For 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

ACID AND ALKALIACID AND ALKALI

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.

ACID

ALKALI

For Strong Acid

For Weak Acid

Back to Menu

HNO3

HClH2SO4

Back

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-

HCOOHH3PO3

Back

Next

Back

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+

STRONG ALKALI AND WEAK STRONG ALKALI AND WEAK ALKALIALKALI

Back to Menu

For Strong Alkali

For Weak Alkali

Next

Back

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-

NaOH KOH

Back

Next

Back

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-

NH4OH

CH3OH

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

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

Concentration

Concentration = mass ( g )

volume ( dm 3 )

Unit : g / dm 3

Concentration = number of mole

volume ( dm 3 )

Unit : mole / dm 3

Formula :

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 :

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

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

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 .

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

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 :

2 ) Calculate the number of mole of H2SO4 in 25 cm 3

H2SO4 aqueus 0.5 mole / dm 3 .

Answer

( 1 ) i. Concentration = mass ( g )

volume ( dm 3 )

= 5.0 g

10.0 dm 3

= 0.5 g / dm 3

Answer :

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

ANALYSINGANALYSINGCONCENTRATIONCONCENTRATION

OF ACIDS & OF ACIDS & ALKALISALKALIS


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

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

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.

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

Number of mole = concentration x volume

1000

= MV / 1000 1

= mass(g)

J.M.R (g/mole) 2

STANDARD SOLUTION

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

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

DILUTION

preparation

formula

meaning

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

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) V2V2 = (10mg/L) (100mL)

(50mg/L)V2 = 20.0mL

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.

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.

ACID & ALKALI

Related pH value with molarity of acid and alkali.

Solve numerical problems involving molarity of acids and alkalis.

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

pH scale

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

pH meter

The pH value of an acid or alkali depends on 2 factors:

a) degree of ionisation / dissociation

b) molarity of the solution

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

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

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-

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

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 molarityM2 = Final molarityV1 = Initial volumeV2 = 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

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

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

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

LEARNING OUTCOME


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)

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 )

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 )

LEARNING OUTCOME


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)

Indicator

Buret fill with acid or base

Erlenmeyer flask fill with acid or base

Apparatus position in Titration

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:

The end point

Na+

Cl-

Na+

Na+

Cl-

Cl-

H+

OH-

OH-

OH-

H+

H+

Sodium hidroxide solution

Hidroclorix acid solution

Contain 3 drops of Phenolphthalein

acidbase

INSTRUCTION ON HOW TO USE A RED CABBAGE AS INDICATOR

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.

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?

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

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

Assignment 4 power point

Technology

Transcript of Assignment 4 power point