ACID AND BASELearning 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 BASEShttp://www.youtube.com/watch?v=RF40cI2O16U
ACID ALKALIChemical substance that produce Chemical substance thathydrogen 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
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 alkalisoap, vitamin everyday?C, pineapple, orange, toothpasteand shampoothat based on acid, base and alkali
Uses of acid in daily lifeUses of acid in daily life
ACIDS AND BASES ARE EVERYWHERE ACIDS AND BASES ARE EVERYWHEREEvery liquid you see will probably be eitheran 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 HWhen the hydrogen chloride, HCl is dissolve inwater, one of the hydrogen ion H+ or proton is donatedto water molecule to form hydroxonium ion.
The role of waterIf you have an ionic compound and you put it in water, itwill 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 OFACIDS AND ALKALIS LEARNING OUTCOMEAt 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 PROPERTIESSource: http://www.youtube.com/watch?v=tjewLktzy9k
DISSOCIATION OF BASE SOLID AQUEUSNaOH (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 AQUEUSHCI (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 paperNaOH + water Red to blue NaOH paper changes from blue to red Na+ + OH-
MATERIAL LITMUS DISSOCIATION ACID CONCLUSION PAPER EQUATIONTamarind No change - Blue to redAsam gelugor + waterLemon No change - Lemon+ water Blue to red - Hydrogen ionEtanoik acid No change caused the litmus paperEtanoik 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 changesHCl ( 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 changesCa ( OH )2 + H2O Ca+2 + Red to blue2 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 ofbases 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 OUTCOMEAt 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 SCALEBases solution blue Acidic solution red
pH METERSThere are two different apparatus that use tomeasure 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 pHvalue?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 pHvalues
Synthesizing the concept of strong acid, weak acid, strong alkalis and weak alkalis. LEARNING OUTCOMESRelate 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.
LEARNING OUTCOMEAt 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 ofmoles 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 3Concentration = 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. Answerii . 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, VNumber 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 volume of solution ( dm 3 )Relative molecular mass = 23 + 35.5 = 58.5Number of mole,n = 2.0 g 58.5 g / mole = 0.03 moleMolarity, 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. Answerii . 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 volume of solution ( dm 3 )Relative molecular mass = 23 + 16 + 1 = 40Number of mole,n = 5.0 g 40 g / mole = 0.125 moleMolarity, 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
LEARNING OUTCOME At the end of the lesson, students should be able to: describe the methods of preparingstandard solution describe the preparation of a solution with a specified concentration using dilution method relate pH value with molarity of acid andalkali 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 SOLUTIONc) 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 SOLUTIONNumber 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 1000MASS (g) NaOH = MV x (23+16+1) 1000 = 0.1 x 100 x 40 =0.4g 1000
(B) PREPARATION SOLUTION NaOH2) 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
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.
? FORMULAThe total amount of = The total amount of solutesolute before dilution after dilution M1V1 = M2V2 1000 1000 M1V1 M2V2
EXAMPLEGiven that the concentration of dilutedstandard is 100mg/L fluoride is diluted bydispensing 10mL of standard into a1000mL volumetric flask and filling it tothe line with distilled water. Calculate theconcentration of the diluted standard.
TRY THISGiven that the volume of a standard required fordilution 10mg/L nitrate standard is required fortesting. The lab has a 50mg/L nitrate, somepipettes, and a 100mL volumetric flask. Calculatethe volume of 50mg/L standard to prepare100mL of a 10mg/L standard.
PREPARING DILUTIONSteps 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 DILUTION3) 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.
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 = M2V2can be used to find the new molarity
EXAMPLE 1Find the volume of distilled water that is added to100cm3 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 obtainFind final volume 250 cm3 acidV2 = M1V1 = 250 – 100 M2 = 150 cm3= 0.5 x 100 0.2= 250 cm3
EXAMPLE 2Find 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 molarityHCl = number of moles= (2+0.5) = 2.5 mole dm-3 volume = 2.5Total 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 OUTCOMEAt the end of the lesson, students should beable to: explain the meaning of neutralisation explain the application of neutralisation in daily life
Is a reaction between acid and alkali andit’s produce salt and water.- So, the acid and alkali will lost it’sproperties
The Ion EquationNaOH(aq) + HCl(aq) NaCl(aq)+H2O(aq)# H+ (aq) + OH- (aq) H2O(l)So, in neutralisation ion hidrogen(H+) from acid will combine withhidroxide ion (OH-) to produce watermolecule.
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 OUTCOMEAt 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 duringneutralization.can solve numerical problems involving neutralizationreaction to calculate either unknown concentration orunknown volume.
Definition of titration:Titration is a neutralization of an acid with a base. It is usedto find concentrations of unknown solutions.Definition of end point:The end point is found by when the indicators change color. Ithappen when the mol hydrogen ion from acid is equal to themol of hydroxide ion from the base solution. It shows whenthe indicator change the color. MaVa = MbVb (acid) (base)
Apparatus position in Titration Buret fill with acid or baseErlenmeyer Indicatorflask fill withacid or base
Acid base indicator:Medium Orange methyl Phenolphthalein Litmus color color paperAcid Red None Blue – redNeutral Orange None NoneBase Yellow Pink Red –blue
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 H2OFrom the equation, 2 mol NaOH react with 1 mol H2SO4.The number of NaOH =2The number of Sulfuric Acid 1MBVB =2MAVA 1where MB= molarity of NaoHMB(25.0) =2 , MB = 2 x 26.5 x 1.01.0(26.5) 1 1 25.0
BROMIN GROUPThe equation shows the reaction between sulphuric acid andsodium hydroxide.H2SO4 + 2NaOH Na2SO4 + H2OWhat is the volume of 1.0mol dm-3 sodium hydroxide solutionwhich can neutralize 25 cm3 of 1.0 mol dm-3 sulphuric acid?
FLUORIN GROUPFigure above shows the set-up of apparatus for the titration ofpotassium hydroxide solution with sulphuic acids. What is thetotal volume of the mixture in the conical flask at the end point ofthe titration?
CHLORINE GROUPDilute sulphuric acid added exessly to 100 cm3potassium hydroxide solution 0.1mol. calculatethe mol dm-3 number of potassium sulphate thatproduce.
Which of the acid base pair that produce orange colour when usingorange methyl indicator in titration? Acid Base IODINE GROUPI 25cm3 hydrocloric acid 1.0 25 cm3 sodium hidroxide mol dm-3 solution 1.0mol dm-3II 25cm3 hydrocloric acid 1.0 25 cm3 ammonia aques mol dm-3 1.0mol dm-3III 25cm3 sulphuric acid 1.0 mol 25 cm3 sodium hidroxide dm-3 solution 1.0mol dm-3IV 25cm3 nitric acid 1.0 mol dm-3 25 cm3 ammonia aques 1.0mol dm-3