Practical General Chemistry - PowerPoint Presentation

1. Practical General Chemistry Experimental No. 1 Titration of Sodium Hydroxide by Hydrochloric acid

2. IntroductionWhat is a titration?The act of adding standard solution in small quantities to the test solution till the reaction is complete is termed titration.What is a standard solution?A standard solution is one whose concentration is precisely known.What is a test (unknown) solution?A test solution is one whose concentration is to be estimated

3. IndicatorsIndicators are chosen, such that they change colors at the range of the pH of interest.

4. A Titration curve shows the change in pH versus volume of titrant as the titration proceeds

5. Aim of the experimentDetermination the Molarity, Normality and strength of NaOH using 0.1 N HClPractical handling and establish a titration system. How the student able to determine the end point of the reaction using an indicator

6. Reagents and materials: 0.1 N HCl solution Unknown NaOH solution Phenolphthalien (ph.ph) indicatorMethyl orange (M.O) indicatorGlass ware and apparatus: 250 mL conical flask 250 mL glass beaker 10 mL pipette 50 mL burette Washing bottle Dist. water

7. Conical flaskbeakerPipetteWashing bottle

8. Procedure1- Wash all the glassware with tap water and then with dist. water2- Put 0.1 N HCl soln in the burette using funnel and beaker3- Adjust the reading of the burette at exactly 0.0 mL4- Take 10 mL of unknown NaOH with pipette and quantitatively in the conical flask5- Add two drops of ph.ph to NaOH in conical flask, then the color or soln convert into pink6- Titrate NaOH in conical flask by 0.1 N HCl from the burette 7- Determine the end point of the reaction when the color change from pink to colorless8- Record the volume of 0.1 N HCl.9- Repeat the above steps using M.O indicator till the color change from yellow to red10- Repeat three times in each indicator

9. 1- Wash all the glassware with tap water and then with dist. water

10. 2- Put 0.1 N HCl soln in the burette using funnel and beaker3- Adjust the reading of the burette at exactly 0.0 mL

11. 4- Take 10 mL of unknown NaOH with pipette and quantitavelly in the conical flask

12. 5- Add two drops of ph.ph to NaOH in conical flask, then the color or soln convert into pink6- Titrate NaOH in conical flask by 0.1 N HCl from the burette 7- Determine the end point of the reaction when the color change from pink to colorless

13. 8- Record the volume of 0.1 N HCl.

14. 9- Repeat the above steps using M.O indicator till the color change from yellow to red

15. Results Put the obtained results in the following Table:1- in case of ph.ph indicator: ExpStartEndDifferenceAverage 1------V1V=(V1+V2+V3)/32------V23------V3Also in case of M.O construct another Table

16. CalculationsAt the end point the following equation used:NaOH N x V = N‛ x V‛ HClSo, N x 10 = 0.1 x V average in case of ph.phStrength of NaOH = N x equivalent weight of NaOH g/L equivalent weight of NaOH = (23 + 16 + 1) / 1 = 40

17. HOMEWORKChoose the best answer:A- The color of M.O indicator change from ….. to …..1- colorless to red 2- yellow to red 3- pink to redB- Both NaOH and HCl are:1- Acids 2- Bases 3- Strong C- one of the following does not used during titration:1- pipette 2- Balance 3- buretteD- both molecular weight and equivalent weight of NaOH 1- are the same 2- different E- Methyl orange indicator is type of 1- azo dye 2- inorganic compound 3- Strong base

18. 2) What is the molarity of a nitric acid (HNO3) solution if 43.33 mL of 0.1000 M KOH solution is needed to neutralize 20.00 mL of the acid solution?HOMEWORKAnswer: A- 0.12 MB- 4.33 MC- 0.21 MAt the end point the following equation used:KOH M x V = M‛ x V‛ HNO3So, M x 20 = 0.1 x 43.33

19. 3) What is the concentration of HCl if 30.0 mL of 0.10 M NaOH neutralizes 50.0mL HCl? NaOH + HCl  H2O + NaClHOMEWORKAnswer: A- 0.06 MB- 6.33 MC- 0.60 MAt the end point the following equation used:HCl M x V = M‛ x V‛ NaOHSo, M x 50 = 0.1 x 30

20. Practical General Chemistry Experimental No.2Titration of acetic aid by Sodium Hydroxide

21. Aim of the experimentDetermination the Molarity, Normality and strength of CH3COOH using 0.1 N NaOH How the student able to determine the end point of the reaction between weak acid and strong base using an indicator

22. Reagents and materials: 0.1 N NaOH solution Unknown CH3COOH solution Phenolphthalien (ph.ph) as indicatorGlass ware and apparatus: 250 mL conical flask 250 mL glass beaker 10 mL pipette 50 mL burette Washing bottle Dist. water

23. Procedure1- Wash all the glassware with tap water and then with dist. water2- Put 0.1 N NaOH soln in the burette using funnel and beaker3- Adjust the reading of the burette at exactly 0.0 mL4- Take 10 mL of unknown CH3COOH with pipette and quantitatively in the conical flask5- Add two drops of ph.ph to CH3COOH in conical flask, then the color of soln still colorless6- Titrate CH3COOH in conical flask by 0.1 N NaOH from the burette 7- Determine the end point of the reaction when the color change from colorless to pink8- Record the volume of 0.1 N NaOH.9- Repeat the above steps three times.

24. The Equation of Experimental Reaction CH3COOH + NaOH CH3COONa + H2O ReactantsProductsWeak AcidStrong base

25. A Titration curve for the reactionNOT USED

26. 1- Wash all the glassware with tap water and then with dist. water

27. 2- Put 0.1 N NaOH soln in the burette using funnel and beaker3- Adjust the reading of the burette at exactly 0.0 mL

28. 4- Take 10 mL of unknown CH3COOH with pipette and quantitatively in the conical flask

29. 5- Add two drops of ph.ph to CH3COOH in conical flask, then the color of soln still colorless6- Titrate CH3COOH in conical flask by 0.1 N NaOH from the burette 7- Determine the end point of the reaction when the color change from colorless to pink

30. 8- Record the volume of 0.1 N NaOH.9- Repeat the above steps three times.

31. Methyl orange (M.O) is not suitable indicator in titration of weak acid with strong base

32. Results Put the obtained results in the following Table:1- in case of ph.ph indicator: ExpStartEndDifferenceAverage 1------V1V=(V1+V2+V3)/32------V23------V3

33. CalculationsAt the end point the following equation used:CH3COOH N x V = N‛ x V‛ NaOHSo, N x 10 = 0.1 x V average in case of ph.phStrength of CH3COOH = N x equivalent weight g/L equivalent weight of CH3COOH = (12 + 3x1 + 12 +16 + 16 + 1) / 1 = 60

34. HOMEWORKChoose the correct answer:A- Acetic acid considered as: 1- strong acid 2- weak acid 3- an electrolyte 4- both 2 and 3B- the equivalent weight of CH3COOH equals:1- 60 2- 30 3- 40 4- none of theseC- A suitable indicator for titration of strong base with weak acid is: 1- ph.ph 2- pH 3- titration curve 4- M.OD- During titration of CH3COOH by NaOH, the color of the indicator changes from …….. to ……..1- Yellow to red 2- Pink to yellow 3- Red to pink 4- Colorless to pink

35. Practical General Chemistry Experiment No.3 DENSITY

36. The principle of density was discovered by the Greek scientist Archimedes.Density Definition In general, density is defined as the mass of substance per unit volume The density (usually represented by the Greek letter "ρ") of an object can be calculated by dividing the the mass (m) and by the volume (v) of the object.ρ = m / vThe SI unit of density is kilogram per cubic meter (kg/m3)It is also frequently represented in the cgs unit of grams per cubic centimeter (g/cm3)

37. Why is density important:Density show how different materials interact when mixed together. Examples:Wood floats in water because it has a lower density. Helium balloons float because the density of the helium is lower than the density of the air.Density is a key concept in analyzing how materials interact in fluid mechanics, weather, geology, material sciences, engineering, and other fields of physics

38. to measure the density of a liquid and solid using different methods.Each method has different precisions of measurement. You will need to keep this in mind when recording your data.Part A: Density of a liquid Graduated Cylinder Method: Choose a liquid, record its identity. Volume: Read to ± 0.1 mL Mass: Read to ± 0.01g (top loading balance) calculate the density of the liquid.The objectives of this laboratory are:

39. 39Pycnometer Method: A “pycnometer” is a flask, usually made of glass, with a close-fitting ground glass stopper so that air bubbles may escape from the apparatus. This enables the density of a fluid to be measured precisely accurately.Step 1: Rinse and dry your pycnometer before using. Calibrate the flasks volume using water, wipe all excess water off of the pycnometer before weighing!All masses are recorded to ± 0.0001g using the analytical balance.Step 2: Using the same liquid as before, determine the density with your calibrated pycnometer as follows: pycnometer

40. 40Get the Mass of water contained by the pycnometerCalculate the volume of water contained by the pycnometerBy using density  Volume of the pycnometer

41. 41Part B: Density of a SolidObtain a solid sample from your. Use the larger of the two samples. Record the identity.Water Displacement MethodGet the Mass of the solid: Read to ± 0.01g (top loading balance)2. Using the mass , determine the volume of your solid via water displacement. 3. Calculate the density using the mass-volume relation.

42. 42The level of the liquid rises due to displacement.The difference in volume is the volume of the object.6.0 mL9.0 mL 9.0 mL 6.0 mL3.0 mLThe volume is accurate because the liquid fills in completely around the irregular shape!All volumes using this method must be reported to  0.5 mL.

43. 43Pycnometer Method:Get the mass of the solid.Place the solid in your pycnometer and weight to ± 0.0001g on the analytical balance. Fill the pycnometer with water making sure there are no bubbles when the stopper is replaced. Record the mass.

44. 44 Mass Pycnometer + solid + water- Mass Pycmometer + solid Mass water surrounding the solidMass water surrounding the solidVolume of water surrounding the solidVolume of solidDensity of solidusing densitySubtract the volume of water from the total pycnometer volumeusing the solid massCalculations:

45. Answer the following questions:1.If 25 g of a liquid occupies 20 cm3 in a measuring cylinder, what is the density of the liquid?a) 0.25 g cm-3b) 0.8 g cm-3c) 1.25 g cm-3d) 5 g cm-32.You have a rock with a volume of 15cm3 and a mass of 45 g. What is its density?Ans: 3.0g cm-33. If we use the units of grams (g.)for mass and cubic centimeters (cm3) for volume, then the units for density will bea) gramsb) cm 3  c) g - cm 3d) g/cm 34.Which one of the following is not a unit of density?   a)g/m³ b) kg/m³ c) kg/m d) g/cm³

46. 5.Oil floats on water. The most accurate reason for this is a) oil is less dense than waterb) oil is immiscible (does not dissolve) in waterc) oil is both less dense and immiscible with waterd) water is heavier than oil6.Liquid water is more dense than ice becausea) A liquid H 2O molecule has more mass than an ice H 2O moleculeb) A chemical change occurs when ice melts that causes the mass of water to increasec) When ice melts there is an increase in the amount of water moleculesd) there are a greater number of H 2O molecules per unit of volume in liquid water than ice

47. Practical General Chemistry Experiment No.4 Viscosity

48. IntroductionDefine Viscosity?Viscosity is defined as the resistance of a fluid to flow either liquid or gas, in the minds of many students it is often thought of as "thickness“.Viscosity is primarily due to intermolecular forces which result from asymmetrical distribution of electrons around moleculesWhat causes Viscosity?

49. Viscosity has dimensions of mass x length-1 x time-1 Note: It is usually expressed as force x time per area One unit that has the correct dimensions is the poise The viscosity of water at 25 °C is 0.008904 poise, or 0.89 centipoise viscosity decrease as temperature increase

50. The aim of this experiment is to: determine the viscosity of a number of liquids by means of Ostwald viscometer.Viscosity can be measured using a viscometer. Ostwald viscometer is a commonly used viscometer, which consists of a U-shaped glass tube held vertically. For more accurate measurements it is held in a controlled temperature bath. It is also known as a glass capillary viscometer. A liquid is allowed to flow through its capillary tube between two etched marks and the time of flow of the liquid is measured using a stopwatch

51. Viscosity MeasurementsCapillary ViscometersIt gives the ‘kinematic viscosity’ of the fluid. It is based on Poiseuille’s law for steady viscous flow in a pipe. Procedure:Clean the viscometer with acetone and dry it.The liquid under test is introduced into the bulb; liquid is drawn into the upper bulb by suction, and then allowed to flow down freely through the capillary into the lower bulb. Two marks (one above and one below the upper bulb) indicate a known volume. The time taken for the level of the liquid to pass between these marks is measured with a stop watch and recorded. The tube is then cleaned again and the procedure repeated. You should repeat the test on each liquid three times and determine the mean time for each liquid. You also need to do test using pure distilled water.

52. Calculations:Values of the fluid viscosity are determined from the viscometer observations by means of the relation ship:=s (t) / (tss)where : is the viscosity of the fluids viscosity of water t time taken for the level of the liquid to pass between the marksts time taken for the level of water to pass between the marks density of the liquids density of water

53. Report your data in the following table for the first unknown liquidSample namedensitytimeviscosity123AverageRepeat for each liquidDo not forget to measure water

54. Answer the following questions:1- What happens to the viscosity of a liquid when its temperature is raised?The viscosity of the liquid increases.The viscosity of the liquid stays the same.The viscosity of the liquid decreases.2- What are fluids?substances that flowLiquids, and gasesa&bnon of the above3- The ..................... the attraction between molecules, the greater the viscosity of the fluid.strongerweakermore red4- What measures a materials resistance to flow?MatterVolumeHardnessTensile Strengthviscosity

55. Complete the following sentences:Water has a very low viscosity equal to 0.008904 poise at 25 ºCAn instrument that measures viscosity is called viscometerSubstances have different viscosities because they have different flow rates and thicknesses

56. Practical General Chemistry Experiment No.5 Surface Tension

57. Introduction Interface is the boundary between two or more phases exist together The properties of the molecules forming the interface are different from those in the bulk that these molecules are forming an interfacial phase. Several types of interface can exist depending on whether the two adjacent phases are in solid, liquid or gaseous state.SURFACE TENSION [γ ] is the force per unit length that must be applied parallel to the surface so as to counterbalance the net inward pull and has the units of dyne/cm

58. 1. If two liquids are completely miscible, no interfacial tension exists between them.2. Greater surface tension reflects higher intermolecular force of attraction, thus, increase in hydrogen bonds or molecular weight cause increase in surface tension Notes:Methods for measuring surface and interfacial tension 1- Capillary rise method 2- Ring (Du Nouy) tensiometer 3- Drop weight method (Stalagmometer)

59. Capillary Rise MethodWhen a capillary tube is placed in a liquid, it rises up the tube a certain distance. By measuring this rise, it is possible to determine the surface tension of the liquid. Cohesive force is the force existing between like mole­cules in the surface of a liquid Adhesive force is the force existing between unlike molecules, such as that between a liquid and the wall of a glass capillary tube When the force of Adhesion is greater than the cohesion, the liquid is said to wet the capillary wall, spreading over it, and rising in the tube.

60. If a capillary tube of inside radius =r immersed in a liquid that wet its surface, the liquid continues to rise in the tube due to the surface tension, until the upward movement is just balanced by the downward force of gravity due to the weight of the liquidr is the capillary tube radiush is the height of of the liquidg is the acceleration of gravity γ = 1/2 r h p g ProcedureDip vertically the capillary tube of the liquid (water, ethanol, …..)Measure the height (h).Repeat the above steps with different liquids.

61. CALCULATIONS Radius of the capillary tube =……….m Height of water in the capillary tube =…………….m Surface tension= RESULTSThe surface tension of water =…………..Nm-1. PRECAUTIONS 1. Capillary tube and water should be, free from grease -. 2. Capillary tube should be set vertical. 3Temperature of water should be noted. SOURCES OF ERROR Water. and capillary tube may not be free from grease.

62. Report your data in the following table for the first unknown liquidSample namehSurface tension123Average

63. Answer the following questions:1- Rain drops are spherical in shape because ofSurface tension.Capillary.The viscosity of the liquid decreases.Electrophoresis2- Unit of surface tensiondyne/cm2dyne/cmdyne/cm3second3-  Plants get water through the roots because of(a)    Capillarity              (b)      Viscosity(c)    Gravity                  (d)      Elasticity4-Surface tension isthe force per unit length that must be applied parallel to the surfacethe mass of substance per unit volumethe resistance of a fluid to flow5- If two liquids are completely …….., no interfacial tension exists between thema. Immisicible b. soluble c. acidic d. miscible

64. Practical General Chemistry Experimental No. 6 CalorimetryDetermination of the Heat of Reactions

65. Calorimetry is a process of measuring the amount of heat involved in a chemical reaction or other process. In this experiment, thermometer is used to see if the chemicals is gaining or losing heat energyA calorimeter is a device used to measure the quantity of heat transferred to or from an object such as coffee cup calorimeter.Introduction

66. The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the following reaction: AgNO3(aq) + HCl(aq) = AgCl(s) + HNO3(aq)When 50 mL of 0.1 M AgNO3 is combined with 50 mL of 0.1 M HCl in a coffee-cup calorimeter, the temperature changes from 23.40°C to 24.21°C. Calculate ΔHrxn for the reaction as written. Use 1.0 g/mL as the density of the solution and C = 4.18 J/g°C as the specific heat capacity. Example

67. Heat capacity (C): is the ratio of the heat added to (or removed from) an object to the resulting temperature change. The SI unit of heat capacity is joule per kelvinSpecific Heat capacity (Cs): is the heat capacity per gram of the liquid. The units for C are, J/g °C or J/g K. The heat transferred is: q = Cs × m × ΔtEnthalpy (H): is the heat flow into or out of a system in a constant- pressure process. The enthalpy of a reaction (∆H): is defined as the difference between the enthalpies of the products and reactants. Definitions

68. Objective To find out the heat released or changed in a reaction.ExperimentCalculation of heat of a reaction between sodium hydroxide and hydrochloric acid (Heat ofNeutralization)NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

69. BalanceCalorimeterThermometerTwo Beakers 250 mLCylinder 100 mL0.5 , 1 , 2 moles of HCl0.5 , 1 , 2 moles of NaOHMaterials

70. 1. Weight the calorimeter (mcal).2. Prepare Sodium Hydroxide and Hydrochloric Acid solutions.3. Take 100 ml of both NaOH and HCl with concentration of 0.5 M. Put them in two separated beakers and measure the temperatures of both solutions (t1 and t2) and calculate the average temperature (tav). 4. Pour NaOH and HCl together to the calorimeter and close the calorimeter and put the thermometer in. 5. Stir quickly and carefully with the thermometer and record the final temperature (tf).6. Repeat steps for 1.0 M and 2.0 M, but make sure you’ve washed the calorimeter before repeating the steps.Procedure

71. The specific heat capacity of the calorimeter (Ccal) is 0.89 J/g°C. The specific heat capacity of the solution (Csol) is the same as that of water, 4.18 J/g°C. The density of the solution is 1.0 g/mL. Therefore,msol = Vsol = 200 1) Assume that Calculations

72. 2) The difference in temperature is:Δt = tf – tav3) The amount of heat released from the calorimeter:qcal = mcal x Ccal x Δt 4) The amount of heat released from the solution:qsol = msol x Csol x Δt 5) The total amount of heat:qT = qcal + qsol Calculations

73. Solved ProblemA large paraffin candle has a mass of 96.83 gram. A metal cup with 100.0 mL of water at 16.2°C absorbs heat from the burning candle and increases its temperature to 35.7°C. Once the burning is ceased, the temperature of the water was 35.7°C and the paraffin had a mass of 96.14 gram. Determine the heat of combustion of paraffin in kJ/gram. (density of water = 1.0 g/mL).Qparaffin = -QwaterThe energy gained by the water in the calorimeter can be determined.Qcalorimeter = m•C•ΔTQcalorimeter = (100.0 g)•(4.18 J/g/°C)•(35.7°C - 16.2°C)Qcalorimeter = 8151 JThe paraffin released 8151 J or 8.151 kJ of energy when burned. This is based on the burning of 0.69 gram (96.83 g - 96.14 g). The heat of combustion per gram, ΔHcombustion - paraffin = (-8.151 kJ) / (0.69 g)ΔHcombustion - paraffin = -11.813 kJ/g.Solution

74. Questions* Calorimetry* Calorimeter* Heat capacity* The enthalpy of a reaction1) Define the following expressions:

75. i) ……………… is a process of measuring the amount of heat involved in a chemical reaction or other process.a. Calorimetry b. Heat capacity c. The enthalpy of a reactionii) ……………… is a device used to measure the quantity of heat transferred to or from an object. a. Heat capacity b. Calorimeter c. The enthalpy of a reactioniii) …………….. the difference between the enthalpies of the products and reactants. a. Calorimetry b. Heat capacity c. The enthalpy of a reaction2) Choose the correct answer of the following:

76. Practical General Chemistry Experimental No.7 Molecular Formula

77. IntroductionThe molecular (true) formula for a substanceIt is the formula that show the true number of atoms in the compound.The empirical (simplest) formula for a substanceIt is the formula that show the ratio of the atoms in the compound.

78. ExampleIn acetylene C2H2, the molar mass for acetylene is 26 g/mol, In benzene C6H6 the molar mass of benzene is 78 g/mol. BUTThe empirical formula of acetylene and benzene are CH

79. To determine the molar mass of a gaseous substance.Use molar mass value to find the molecular formula of the substance.Aim of the experiment

80. Safety ConsiderationsWear protective glasses and an apron at all times. Avoid skin contact with solids and solutions. There should be no flames in the laboratory during this laboratory activity. Dispose of all solutions in the containers provided by your teacher. Wash your hands before leaving the laboratory.

81. Wash all the glassware with tap water and then with dist. WaterHandle the flask at all times with a paper towel.Determine the mass of a stoppered 250 mL flask of air.In an operating fume hood, place the tubing outlet from an unknown gas supply completely into the flask. Allow the gas to flow for at least 30 s to replace the air. Stopper tightly. If your teacher suggests the gas is less dense than air, invert the flask for filling.Procedures

82. Find the total mass of the flask, stopper, and unknown gas. Repeat Steps 4, 5 and 6.Place a mark on the outside of the flask neck at the bottom of the stopper. In an operating fume hood, fill the flask to the mark with water.Measure the volume of water to the nearest milliliter with a graduated cylinder.Determine the temperature and air pressure in the laboratory.Wash hands thoroughly before leaving the laboratory.

83. Data and Calculationsa Mass of the flask (empty) + cap =……………gm b Mass of the flask + Cap + unknown gas (step No. 8) = .....gmV volume of the flask (from step No. 10) = ....... lP atmospheric pressure = 1 atmT Lab. Temperature = 25+273.15= 298.15 K

84. From the ideal gas Law:PV=nRTWhere:P 1atm.T 298.16KR= 0.082 l.atm./mol. KV is known from step 10n = w/MGass mass (step 8)molar mass

85. HOMEWORKThe empirical formula of acetic acid CH3COOH is:C2H4O2 b) COOH c) CH2O2) If the empirical formula of benzene is CH its molecular formula is?a)C7H8 b) C2H2O2 C) C6H6

86. 3) The empirical formula is ………. the molecular formula a) higher than b) smaller than c) equal

87. Practical General Chemistry Experimental No.8 Acidic and Basic Radicals

88. Aim of the experimentTo verify the presence of a certain ion (cation or anion) in solution , a qualitative analysis must be performed to the solution

89. The meaning of acidic and basic radicalsAny salt is divided into to radicals:1- Acidic radical with negative charge (Anion)2- Basic radical with positive charge (Cation)Any SaltAcidic RadicalBasic RadicalNa2CO3, KCl, CuSO4, CaCO3, NaNO3

90. The acidic radicals divided into three groups, each one with a specific reagent:1st GROUP2nd GROUP3rd GROUPdil HClconc H2SO4BaCl2 solnCO32-CarbonateCl-ChlorideSO42-Sulphate HCO3-BicarbonateBr-BromideB4O72-Borate S2-SulfideI-IodidePO43-Phosphate SO32-SulfiteNO3-nitrateS2O32-ThiosulphateNO2-Nitrite

91. The basic radicals divided into six groups on the basis of solubility product, each one with a specific precipitant:1st Group2nd Group3rd Group4th Group5th Group6th Groupdil HCLdil HCl + H2SNH4Cl + NH4OHNH4Cl + NH4OH + H2SNH4Cl + NH4OH + (NH4)2CO3------Ag+Cu2+Al3+Zn2+Ba2+K+Pb2+Hg2+Cr3+Mn2+Ca2+Na+Hg22+Cd2+Fe3+Ni2+Sr2+Mg2+Bi3+Co2+NH4+ChloridesSulfides in acidic mediumHydroxides Sulfides in basic mediumCarbonates -------

92. Physical properties of the unknown salt1- Color (Any color) 2- Shape (Crystal, Powder, Sheets)3- Solubility (sol, insol) Before the chemical test on the unknown, students must investigate the salt physically:

93. Solubility tests:Try to dissolve the solid material in the following solvents at the order of:Water, Hot waterHCl dil , HCl dil + heatingHCl conc , HCl conc + heatingHNO3 dil , HNO3 dil + heatingHNO3 conc , HNO3 conc + heatingAqua regia (3V of HCl conc + 1V of HNO3 conc)

94. Test TubesTest Tube holderTest Tube Brush

95. Plastic PipetteBunsen burner

96. Flame Test

97. Chemical Properties:ConclusionsObservationsExperiment1) Solid salt solution + Ammonium chloride solution + NH4OH + (NH)2CO32)Solid salt solution + Calcium sulphate solution BaCl2 + CaSO4Barium chloride + calcium sulphate solution3) Salt solution + Sodium phosphate solution 3BaCl2 + 2Na2HPO4Barium chloride + disodium hydrogen phosphate

98. IronFerrous, Fe2+Ferric, Fe3+CopperCuprous, Cu+Cupric, Cu2+TinStannous, Sn2+Stannic, Sn4+LeadPlumbous, Pb2+Plumbic, Pb4+MercuryMercurous, Hg+Mercuric, Hg2+Cations with different charge

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109. HOMEWORKChoose the best answer:A- Investigation a type of both acidic and basic radicals is considered as ….1- Organic analysis 2- Qualitative analysis 3- Quantitative analysisB- The basic radical has ……. charge 1- positive 2- negative 3- neutralC- Basic radicals are divided into …….. Groups1- six 2- three 3- sevenD- Borate radical has a symbol 1- B2O42- 2- B2O22- 3- B4O72-

110. Practical General Chemistry Experimental No.9Colligative Properties

111. BackgroundColligative properties: are properties of a solvent, which depend on the concentration of solute particles dissolved in the solvent. freezing point depression and boiling point elevation,Such as:

112. The decrease in freezing point, Δ Tf (freezing point depression) for a near ideal solution can be described by the equation: ΔTf = kf · m kf is the molal freezing point depression constant of the solvent with units °C · kg solvent/mole solute. m is the molal concentration of the solute dissolved in the solvent expressed as moles of solute/kg solvent.Where:

113. Scince:moles = g/M where M is the molecular weight.By substituting into the unit definition of molality yields: m = g/(M · kg solvent)Then: Δ Tf = (kf · g)/(M · kg solvent)

114. M = (kf · g)/(ΔTf · kg solvent)We can rearrange and solve for the molar mass, mol wt, as is shown below. Therefore, if we know the mass of unknown compound added to a known mass of solvent and determine the change in freezing point of the solution, relative to pure solvent, we can use this equation to determine the molar mass of the unknown compound.

115. Aim of the exoerimentIn this experiment you will determine the freezing point of pure tertiary butyl alcohol (t-butanol)Then the freezing point of t-butanol with an unknown solute dissolved in it. From these freezing point measurements you will be able to calculate the molar mass of the unknown solute.

116. (A)Freezing point of pure t-butanol:Place a 150 ml beaker on a top loader balance and tare it. Place a clean, dry 25 x 150 mm test tube in the beaker and record the mass in the data table, Fill the test tube about half full with t-butanol, reweigh and record this mass in the data table, Place the beaker and test tube aside for now.Place about 250 ml of hot tap water in one 400 ml beaker and place the test tube containing the t- butanol in the hot water; we want to warm the t-butanol to about 40 °C. You may use a tripod and Bunsen burner to warm the t-butanol in the waterProcedure

117. Insert the stirring loop into the test tube and then insert the thermometer so that the loop of the stirrer surrounds the thermometer. Periodically stir the t-butanol with the stirring loop by an up and down motion while warming it.Place about 250-300 ml of ice in the other 400 ml beaker and enough cold tap water to just cover the ice. Once the temperature of the t-butanol has warmed to about 40 °C, transfer the test tube to the ice- water bath making sure that most of the t-butanol is below the surface of the ice-water.

118. Immediately begin to take temperature readings and record them in the table every 15 seconds while continually stirring the t-butanol with the stirring loop in an up and down motion. Continue to stir and take temperature readings until the t-butanol has solidified. When the t-butanol has solidified so that the stirring loop will no longer move, stop trying to stir, but continue to record the temperature every 15 seconds for one more minute.

119. Place the test tube in the 150 ml beaker. Place the beaker with test tube on a top loader balance and tare the balance. With a disposable pipette add about 0.5 g of our unknown to the test tube and record the mass, Reheat the test tube as before to about 40 °C, Try to make sure all your unknown is dissolved in the t-butanol. Use the stirring loop to aid the dissolution of the solute, if needed. As before, once the temperature is about 40 °C, transfer the test tube to the ice-water bath. (B) Freezing point of solutions:

120. Begin stirring and take temperature readings every 15 seconds until the solution has solidified. continue to record the temperature every 15 seconds for one more minute.Return the test tube to the 150 ml beaker and place the beaker with test tube on a top loader balance and tare the balance. With a disposable pipette add about another 0.5 g of the unknown to the test tube and record the mass, Repeat the melting and temperature recording steps(B) Freezing point of solutions:

121. Data and Calculations1) Mass of test tube:2) Mass of test tube and t-butanol:3) Mass of t-butanol in kgs:4) Mass of first sample of unknown:5) Mass of second sample of unknown:6) Total mass of unknown in second solution freezing point

122. Data and CalculationsData Table:t-butanol, pure solventt-butanol plus first sample portion, solution 1t-butanol plus second sample portion, solution 2time temperatureTime temperature

123. Data and CalculationsUse graph paper and plot temperature vs. time for the pure t-butanol and for each solution analyzed; make 3 different graphs. determine the freezing point, Tf, of t-butanol and of each solution. Determine the ΔTf of solution 1 by subtracting the Tf of solution 1 from the Tf of pure t-butanol.

124. Data and CalculationsDetermine the ΔTf of solution 2 by subtracting the Tf of solution 2 from the Tf of pure t-butanol.Use the ΔTf for solution 1 along with the mass of unknown in solution 1, mass of solvent, t-butanol, and the kf of t-butanol, 9.10 oC·kg solvent/mol solute, to determine the molar mass, M, of your unknown compound.Repeat the calculation above for solution 2

125. HOMEWORK1. A homogeneous mixture of two substances is acolloid. b. suspension. c. solution. d. compound.2. The dissolving medium in the solution is the a. solute. b. solvent c. solution. d. mixture.What is the solvent in 70% alcohol solution? a. alcohol b. water c. Sugar d. kerosene

126. The number of moles of solute divided by the number of kilograms of solvent isa. Percent by mass b. Molality c. Molarity d. Mole fractionWhat is the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent?a. 1.28 m c. 2.18 mb. 1.58 m d. 5.18 m

127. Practical General Chemistry Experimental No. 10 Chemical Reactions

128. Changes go on about you all the time. Some changes are chemical changes, such as gasoline burning or a nail rusting. But what is happening when a chemical change occurs? What is the nature of a chemical reaction? Introduction

129. ObjectiveTo examine the behavior of matter in a chemical reaction.Focusing on the behavior of the individual particles of each substance involved.

130. Procedure Part I Three reactions will demonstrate. Record your observations and answer the following questions: • What evidence for a chemical reaction did you observe in each demonstration? • Define "precipitate".

131. Procedure Part IIAssemble the following materials: petri dish, vials containing crystals of lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI, spot plate, small test tube, and two labeled, thin-stem pipets containing solutions of Pb(NO3)2 and KI. Record all your observations and answer the questions after the following procedure:

132. Procedure 1) Grinding of solids: Place a few crystals of solid lead(II) nitrate and solid potassium iodide in the same well of a spot plate. Using the small test tube, carefully grind the solids using a circular motion. Do not press too hard or you may break the test tube. 2) Prepare petri dish: Cover the bottom of the petri dish with distilled water. Place on a white sheet of paper.

133. Procedure 3) Simultaneous addition of solutions:Using the thin-stem pipets, simultaneously add 10 drops of lead(II) nitrate solution and 10 drops of potassium iodide solution on opposite sides of the petri dish as shown in the diagram below. Do not move or shake the dish

134. Procedure 4) Delayed addition of potassium iodide:Repeat Steps 2 and 3 but wait five minutes before adding the potassium iodide solution. 5) Dump all solutions into the waste container provided: Using a small amount of water, wash the solid from Step 1 into the waste container. 6) Wash hands thoroughly before leaving the laboratory.

135. Data Analysis, ConceptHow do you know a chemical reaction occurred? How were the results in Step 1 and Step 3 similar? How were they different?Do you think the chemical reactions were the same in both Step 3 and 4? Why?How is the observation in Step 3 different from that in Step 4? What do you think caused this difference?Why did it take longer in Step 3 for an observable change?

136. Data Analysis AnswersA yellow solid forms; change of color.Both systems form a yellow solid. The reaction takes longer to form in the solution. The solid formed is not where the reactants were added.Yes, the color formed was the same.The color appears in a different spot. The lead(II) nitrate has had more time to move around.The materials had to move in the water before they got together.

137. Other Experiments

138. Reagents0.1 M sodium carbonate, Na2CO3, 4 g per 500 mL 0.1 M calcium nitrate, Ca(NO3)2, 4 g per 250 mL 0.02 M iron(III) chloride, FeCl3, 0.8 g per 250 mL 0.001 M pot. thiocyanate, KSCN, 0.03 g per 250 mL 3 M hyd. acid, HCl, 25 mL; 6 mL conc. HCl per 25 mL 0.25 M lead(II) nitrate, Pb(NO3)2, 8.3 g per 100 mL 0.50 M potassium iodide, KI, 8.3 g per 100 mL

139. ProcedureReaction 1: Pour 200 mL of 0.1 M Na2CO3 solution into a 500 mL beaker. Now add 200 mL of 0.1 M Ca(NO3)2 solution. Observation:Formation of a white precipitate.

140. ProcedureReaction 2: Pour 200 mL of 0.02 M FeCl3 solution into a 500 mL beaker. Add 200 mL of 0.001 M KSCN solution. Observation:Formation of a red-colored solution.

141. ProcedureReaction 3: Pour 200 mL of 0.1 M Na2CO3 solution into a 500 mL beaker. Add 25 mL of 3 M HCl. Observation:Evolution of a gas (CO2).

142. QuestionsWhat is your suggested observation upon:1) Addition of Na2CO3 solution to Ca(NO3)2 solution. Formation of a white precipitate. 2) Addition of FeCl3 solution to KSCN solution.Formation of  a red-colored solution.3) Addition of Na2CO3 solution to HCl solution.Evolution of a gas

143. Practical General Chemistry Experimental No.11 Oxidation Reduction Titration

144. Introduction

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152. Ex. Standardization of potassium permanganate using oxalic acid(Reduction-Oxidation Reaction)Aim: Calculate NKMnO4Theory: KMnO4 is a 2nd standard substanceKMnO4 + C2O4H2 + H2SO4 = K2SO4 + MnSO4 + CO2 + H2O

153. Conditions for KMnO4 titration:Acidic medium using 20 ml 2N H2SO4 Why should you avoid HNO3 and HCl???Heating to 70 oC (not boiling, why??)Define: Oxidation, Reduction, Oxidizing agent, Reducing agent??? How to balance the redox. Equation?????

154. ProceduresTransfer 10 ml oxalic or oxalate soln. in to C.F.Add 20 ml 2N H2SO4Heat to 70 oCTitrate with KMnO4 gradually with stirring till pinkCalculate (N x V)KMnO4 = (N x V)oxalic

155.

156. Home WorkOxidation process involved…………Loss of electronsGain of electronsIncrease of pHDecreasing in acidity2. Chemical structure of potassium dichromate is………K2CrO4K2Cr2O7KMnO4K2CO33. Reduction process has been done by………….Reducing agentOxidizing agentReference electrodeCalomel electrode

157. 4. KMnO4 is oxidizing agent and has been used for redox titration in acidic medium of …………H2SO4HClHNO3CH3COOH5. Potassium permanganate is………K2CrO4Self indicatorReducing agentStandard material

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159. Practical General Chemistry Experimental No.12 pH Measurements

160. Introduction The numerical scale called pH indicates how acidic or basic an aqueous (water) solution is, or whether that solution is neutral. Many products we use daily for personal hygiene, home and auto care, or eating and drinking are suitable for pH testing. For this laboratory activity, you may bring to class as many products, in their original, closed containers, as you wish to test.

161. PurposeTo determine the pH of common household products and classify each based on its acidic or basic properties using the pH scale.ProcedureUse the data table provided, or construct one.• Obtain several strips of wide-range pHydrion paper and several strips each of various narrow-range papers (4-6, 6-8, etc.).

162. • Arrange the strips so they are convenient to use during testing.• Set up an area for temporarily discarding the used test papers. (Not the sink!)• First use the wide-range paper, then the narrow-range paper to test several products, one at a time. Transfer a few drops of the product to be tested to a 50 mL beaker or watch glass. Dip a clean glass stirring rod into the product.

163. • Observe the color change. Then match the final color to those on the chart to estimate the pH value of the product. • If you are unsure of the color, use two narrow-range papers that either overlap or follow. (Example: If pH approximates 8.0, use 6-8 followed by 8-10.) • Rinse and dry the stirring rod before each test. • Record the pH of each product. • Wash hands thoroughly before leaving the laboratory.

164. Data Analysis - Trade data with classmates who tested different products. Gather data on at least 20 products. - Formulate at least three distinct conclusions from your data. Look for any similarities or differences among product pH values.

165. Imply, Apply 1. Are the pH values of the products tested consistent? Do they make sense? Why or why not? 2. State which, if any, pH values surprised you. Why were you surprised? 3. Look up the pH value or range found on human skin. Also try finding the meaning of "pH balanced". (A skin-care book may be helpful.) Were any products "pH balanced?" Which one(s)? Cite any reference(s) used. 4. Predict the mouth - pH range most likely to promote tooth decay. Extra Credit: Provide an answer from your dentist or one obtained

166. through library research. 5. What did you like about this laboratory activity? Why? Did you dislike anything about this activity? 6. Suggest improvements for this laboratory activity. 7. Do you have any questions about pH? If so, list them on a separate sheet of paper and submit them to your teacher.

167. Increasing the hydronium ion concentration causes…………Increase in pH valueDecrease in pH valueNot effective of pH valueDecreasing in acidity2. pH 3 means that the medium is……AcidicNeutralBasic3. pH value can be calculated from the equation below:pH = log [H+]pH = -log [H+]pH = -log [OH-]pH = log[OH-] HOMEWORK

168. 4. pH 9 means that the medium is……AcidicNeutralBasic5. pH 7 means that the medium is……AcidicNeutralBasic