CHM113/Chemistry Lab 2 (UM)
CHM113/Chemistry Lab 2 (UM) CHM113
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Date Created: 09/28/14
CHM113 Chemistry 1 Lab Exam Lab Safety Instructions Approved eye protection worn at all times Full coverage amp closed toe shoes are required at al times nothing below ankle visible Eating and drinking not allowed May not enter lab without TA present only work on authorized experiments Clothing cover entire body don39t wear good clothing to get dirty Learn locations of and how to operate eyewash stations safety showers fire extinguishers and fire alarm In case of injury or accident no matter how small take appropriate first aid action and tell another student to notify TA No visitors horseplay pranks Hotplates never to be turned up past 4 for most purposes keep at 3 can go over 550 When working with chemicals reagents and lab equipment 0 O O O Unplug hotplates when not in use Read labels on chemical and reagent bottles carefully using the wrong chemical can create a serious hazard Prevent skin Contact with solvents and chemicals gloves always use spatula or scoopula when transferring solids if you do touch something wash area for several minutes with cold water Avoid breathing fumes of any kind and evaporate ORGANIC solvents in the hood Always use suction bulb to fill pipets never by mouth All chemical wastes pour into appropriately marked waste bottles if you don39t know where ask Toxic corrosive or ammable chemicals acids bases solvents should be kept in and dispensed inside the hood Never return excess chemicals to the containers you got them from Put lids and caps back on chemical and reagent bottles immediately after use Clean up all spills immediately 0 O Spills on floor someone can slip and don39t know if its acid or water Use sodium bicarbonate to clean up acid splatters Base spills use boric acid to water with water All students are responsible for cleaning up work areas as well as common areas Lab subjective grade will suffer if you dot clean up after yourself Introduction to Computers and LoggerPro LoggerPro designed to make lab measurements simpler and more accurate by using computer to aid in data collection 0 Good practice to let a temperature probe sit on the bottom of a beaker especially if the beaker is on a hot plate easiest to do using a ring stand and a clamp designed to hold the temp probe Possible probes pH conductivity pressure temperature Autoscale linear fit Melting point of lauric acid LabPro Uses temp probe electric thermometir Set data collection parameters 180 seconds as length of time Dip temp probe in alcohol click collect immediately Highlight at portion of graph portion of curve after the temperature has dropped during the evaporation of alcohol and has returned to room temperature linear fit 0 Y intercept value room temperature 16 C Freezing point of Lauric Acid Put test tube of lauric acid into warmed water on hot plate While it melts set up labpro o 200mL cold water in 250mL beaker 0 clear latest run set data parameters for 600 seconds when lauric acid has melted take it off the hotplate and put temp probe in it and the test tube in the cold water collect to get good results make sure the tip is in molten lauric acid constantly stir with probe until solidifies after temp readings stops dropping you can stop stirring and add a few ice chips to the cold water bath collect data until the temp of lauric acid begins to decrease again 10 mins at portion of graph melting point 0 highlight and linear fit 0 y intercept freezing point free probe melt again if need be known melting point of lauric acid 435 C if the temperature of lauric acid dips below the freezing point temperature then rises back up to freezing point just before the at portion super cooling Post Lab temperature V Time plot of evaporation of alcohol o the temp decreases in the presence of alcohol from 21 degrees to 162 degrees removing energy until it plateaus and increases gradually back to room temperature at 183 degrees temperature V time plot for the freezing point of lauric acid 0 section 1 when the acid is cooling down dramatically in room temperature water to become a solid state 0 section 2 the cooling down plateaus and the liquid acid is nearly completely a solid 0 section 3 the ice is added to the water and completely freezes the liquid back to solid state Density density the ratio of a substance s mass to its volume 0 physical properties of a substance independent of the mass of the substance and volume 0 density intensive property 9 independent of size of sample 0 mass and volume extensive properties 9 depend on size of the sample taken since the density of a pure substance is an intensive property of that substance we can use the density to aid in identifying the substance 0 although density intensive property does not mean that extensive properties cannot affect it balloon put in freezer density of substance relates information on how the atoms or molecules that make it up are arranged in space 0 gases low density molecules take up 500 to 1000 times more space than the corresponding liquid or solid 0 liquids less dense than corresponding solids means atoms or molecules are more tightly packed in the solid state than in the liquid 0 tighter packing generally implies stronger interactions between the entities I 2 exceptions water density also conversion factors from a mass basis to a volume basis specific gravity of a substance is the ratio of the density of the substance to the density of water 0 since the density of most everything depends on temp 9 specific gravity must also have the information about the temp of the measurement and temp of the water to which it is referred o Usual reference temp for water density 4 C temp of max density of water Another place density is important is earth earth has been density segregated over the eons 0 Atmosphere is on top then oceans light rocks then heavy rocks then molten heavy stuff in centre 0 Geologists use density to classify rock formations Procedures Direct amp indirect methods to determine densities of objects of various sizes and shapes 0 Difficult to measure density directly Find density of 4 objects with 3 different methods Density of block of wood Measure dimensions of block cm and determine volume Weight the mass of block 9 density How many cubic centimeters in one liter 1000 cm3 1 L Density of a metal cylinder Weigh cylinder on balance record its mass Volume of cylinder Vcyl H r2 h Calculate density and specific gravity of cylinder Density of a rock Irregular object Weigh rock in the air Weigh rock while totally submerged in water Water has buoyancy property 9 water exerts force on the object that is in the oppose direction of gravity 0 The rock weighed less in water because water pushing back on rock 0 Difference in 2 masses mass of the water that the rock displaced Density of Plastic PVC Too small to measure directly must measure density indirectly by levitating it in a solution of the same density then determine density of solution Only works for water insoluble materials if we use aqueous solution 9 method of oatation 0 When the resin is more dense that the solution the resin will sink 0 But when the resin and the solution have the same density the resin will levitate in the middle of the solution Use 2 solutions of calcium chloride that we will use 0 Concentrated calcium chlorides density greater than plastic Dilute calcium chloride density less than plastic Add 3 mL of each olution to separate test tubes Add plasic to dilute solution first note what happens Use spatual to fish out piece of platic and add to concentrated solution note what happens 0 Add small amounts of dilute into concentrated and mix after each addition until the piece levitates I If oats add more dilute I If sinks add more concentrated 0 Important to mix them very well they have different densities will not mix together readily Once they are at the dame density 9 10 mL graduated cyclinder 0 Weigh record its mass add 8 mL inside wipe off all outside solution 0 Measure record the exact volume of solution Weigh the mass of the graduated cylinder with the solution 0 Difference between mass of graduated cylinder with solution and withought mass of solution I Determine the density of solution and thus the density of the plastic 0 O O O 0 Post Lab Units of density gcm3 OR gmL Density of objects substances depends on the state of that substance at a particular temperature which is the measurement about how the atoms and molecules are arranged in space Tighter packing generally implies higher density Density is an important because as an intensive physical property it can aid in the identification of a substance It can also be useful in classifying entities on earth such as rocks Highest error 9 the measurements of the pieces of wood due to the reading of the rules measurements which were only to 2 sigfigs The lowest error 9 weighing the substances due to the electronic values gives as objective unarguable measurements to 3 sigfigs Why ice cubes oat in water o The density of ice is less than water as a liquid Water is one of the few substances that has a solid state that is less dense than its liquid state Structure of ice and water 0 The structure of ice and water differ in the way that their molecules are arranged in space o The expansion of water when it freezes increases the space between the molecules thus rendering it less dense o In the liquid form the atoms are more tightly packed and therefore more dense than its solid form Notes Density intensive property Mass and volume extensive property Which is less dense 0 Water ice ice 0 Gold powder gold nugget same 0 Ocean water distilled water distilled water Rank in increasing density 0 Gases lt liquids lt solids Why does the marble weight less in water than in air 0 Because water is more dense than air and water has buoyancy 9 it exerts a force on the object that is in the opposite direction of gravity Purification of Adipic Acid H2C5H304 Part of crude adipic acid needs purification came from waste pile in Odessa Texas Used in food avorings economically important in synthesis of nylon 66 9 used in upholstery carpet fibers clothing tire reinforcement auto parts Worldwide production of adipic acid 22 million metric tons per year 0 Problem comes in how this rather large quantity of adipic acid is synthesized or made I Relies on use of nitric acid oxidation of compounds such as cyclohexanol and cyclohexanone which must be synthesizes from benzene toxic I Use of nitric acid 9 results in waste emissions of nitrous oxide N20 I Nitrous oxide greenhouse gas contributes to global warming ozone depletion smog and acid rain 0 Even with efforts to recover and recycle N20 around 400 thousand metric tons of it are still emitted into atmosphere each year 0 Estimated to be between 5 8 of worldwide emissions of N20 A greener approach would be better Adipic acid 9 very soluble in boiling water slightly soluble in cold water 0 Temp dependence of the solubility is very favorable for recrystallization from water 0 Recrystallization9 one of best ways to purify compounds I When crystals form the impurities are left of the crystal because they are in solution Brown color of crude material 9 due to oxidized compounds and fine carbon particles 0 May be removed by absorption on to activated charcoal Separation of adipic acid from impurities done in 2 stages 0 Day 1 get adipic acid into solution and filter out impurities then when crystallization beings separate crystals from water amp put in locker to dry 0 Day 2 weigh the crystals and determine their purity using a MelTemp or other apparatus used to determine the melting point of a substance Procedure First Lab Period Make boat to weight out sample 9 weigh 20 g crude adipic acid record to 3 dp Add crude adipic acid to 250 mL Erlenmeyer Flask and 75 mL of distilled water Bring to boil on hotplate stir occasionally After bult of crude acid is dissolved add small scoop of activated charcoal to boiling solution leave to boil for 3 5 mins Prepare uted filter paper using plastic funnel turn off hot plate Pour solution into uted filter which has a 150 mL beaker as a receiver 0 Pour down the side of paper or center will break do quickly so it doesn39t begin to crystallize before it gets into the beakers Allow filtrate to cool colourless 15 mins place in ice bath for 1015 mins 0 Do not stir or crystals will be small let larger ones form Prepare clean uted filter paper and place in a funnel Weigh filter paper record mass to 3 dp Filter crystals of purified adipic acid 9 crystals should be colorless o The more folds in the filter paper increase surface areas for contact with solution leading to more rapid rate of filtration less time for the cool solution to warm Second Lab Period When crystals are dry weigh the in boat Determine purity based on weight of crude adipic acid at start of procedure Determine melting point of pure adipic product o Compare it to that you started with Recovery weight of pure adipic acid weight of crude adipic acid x 100 Taking melting point is good way to determine purity of a product o Melting point of pure produce closer to literature value o Range of temps from started to finished very narrow When impurities present 9 melting point lower and broader o Impurities interrupt the normal intermolecular forces that keep the crystal together MelTemp to determine melting point 0 Obtain glass capillary tube put adipic acid all the way to the bottom of tube push open end down on sample and tap closed end gently on bench 0 Put it in the slots in the window of the apparatus o Adjust temp controls to give you a final temp just below literature value of adipic acid I When it reaches set temp it will beep I Then start heating sample at rate 1 degrees C per minute by preson go to button o Record temp of final melting point Post Lab Impurities of crude adipic acid 0 Carbon compounds o Nitrous oxide compounds oxidized compounds 9 N20 Purpose of adding carcoal o Insoluble impurities are attracted to charcoal form an aggregate which can be separated by filtration Heating adipic acid solution o In order to filter the solution and allow soluble adipic acid to filter through leaving insoluble impurities on filter paper the acid is soluble in boiling water Solution cooled 0 In order to collect the pure adipic acid crystals and be separated from the soluble impurities through filtration Using uted filter paper 0 Numerous folds in filter paper gives a larger surface area for contact with the solution leading to a more rapid rate of filtration less time for cool solution to warm up The crystals formed 0 Were solid and white and formed in clusters were fragile and compacted powder like What happened to the impurities 0 Both soluble and insoluble impurities are separated through filtration Purpose of this experiment 0 Purpose is to purify the solution of adipic acid by removing the impurities by boiling as then the acid becomes soluble and can be filtered Chemical principle illustrated 9 solubility Why is there a difference between the 2 melting points of crude and pure adipic acid 0 Due to the impurities o The crude has a lower melting point as the impurities are present and break down the covalent bonds quicker by weakening them Conclusions 0 Impurities interfere with packing of molecules in a substance lowing the melting point 0 Removing the impurities and allowing the substance to reform will allow the molecules to pack more tightly raising the amount of energy needed to melt the substance Other physical properties of pure compounds 0 Pure compound color will change with the presence of impurities 0 Density and volume will also be effected Notes Compared to crude adipic acid 9 pure adipic acid has HIGHER melting point NARROWER range Dissolve crude in hot water and filter crystals in low temperature Recrysallization is one of the best was to purify acid because its only slightly soluble in cold water rand very soluble in boiling water Important not to touch crystals in ice bath 9 so fully form into larger crystals Important to record the range of temperatures at which sample beings to melt into a liquid 0 If there are impurities the range is broader than that of the pure substance Production and release of nitrous oxide 0 O O Depletes ozone layer Acid rain smog Contributes to global warming Characterize sample is pure 0 Compare literature melting point to the one obtained Brown color 9 due to oxidized compound 0 Removed by absorption on activated charcoal Right order of steps 0 OOOOOOOO O Weigh crude adipic acid dissolve crude adipic acid with cold water boil and stir the solution add pea sized amount charcoal filter the hot solution cool the filtrate to room temperature cool filtrate to zero degrees filter crystals wash crystals with cold water dry crystals in locker 3 ways that recovery may be lowered O O O Molecular Modeling VSEPR Theory amp Shapes Building molecules using Spartan PC Pro 0 Build models of molecules to see where the atoms and electron pairs go then they are bound in molecules Molecule with a central atom and 2 substituents 9 linear o Substituents are on opposite sides of central atom 0 Eg Carbon dioxide Molecule with central atom and 3 substituents 9 trigonal planar o Are all in the same place 120 degrees from each other 0 Not may molecules take this shapes because only boron has a valence of 3 0 Usually assumed by ions like N03 Molecules with 4 substituents 9 tetrahedral o All are 1095 degrees from each other 0 Can be visualized drawing a cube 0 Eg Carbon is most common atom to assume this shape 0 When one of the substituents on a tetradhedal array is in the form of a lone pair of electrons we see a different shape I Can be visualized by simply taking one of the substituent off of the tetrahedron trigonal pyramid 0 When 2 of the subtituents electron pairs 9 bent shape 0 Eg Water Molecule with 5 substituents 9 trigonal bipyramidal o 2 trigonal pyramids placed base to base 0 2 substituents on opposite sides are called axial while 3 in opposite plane are called equatorial 0 When 1 substituent electon pair 9 seesaw shape I First substituent that must be removed is an equatorial one if an axial one is removed the substituents are too close and a high energy molecule results 0 When 2 lone pairs 9 t shaped because another equatorial substituent is removed Molecule with 6 substituions 9 octahedral o Visualized with cube again 0 Eg Transition metal complexes are the most common octahedral arrays but sulfur can do it too Conductimetric Titration and Gravimetric Determination Analyzing solution of BaOH2 of unknown concentration 2 basic questions must be answered 0 What compounds or elements are present in the substance I Qualitative analysis or qual 0 Exactly how much of a particular compound or element is present in a sample I Quantitative analysis or quant Goal to learn the careful handling isolation and analysis of a chemical sample Perform quantitative analysis of BaOH2 in 2 ways 0 Conductimetric titration o Gravimetric dermination Conductimentric Titration Monitor conductivity during reaction between sulfuric acid H2504 and barium hydroxide BaOH2 9 determine the equivalence point From this info find the concentration of the BaOH2 solution The reaction between sulfuric acid and barium hydroxide yields an insoluble product barium sulfate and water Ba 2 aq 2 OH aq 2H 39 aq S04 239 aq 9 BaS04 s 2 H20 1 In reaction total number of dissociated ions in solution is reduced dramatically during the reaction as a precipitate is formed 01 M H2804 slowly added to BaOH2 of unknown concentration in conductivity of solution monitored conductivity probe 0 when probe is placed in solution with ions has ability to conduct electricity electric circuit is completed across electrodes that are located on either side of the hole near the bottom of the probe 0 UNITS of conductivity 9 microsiemes per cm uS cm As reaction proceeds 9 solid BaOH2 precipitates 0 Collecting this barium sulfate salt when dry and weighing it will allow you to calculate amount barium ion in original sample of unknown concentration determining the concentration 0 Depends on weighing called gravimetric analysis 9 most accurate and reliable procedure available For gravimetric analysis to work requirements 1 Compound formed must be pure amp known stoichiometry 2 Precipitate reaction must be virtually complete 9 yield of solid should be 999 or better 3 precipitate reagent should be specific for the sample being determined 9 interference by other compounds forming precipitates should be minimal 4 precipitated solid 9 should be reasonably large well formed crystals necessary so precipitate can be easily filtered amp compound has highest possible purity 5 molecular weight of precipitated solid should be high enough that a reasonable weight of precipitate is generated even when the weight of substance being determined is low in the unknown sample To make sure to get complete precipitate 9 occulate cause to aggregate into a mass by heating solution of barium sulfate just to boiling Most common cations sodium magnesium aluminum do not interfere by producing precipitates with sulfate anion O Strontium mercury I lead 11 ions could interfere with reaction and co precipitate but none of these cations will be present in sample To determine the concentration of barium hydroxide in unknown by gravimetric determination 9 will need to know 2 experimental quantities accurately O O 1 Volume of unknown sample 2 Weight of barium sulfate obtained from that sample weight must be obtained to nearest mg and avoid sample loss or contamination Procedure before exp take out filter crucible from drying oven and place in desicooler to cool while completing titration use 10mL graduated cylinder of barium sulfate into 100mL beaker add 30mL distilled water 0 barium hydroxide is caustic do not spill on skin or clothes connect drop counter to LabPro of Vernier computer interface lower it onto ring stand conductivity probe must be at 02 0000 range connected to interface In hood 60ml of 0100 M hydrogen sulfate into 250 mL beaker 0 Hydrogen sulfate strong acid handle with care 60 mL of reagent reservoir close both valves by turning handles HORIZONTAL OOOOOO fill reservoir with water check for leaks rinse it with few mL of hydrogen sulfate attach it to the ring stand with clamp fill it with 40 mL of 01M hydrogen sulfate solution place 250mL beaker with hydrogen sulfate rinse below it drain a small amount from reservoir to beaker to fill reservoir tip both values VERTICAL then back to horizontal open TOP value VERTICALLY and slowly BOTTOM value open until drop rate of about one drop per second I do not touch it again for the rest of the experiment I close top valve horizontal and just use top valve to start and stop liquid ow as needed I discard drained solution in 250 beaker SUCCESS relies on having drops from reservoir fall directly through middle of drop counter AND keep drop rate below 1 drop per second the counter will miss the drops if its any faster Calibrate volume of drops that will be delivered form reservoir o Calibrate to automatic and put volume of dripped solution as 56 mL and record the drop count calculated in case you mess up and don39t need to recalibrate Assemble apparatus o Conductivity sensor in large hole in drop counter line up drop counter and reservoir in center of magnetic stirrer o Magnetic stir bar in solution and adjust tip to be just above drop counter Conduct titration o Collect and monitor conductivity fully open TOP valve o When conductivity drops below 100 uS and then rises again I Titration curve will be V SHAPED o When reaches about 500 uS stop Examine the data to find EQUIVALENT POINT volume when the conductivity value reaches a minimum o Lightly linear portion of data where conductivity is decreasing only 9 linear fit o Same for linear portion where conductivity is increasing past the equivalence point o Highlight area 2 linear fit lines intersect volume equivalence point volume for the titration Gravimetric Determination of Barium Sulfate Beaker with titrated solution on hot plate heat to near boil While heating occulte barium sulfate o Weight cool filter crucible to near mg 0 Place vacuum filtration apparatus securely to from good seal between crucible and rubber cup Remove from hotplate after 2 mins allow to cool to room temp o Before filtering squirt bottle with methanol and spray mL of it into crucible to wet filter disk so it sticks to the bottom and doesn39t let the solids from going down into beaker o Stir solution to suspend solid in solution and immediately begin filtering solution do not overfull crucible or spill o Do not tilt beaker steeply to allow magnetic stir bar to enter crucible it should stay in beaker Place crucide and precipitate in drying over for 1520 mins o Place in desicooler o Record weight of crucible with barium sulfate o Determine mass of barium sulfate collected I Calculate the concentration of the barium hydroxide solution Post Lab Suppose a small amount of Lead 11 chloride got into your BaOH2 sample before you titration contaminating it How do you think the contamination would affect your calculated concentration for the sample What about the same amount of FeCl3 0 The small amount of lead 11 chloride could interfere with the reaction and co precipitate leading to an inaccurate measurement of yield of barium hydroxide leading to an inaccurate calculated high concentration 0 Most common cations do not interfere by producing precipitates with the sulfate anion so a small amount of iron 11 chloride would probably not affect the calculated concentration Suppose the barium sulfate weighed in the crucible above was not completely dry How would this affect the concentration calculated for the barium hydroxide solution 0 The concentration calculated in this instance would be higher than the true value because the added moisture would add to the weight Purpose of occulating the barium sulfate precipitate by heating the solution to boiling 0 So it will aggregate If you forget to put crucible in over 9 moisture of solute along with barium sulfate would be present in sample 0 Increase moles of sample 0 Increase concentration of barium sulfate Drops from reagent reservoir fall directly through middle of drop counter 0 Keep drip rate below 1 drop sec 0 Any faster misses drops 4 observations 0 color change 0 bubbles 0 change in conductivity 0 change in temperature 2 properties being measured 0 temperature 0 conductivity conductivity is the reciprocal of resistance of ow of current Types of Reactions Examine reactions acid base precipitation gas evolution combustion redox simple dissolution 0 Easiest to see I gas evolution bubbles and effervescence apparent I precipitation reactions formation of insoluble substance that falls to bottom of vessel perform theoretical and real oxidation reduction reaction 0 to visualize the chemical reactions is to find a molecule that changes color in different chemical environments 9 indicators 0 today use instruments to monitor conductivity and temperature conductivity ability of substance to conduct electricity 0 ions in solution carry charge that ow of charge current can be measured 0 reciprocal of resistance ohms of the ow of current the ow of ions in the solution 0 higher concentration of ions in solution higher conductivity 0 smaller ions hydrogen good carried of charge because they can run through the solution faster than the bigger ions barium perchlorate I smaller ions also have higher charge to mass ratio should be slightly better conductors than larger ions 0 higher charged ions carry more charge attract a lot of water molecules are solvated I higher charged ions are more polarizing and therefore carry or drag more water with them the efficiency of higher charged ions current carriers is reduced somewhat would expect Mg 2 to be a better carrier of charge but not twice as conductive as Na why measure temp 0 measurement of temp change can be used to determine the presence of a chemical reaction 0 EVOLVES energy EXOTHERMIC o ABSORBS energy ENDOTHERMIC Procedure Conductivity probe in channel 1 O 2000 temp probe channel 2 Change data collection parameters events with entry mode drops as units Collect after each drop click blue keep wheel and enter drop number for total drops added After drop swirl reagent and wait for it to come to rest Before every reaction rinse all probes and beakers with deionized water Reactions 1 9 Add 20 drops of 01 HCL to the 50mL deionized water a Addition of HCl to water increases conductivity of the solution b Conductivity increases with each drop because of the increased concentration of hydrogen ions excellent conductors To the above solution add 30 drops of 01 M NaOH dropwise a Adding NaOH decreases conductivity b Continues to decrease until all H 39 has been neutralized with OHquot to form water which does not conduct electivity and NaCl is not as good a conductor as HCl in aqueous solution c Conductivity begins to go back up as excess NaOH is added giving Na and OHquot in solution d Intersection of 2 linear portions of graph of conductivity vs number of drops stoichiometry equivalence point Add 20 drops of NH3 to 50 mL deionized water then 8 drops of CuNO32 Add 50 mL of the dilute Na2CO3 solution to a beaker dropwise add 45 drops ofO1 M HCl VISIBLE REACTION a Take small test tube add 1mL of 10 M HCl add 1 mL of saturated Na2CO3 observe results AS you add Na2CO3 b Immediately pour mixture into sink and observe again as solution hits the bottom of the sink ush with water Add 20 drops of NACl to the 50 mL of deionized water Dropwise add 20 driops of KNO3 Add 20 drops of 05 M Na2SO4 to the 50 mL deionized water and dropwise 30 drops of 01M BaNO32 VISIBLE REACTION a Take small test tube add 1 mL saturation Na2SO4 then add 10 drops BaNO32 b Observe results as you add BaNO32 Give a balanced chemical reation for the addition of FeCl3 to SnCl2 indicating which oxidation states and oxidized and reduced species no net reation Add one small piece of Mg turnings solid to 50 mL of deionized water in a 100 mL beaker add 20 drops of 1 M HCl over 4 minutes then monitor reaction visually and with LabPro for 4 minutes a Means you need to add fake drops to get LabPro to record data every 10 seconds b Involves 3 types of reactions Data analysis each portion of each experiment highlight and linear fit 0 Slope of resulting line relative change in conductivity 0 Steeper the slope means conductivity changed more rapidly Post Lab How does the size and charge of ions affect the conductivity of a solution 0 Any ions in a solution carry change and it is those charges that are measured as conductivity 0 The higher the concentration of ions in a solution 9 the higher the conductivity 0 Smaller ions are slightly better conductors than larger ions 0 Higher charged ions are not as efficient because they attract more water polar slowing them down but they carry more charge What would you expect to happen to the observed temperature in an endothermic reaction 0 Expect a drop in temperature because the system is absorbing heat leading the surroundings colder ENDO 9 heat absorbed temperature surroundings decreases EXO 9 head released temperature of surroundings increases Conductivity and bubbles are visible properties that tell us chemical reaction is taking place 9 False Reaction 1 amp 2 9 acid bas o HCl NaOH 9 H20 Na Cl 0 H30 OH 9 2H2O Reaction 3 9 precipitation o Ca2 20H 9 CuOH2 s Reaction 4 9 gas evolution 0 H C032 9 H2C03 O H2C03 69 H20 CD2 Reaction 5 9 double displacement o NaClKNO39NaClKNO3 Reaction 6 9 precipitation 0 Ba 2 SO42 9BaSO4 s Reaction 7 9 redox o Fe3 Sn2 9 Fe2 Sn3 0 Fe reducing agent 0 Sn oxidizing agent Reaction 8 9 redox gas evolution acid base 0 Mg s 2Hquot 9 Mg2 H2 Post Lab Precipitation one that leads to the formation of an insoluble substance that falls to the bottom of the reaction vessel Gas evaporation one in which gas bubbles are observed out of the reaction vessel This effervescence is apparent The purpose of measuring conductivity and temperature in experiment o not all the chemical processes have visual indicators therefore they are other properties that may change Acids amp Bases Acid proton donor 0 Sour 0 Turns litmus RED 0 pHlt7 Base proton acceptor o Bitter 0 Turns litmus BLUE 0 pHgt7 Litmus test natural dye stuff which is sensitive to acids and bases 0 Universal indicator allow materials tested to be rated on their strength as acids and bases pH the negative log of the hydrogen ion concentration 0 p function is a mathematical function described as the negative log of a desired value pH Colors 4 Red 45 Red orange 5 Orange 55 Orange yellow 6 Yellow 67 Yellow green 7 Green 8 Green blue 85 Blue 9 Violet Procedure to test materials need dilute indicator solution 10 mL of universal indicated dilute it with 30 mL of distilled water 0 place materials on spot plates and test with 24 drops of dilute indicator solution testing vinegar starch sugar sat ammonia borax club soda tomatoes tap water baking soda sea water Epsom salt apples onion washing soda red cabbage grapes beets contain natural dyes or pigments that also change color with acids and bases 0 shred leaf into 150mL of water and bring to boil 0 after boil take it off hotplate to steep if you boil too long will denature the pigments rendering them useless just need to extract the dye prep diluted grape juice mix 10mL grape juice with 40 mL of distilled water 0 test 10 mL of red cabbage juice and diluted grape juice with o vinegar borax baking soda ammonia washing soda since used universal indicator you know their respective pH s use pH meter to test baking soda borax washing soda 0 add small amount of solid into 100 mL water for each case 0 for vinegar ammonia add few drops of each liquid in 100mL of water test Coca Cola amp Mug Root Beer 0 Coke phosphoric acid 0 Root Beer citric acid Post Lab In carbonated beverages like coke carbon dioxide gas is dissolve din what to form the carbonic acid present in soda 0 Water Conclusions about observations 9 red cabbage extract 0 It is clear that red cabbage extract can be used to determine relative pHs 0 The more acidic substances were pink whereas the extract turned blue then green the more basic the substances were Conclusions 9 grape juice 0 grape juice and cabbage produced similar colors when indicating pHs 0 pink grape juice went to pink for more acidic substances and green for more basic substances conclude about acids and bases 0 universal indicator good method of getting relative pHs because the more acidic a substance the redder it is and as pH goes up the indicator goes through the colors of the rainbow o acids have a pH before 7 bases above 7 and neutral 7 o salts are generally neutral because they are a product of neutralization reactiosn types of techniques 0 litmus test 0 universal indicator 0 pH meter 0 grape juice cabbage juice solids cannot go down the sink Gas Laws 1 Boyle39s amp Charles Law Objectives Determine the effect of pressure on the volume of gas Determine the effect of temperature on the volume of gas Graphically illustrate the relationship between PampV TampV Compare your relationship to Boyle39s Law and Charles Law Of all states of matter gases are most affect by their environment Solids maintain their shape and volume with only minor changes resulting form changes in environment temperature pressure shape of container etc Liquids assume shape of their container but tend to maintain their volume with changes in the environment Pressure does not appreciable affect matter in either solids or liquids Gases 0 Significantly affected by changes in the environment o A given amount of gas will expand to fill the container completely and uniformly 0 Volume of gas sensitive to the pressure and temperature applied Boyle39s Law Relationship between volume and pressure at a constant temperature a fixed amount of gas occupies a volume inversely proportional to pressure exerted on it 0 As pressure exerted on a gas increases volume decreases 0 Pressure vs volume hyperbola 0 Pressure vs 1volume linear constant P1V1 P2V2 Charles Law Relationship between volume and temperature at a constant pressure the volume occupied by a fixed amount of gas is directly proportion to the absolute temperature o VT constant o Temperature vs volume linear I In celcius y intercept 273 degreee I In Kelvin y intercept passes through origin V1T1 V2T2 Procedure Using LabPro to determine the effects of pressure and temperature on the volume of a gas o The 60 mL syringe setup used to determine the volume and to generate pressures needed Effect of Pressure on Gas Volume Boyle39s Law Use pressure sensor on LabPro port connect end of tube of known volume 5mL to pressure port of the gas pressure sensor do not attach syringe yet Change units y axis of pressure from kPa 9 mmHg torr set up sensors Set up x axis to read volume in mL Set plunger in 60mL syringe to 25 mL exactly total volume of 30mL with 5 mL of tubing 0 Attach syringe to the tubing coming from the gas pressure sensor 0 At this point the pressure in the syringe should be the atmospheric pressure you recorded from the barometer since you pre set the plunger to 25 mL before attaching the syringe to the tube giving a total of 30 mL in the syringe 0 You must remember to always add the 5 mL due to the tubing volume to the reading you see on the syringe in this experiment One adjusts the syringe plunger to appropriate mL setting and hold it steady the other clicks blue keep wheel 9 enters volume data 0 Must have the syringe plunger to correct volume before clicking the blue keep on each data entry Syringe should still be attached to the tube and pressure sensor with the plunger at the 25 mL mark adjust the plunger to 10mL 0 Enter 15 volume of the syringe and the tubing Repeat from 15 55 mL always adding 5 mL to values entered in Logger Pro Data Analysis Select Autoscale axes Pressure mmHg vs Volume mL 0 For each set of volume pressure measurements calculate the product of pressure and volume PV 0 Set up new calculated column 0 Remember P x V should be constant 9 your standard deviant low 0 new calculated column 1volume I correlation coefficient reported is an indication of the accuracy of the data I the closer to 1 the better the data I good data for this lab will be a correlation coefficient of at least 099 I NB We have assume temperature was constant in reality compressing a gas heats it up and conversely expanding a gas cools it 9 quite small therefore negligible there can assume its constant Effect of Temperature on Gas Volume Charles Law Obtain glass Volumetric pipet sealed at one end with a small plug of glycerin somewhere in the middle of the range market with volumetric gradation o Plug will trap a small sample of air between the plug and bottom of pipet o Plug of constant weight plus atmospheric pressure will maintain a constant pressure on trapped air 0 Pipet can be attached to a fudge stick Small plug in pipet is glycerin 9 highly Viscous liquid 0 Since weight of this small plug is very small compared to atmospheric pressure we can assume that the air trapped below the plug is constantly at atmospheric pressure o We want to know the volume of air trapped under the glycerin plug o Since the pipet is calibrated to deliver the scale is reversed 9 must assume that the tip of pipet is 00 mL not 05 mL as indicated count UP using the gradation marks to determine the volume of air under the plug Collecting the data Restad loggerPro with temp probe Change the y axis to pressure and x axis as volume 9 plot T vs V Place pipet into a 400 mL beaker of cool water add chips of ice making sure liquid plug in pipet is submerged below the water o Temperature sensor goes next to the pipet in the beaker of water carefully clamped in vertical position o Be sure that neither the thermistor or the pipet touches the bottom for the beaker by marking the Popsicle stick extend a centimeter blow the bottom of the pipet and thermistor Put beacker on hotplate as heat water bath glycering plug rises Use hotplate to heat water bath until its 7075 degrees Celcius 0 Make sure that the temperature of the pipetwatertemperature probe are in equilibrium and record the temperature of the water bath and the volume of the air column in the pipet to at least 0005 mL o When you record the initial Volume LabPro will record the temperature 0 NB IMPORTANT that you take the time to record the volume of air under the plug as accurately as possible for this experiment to succeed do not rush To ender data one partner should determine the volume of air under the glycerin plug THEN as soon as the volume has been determined the other partner should click blue bottom and enter that Value o Input volume of air in pipet into your labpro program at 5 degree increments as the water cools until reaches 50C o Program should automatically send a temperature reading to the data column after you enter volume 0 Add ice stir constantly to maintain equilibrium and take more reasings until the temperature settles between 0 5 degrees and keep temp dropping Linear fit data y intercept value of absolute zero 0 Record regression data ymxb and correlation coeffecient Post Lab Absolute zero in Kelvin as 270 K or 0 Degrees Celsius o Temperature that water freezes and SI unit used for temperature is calculated in Kelvin Which state of matter is most affected by their environment 0 Gas Boyles Law During Boyle39s law experiment after finding best fit line what statistical measurement will you use to determine accuracy o Fractional analysis correlation coefficient 0 Should be at least 099 Boyle39s law experiment don39t forget to 0 Add 5 ml to tubing volume to the reading on the syringe 0 Change units from kPa 9 mmHg torr 0 Assume temperature is constant change is negligible I Reality when compressing gas it heats and expanding 9 cools Assumption in Boyle39s law experiment o Some error in assumption because syringe is not at room temperature and the temperature of the gas increases as it is compressed o This assumption is devalued because the temperature is negligible Suppose 2 identical balloons are filled with 01 mole of helium one in Miami and one Denver which would be bigger o Balloon in Denver would be bigger because air pressure is lower with higher altitude and the lower the pressure the greater the volume What would happen to the pressure of a gas sample if volume were increased to infinity o If volume was increased to infinity the pressure would approach zero Likely source of error assuming temperature constant Charles law Charles Law experiment 0 Measure the air trapped BLOW glycerin plug 9 volume of air 0 As you heat water glycerin will move up o Pipet is calibrated to deliver scale is reversed I Must assume tip of pipet is 00 mL not 05mL and count UP using gradation marks to determine the volume of air under the plug Assume that pressure is kept constant therefore the ratio of volume and temperature is constant 0 This is a valid assumption because high viscosity of glycerin reduces friction that could cause an increase in temperature o Also because the weight of the plug is very small compared to atmospheric pressure there is hardly a difference in the pressure of the air below the plug Likely source of error reading the glycerin plug wrong assuming pressure constant Gas Laws 2The Ideal Gas Equation Determination of the Equivalent Weight of Magnesium Determination of the Equivalent Weight of Magnesium The equivalent weight the weight of the substance which will react with or produce one mole of hydrogen 0 Can also be discussed in terms of unit loss or gain of electrons for oxidation reduction reactions or of one mole of H for acid base reaction I eg Equivalent weight of MgOH2 hat would react with 1 mole of H would be the molecular weight of MgOH2 divided by 2 since there are 2 moles of OH for every mole of MgOH2 Mg ribbon is reacted with aqueous HCl in such a fashion that all the H2 produced is collected in a buret 0 From weight of Mg used and from the corrected gas volume the equivalent weight of Mg is calculated for this ractio o How much Mg is required to produce 1 mole of H atoms Mg 5 2 HCI aq 9 MgCl2 aq H2 g The ratio of Mg to H atoms 9 12 Mg ribbon cut into 30mm pieces 9 use steel wool to clean surface of oxide coating or impurities I When Mg metal left in air reacts with oxygen to form MgO which will not react with HCL that wont react 0 Record its mass of cleaned Mg 0 Wrap copper wire around it Place 10 mL of 3 M HCl solution into a measuring cylinder under hood and transfer it to a gas buret 0 Add water until level is 1 inch from top 0 Put copper wise and Mg in mouth of buret and invert buret and submerge the open en din water in beaker do not release thumb before complete submersion 0 If you leave too much space at the top a large bubble will ow up into the buret when you turn it over mixing the HCl and water along the way ruining the 2 layers Even though not meant to never add water to acid because it will boil and splatter and cause excruciating pain 0 Since using concentrated HCl which is more dense than pure water and carefully pouring water down the side of buret we form 2 layers 0 Will create a delay for us to get the buret set up in the beaker initially HCl will be trapped at the bottom of the buret after you invert it As concentrated HCl slowy diffuses down the buret and reaches the Mg ribbon it will begin to react o Clamp it immediately and read and record the water level in the buret I Volumes should be read to nearest 001 mL I HCL will diffuse downward and reaction will begin I All H2 must be trapped in buret or serious error will result 0 If you don39t collect all gas on first try or your copper wire falls out try again I Reactions stops when there are no more bubbles tap buret to make sure all bubbles come out and let system stand for 510 mins for temperature equilibration I Read and record the final buret volume and temperature for water in beaker 9 appropriately the tmperature of the gas in buret I Measure higher of water in buret Ah with ruler meter stick From data calculate equivalent weight of Mg using accepted value for the atomic weight compute the oxidation state of the reacted magnesium valence Since gas is not easily weighed we will calculate the amount in a volume o Using ideal gas law assume standard conditions o If 2 moles of hydrogen atoms on each side I One on left in form of HCl and on right of H2 Equivalent weight is calculated with the initial mass of the Mg ribbon Post Lab Errors inherent in this experiment 0 Some hydrogen is lost throughout the reaction leading to an incorrectly high equivalence weight and a lower oxidation number o Another error that could have the same erroneous results would be the persistence of oxide on the Mg ribbon If a think film of MgO was present on the surface of the Mg ribbon the weight recorded fro the ribbon would be higher than in actuality leading to an incorrect equivalent weight than would be too high and an incorrect oxidation number that would be too high Why vapor pressure is important o Because it must be subtracted from barometric pressure along with the pressure exerted by liquid water to obtain the pressure of dry hydrogen gas What is meant by dry gas o Refers to the hydrogen gas generated by the reaction once the water vapor has been taken into account and the corrected volume of hydrogen gas is obtained by the combined gas law Why is the height of the water delta h used why is it divided by 136 o Its sued because it is the pressure of the water exerted on the system and must be subtracted form the barometric pressure along with vapor pressure of water to obtain the pressure of the dry hydrogen gas o Change hydrogen is divided by 136 to convert the units of pressure from mmHg to torr Why is it important to calculate the volume of dry gas o So that water vapor is not taken into account when measuring the hydrogen gas volume Value of equivalence weight would be affect if some hydrogen gas produced during the reaction of Mg s and acid escaped from gas buret o Calculated value of equivalence weight would be too high 3 types of reactions for which equivalence weight can be calculated 0 gas evolution this exp 0 acid base o redox bending the rules of adding acid to water 0 can bend the rules because concentration of HCl is more dense than water so the point is not to mix them but to create 2 separate layers volume of 1 mole of ideal gas at STP 0 224 L Freezing Point Depression Constant of Lauric Acid When substance freezes or solidifies molcules come into a certain arragnemnets 0 Some cases molecules become very closely packed and form a dense solid 0 With water hydrogen bonding takes over when the molecules have such low Kinetic energy 0 Hydrogen bonds are longer than normal distance between water molecules in liquid state 9 water molecules will be father away from each other in solid state I Solid less dense than liquid In process of freezing 9 impurities affect normal molecular interactions 0 When this occurs more energy must be taken out of system for system to solidify I Temperature must be lower I Eg Salt on roads lowers the freezing point of water ice on roadways and preents ice sheets from forming on the road or melts them Using LabPro to determine freezing point depression constant Kf for lauric acid which is a non electrolyte The amount that the freezing point is lowered is given by O A T Kf m I A T lowering of freezing point in Celsius for solution I Kf freezing point depression constant for the solvent I m is the molality of the solution solution will be a solute benzoic acid in the solvent lauric acid 0 solution will have specific molality o molality moles solute mass solvent I compared to molarity moles divided by volume of solution Volume is temperature dependent so at different temperatures the volume of a solution will change which would also affect the molarity I molality is not effected by temperature Freezing Point of Lauric Acid temperature probe used to determine freezing point of pure lauric acid 0 lauric acid melts at temp slightly above room temp hot plate with 400 mL beaker of water about 6070 degrees Celsius 0 test tube of pure lauric acid in hot water until melts put temp probs in melted lauric acid and begin stirring the melting lauric acid 0 5 secs then collect 0 put in room temperature water stir constantly and keep lauric acid in test tube 0 once you reach at area of cooling curve stop stirring 0 add ice to aid in cooling and record data until temperature starts to trail off and drops below 30 degrees Celsius o highlight lat portion of curve linear fit 0 y intercet freezing point 0 store latest run with now frozen lauric acid re melt acid completely and repeat experiment 9 determine average melting freezing point Data Analysis freezing point of lauric acid will be at portion of curve sometimes as we remove heat form liquid we can temporarily cool it below its freezing point without forming a solid 9 super cooling eventually stirring and agitating will cause liquid to freeze temperature will return to that of the freezing point 0 at portion occurs immediately after this 0 during freezing process the at part o the cooling curve represent the heat of fusion 0 the temperature does not change during the heart of fusion because of energy lost during the freezing process is used in packing the lauric acid molecules into the dense solid state 0 temp continues to fall after at portion due to further thermal cooling of solid lauric acid know the freezing point of lauric acid 9 begin to determine the freezing point depression constant 8g Lauric Acid O75g Benzoic acid solute amp 150g benzoic acid solute melt mixture in hot water 9 take out and use temp probe to collect data find freezing curve for solution calculate A T and molality of solution use molecular weight ofbenzoic acid 12212 9 moles of solute 9 divide by kg of solvent molality Purpose The purpose of the experiment was to nd the molarity M of a Sample of BaOH2 using two methods of quantitative analysis conductimetric titration and gravimetric determination The morality of a solution or concentration is the amount of moles of substance per liter of solution Conductimetric titration is performed by removing ions form solution until the conductivity of the solution reaches zero labeled the equivalence point 0 This is the point at which the conductivity of the solution is at its lowest indicating the formation of the products 0 This type of reaction in which ions are removed from a solution is a metathesis reaction however it is also a precipitation reaction A precipitation reaction is one in which anions and cations in aqueous solution combine to form a solid precipitate o The amount of added H2SO4 required to reach the equivalence point allow for the calculation of the concentration of BaOH2 in the original solution 0 In this case the ions present in aqueous BA and aqueous H2SO4 combine to form solid BaOH2 and water The formation of a precipitate is what allows us to measure the concentration of BaOH2 in the experiment gravimetric determination Gravimentric analysis depends on weighing and is one of the most accurate forms of quantititative analysis 0 Several requirement must be must in order for it to work Lab Manual p44 the compound formed must be of pure and known stoichiometry the precipitation reaction must be virtually completely that is the percent yield of the solid product should be 999 or better the precipitation reagent should be speci c for the sample being determined interference by other types of compounds forming precipitates should be minimal the solid that is precipitated should be in the form of reasonably large wellformed crystals and the molecular weight of the precipitated solid should be high enough that a reasonable weight of precipitate is generated even when the weight percent of substance being determined is low in the unknown sample 0 The solid BaSO4 precipitate formed in the conductimetric titration portion of the experiment is collected dried and weighed The weight of the BaSO4 collected allows for the calculation of the concentration of the original BaOH2 using the stoichiometric equation as a method of establishing quantitative relationships between the compounds and ions that partake in the reaction Procedure Conductimetric titration l 2 L11 Place a lter crucible from the drying over in a desicooler to cool while the titration is completed Transfer l00mL of BaOH2 solution into a l00mL beaker using a lOmL graduated cylinder Add 30mL of distilled water to the BaOH2 solution using the same graduated cylinder Connect the Drop Counter and Conductivity Sensor to the LabPro interface and open the l6b Conductivity Titration Drop Count from Advanced Chemistry with Vemier folder Measure out approximately 60mL of 0100 M H2SO4 into a 25 0mL beaker Obtain and set up the reagent reservoir for the titration making sure that the drops are going through the middle of the drop counter and that the drop rate is below one drop per second to ensure an accurate drop count by the drop counter Calibrate the volume of the drops that will be delivered form the reagent reservoir using the H2SO4 solution Assemble the apparatus by inserting the Conductivity Sensor through the large hole in the Drop Counter Make sure the Drop Counter and reagent reservoir are lined up with the centre of the magnetic stirrer Place a magnetic stir bar in 10 the breaker of BaOH2 solution place the breaker onto the magnetic stirrer and lower the conductivity sensor into the breaker Tum the stirrer on Conduct the titration Click collect on the Logger Pro program so it begins monitoring conductivity Open the top valve of the reagent reservoir so the H2SO4 begins dropping into the BaOH2 Observe the graph the conductivity will drop and then rise again after it rises again to 5000 us close the top value of the reagent reservoir to stop the H2SO4 from dropping into the solution Examine the data on the graph to find the equivalence point which is the volume of H2SO4 added when the conductivity value reaches a minimum Gavimetric Determination 1 Place the beaker with the titrated solution on a hot plate and heat to a nearto low boil 2 While the solution is heating to occulate the barium sulfate weigh the filter crucible to the nearest milligram 3 Place the crucible in the vacuum filtration apparatus securely to form a good seal between the filter crucible and the rubber cup 4 Once the solution has been near boiling for a minute or two remove it from the hot place and allow it to cool to about room temperature 5 Before ltering the solution spray about a milliliter of methanol using a squirt bottle into the crucible to wet the lter disk 6 Tum on the vacuum apparatus to seat the filter disk and prevent the loss of solid BaSO4 7 Quickly stir the solution to suspend the BaSO4 solid in the liquid and immediately begin filtering the solution being careful not to spill the solution or overfill the crucible Do not tilt the breaker steeply enough that the magnetic stir bar enters the crucible it should stay in the beaker 8 Once all the liquid is out of the beaker rinse any BaSO4 left into the crucible using the methanol squirt bottle 9 Once all the precipitate has been transferred into the crucible wash the precipitate with some methanol and allow the crucible to sit on the vacuum for a minute 10 Place the dry crucible with the precipitate into a breaker and into the drying oven and let it dry for 1520 minutes 11 Carefully remove the breaker with the crucible and cool it in the desicooler 12 Once cool record the weight of the crucible with the BaSO4 and determine the mass of BaSO4 collected 13 At the end of the experiment discard the BaSO4 and clean the crucible and place it back to the drying oven 14 Rinse conductivity probe tip with distilled water and the solvent reservoir with distilled water Discussion Using conductimetric titration and gravimetric determination allowed for the calculation of barium ions in the original BaOH2 solution Given that there are the same number of moles of barium ions as there are moles of BaOH2 the calculated concentration is the same for both The calculated concentration of BaOH2 and by extension barium ions in the conductimetric titration portion of the lab gave a concentration of 0168 M The gravimetric determination of BaOH2 led to a calculated concentration of 0228 M The difference in the two values is the result of various sources of error A source of error that I observed was during the condutimetric titration as the H2SO4 was dropped into the BaOH2 solution drop of BaOH2 bounced out of the beaker o This did not affect the concentration calculate in this instance as the drop counter still recorded an accurate amount of HSO entering the breaker but did affect the amount of BaSO4 weighed in the gravimetric determination giving an inaccurate concentration as some BaSO4 was lost The concentration calculated by gravimetric determination was also affected by weighing the precipitate before it was completely dry giving a higher weight and thus a higher concentration We weighed the precipitate before a full 15 minutes due to time constraints This gave us a higher calculation of BaOH2 than in the conductimetric titration even though it should have been lower due to the loss of BaOH2 during the titration The experiment show that these methods of concentration calculation with those two courses of errors eliminated would be very accurate in determining the concentration of barium ions in the original BaOH2 solution or any other sample undergoing quantitative analysis The deductions we made in equating the number of moles of BaSO4 to the number of moles of BaOH2 along with the deciding we were able to making equating the number of moles of H2SO4 to moles of BaOH2 would not have been possible without having a known stoichiometric equation showing that neither conductimetric titration nor gravimetric determination would have been possible without it The computer software that allowed us to visualized the equivalence point drive home the concept of having the least amount of conductivity when the least amount of ions were present Through this experiment I now understand the process of conductimentric titration and gravimetric determination as forms quantitative analysis The experiment also allowed me to see and understand first hand a precipitation reaction If I were to perform this experiment again I would try to reduce the amount of BaOH2 that was lost by having the reagent reservoir closer to the breaker of solution and allow the precipitate more time to dry in the drying over before weighing it Ka of Weak Acid use LoggerPro and LabPro to follow pH changed during acid base titration obtain Ka of weak acid being titrated acetic acid vinegar 9 HC2H3O2 in water solutions weak acids react with water to establish equilibrium 0 HA H20 69 H30 A39 Equilibrium represented by acid dissociation constant Ka o Ka is constant for given acid at given temperature 0 Value determined by 2 methods used in this experiment 1 measurement of pH of a solution containing a known concentration of a weak acid 2 measurement of the pH of the half neutralization point in the titration of the weak acid with a strong base Titration Curves graph should like this pH pKa log A39 HA equation above gives pH of any A39 HA ratio 9 the ratio changes as base is added to acid 0 because the reaction is pushed farther towards the right with each addition of the base 0 at beginning only acid HA and small concentration of H3O and A from ionization are present 0 as base added acid is neutralized decreasing concentration of HA I at the same time salt formation increased A39 throughout the titration HA drops and A39 increases until HA is neutralized the equation shows as tthis happens and ration A39 HA increases from low to high value 9 pH of solution changes according to graph 0 easy calculate concentration of A39 HA at various stages I substituting concentrations and pKa into equation NB First addition of base produced significant rise in pH 0 Follow by region where pH only changes slightly 9 solution is buffered by presence of both weak acid and its salt 0 As you add base the acid concentration drops so much that the solution is no longer buffered 0 Now pH rises rapidly through neutralization equivalence point and slightly beyond 0 Beyond this point acid has been neutralized and pH of solution changes only slightly as more base is added 0 Rise of pH beyond equivalence point is due to addition of base to a relatively large volume of solution Rapid change in pH near equivalence point makes a quantitative titration of acid by base a feasible experiment o 2 drops of base solution near equivalence point 9 changes pH of 170 units in solution 0 sufficient change for color change in indicator from acid to base 0 rapid change near equivalence point 9 react tritrant must be added in smaller and smaller amounts as it is approached calculating Ka o because of equilibrium is reestablished after each addition of base during titration a value of Ka could be obtained from data corresponding to any point before the equivalence point on titration curve I using data from two particular points first point on curve and half neutralization point o first point on curve is pH of acid solution before base is added I knowing concentration of acid in solution of pH should be able to calculate the Ka o when acid is exactly half neutralization pH of solution pKa of acid I when half of acid is neutralized present as A the other half is not HA the concentrations of both are equal I log function 1 pH12 pKa several ways to identify half neutralization point o easiest way 9 select point on curve that corresponds to half the number of mL of base needed to reach equivalence point o ka of acid can be obtained form pH corresponding to this point by taking antilog of pH negative log Procedures 1 Add 200 mL of acetic acid solution to a 100 mL beaker Do not dilute 2 Attach the Drop Counter and pH Sensor to the computer interface of the Vemier software 3 Run the Logger Pro program Open the 07b AcidBase Drop Count le from the Advanced Chemistry with Computers folder 4 Set up the reagent reservoir for the titration see p 94 of the lab manual 5 Calibrate the drop Volume for the drop counter see p 94 of the lab manual 6 Assemble the apparatus see p 94 of the lab manual 7 Tum on the magnetic stirrer so that the stir bar is stirring at a moderate rate 8 Click Collect and allow the NaOH to drop into the acetic acid 9 Watch the graph to see when a large increase in pH occurs stop adding the NaOH when the pH levels out after this large increase 10 Clean up Notes The purpose of this lab is to determine the Ka Value of a weak acid in this case acetic acid As opposed to strong acids that completely dissociate into ions in solution weak acids only partially dissociate The strength of an acid depends on whether the equilibrium favors the reactants or the products The formula for the equilibrium of the reaction of a weak acid HA in water is HAaq l H20 H30 aq l A aq A represents the salt of the weak acid The relation strength of a weak acid can be quantified using the acid dissociation constant Ka Strong acids have a Ka greater than 1 Weak acids have a Ka less than 1 The Ka of an acid is the equilibrium constant for the dissociation reaction of a weak acid The smaller the Ka the less the acid dissociates o For example the strong acid HCL hydrochloric acid that completely dissociates in water has a Ka of l3xl06 whereas a weak acid such as H2S hydrosulfuric acid has a Ka of llxl0397 The Ka of an acid is a constant at a given temperature Two methods of determining Ka are used in this experiment First the Ka is determined by measuring the pH of a solution with a known concentration of a weak acid The equivalence point at which the moles of base present equal the moles of acid initially present in the solution determines the concentration of the weak acid The pH of a solution is the negative log of the hydrogen ion concentration a pH below 7 indicated an acidic solution a pH about 7 indicates a basic solution and a pH at 7 indicated a neutral solution Second the Ka is determined by measuring the pH of the solution at the half equivalence point in the titration of the weak acid with a strong based in this case NaOH As each drop of NaOH is added to the solution the equilibrium reestablishes itself according to the relationship established by the Ka This means that the concentration of the weak acid decreases as the strong base is added which increases the concentration of Aquot until all of the weak aid in neutralized Discussion The Ka was determined using tow methods First by measuring the pH of the solution and calculating the concentration of the weak acid using the equivalence volume of NaOH Second the Ka was determined by measuring the pH at the halfequivalence point of the titration At the halfequivalence point equal amounts of the acid and its salt are present creating a buffer solution which explains why the pH does not change much in this portion of the graph This is the point at which pH pKa At the equivalence point however all of the weak acid has been converted to its salt but because of equilibrium some sat backreacts to reform its acid along with OHquot This is shown by the equations A aq 1 H20 1 69 HAaq 1 OH OH39 from the backreaction plus the OHquot from the NaOH means that the solution will be basic at its equivalence point which was supported by our data The book value of the Ka of acetic acid is l8xl0395 Using the data we obtained two different Ka values were determined indication that there was error in our experiment The Ka value determined by the first method was 1167 X 105 whereas the Ka determined by the second method was 1995 X 10 395 The Ka obtained using the first method has a 352 error however the second method only had a 108 error One possible source of error could have bee that the magnetic stir bar did not mix the solutions well enough so that the pH recorded was inaccurate particularly when the volume of the solution was higher at the end and the base added at the top of the solution did not get mixed quickly enough by the stir bar which was located at the bottom of the solution 0 This explains why the first method which used the equivalence volume to calculate the initial concentration of HA was les accurate than the second method which calculated Ka at the equivalence point using half the volume Another source of error could have resulted form our assumption that there was no change in temperature Ka is constant at a given temperature therefore if the temperature of the solution changed throughout the titration which it probably did then the Ka would also change introduction of solute into a solution changes the freezing point because the solute may interfere with the normal molecular interactions Because of this interference more energy must be taken out of the system in order for the system to solidify resulting in a lower freezing point
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