CHM 105 & 106 UNIT 2, LECTURE SEVEN 1 CHM 105/106 Program 14: Unit 2 Lecture 7 IN OUR PREVIOUS LECTURE WE WERE TALKING ABOUT THE DYNAMICS OF DISSOLVING AND WE WERE TALKING ABOUT PARTICLES GAINING ENERGY AND LEAVING THE SOLID SURFACE MOVING INTO SOLUTION AND WE ALSO WERE TALKING ABOUT THEN THE REVERSE PROCESS, THE PROCESS OF PARTICLES THAT WERE ALREADY DISSOLVED MOVING BACK OUT OF SOLUTION AND STICKING TO THE SOLID, AND OF COURSE WE CALL THE TWO PROCESSES DISSOLVING FOR LEAVING AND PRECIPITATION FOR RETURNING TO THE SURFACE. WE ALSO TALKED ABOUT THE FACT THAT IF E HAVE ENOUGH SOLID IN THE SYSTEM THAT SOONER OR LATER WE LL REACH A POINT WHERE THE RATE AT WHICH THE PARTICLES ARE LEAVING THE SURFACE BECOME EQUAL TO THE RATE AT WHICH PARTICLES RETURN TO THE SURFACE ONCE AGAIN, AND WHEN THIS HAPPENS WE SAID THAT WE HAVE A SOLUTION WHICH WE REFER TO THEN AS A SATURATED SOLUTION. SATURATED SOLUTION THEN OCCURS, GET A LITTLE BETTER FOCUS HERE, SATURATED SOLUTION OCCURS THEN WE SAID WHEN THE RATE OF DISSOLVING BECOMES EQUAL TO THE RATE OF PRECIPITATION. NOW ONE OF THE THINGS THAT WE SHOULD POINT OUT HERE IS THAT WHEN THIS OCCURS WE REFER TO THIS AS AN EQUILIBRIUM BECAUSE TWO OPPOSING THINGS HAVE BECOME EQUAL BUT THIS PARTICULAR TYPE OF EQUILIBRIUM IS A DYNAMIC EQUILIBRIUM, MEANING THAT THE PROCESSES, THE DISSOLVING AND THE PRECIPITATION BOTH ARE STILL GOING ON. IT S NOT THAT THEY HAVE STOPPED, IT S JUST THE MATTER THAT THEY HAVE BECOME EQUAL. SO ON A MACROSCOPIC SCALE SUCH THINGS AS COLOR OR TASTE OR CONDUCTIVITY, WHATEVER THOSE THINGS WERE THAT WE COULD MEASURE, THEY WON T BE CHANGING, BUT ON A MICROSCOPIC LEVEL IT S CONTINUALLY CHANGING. IN OTHER WORDS, THE ACTUAL PARTICLES THAT ARE IN SOLUTION ARE NOT THE SAME MINUTE AFTER MINUTE AFTER MINUTE. THEY ARE CONSTANTLY UNDERGOING THIS CHANGE, AND SO WE CALL THAT A DYNAMIC EQUILIBRIUM. OF COURSE, WE ALSO SAID THEN IF WE DON T HAVE ENOUGH SOLID WE HAVE THE SITUATION OF AN UNSATURATED SOLUTION. WE RE NOT GOING TO GET TO THE CONDITION WHERE THE TWO OPPOSING RATES BECOME EQUAL BECAUSE WE DON T HAVE ANYTHING LEFT TO DISSOLVE,
CHM 105 & 106 UNIT 2, LECTURE SEVEN 2 AND SO WE REFER TO THAT TYPE OF SOLUTION MERELY AS SATURATED. WE ALSO DID TALK ABOUT AND LOOKED AT AN EXAMPLE OF WHAT WE CALL A SUPERSATURATED SOLUTION, AND OF COURSE IN THE SUPERSATURATED SOLUTION WE INDICATED FIRST OF ALL THAT THEY RE VERY UNSTABLE AND WE CAN T MAKE UP A SUPERSATURATED SOLUTION DIRECTLY. WE HAVE TO FIRST OF ALL MAKE UP A SATURATED SOLUTION AT HOT CONDITIONS, HAVE HOT CONDITIONS AND THEN WE COOL THE SOLUTION DOWN. WHEN WE DO SO WE CAN SOMETIMES, FOR CERTAIN CHEMICALS, PRODUCE A SUPERSATURATED SOLUTION. AS WE SAW THE OTHER DO, DOESN T, SUPERSATURATED SOLUTION IS NOT VERY STABLE. IF WE JAR IT, IF WE DROP A PIECE OF DIRT IN IT OR SOMETHING, THE EXCESS SOLUTE IMMEDIATELY BEGINS TO PRECIPITATE OUT. NOW THESE TERMS OF COURSE SAY SOMETHING ABOUT THE AMOUNT OF SOLUTE DISSOLVED, BUT THEY RE RATHER QUALITATIVE TERMS, AND AS I MENTIONED IN THE PREVIOUS LECTURE, TODAY WE WERE GOING TO LOOK AT SOME MORE QUANTITATIVE TYPES OF RELATIONSHIPS. WELL TWO OF THE TERMS LET ME QUICKLY MENTION HERE BEFORE WE GO ON AND LOOK AT SOME VERY VERY QUANTITATIVE TERMS. WE TALK ABOUT CONCENTRATED SOLUTIONS AND DILUTE SOLUTIONS. THOSE ARE TWO OTHER WAYS OF EXPRESSING THE AMOUNT OF SOLUTE THAT WE HAVE IN THE SOLUTION. FOR THESE TWO, AGAIN, WE HAVE SOME SORT OF DEFINITION. CONCENTRATED IN GENERAL APPLIES THAT WE HAVE AT LEAST SIX MOLES. WE TALKED ABOUT THE MOLE AS OUR BASIC UNIT OF MEASUREMENT, SIX MOLES OF SOLUTE IN A LITER OF SOLUTION. IF WE HAVE AT LEAST THAT AMOUNT WE SAY THAT THE SOLUTION IS CONCENTRATED. ALL THAT REALLY MEANS IS THAT WE HAVE QUITE A BIT OF SOLUTE DISSOLVED. IF IT S LESS THAN SIX MOLES OF SOLUTE PER LITER OF SOLUTION, BY THE WAY, THIS RELATIONSHIP HERE WE RE GOING TO TALK ABOUT IN JUST A MINUTE IS ALSO KNOWN AS MOLARITY, ANOTHER CONCENTRATION UNIT. IF IT S LESS THAN SIX MOLES PER LITER, THEN OF COURSE WE REFER TO IT AS DILUTE. SO ONCE AGAIN, THESE ARE TWO TERMS THAT SAY SOMETHING ABOUT THE AMOUNT OF SOLUTE, BUT CERTAINLY AREN T VERY QUANTITATIVE. SO LET S LOOK AT A COUPLE OF TERMS HERE NOW THAT ARE MUCH MORE QUANTITATIVE. FIRST OF ALL, PERCENT BY MASS. WHEN WE SAY, PERCENT BY MASS THEN WE ARE TALKING ABOUT HE MASS OF SOLUTE DIVIDED BY THE MASS OF SOLUTION,
CHM 105 & 106 UNIT 2, LECTURE SEVEN 3 MULTIPLIED BY 10 TO T HE SECOND (10 2 ) TO EXPRESS IT IN TERMS OF PERCENT. SO IT S GOING TO BE GRAMS OF SOLUTE OVER GRAMS OF SOLUTION TIMES 10 2 ). IF I WERE TO GO TO THE GROCERY STORE FOR INSTANCE AND PICK UP A BOTTLE OF BLEACH AND LOOK AT THE LABEL, IT WOULD SAY IT CONTAINS A 5.25% SOLUTION OF SODIUM HYPOCHLORITE. WHAT THAT TELLS ME IS THAT WE HAVE 5.25 GRAMS OF SODIUM HYPOCHLORITE PER 100 GRAMS OF SOLUTION. THAT S VERY DEFINITE. THAT S VERY QUANTITATIVE. WE KNOW EXACTLY HOW MUCH WE RE TALKING ABOUT IN THAT CASE. NOW ALONG WITH PERCENT SOMETIMES IF WE RE WORKING WITH TWO LIQUIDS MAKING UP SOLUTION WE HAVE A PERCENT BY VOLUME, AND SO WE CAN WRITE THAT DOWN HERE QUICKLY. PERCENT BY VOLUME THEN, INSTEAD OF BY MASS, IS MILLILITERS OF SOLUTE. IF IT S A LIQUID IT S GOING TO BE EASIER TO MEASURE OUT OF VOLUME OF THE LIQUID THAN TO WEIGH THE MASS, DIVIDED BY MILLILITERS OF SOLUTION, MULTIPLIED BY 10 2, THAT S PERCENT BY VOLUME. IF WE GO TO A LIQUOR STORE AND LOOK AT A BOTTLE OF WINE FOR INSTANCE IT WOULD SAY 10% BY VOLUME ETHANOL, OR ETHYL ALCOHOL, AS WE WOULD CALL IT. SO IT TELLS US THAT THERE S 10 MILLILITERS OF ETHYL ALCOHOL FOR 100 MILLILITERS OF THE WINE. OR IF I GO TO A PHARMACY AND PICK UP A BOTTLE OF HYDROGEN PEROXIDE, AND IT MIGHT SAY 3% HYDROGEN PEROXIDE. THAT WOULD MEAN THAT IT HAS 3 MILLILITERS OF HYDROGEN PEROXIDE LIQUID IN 100 MILLILITERS OF SOLUTION. SO AGAIN, PERCENT BY VOLUME. SO THOSE ARE TWO VERY QUANTITATIVE WAYS THAT WE CAN EXPRESS CONCENTRATION OF SOLUTION. THE LAST ONE I WANT TO JUST MENTION HERE VERY QUICKLY AND THEN WE LL GO ON TOA SPECIFIC EXAMPLE PROBLEM DEALING WITH PERCENT, BUT MOLARITY. THIS IS THE CONCENTRATION UNITS THAT IS USED MOST EXTENSIVELY IN CHEMISTRY AND IN SCIENCE PRIMARILY, OKAY, MOLARITY. NOTICE THE UNITS, MOLES OF SOLUTE DIVIDED BY LITERS OF SOLUTION. MOLARITY, VERY SPECIFIC, AND OF COURSE IT INVOLVES THE BASIC UNIT OF MEASUREMENT THAT WE VE TALKED ABOUT PREVIOUSLY, THE MOLE. OKAY, SO BEFORE WE GO ON AND TALK ANY FURTHER ABOUT MOLARITY THEN, LET S TURN THEN TO A COUPLE OF EXAMPLE PROBLEMS INVOLVING THE PERCENT COMPOSITION AS EXAMPLES. SO THE FIRST PROBLEM THAT WE WILL TAKE A LOOK AT HERE STATES THAT WE HAVE 735 GRAMS OF SOLUTION, WHICH CONTAINS 11.9 GRAMS OF
CHM 105 & 106 UNIT 2, LECTURE SEVEN 4 SODIUM PHOSPHATE. WHAT IS THE PERCENT BY MASS OF THIS SOLUTION? SO WE WANT OT CALCULATE THE PERCENT BY MASS, SO PERCENT BY MASS THEN AGAIN WILL BE THE GRAMS OF SOLUTE DIVIDED BY THE GRAMS OF SOLUTION, TIMES 10 2. IN THIS PARTICULAR CASE THE NUMBER OF GRAMS OF SOLUTE, REMEMBER BY DEFINITION THE SOLUTE IS THE COMPONENT OF THE SOLUTION IN THE LEAST AMOUNT. SO OBVIOUSLY THIS IS THEN THE LEAST AMOUNT, SO WE HAVE 11.9 GRAMS DIVIDED BY 735 GRAMS, NOTICE THAT S GRAMS OF SOLUTION, MULTIPLIED BY 10 2, AND SO IF WE CALCULATE THAT OUT WE FIND THAT WE WILL HAVE 11.9 DIVIDED BY 735. WE WOULD HAVE 1.62% BY MASS SOLUTION OF SODIUM PHOSPHATE. NOW NOTICE THAT WE DIDN T PUT IN THE UNITS GRAMS OF SODIUM PHOSPHATE HERE BECAUSE IT S GRAMS OF SOLUTE, AND IT REALLY DOESN T MAKE ANY DIFFERENCE WHAT THE SOLUTE IS. IT S JUST GRAMS OF SOLUTE, SO IF WE RE 11.9 GRAMS OF SODIUM CHLORIDE IT WOULD STILL BE A 1.62 PERCENT SOLUTION. IT WOULD BE A 1.62% SOLUTION OF SODIUM CHLORIDE, BUT IT WOULD BE EXACTLY THE SAME NUMERIC VALUE OKAY? SO WHAT THE SOLUTE IS DOESN T MAKE ANY DIFFERENCE AND SO WE DIDN T DISREGARD THE UNITS IN CALCULATING MOLES AND MOLAR MASS ET CETERA, WE WERE VERY CAREFUL TO MAKE SURE WE SAID WHAT IT WAS. HERE IT REALLY DOESN T MAKE ANY DIFFERENCE. JUST GRAMS OF SOLUTE, WHATEVER IT IS, OVER GRAMS OF SOLUTION. WELL LET S LOOK AT ANOTHER PROBLEM. THIS TIME THE QUESTION IS IF 17.3 GRAMS OF SODIUM CHLORIDE ARE MIXED WITH 62.7 GRAMS OF WATER, WHAT IS THE PERCENT BY MASS OF THIS SOLUTION? SO PERCENT BY MASS WOULD BE EQUAL TO, OBVIOUSLY THIS IS MY SOLUTE, IT IS THE ONE IN THE LESSOR AMOUNT, SO WE HAVE 17.3 GRAMS OF SOLUTE DIVIDED BY 72.7 GRAMS, TIMES 10 2. ALRIGHT, SO IF WE WERE TO WORK THAT OUT THEN WE WOULD GET OUR PERCENT BY MASS RIGHT? SOMEBODY HERE WAS SHAKING THEIR HEAD NO. WHY NOT? (STUDENT RESPONSE NOT AUDIBLE) YES, BECAUSE REMEMBER THE BOTTOM UNIT IS NOT SOLVENT, THE BOTTOM UNIT IS GRAMS OF SOLUTION, AND WHAT ARE THE COMPONENTS MAKING UP A SOLUTION? SOLUTE PLUS SOLVENT. SO THIS IS NOT COMPLETE. WE WOULD HAVE TO ADD 17.3 GRAMS HERE SO THAT THIS PLUS THIS GIVES ME THEN GRAMS OF SOLUTION. REMEMBER, IT S GRAMS OF SOLUTE OVER GRAMS OF SOLUTION, NOT SOLVENT. BOTTOM TERM IS ALWAYS SOLUTION. WELL, SO IN THIS
CHM 105 & 106 UNIT 2, LECTURE SEVEN 5 PARTICULAR CASE THEN WE D HAVE 17.3 GRAMS, NOW DIVIDED BY 90.0 GRAMS TIMES 10 2, AND THEN WE WOULD GET A FINAL ANSWER, 17.3 DIVIDED BY 90, AND WE HAVE A CONCENTRATION OF 19.2% BY MASS, SODIUM CHLORIDE. ANY QUESTION ON EITHER STEP? OR EITHER PROBLEM IN THIS CASE? AS WE SAID, ONE CAN DO THE SAME TYPE OF CALCULATION IN TERMS OF PERCENT BY VOLUME, THE ONLY DIFFERENCE WILL BE IS THAT IT WOULD BE MILLILITERS OF SOLUTE OVER MILLILITERS OF SOLUTION, RATHER THAN GRAMS OF SOLUTION. WELL ALTHOUGH THAT S COMMERCIALLY THE MOST COMMON CONCENTRATION UNIT, IT IS NOT THE CONCENTRATION UNIT THAT WE USE PRIMARILY IN CHEMISTRY. THE PRIMARY CONCENTRATION UNIT THAT WE USE IN CHEMISTRY IS CALLED MOLARITY, AND IT IS GIVEN THE SYMBOL UPPER CASE M, WHEN WE WRITE IT OR ABBREVIATE. SO IF YOU SEE A LARGE M AFTER A NUMBER IN A SOLUTION THAT INDICATES ITS MOLARITY. FOR THOSE OF YOU INVOLVED IN THE LABORATORY PORTION OF THE COURSE, WHEN YOU GO TO THE LAB AND CARRY OUT THE EXPERIMENTS YOU LL OFTEN HAVE BOTTLES OF CHEMICALS ON THE LAB BENCH THERE AND IF YOU LOOK YOU LL SEE SOMETHING LIKE 6.0M, AND THEN WHATEVER THE COMPOUND IS. MEANING 6 MOLAR NITRIC ACID OR 6 MOLAR SODIUM HYDROXIDE, WHATEVER THE CHEMICAL IS, AND SO MOLARITY IS THE UNIT THAT WE FREQUENTLY, NOT ONLY FREQUENTLY, BUT PRIMARILY, USE IN CHEMISTRY. NOTICE THE UNITS HERE. MOLES OF SOLUTE PER LITER OF SOLUTION. NOTICE THAT IT S THE SAME IN TERMS OF THE PERCENT BY MASS, IT S SOLUTE ON TOP AND SOLUTION ON THE BOTTOM. THE ONLY DIFFERENCE IS THIS TIME WE RE EXPRESSING IT IN MOLES RATHER THAN GRAMS FOR THE TOP UNIT, AND WE RE EXPRESSING THE SOLUTION IN VOLUME RATHER THAN IN TERMS OF MASS IN THIS PARTICULAR CASE. WELL LET S TAKE A LOOK AT SOME CALCULATIONS THEN INVOLVED WITH MOLARITY, AND AS I SAID, THIS IS THE UNIT THAT WE RE GOING TO USE NOT ONLY IN THIS CHAPTER, BUT EXTENSIVELY THROUGH OTHER CHAPTERS AS WELL. THE FIRST PROBLEM THAT I HAVE TO LOOK AT ASKS THE QUESTION: WHAT IS THE MOLARITY OF A SOLUTION IF I WERE TO DISSOLVE 41.8 GRAMS OF LITHIUM CHLORIDE IN 785 MILLILITERS OF SOLUTION? IN OTHER WORDS, WE HAVE A VOLUME OF SOLUTION THAT WE RE GOING TO END UP WITH. IT DOESN T SAY HOW MUCH WATER IT S GOING TO TAKE, BUT WE RE GOING TO END UP WITH 41.8 GRAMS OF LITHIUM CHLORIDE, THE
CHM 105 & 106 UNIT 2, LECTURE SEVEN 6 SOLUTE, IN 785 MILLILITERS OF SOLUTION. NOW, IN SETTING THE PROBLEM UP THEN WE WANT TO CALCULATE MOLARITY AND SO WE KNOW THAT WE NEED TO END UP WITH UNITS OF MOLES OF SOLUTE OVER LITER OF SOLUTION. THAT S WHAT WE HAVE TO HAVE FOR UNITS WHEN WE FINISH UP. SO LET S GO AHEAD THEN AND START. WE HAVE 41.8 GRAMS OF LITHIUM CHLORIDE OVER 785 MILLILITERS OF SOLUTION. NOW LOOKING AT THAT THEN, WE RE GOING TO ASK THE QUESTION: WHAT DO WE NEED TO DO TO CHANGE THEN FROM THE UNITS OF GRAMS OF LITHIUM CHLORIDE TO MOLES, AND WHAT DO WE NEED TO DO TO CHANGE MILLILITERS OF SOLUTION TO LITERS? SO WE RE GOING TO BE LOOKING FOR THEN TWO CONVERSION FACTORS HERE TO ALLOW US TO MAKE THAT CHANGE. WELL LET S DO THE BOTTOM ONE FIRST. IF WE MULTIPLY NOW BY 1000 MILLILITERS PER 1.00LITERS THEN NOW WE RE DOWN TO LITERS OF SOLUTION IN THE BOTTOM TERM. I SHOULDN T HAVE SCRATCHED THROUGH THE SOLUTION PART, SHOULD VE JUST CANCELLED OUT THE UNIT MILLILITERS. SO WE VE TAKEN CARE OF THE FIRST PART, WE NOW HA VE LITERS OF SOLUTION, BUT NOW WE NEED TO CHANGE FROM GRAMS OF LITHIUM CHLORIDE TO MOLES, AND SO THE QUESTION IS: ONE MOLE OF LITHIUM CHLORIDE HAS WHAT NUMBER OF GRAMS OF LITHIUM CHLORIDE? AND OF COURSE THAT MEANS THAT WE JUST NEED TO CALCULATE OUR MOLAR MASS, AND SO WE WOULD REFER TO THE PERIODIC TABLE OR THE TABLE AT THE BACK OF THE BOOK. WE WOULD LOOK FOR LITHIUM, AND WE FIND LITHIUM TO BE 6.94, AND WE LOOK FOR CHLORINE AND WE FIND IT TO BE 35.45, USING TWO PLACES AFTER THE DECIMAL, AND SO WE HAVE A MASS OF 9, 3, 2, 4, 42.39 GRAMS OF LITHIUM CHLORIDE, AND WE NOW HAVE MOLES OVER LITERS, AND THOSE ARE THE UNITS THAT WE WANT, SO WE SHOULD HAVE THE PROBLEM SET UP CORRECTLY. WELL LET S SEE THEN, WE HAVE 41.8 DIVIDED BY POINT, OR 785, MULTIPLIED BY 1000, AND THEN DIVIDED BY 42.39, AND WE END UP WITH A MOLARITY OF THIS SOLUTION EQUAL TO 1.2, AND LET S SEE, I GUESS WE RE LIMITED TO THREE SIGNIFICANT FIGURES, 1.26 MOLAR SOLUTION IS OUR ANSWER. ANY QUESTIONS ON ANY STEP OF THAT PARTICULAR PROBLEM? ALRIGHT, WELL LET S LOOK AT ANOTHER ONE, JUST FROM A LITTLE DIFFERENT ANGLE IN OUR QUESTION. LET S SUPPOSE THAT WE HAVE A VOLUME OF SOLUTION. THE VOLUME THAT WE HAVE IS A BOTTLE CONTAINING 1.75 LITERS, AND ON THE LABEL IT INDICATED THAT IT HAS.650 MOLAR SOLUTION
CHM 105 & 106 UNIT 2, LECTURE SEVEN 7 OF SODIUM NITRATE. THAT QUESTION WOULD BE: HOW MANY GRAMS OF SODIUM NITRATE THEN ARE THERE CONTAINED IN THERE? OR, ASKING IT IN A DIFFERENT WAY, HOW MANY GRAMS OF SODIUM NITRATE WOULD WE NEED IN ORDER TO PREPARE THAT AMOUNT OF SOLUTION OF THAT CONCENTRATION? WELL WE CAN THINK OF IT SORT OF IN A TWO-STEP FASHION. TO CALCULATE NUMBER OF GRAMS OF SODIUM NITRATE WE NEED TO KNOW HOW MANY MOLES OF SODIUM NITRATE THAT WE HAVE. ONCE WE KNOW HOW MANY MOLES OF COURSE, WE CAN CONVERT TO GRAMS. WE VE DONE THAT NOW MANY TIMES IN MANY TYPES OF PROBLEMS. SO WE ASK THE QUESTION: HOW ARE WE GONNA DETERMINE THE NUMBER OF MOLES IF WE ARE STARTING WITH A SOLUTION THAT HAS 0.650 MOLES OF SODIUM NITRATE PER LITER OF SOLUTION, AND WE NOTICE THAT WE HAVE 1.75 LITERS OF THAT SOLUTION. IF WE MULTIPLY THOSE TWO, WHAT DO WE HAVE LEFT FOR UNITS? MOLES OF SODIUM NITRATE, WHICH IS WHAT WE WANTED TO CALCULATE. AS A MATTER OF FACT, IF WE LOOKED AT THIS, WE COULD REALLY SAY, AND ALL THOUGH I DON T LIKE TO MEMORIZE EQUATIONS, BUT SOMET IMES THEY RE USEFUL TO KEEP IN MIND. MOLES OF X WILL BE EQUAL TO THE MOLARITY OF X MULTIPLIED TIMES THE VOLUME IN LITERS. SO ANY TIME WE MULTIPLY MOLARITY TIMES THE VOLUME EXPRESSED IN LITERS WE RE GOING TO HAVE THE NUMBER OF MOLES OF SOLUTE CALCULATED. SO IF WE CARRY OUT THIS STEP AND DETERM INE THE NUMBER OF MOLES OF SOLUTE THEN OBVIOUSLY WE CAN GO FROM THERE TO DETERMINE THE NUMBER OF GRAMS OF SOLUTE THAT WE HAVE IN THE SYSTEM OR THAT WE NEED TO PREPARE THE SYSTEM, EITHER WAY WE WANT TO LOOK AT IT. WELL LET S SEE, IF WE MULTIPLY THAT OUT HERE QUICKLY THEN WE HAVE 1.75 MULTIPLIED BY.650. IT SAYS THAT WE HAVE 1.14 MOLES OF NANO3, AND THEN WE CAN ASK THE FINAL QUESTION, HOW MANY GRAMS OF NANO3 BECAUSE WE HAVE 1.14 MOLES NANO3 MULTIPLI ED BY THE NUMBER OF GRAMS PER ONE MOLE, AND OF COURSE WE LL HAVE TO CALCULATE THAT. SO CALCULATING THEN THE MOLAR MASS OF SODIUM NITRATE SO THAT WE CAN PLUG THAT IN HERE, WELL LET S SEE, WE LL JUST DO THAT DOWN HERE IN THE CORNER. SODIUM 22.99, LOOKING AT OUR PERIODIC TABLE, NITROGEN 14.01, AND 3 OXYGENS, OXYGEN 16 X 3 WOULD BE 48, AND SO WE END UP WITH 0037534 85.00 GRAMS PER ONE MOLE. IF WE NOW MULTIPLY THAT OUT, MULTIPLIED BY 85, WE GET AN ANSWER THEN OF 9. WHOOPS,
CHM 105 & 106 UNIT 2, LECTURE SEVEN 8 96.7GRAMS OF SODIUM NITRATE. NOW LET ME JUST MENTION THAT IF IN FACT YOU STEP-WISE WROTE DOWN THE 1.14, AND THEN DID THIS AND CALCULATED, YOUR ANSWER PROBABLY COMES OUT TO BE 96.8. THE ANSWER 96.7 THAT I OBTAINED WAS THAT I JUST KEPT THE ANSWER FOR THE FIRST STEP ONO THE CALCULATOR. I THEN PLUGGED IN THIS SECOND NUMBER AND THEN RECORDED THE FINAL ANSWER IN THE CORRECT NUM BER OF SIGNIFICANT FIGURES, AND I MENTIONED THIS BEFORE AND I MENTION THIS AGAIN, WHEN YOU HAVE AN ANSWER THAT IS SLIGHTLY DIFFERENT IN THIS LAST DIGIT IN A PROBLEM TWO OR THREE PLACES DIFFERENT, THAT S NOT ANYTHING TO BE CONCERNED ABOUT. IT MAY MERELY BE THE WAY THAT WE DID IT. IF YOU WRITE DOWN THE INDIVIDUAL NUMBERS ROUNDING OFF EACH TIME TO THE CORRECT NUMBER OF SIGNIFICANT FIGURES YOU RE GOING TO GET A SLIGHTLY DIFFERENT ANSWER IN SOME CASES AT LEAST THAN IF YOU CARRY IT CONTINUOUSLY THROUGH ON THE CALCULATOR. RIGHT? ANY QUESTIONS ON ANY STEP OF THIS ONE? ALRIGHT. NOW OFTEN TIMES WHEN WE ARE GOING TO PREPARE SOLUTIONS, OBVIOUSLY THERE S CERTAIN WAYS WE CAN DO IT IF WE RE PREPARING A SOLUTION THAT INVOLVES A SOLID SOLUTE WE CAN GO UP AND MEASURE WE CAN MEASURE OUT THE 96.7 GRAMS OF SODIUM NITRATE, DUMP IT INTO A CONTAINER, ADD ENOUGH WATER TO BRING IT UP TO THE VOLUME THAT WE ASK FOR, 1.75 LITERS, AND AT THAT POINT WE WOULD HAVE THE CONCENTRATION THAT WE WANT. BUT, IF WE RE NOT WORKING WITH A SOLID OF COURSE, THEN WE HAVE TO MEASURE OUT A VOLUME, AND WE WOULD HAVE TO KNOW WHAT ITS DENSITY WAS. OR MORE TIMES THAN NOT WHEN WE PREPARE SOLUTIONS WE ARE STARTING WITH A SOLUTION THAT S ALREADY PREPARED AND ARE GOING TO PRODUCE SOLUTIONS THAT WE USE IN THE LAB FROM THE CONCENTRATED ONES. THIS INVOLVES THE PROCESS THEN THAT WE CALL DILUTION, AND SO IF I WERE TO GO UPSTAIRS TO PREPARE SOME SULFURIC ACID FOR INSTANCE WHAT I WOULD DO IS I WOULD GO TO THE STOCK ROOM AND GET A BOTTLE OF CONCENTRATED SULFURIC ACID, WHICH IS ABOUT 18 MOLAR, AND LET S SUPPOSE FOR MY CHEMICAL LAB THIS WEEK THAT I NEEDED SOM E 1 MOLAR SULFURIC ACID. WHAT I M GOING TO DO THEN IS PREPARE THAT 1 MOLAR BY TAKING THE CONCENTRATED, ADDING MORE SOLVENT, DILUTING IT DOWN TO GET IT TO THE CONCENTRATION THAT I WANT. DILUTING
CHM 105 & 106 UNIT 2, LECTURE SEVEN 9 PROCESS, AND THIS IS THE WAY THAT WE PREPARE MOST OF OUR SOLUTIONS THAT AT LEAST INVOLVE LIQUID/LIQUID SOLUTIONS. LET S USE THIS HERE AS THEN AN EXAMPLE OF WHAT WE RE TALKING ABOUT. HERE WE HA VE A SOLUTION OF LET S SUPPOSE THAT THIS PARTICULAR SOLUTION WE RE LOOKING AT DOWN HERE, AND THERE ARE 25 MILLILITERS, IT S A LITTLE HARD TO READ BUT YOU CAN SEE THE 30 THERE, AND THE 20 IS DOWN HERE, IT IS SET AT 25. THIS PURPLE COLORED SOLUTION WHICH IS A COPPER SOLUTION. SO WE HAVE 25 MILLILITERS OF THIS, AND LET S ASSUME THAT ITS CONCENTRATION IS 1 MOLAR. SO THE QUESTION IS HOW MANY MOLES OF SOLUTE DO I HAVE IN THIS GRADUATED CYLINDER? WELL REMEMBER WE SAID MOLES OF SOLUTE IS WHAT? MOLARITY TIMES VOLUM E IN LITERS. SO THE MOLARITY I SAID WAS 1 MOLAR, THAT S ONE MOLE PER LITER, MULT IPLIED TIMES.025 LI TERS, 25 MILLILITERS, AND SO WE HAVE A TOTAL OF.025 MOLES OF SOLUTE AT THAT MOMENT. NOW I M GOING TO TAKE SOM E OF THIS WATER IN THE BEAKER NEXT TO IT AND I M GOING TO ADD THAT NOW TO THE GRADUATED CYLINDER, AND I M GOING TO POUR IN ENOUGH THAT I NOW HAVE 100 MILLILITERS, AGAIN, A LITTLE BIT DIFFICULT TO READ THE PRINT ON THERE. LET S SLIDE IT DOWN HERE JUST A SECOND. WE HAVE DILUTED UP TO THE 100 MILLILITER MARK, AND THE QUESTION NOW IS? HOW MANY MOLES OF SOLUTE DO I NOW HAVE? ANYONE? HOW MANY MOLES OF SOLUTE DO WE NOW HAVE IN THAT GRADUATED CYLINDER? HOW MUCH? (STUDENT RESPONSE NOT AUDIBLE).1. ALRIGHT, AND WHY DO YOU SAY.1? (STUDENT RESPONSE NOT AUDIBLE) OKAY WE HAVE 4 TIMES THE VOLUME RIGHT? OKAY, WE HAVE 4 TIMES THE VOLUME, SO.1 SORT OF MAKES SENSE. NOTICE I HINT A LITTLE BIT, I SAID SORT OF MAKES SENSE. BUT HOW DID WE GET THAT OTHER.75 MOLES OF SOLUTE? WHERE DID THAT COME FROM? DID IT COME FLYING IN FROM SPACE? WATER? THERE WASN T ANY SOLUTE IN THE WATER THOUGH WAS THERE? WATER IS WATER. WHEN I ADDED THE WATER TO THE PREVIOUS GRADUATED CYLINDER, DID ANY SOLUTE JUMP OUT OF THE GRADUATED CYLINDER? DID ANY SOLUTE JUMP INTO THE GRADUATED CYLINDER? NO. HOW MANY MOLES OF SOLUTE DO I HAVE IN THE GRADUATED CYLINDER? THE SAME AMOUNT. NOTHING CHANGED. I DIDN T DO ANYTHING TO THE SOLUTE. I
CHM 105 & 106 UNIT 2, LECTURE SEVEN 10 ADDED SOLVENT. THERE IS STILL 0.025 MOLES OF SOLUTE IN THAT GRADUATED CYLINDER. OH YES, IT S OVER A GREATER VOLUME, AND IF IT S OVER A GREATER VOLUME THEN THE CONCENTRATION HAS CHANGED, BUT THE MOLES OF SOLUTE DIDN T CHANGE, AND THAT S OUR REFERENCE POINT WHEN WE RE TALKING ABOUT DILUTING, IS THAT THE MOLES OF SOLUTE, LET ME USE A DIFFERENT ONE TO WRITE ON HERE. IN A DILUTION PROCESS THE MOLES OF SOLUTE INITIALLY MUST NE EQUAL TO THE M OLES OF SOLUTE IN THE FINAL CONDITION. WE AREN T ADDING ANY. WE RE NOT TAKING ANY OUT. WE RE MERELY SPREADING IT OUT OVER DIFFERENT VOLUMES. WE RE ADDING SOLVENT. SO THEREFORE, IF THE MOLES OF SOLUTE INITIALLY IS EQUAL TO MOLES OF SOLUTE FINAL, OF COURSE THAT CAN BE OUR FOCUS POINT FOR CONVERSION FACTORS, OR KEEPING IN MIND THAT WE SAID MOLA RITY TIMES LITERS MUST BE EQUAL THEN TO THE MOLARITY TIMES THE LITERS. SO THE INITIAL MOLARITY TIMES THE INITIAL VOLUME MUST BE EQUAL TO THE FINA L MOLARITY TIMES THE FINAL VOLUME IN ANY DILUTION INVOLVING M OLARITY. NUMBER OF MOLES OF SOLUTE DOES NOT CHANGE. SO WE MIGHT GO BACK TO OUR GRADUATED CYLINDER AND WE MIGHT ACTUALLY SAY: WHAT IS THE MOLARITY OF THIS NEW SOLUTION. WELL I THINK IT WOULD BE RATHER OBVIOUS THAT THE MOLARITY OF THIS SOLUTION BETTER BE LESS THAN THE MOLARITY OF THE SOLUTION WE STARTED WITH. IF WE VE ADDED SOLVENT AND MADE THE VOLUME BIGGER THE CONCENTRATION MUST BE DECREASING, IN OTHER WORDS, DILUTING IS GOING TO CAUSE THE CONCENTRATION TO DECREASE, SO THE MOLARITY IS MOLES OVER LITERS, AND WE JUST CALCULATED THAT WE STILL HAVE.025 MOLES, BUT NOW WE HAVE IT OVER 1.00 LITERS. W E COULD HAVE IT OVER 100 MILLILITERS, AND THEREFORE THE NEW CONCENTRATION IS 0.25 MOLES MOLAR. WHAT WAS THE INITIAL CONCENTRATION? 1 MOLAR W E SAID. IT WAS 1 MOLAR. WE VE NOW ADDED SOLVENT AND WE RE AT A CONCENTRATION OF.250 MOLAR. OUR ANSWER MAKES RELATIVE TO THE FACT THAT WHEN WE DILUTE SOMETHING THE MOLARITY SHOULD DECREASE. OKA Y, WELL LET S TAKE A LOOK AT A COUPLE OF PROBLEMS INVOLVING THEN DILUTION. ONCE AGAIN, AS I POINTED OUT, ONE CAN USE THIS TO THEN SOLVE FOR ANY ONE OF THOSE FOUR ITEMS, IF YOU KNOW THE OTHER THREE. ALRIGHT, AND SO THEREFORE WE CAN GO AHEAD AND TAKE A LOOK, BUT WE RE GOING TO DO IT FOLLOWING UNITS RATHER
CHM 105 & 106 UNIT 2, LECTURE SEVEN 11 THAN MEMORIZE THE EQUATION. SO WE HAVE THE QUESTION 185 MILLILITERS OF A SOLUTION, 1.56 MOLAR IN CONCEN TRATION, IS DILUTED TO 1000 MILLILITERS. WHAT IS THE NEW FINAL CONCENTRATION? NOW THE WAY WE LL LOOK AT IT FIRST IS WE RE GOING TO SAY, ALRIGHT, HOW MANY MOLES OF SOLUTE DID I HAVE INITIALLY? AND I HAVE 1.56 MOLES OF SOLUTE, AND IT DOESN T MAKE ANY DIFFERENCE WHAT IT IS, PER LITER, THAT WAS MY CONCENTRATION, MULTIPLIED TIMES 1.000 LITERS. AND SO THEREFORE WE HAVE A TOTAL OF 1.56 MOLES. OH, I M SORRY, X THAT OUT. WE LL START AGAIN. THE MOLES OF SOLUTE IN INITIALLY IS 1.56 MOLES PER LITER, MULTIPLIED BY 185 MILLILITERS. THIS IS WHAT WE RE STARTING WITH, I GOT OVER ON THIS WRONG SIDE. THAT S WHAT WE RE GONNA END UP WITH. THIS IS WHAT WE RE STARTING WITH. SO INITIALLY THEN I HAD THAT CONCENTRATION AND THAT VOLUME, AND OF COURSE THEN 1 LITER OVER 1000 MILLILITERS TO GET IT INTO THEN MOLES SO THAT OUR UNITS ALL CANCEL, AND WE WOULD HAVE THEN LET S SEE, 1.56 X.185, AND WE HAVE 0.289 MOLES. WHAT WE WANTED TO CALCULATE WAS THE FINA L MOLARITY. WE KNOW THAT MOLARITY IS MOLES OVER LITERS, AND WE HAVE HOW MANY MOLES? 0.289 MOLES AND THEN WHAT FINA L VOLUME IS THAT GOING TO BE? 1.000 LITER, THE FINAL CONCENTRATION IS 0.289 CAPITAL M. IS THAT A SMALLER VA LUE THAN WHAT WE STARTED WITH? CERTAINLY IS. WE STARTED WITH 1.56. WE WOULD PRED ICT THAT THE CONCENTRATION SHOULD DECREASE IF WE RE INCREASING THE VOLUM E, AND SO OUR ANSWER MAKES SENSE. ALRIGHT, ANY QUESTIONS ON ANY STEP OF THAT ONE? NOW AS I SAY, ONE COULD USE THE MEMORIZED EQUATION. OUR UNKNOWN IN THIS PART ICULAR CASE WAS FINA L MOLARITY, AND SO W E COULD HAVE REARRANGED THE EQUATION THAT WE HAD AND SOLVED ALGEBRAICALLY FIRST OF ALL FOR THE FINAL M AND THEN PLUGGED IN THE NUMBERS ACCORDINGLY. BUT I PREFER TO LOOK AT THE PROBLEMS INSTEAD IN TERMS OF THE UNITS AND KEEP IN MIND THE MOLES BECAUSE THE MOLE PART IS GOING TO BE IMPORTANT TO US AS WE GO INTO SOLUTION STOICHIOMETRIC PROBLEMS. WELL LET S LOOK AT ANOTHER EXAMPLE HERE. THIS PARTICULAR ONE ASKS THE QUESTION: HOW MANY MILLILITERS OF THIS MORE CONCENTRATED SOLUTION, 5.46 MOLAR SODIUM CHLORIDE, HOW MANY MILLILITERS OF THAT SOLUTION WOULD I NEED TO TAKE TO PREPARE 785 MILLILITERS OF.250 MOLAR SODIUM CHLORIDE? NOW, THIS IS WHAT I WANT TO
CHM 105 & 106 UNIT 2, LECTURE SEVEN 12 END UP WITH MOLE-WISE, AND SO I CAN SAY: HOW MANY MOLES OF NACL DO I NEED? I NEED 0.250 MOLES PER LITER, MULTIPLI ED BY 0.785 LITERS. ALRIGHT, I M HOPING AT THIS POINT NOW THAT MOST EVERYBODY HAS WORKED ENOUGH NOW WITH THE AT LEAST THOUSAND UNIT, AND WE CAN JUST GO AHEAD AND MOVE THE DECIMAL OVER AND NOT PUT IN THIS METRIC CONVERSION EACH STEP AS WE GO THROUGH. SO LITERS CANCEL OUT AND IT TELLS ME THAT THE FINAL SOLUTION THAT WE ARE GOING TO PREPARE CONTAINS.25 X.785, CONTAINS 0.196 MOLES OF SOLUTE. THE QUESTION THEN IS: WHAT VOLUME OF THIS SOLUTION WOULD I NEED TO TAKE TO GET THAT NUMBER OF MOLES OF SOLUTE? SO THE QUESTION WE RE REALLY ASKING IS LITERS OF THE MORE CONCENTRATED 0.196 MOLES IS WHAT WE NEED, AND OUR CONCENTRATION TELLS US THAT ONE LITER OF OUR STARTING SOLUTION CONTAINS 5.46 MOLES. REMEMBER, ANY CONVERSION FACTOR CAN BE USED EITHER WAY. IF WE SAY ONE MOLE PER LITER, WE CAN SAY ONE LITER PER ONE MOLE. SO IN THIS PARTICULAR CASE W E WANT THE MOLARITY NOT AT MOLES OVER LITERS, BUT AT LITERS OVER MOLES, SO THIS CANCELS. AND WE WOULD HAVE AN ANSWER IN LITERS, BUT, NOTICE THAT THE QUESTION WAS HOW MANY MILLILITERS DO WE WANT TO TAKE? WELL IF WE WANT TO GO AHEAD AND DO THAT WE LL JUST PUT MILLILITERS HERE. WE VE GOT LITERS. NOW WE LL MULTIPLY THEN BY 10 3 MILLILITERS PER LITER, AND WE RE GOING TO END UP WITH WHAT WE NEED IN TERMS OF MILLILITERS. SO.196 DIVIDED BY 5.46 X1000, WE WOULD NEED TO TAKE 35.9 MILLILITERS OF THE 5.46 MOLAR SOLUTION. IN OTHER WORDS THEN, IF I WERE TO TAKE AND MEASURE OUT 35.9 MILLILITERS INTO A GRADUATED CYLINDER, I DUMP IT INTO A CONTAINER, A LARGER GRADUATED CYLINDER, AND I A DD WATER UNTIL IT REACHES 785 MILLILITERS. AT THAT POINT I HAVE A CONCENTRATION OF SOLUTE OF.250 MOLAR CONCENTRATION. OKAY? SO OFTEN TIMES THAT S THE APPROACH WE HAVE. HOW MUCH DO WE NEED OF THE CONCENTRATE TO PREPARE WHAT IT IS THAT WE RE MAKING? W ELL LET S LOOK AT ONE MORE EXAMPLE HERE OF DILUTION. WHAT VOLUME OF 3 MOLAR NITRIC ACID COULD BE PREPARED BY DILUTING 25 MILLILITERS OF 15.4 MOLAR NITRIC ACID? OKAY, WE WANT TO KNOW WHAT VOLUME, HOW MUCH OF THIS 3 MOLAR SOLUTION CAN I MAKE BY DILUTING 25 MILLILITERS OF 15.4 MOLAR? BY THE WAY, THIS WOULD BE CONCENTRATED NITRIC ACID, THAT S THE MOLARITY OF IT. THIS
CHM 105 & 106 UNIT 2, LECTURE SEVEN 13 TIME LET ME GO AHEAD AND AS I SAY, I DON T PREFER TO MOST PROBLEMS THIS WAY, BUT LET S GO AHEAD AND USE THEN THE EQUATION THAT WE HAD JUST A MOMENT AGO. WE SAID MOLARITY INITIAL TIM ES LITERS INITIAL EQUALS MOLARITY FINAL TIMES LITERS FINAL. BUT WE CAN MODIFY THAT A LITTLE BIT BECAUSE AS LONG AS WE DO THE SAME THING TO EACH SIDE OF THE EQUATION WE WON T CHANGE IT ANY. WE COULD SAY MOLARITY INITIAL TIMES MILLILITERS INITIAL IS EQUAL TO MOLARITY FINAL TIMES MILLILITERS FINAL. NOW, LOOKING AT THIS THEN WHAT IS IT OUT OF THAT FOUR THAT WE RE LOOKING FOR? WE SAY WHAT VOLUME, WHAT FINAL VOLUME WILL WE BE ABLE TO OBTAIN? SO THIS IS WHAT WE RE LOOKING FOR RIGHT HERE. W E RE LOOKING FOR THE MILLILITERS THAT WE CAN PREPARE, AND SO IF WE ALGEBRAICALLY DO THIS WE WOULD DIVIDE THEN THIS SIDE BY M FINAL, SO THAT WE CAN GET RID OF IT, AND IF WE DO THAT WE MUST DIVIDE THIS SIDE BY M FINAL, AND THEN WE RE READY TO GO. MILLILITERS FINAL IS EQUAL TO THE MOLARITY INITIALLY, 15.4 MOLAR TIMES THE MILLILITERS INITIALLY 25.0 MILLILITERS, ALL DIVIDED BY THE FINAL MOLARITY 3.00 CAPITAL M. MAKE SURE OUR UNITS CHECK, AND THEY DO IN THIS CASE. WE STILL ARE USING AN ANALYSIS, WE HAVE MILLILITERS ON EACH SIDE, AND OUR FINAL ANSWER THEN IS GOING TO BE ABOUT 5 TIMES THIS, SO APPROXIMATELY 128 OR 130 MILLILITERS IS WHAT WE COULD PREPARE OF 3 MOLAR SOLUTION. ANY QUESTION ON THAT? ALRIGHT, IN OUR NEXT LECTURE WE LL TAKE A LOOK AT HOW WE CAN NOW USE THIS QUANTITY OF MOLARITY TO WORK STOICHIOMETRIC PROBLEMS INVOLVING SOLUTIONS. WE LL ALSO LOOK AT SOME CONCENTRATION UNITS TO DEAL WITH SOLUTIONS THAT ARE VERY DILUTE. MOLARITY IS USUALLY USED WHEN WE RE TALKING ABOUT AN APPRECIABLE AMOUNT OF SOLUTE, BUT IF WE RE TALKING ABOUT ENVIRONMENTAL IMPACTS OF THINGS WE NEED SOME CONCENTRATION UNITS THAT WILL ALLOW US TO TALK IN VERY SMALL AMOUNTS OF SOLUTE. WE LL LOOK AT THAT IN OUR NEXT LECTURE.