CHM 105 & 106 UNIT TWO, LECTURE EIGHT 1 IN OUR PREVIOUS LECTURE WE WERE LOOKING AT CONCENTRATION UNITS FOR SOLUTIONS

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1 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 1 CHM 105/106 Program 15: Unit 2 Lecture 8 IN OUR PREVIOUS LECTURE WE WERE LOOKING AT CONCENTRATION UNITS FOR SOLUTIONS AND WE HAD LOOKED AT PERCENT BY MASS AND PERCENT BY VOLUME, AND WE ALSO LOOKED THEN AT WHAT WE CALL THE PRIMARY CONCENTRATION UNIT FOR CHEMISTRY WHICH IS MOLARITY, AND WE SAID THAT MOLARITY IS DEFINED AS THE MOLES OF SOLUTE PER LITER OF SOLUTION. AND, AS I INDICATED, THIS IS OUR PRIMARY UNIT AND OF COURSE IT INVOLVES MOLES, WHICH TAKES US BACK TO THE BASIC MEASURING UNIT OF CHEMISTRY, WHICH OF COURSE MEANS A MOLAR MASS OF A SUBSTANCE, OR AVOGADRO S NUMBER OF PARTICLES, MOLECULES. SO MOLES PER LITER OF SOLUTION. LET S JUST TAKE ONE MORE QUICK EXAMPLE HERE BEFORE WE GO ON TO TAKE A LOOK AT A COUPLE OF OTHER CONCENTRATION UNITS. LET S SUPPOSE THAT WE DISSOLVED 322 GRAMS OF SODIUM HYDROXIDE IN WATER AND PREPARED 3.75 LITERS OF SOLUTION, AND WE WANT TO KNOW WHAT THE MOLARITY IS OF THAT SOLUTION, SO AGAIN, THEN, SETTING UP, KEEPING IN MIND OUR UNITS THAT WE NEED TO END UP WITH, MOLES OVER LITER OF SOLUTION WE WOULD SAY THAT MOLARITY IS EQUAL TO 322 GRAMS OF NAOH OVER 3.75 LITERS, IN THIS PARTICULAR CASE THE BOTTOM UNIT IS ALREADY CORRECT, SO WE DON T NEED TO MAKE ANY CONVERSION. THE TOP UNIT, HOWEVER, IS NOT IN THE UNITS OF MOLES, SO WE NEED TO MAKE A CORRECTION FACTOR. SO WE NEED TO BRING IN THE MOLAR MASS THEN OF SODIUM HYDROXIDE, AND ONE MOLES OF SODIUM HYDROXIDE THEN IS EQUAL TO, AND WE WOULD HAVE TO DETERMINE THE MOLAR MASS. WELL SODIUM IS 22.99, AND HYDROGEN 1.01, SO THAT WOULD GIVE US 24, AND 16 FOR THE OXYGEN, WOULD GIVE US A MASS OF 40 GRAMS FOR THE MOLAR MASS. OUR UNITS OF GRAMS CANCEL. WE SEE THAT WE RE NOW LEFT WITH MOLES OF SOLUTE PER LITER OF SOLUTION, AND SO WE HAVE THE PROBLEM SET UP CORRECTLY. IT THEN WOULD BE A MATTER OF DETERMINING WHAT THE ACTUAL CONCENTRATION WAS. SO TAKING 322, DIVIDE BY 3.75, AND DIVIDE THAT NEW TOTAL BY 40, AND WE END UP WITH A SOLUTION WHICH IS 2.15 MOLAR IN CONCENTRATION. SO AS I INDICATED PREVIOUSLY, WE RE GOING TO UTILIZE OF COURSE THIS MOLARITY AS WE GET INTO CALCULATIONS INVOLVING SOLUTION STOICHIOMETRY. BUT BEFORE WE DO THAT LET S TAKE

2 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 2 A LOOK AT A COUPLE OF OTHER CONCENTRATION UNITS WHICH DEAL WITH VERY DILUTE SOLUTIONS, VERY LOW AMOUNTS OF SOLUTE IN SOLUTION, AND THESE ARE UNITS THAT ARE TYPICALLY USED WHEN WE TALK ABOUT ENVIRONMENTAL HAZARDS, WHEN WE TALK ABOUT LEAD IN THE ENVIRONMENT OR MERCURY IN THE ENVIRONMENT, OR PESTICIDES IN THE ENVIRONMENT, WE TALK ABOUT IN TERMS OF PARTS PER THOUSAND OR PARTS PER MILLION OR PARTS PER BILLION, ARE THE UNITS THAT WE USE. VERY TOXIC MATERIALS, WE MAY REGULATIONS THAT ALLOW ONLY PARTS PER BILLION BEING PRESENT IN DRINKING WATER, OR FOOD, WHATEVER THE CASE MIGHT BE, OR EVEN IN THE AIR. IF IT ISN T THAT TOXIC, BUT SOMEWHAT HARMFUL, WE MAY HAVE REGULATIONS THAT LIMIT US TO PARTS PER THOUSAND, IN CONTRAST, INSTEAD OF PARTS PER BILLION. SO LET S TAKE A LOOK AT THESE UNITS AND SEE WHAT THEY MEAN WHEN WE TALK ABOUT PARTS PER THOUSAND FOR INSTANCE. PARTS PER THOUSAND IS DEFINED AS THE NUMBER OF GRAMS OF SOLUTE PER LITER OF SOLUTION. NOW, WHEN WE THINK OF PARTS PER THOUSAND, WE OBVIOUSLY WOULD THINK ONE UNIT OUT OF A THOUSAND UNITS, SO THE QUESTION IS: HOW ARE WE EXPRESSING PARTS PER THOUSAND IN TERMS OF GRAMS OVER LITERS OF SOLUTION? WELL FOR MOST DILUTE SOLUTIONS, THE CONCENTRATION OF WATER IS ONE GRAM PER MILLILITER. SO THEREFORE, A LITER OF WATER, OR A LITER OF SOLUTION WHICH IS PRIMARILY WATER THEN, WOULD CONTAIN ABOUT 1000 GRAMS, AND SO THAT IS WHY WE HAVE THE STATEMENT GRAMS PER LITER OF SOLUTION. JUST RE-WRITE THAT OVER HERE, WHAT WE RE SAYING THEN IS THAT WE HAVE GRAMS OF SOLUTE AND BECAUSE THE DENSITY OF VERY VERY DILUTE SOLUTIONS IS ONE GRAM PER MILLILITER, A LITER WOULD THEREFORE BE 1000 GRAMS OF SOLUTION. SO THIS IS THE ACTUAL DEFINITION, WOULD BE GRAMS OF SOLUTE PER THOUSAND GRAMS OF SOLUTION. HOWEVER, AS I SAID, FOR VERY DILUTE SOLUTIONS, 1 GRAMS EQUALS ONE MILLILITER, AND SO THEREFORE 1000 GRAMS WOULD BE 1000 MILLILITERS, WHICH WOULD TAKE US BACK TO ONCE AGAIN A LITER. ALRIGHT, SO GOING BACK TO OUR BASIC DEFINITION THEN FIRST OF ALL, PARTS PER THOUSAND EXPRESSED AS GRAMS OF SOLUTE PER LITER OF SOLUTION. WE ALSO DEFINE VERY LOW CONCENTRATIONS IN TERMS OF UNITS OF PARTS PER MILLION, ABBREVIATED PPM, AND HERE WE SEE THAT WE HAVE UNITS OF MILLIGRAMS OF SOLUTE PER LITER OF SOLUTION. NOW A

3 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 3 PART PER MILLION ONCE AGAIN WOULD SEEM TO BE ONE OUT OF A MILLION UNITS. AGAIN IF WE GO BACK TO TAKE A LOOK AT THIS IN TERMS OF RELATIONSHIP OF MASS, THEN WHAT WE RE SAYING HERE IS A MILLIGRAMS PER LITER. BUT A MILLIGRAMS IS WHAT?.001 GRAM OVER 1000 GRAMS. THIS IS STILL RELATING TO THE FACT THAT VERY DILUTE SOLUTIONS, A MILLILITER IS EQUAL TO A GRAM, AND IF WE WERE TO CONVERT THIS THEN TO 1 INSTEAD OF A 1000 TH, IF WE MULTIPLIED BOTH THE TOP AND BOTTOM BY 1000 WE WOULD NOW BE SAYING 1 GRAM PER 1O 6 GRAMS, OR 1O 6 OF COURSE IS A MILLION. SO WE RE SAYING ONE GRAM OUT OF A MILLION GRAMS WOULD BE 1 PART PER MILLION. BUT FOR PRACTICAL PURPOSES FOR CALCULATING PURPOSES THEN WE GO BACK TO OUR BASIC DEFINITION HERE, MILLIGRAMS OF SOLUTE PER LITER OF SOLUTION. FINALLY, FOR VERY LOW LEVELS OF MATERIALS, OR VERY HIGHLY TOXIC MATERIALS IN THE ENVIRONMENT WE MAY TALK ABOUT THEM IN TERMS OF PARTS PER BILLION. OKAY, SO WE HAD PARTS PER THOUSAND, PARTS PER MILLION, PARTS PER BILLION. EACH OF THESE VARIES BY A FACTOR OF OKAY, SO EACH STEP WE RE GOING IS A THOUSAND CHANGE. SO IN THIS PARTICULAR CASE WE EXPRESS IT AS MICROGRAMS, THAT S 10-6 GRAMS OF SOLUTE PER LITER OF SOLUTION. AGAIN, IF WE WERE TALKING IN TERMS OF JUST GRAMS TO GRAMS RELATIONSHIP WE COULD SHOW THAT. WE WOULD BE SAYING THEN THAT WE HAD 10-6 GRAMS, THAT S A MICROGRAM, PER 1000 GRAMS OF SOLUTION. NOW, IN ORDER TO CLEAR THAT UP THEN SO THAT WE HAVE JUST ONE ON TOP WE D MULTIPLY BOTH THE TOP AND THE BOTTOM BY 10 6, SO WE WOULD END UP WITH ONE GRAM PER 10 9 GRAMS, AND NOW WE RE TALKING ABOUT PARTS PER BILLION. OKAY, SO FOR OUR PURPOSES, FOR OUR CALCULATION PURPOSES THIS IS ONLY TO SHOW THAT WE REALLY TRULY DEFINE INITIALLY PARTS PER MILLION, PARTS PER BILLION, PARTS PER THOUSAND RELATIVE TO GRAMS TO GRAMS. BUT BECAUSE OF THE DENSITY OF VERY DILUTE SOLUTIONS THEN FOR ALL PRACTICAL PURPOSES WE CAN USE THESE AS OUR WORKING DEFINITIONS FOR THESE THREE VERY DILUTE SOLUTIONS. SO LET S TAKE A LOOK AT A COUPLE OF PROBLEMS HERE. WELL, BEFORE I DO THAT LET ME JUST SAY ONE THING, AND THIS IS TO TELL YOU OR ASK YOU TO BE A LITTLE SKEPTICAL WHEN YOU READ NEWSPAPER REPORTS ET CETERA ABOUT HAZARDS IN THE ENVIRONMENT OF SOMEBODY HAS MEASURED SOME TOXIC METALS IN SOME LAND SOMEWHERE. NEWSPAPER

4 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 4 REPORTERS ARE VERY PRONE TO MAKE THE STORY SOUND VERY GOOD, AND SO THEREFORE, BECAUSE MOST PEOPLE DON T REALLY KNOW WHAT THE DIFFERENCE BETWEEN THESE ARE, YOU SEE, LET S SUPPOSE THAT THE FEDERAL GOVERNMENT ALLOWED US TO HAVE IN DRINKING WATER, AND WE LL SAY CADMIUM WHICH IS A METAL WHICH WE ARE CONCERNED ABOUT, LET S SUPPOSE THAT THE FEDERAL GOVERNMENT HAD AS A MAXIMUM LEVEL 10 PARTS PER MILLION OF CADMIUM. AND CADMIUM ACTUALLY IN NATURE EXISTS AS A CHARGED PARTICLE, AN ION AS WE TALKED ABOUT BEFORE. ALRIGHT, SO LET S SUPPOSE THAT SOMEBODY NOW HAD A WATER SAMPLE THEY HAD COLLECTED FROM A RUN-OFF OF A POND SOMEPLACE, AND THEY HAD IT ANALYZED, AND IT TURNED OUT TO BE, LET S SAY 7 PARTS PER MILLION. WELL THAT DOESN T MAKE ANY STORY. BUT YOU SEE IF I WANT TO MAKE A STORY OUT OF THIS AS A NEWSPAPER PERSON I CAN PUBLISH AND JUSTIFIABLY SAY IT, I COULD SAY THAT THERE WERE 700 PARTS PER BILLION CADMIUM. NOW WHEN YOU TELL SOMEBODY THERE S 700 SOMETHING UNITS OUT THERE THAT S A TOXIC THAT S PRETTY SCARY. IF YOU SAY THERE S 7, OR IF YOU SAY THERE S.007 PARTS PER THOUSAND DOESN T HAVE NEARLY THE IMPACT, AND YOU RALLY NEED TO PAY ATTENTION IF YOU RE LOOKING AT STORIES RELATING TO ENVIRONMENTAL PROBLEMS WHAT THEY ARE STATING IN, BECAUSE WE DO HAVE STANDARDS MAXIMUMS THAT WE CAN HAVE, AND OFTEN TIMES THESE NUMBERS THAT ARE BEING MEASURED REALLY FALL BELOW THE ALLOWABLE EPA STANDARDS, BUT YET THE STORY WILL BE REPORTED IN SUCH A WAY THAT IT ACTUALLY MAKES IT SOUND LIKE WE HAVE A HORRIBLE CONDITION EXISTING, AND THIS REALLY CAME HOME OT ME WHEN I WAS IN DALLAS, TEXAS A FEW YEARS AGO AT A MEETING, AND IT JUST HAPPENED THAT PARTICULAR DAY THAT THEY HAD A STORY ABOUT AN AREA OF DALLAS THAT HAD A TRENCH THAT KIND OF WENT THROUGH OR A CRICK THAT WENT THROUGH AND THESE PEOPLE WERE CONCERNED ABOUT POLLUTION IN THE WATER AND THIS LADY WAS STANDING ON THE FRONT STEPS OF THE COURTHOUSE, OR I GUESS IT WAS THE COURTHOUSE WITH A QUART JAR AND THEY HAD A LABEL ON THERE AND THEY HAD ALL THESE THINGS LABELED AND THEY HAD THEM IN THEN PARTS PER BILLION, AND SOME OF THESE THINGS WERE REALLY LARGE NUMBERS. LIKE, IN TERMS OF PARTS PER BILLION, THEY HAD SOMETHING LIKE 246,000 PARTS PER BILLION FOR

5 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 5 SOME SORT OF ORGANIC COMPOUND THAT WAS THERE. WELL IF WE CONVERT THAT BACK TO PARTS PER THOUSAND THAT S.246 PARTS PER THOUSAND AND IT JUST HAPPENED THAT THAT PARTICULAR CHEMICAL THAT THEY HAD LISTED THERE THE EPA ALLOWS 5 PARTS PER THOUSAND, BEFORE IT S EVEN CONSIDERED TO BE A HAZARD. BUT YET HERE S THIS LABEL ON THIS JAR SHOWING THESE TREMENDOUS LARGE NUMBERS INDICATING THIS TERRIBLE POLLUTION IN THIS WATER IN THIS DITCH BEHIND THIS SET OF HOMES. NOW I M NOT SAYING THERE MIGHT NOT HAVE BEEN SOME IN THERE THAT WERE TOXIC OR AT HIGH ENOUGH LEVELS, BUT THE STORY WAS CERTAINLY OUT OF PROPORTION AS COMPARED TO WHAT IT TRULY WAS. YOU SEE, THAT NUMBER LOOKS PRETTY BIG. IF YOU TOLD SOMEBODY THEY HAD THAT MAGNITUDE OF SOME SORT OF PESTICIDE THEY WOULD SAY, OH MY GOODNESS. THIS IS HORRIBLE. IF YOU PUT IT THIS WAY AND THEN TELL THEM THAT, YES, YOU CAN HAVE 5 PARTS PER THOUSAND WITHOUT ANY HEALTH PROBLEMS IT LOOKS CONSIDERABLY DIFFERENT. SO I JUST SAY THAT WHEN YOU RE LOOKING AT REPORTS BE SURE THAT YOU PAY ATTENTION TO HOW THE CONCENTRATION IS EXPRESSED. IN CHEMISTRY, WHERE W E RE USING PRIMARILY MOLARITY, THAT S NOT A PROBLEM, IT ALWAYS MEANS THE SAME THING. A MOLE PER LITER IS A MOLE PER LITER IS MUCH LESS CONCENTRATED THAN 10 2, AND WE KNOW THAT DIFFERS BY 10 POWERS, OR SIX POWERS OF 10. HAD A QUESTION OVER HERE? (STUDENT RESPONSE NOT AUDIBLE) YES, THAT IS CORRECT, THANK YOU. I CAN T EVEN MULTIPLY BY 1000 HERE THIS MORNING. SO THAT MAKES IT EVEN LOOK WORSE DOESN T IT PARTS PER BILLION THEN OF CADMIUM. ALRIGHT, WELL LET S TAKE A LOOK AT CALCULATING A COUPLE OF THESE THEN. AND I HAVE A PROBLEM HERE TO LOOK AT. THE PROBLEM SAYS: HOW MANY PARTS PER MILLION MERCURY, MERCURY IS ONE OF THE CHEMICALS THAT WE ARE CONCERNED ABOUT, AS FAR AS TOXICITY - HOW MANY PARTS PER MILLION MERCURY WOULD THERE BE IN A SOLUTION THAT HAS A MOLARITY OF 3.4 X 10-5 MOLES PER LITER? NOW, KEEP IN MIND HOW WE DEFINE PARTS PER MILLION. WE SAID PARTS PER MILLION WAS EQUAL TO MILLIGRAMS OF SOLUTE, SO IN THIS CASE, MILLIGRAMS OF MERCURY PER LITER OF SOLUTION. SO WE RE GOING TO START HERE WITH 3.4 X 10-5 MOLES OF MERCURY PER LITER OF SOLUTION, AND SO WE SEE THAT THE CHANGE NOW TO PARTS PER MILLION, TO EXPRESS IT IN THAT FASHION, WE WILL

6 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 6 NEED TO CONVERT FROM MOLES TO MILLIGRAMS. OKAY, SO LET S GO AHEAD AND RESET IT DOWN HERE SO WE HAVE JUST A LITTLE BIT MORE WORKING ROOM. SO ONCE AGAIN, 3.4 X 10-5 MOLES OF MERCURY PER LITER, MULTIPLIED BY, ALRIGHT, WE WOULD HAVE TO LOOK AT THE PERIODIC TABLE AND FIND MERCURY. WE SEE THAT IT HAS A MASS OF THAT WOULD BE GRAMS PER ONE MOLE. BUT WE DON T WANT IT IN GRAMS, WE NEED IT IN MILLIGRAMS, SO MULTIPLY BY 1000, 10 3 MILLIGRAMS PER ONE GRAMS. SO NOW THE GRAMS CANCEL, AND NOW WE HAVE MILLIGRAMS PER LITER OF SOLUTION WHICH IS THEN THE CORRECT UNIT FOR PARTS PER MILLION. WELL LET S SEE WHAT WE WOULD END UP WIT THEN. WE HAVE 3.4 X 10-5 X , AND TIMES AND WE HAVE AN ANSWER OF 6.82 PARTS PER MILLION MERCURY. OKAY, NOW AGAIN, YOU SEE, IF I REPORTED THAT AS PARTS PER BILLION NOW AND MAKE THAT 6820 PARTS PER BILLION, ONCE AGAIN THAT SOUNDS LIKE A TREMENDOUS AMOUNT PARTS PER MILLION SOUNDS CERTAINLY MUCH MORE REASONABLE, AND I M NOT EVEN SURE WHAT THE ACTUAL ALLOWABLE LEVEL OF MERCURY IS IN SAY WATER SAMPLES. ALRIGHT, HERE S ANOTHER EXAMPLE. CALCULATE THE PARTS PER BILLION OF POTASSIUM CYANIDE. POTASSIUM CYANIDE IS A POISON, AND IT S A POISON THAT S USED IN LIKE SOME COCKROACH OR ROACH, WHAT DO I WANT O SAY, PESTICIDES. SO IT WOULD BE ONE OF THE THINGS THAT WE MIGHT BE CONCERNED ABOUT, AS FAR AS IN A WATER SAMPLE. THIS TIME W E WANT TO EXPRESS IT IN TERMS OF PARTS PER BILLION, AND WE SAID THAT PARTS PER BILLION IS MICROGRAMS PER LITER OF SOLUTION. SO WE WOULD SET UP THE PROBLEM THEN IN SIMILAR FASHION TO WHAT WE DID BEFORE. PARTS PER BILLION WOULD BE EQUAL TO 2.8 X 10-6 MOLE OF KCN PER LITER OF SOLUTION. NOW THE FIRST ONE WE LOOKED AT WAS OF COURSE PARTS PER MILLION OF AN ELEMENT. HERE WE RE TALKING NOW PARTS PER MILLION OF A COMPOUND. SO THE FIRST THING WE NEED TO DO IS CONVERT THIS TO GRAMS, AND SO WE NEED TO KNOW HOW MANY GRAMS OF KCN WE HAVE PER ONE MOLE, AND SO WE D HAVE TO CALCULATE THAT. WELL LET S SEE, POTASSIUM WOULD BE 39.10, AND WE HAVE ONE CARBON12.01, AND WE HAVE ONE NITROGEN 14. SO WE WOULD END UP WITH GRAMS PER MOLE. SO NOW WE HAVE GRAMS, BUT WE STILL NEED TO GO A STEP FURTHER, AND THAT WOULD BE NOW TO CONVERT TO MICROGRAMS. SO IF WE MULTIPLIED NOW BY 10 6 MICROGRAMS PER ONE GRAM, THEN THE

7 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 7 GRAMS WILL CANCEL AND WE LL NOW BE EXPRESSING UNITS OF MICROGRAMS PER LITER, WHICH WOULD BE PARTS PER BILLION. WELL IN THIS PARTICULAR CASE LET S SEE WE CAN PROBABLY ALMOST EYEBALL THIS. FIRST OF ALL, 10-6 X 10 6, THOSE WILL CANCEL OUT, AND SO WE LL JUST HAVE 2.8 X 65. SO 2.8 X 65.11, AND WE HAVE A FINAL ANSWER THEN OF 182 PARTS PER BILLION KCN. WHICH WOULD ALSO BE EQUAL TO.182 PARTS PER MILLION, AND SO ON DOWN THE LINE. SO YOU SEE AGAIN THE MAGNITUDE CHANGE THAT WE HAVE. ANY QUESTION ON EITHER OF THESE TWO CONCENTRATIONS, OR THREE CONCENTRATION UNITS, I SHOULD SAY? PARTS PER THOUSAND, PARTS PER MILLION, PARTS PER BILLION. IN CASE OF AIR, AND I LL GET YOUR QUESTION IN A SECOND, IN CASE OF AIR OF COURSE, BECAUSE THE DENSITY IS VERY LOW, WE REALLY DO HAVE TO WORK IN TERMS OF GRAMS. OKAY. BECAUSE WE DON T HAVE A DENSITY OF 1. BUT FOR AQUEOUS SOLUTIONS, BECAUSE THE DENSITY IS NEAR 1 GRAM PER MILLILITER, THAT S WHY WE CAN WORK IN UNITS OF LITERS. FOR ATMOSPHERIC OR AIR MATERIALS WE WOULD HAVE TO TALK ABOUT IT RELATIVE TO GRAMS OF AIR. OR, IF WE KNOW THE DENSITY OF AIR WE COULD TALK ABOUT IT, BUT THE VOLUME OF AIR OF COURSE IS GOING TO BE VERY LARGE, SO THE DENSITY IS VERY SMALL. WE HAD A QUESTION HERE? (STUDENT RESPONSE NOT AUDIBLE) IT S, IT WOULD BE ONE MICROGRAM OVER 10-6 GRAMS. THAT WOULD BE CORRECT. SO YOU COULD WRITE IT EITHER WAY. THE QUESTION WAS: WHY DID I EXPRESS THE CONVERSION FACTOR HERE AS 10 6 MICROGRAMS OVER 1 GRAM, SHOULDN T IT HAVE BEEN 10-6? IF WE DO IT THAT WAY, AND THAT S TOTALLY CORRECT, WE WOULD SAY 1 MICROGRAM OVER 10-6 GRAMS. WE KNOW THAT MICRO IS 10-6 OF ANYTHING. SO WE COULD PUT IT IN EITHER THIS WAY OR THIS WAY, AND OF COURSE WE STILL WILL SEE THAT WE LL HAVE THE SAME VALUE WHEN WE GET DONE. SO YOU CAN EXPRESS IT EITHER WAY. ALRIGHT, ANY OTHER QUESTIONS ON THAT? LET S TURN THEN TO USING NOW MOLARITY TO MAKE SOME CALCULATIONS INVOLVING BALANCED CHEMICAL EQUATIONS. SO WE VE LOOKED AT STOICHIOMETRY BEFORE. WE LOOKED AT STOICHIOMETRY IN CHAPTER THREE, AND SO FAR WE VE LOOKED AT PROBLEMS THAT WE CALL MOLE/MOLE PROBLEMS, AND WE TALKED ABOUT MOLE/MASS PROBLEMS, AND WE TALKED ABOUT MASS/MASS PROBLEMS. NOW WE RE GOING TO INCORPORATE INTO THAT SOLUTIONS INTO OUR CALCULATIONS AS WELL. WELL

8 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 8 REMEMBER THE THREE BASIC STEPS THAT WE ALWAYS HAD. THE FIRST STEP IS TO DETERMINE THE NUMBER OF MOLES OF KNOWN SUBSTANCE. THE SECOND STEP IS TO DETERMINE THE NUMBER OF MOLES OF UNKNOWN, USING OUR BALANCED CHEMICAL EQUATION, AND THE THIRD STEP IS TO CONVERT TO WHATEVER UNITS IT IS THAT WE RE ASKED FOR. SO WE RE STILL GOING TO FOLLOW THE SAME STEP-WISE FASHION WE DID BEFORE. THIS PARTICULAR QUESTION THEN SAYS: HOW MANY GRAMS OF SODIUM CARBONATE, THAT S WHAT THIS COMPOUND IS CALLED, COULD BE REACTED WITH BY 48 MILLILITERS OF.465 MOLAR HYDROCHLORIC ACID? HOW MANY GRAMS OF THIS COULD WE REACT IS THE QUESTION. NOW MYSELF, I FIND IT PROBABLY A LITTLE EASIER WHEN YOU RE DEALING WITH SOLUTION STOICHIOMETRY TO ACTUALLY TAKE AND DO EACH STEP INDEPENDENTLY, RATHER THAN SETTING IT UP IN ONE CONTINUOUS FASHION. THERE S NOTHING WRONG WITH SETTING IT ALL UP, BUT YOU SOMETIMES FIND IT A LITTLE BIT EASIER JUST TO GO AHEAD AND DO IT STEP -WISE. SO I M GOING TO DO THESE IN TERMS OF DOING THEM STEP-WISE HERE. MY FIRST QUESTION THEN, STEP NUMBER ONE, IS TO DETERMINE THE NUMBER OF MOLES OF KNOWN SUBSTANCE, WHICH IN THIS PARTICULAR CASE IS, WHOOPS, CHCL. SO HOW MANY MOLES OF HCL DO I HAVE IN THAT? WELL WE VE ALREADY DISCUSSED THAT IN DEALING WITH SOLUTIONS THAT THE MOLARITY TIMES THE VOLUME OF SOLUTION IN LITERS GIVES US THE NUMBER OF MOLES OF SOLUTE. SO IF WE WERE TO TAKE THE MOLARITY, MOLES OF HCL PER LITER OF SOLUTION, MULTIPLY THAT BY.0480 LITERS OF SOLUTION. THEN WE RE GOING TO END UP WITH THE NUMBER OF MOLES OF KNOWN SUBSTANCE. THAT S MY FIRST STEP, AND LET S SEE WHAT WE WOULD GET FOR A NUMBER,.465 TIMES.0480, AND WE HAVE MOLES OF HCL THAT REACTED. STEP NUMBER TWO IS TO DETERMINE THE NUMBER OF MOLES OF UNKNOWN, WHICH IN THIS CASE IS THE SODIUM CARBONATE, AND WE RE GOING TO START WITH OUR KNOWN MOLES OF HCL, AND MULTIPLY BY DIDN T LEAVE MYSELF A WHOLE LOT OF ROOM THERE SO WE LL JUST PUT IT DOWN BELOW HERE MULTIPLIED BY 1 MOLE OF SODIUM CARBONATE PER 2 MOLES OF HCL. SO MULTIPLY BY 1 MOLE OF NA2CO3 PER 2 MOLES OF HCL. SO AT THIS POINT THEN DIVIDING BY 2, MULTIPLYING BY ½ WE WILL HAVE THEN AN ANSWER MOLES, AND THE THIRD STEP IS TO CHANGE TO WHATEVER UNITS IT IS THAT WE RE ASKED FOR. IN THIS

9 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 9 PARTICULAR PROBLEM WE RE LOOKING OF UNITS OF GRAMS OF SODIUM CARBONATE, AND SO GRAMS OF NA2CO3 WOULD BE EQUAL TO MOLES, MULTIPLIED BY THE NUMBER OF GRAMS PER 1 MOLE. AND WE WOULD HAVE TO CALCULATE THAT. SO LET S SEE WE WOULD HAVE 2 SODIUMS, SO X 2 WOULD GIVE US 45.98, AND A CARBON AND 3 X 16 FOR THE OXYGENS, 48, AND IT LOOKS LIKE WE HAVE 60, LOOKS LIKE WE HAVE GRAMS PER ONE MOLE OF SODIUM CARBONATE. SO IF WE NOW MULTIPLY THAT, SO , WE WILL END UP WITH THEN A FINAL ANSWER 1.18 GRAM OF SODIUM CARBONATE. ALRIGHT, NOW AS I SAID, THERE S CERTAINLY NOTHING INCORRECT WITH SETTING IT UP IN A CONTINUOUS FASHION IF YOU WISH TO DO SO. IN THIS PARTICULAR TYPE IT PROBABLY WOULD HAVE BEEN RELATIVELY EASY TO DO SO. WHERE WE GET INTO WHERE IT S A LITTLE MORE DIFFICULT IS IF WE RE REACTING TWO SOLUTIONS AND WE RE ASKING FOR, SAY THE MOLARITY OF ONE OR THE VOLUME OF ONE. THEN IT MAKES IT A LITTLE BIT EASIER TO ACTUALLY BREAK INTO INDIVIDUAL PARTS. ANY QUESTION ON THIS PARTICULAR ONE BEFORE WE GO ON HERE? OKAY, LET S LOOK AT ONE OTHER ONE THEN. IN THIS PARTICULAR PROBLEM IT ASKS THE QUESTION: HOW MANY MILLILITERS OF CALCIUM IODIDE SOLUTION WOULD I NEED? AND IT HAS A CONCENTRATION OF.186 MOLAR CONCENTRATION. HOW MANY MILLILITERS OF THIS SOLUTION WOULD I NEED TO REACT WITH 9.76 GRAMS OF THE OTHER COMPOUND CALLED SILVER NITRATE? SO THIS IS OF COURSE THE KNOWN IN THIS PARTICULAR PROBLEM, AND WE RE LOOKING FOR MILLILITERS OVER HERE. SO LET S GO AHEAD AGAIN AND SET IT UP IN STEP-WISE FASHION. SO THE FIRST THING I M GOING OT ASK HERE IS THE NUMBER OF MOLES OF SILVER NITRATE, AND THAT S 9.76 GRAMS OF SILVER NITRATE TIMES 1 MOLE OF SILVER NITRATE FOR A CERTAIN NUMBER OF GRAMS. AND WE LL HA VE TO CALCULATE THAT AGAIN, LET S SEE. SO WE WOULD HAVE FOR SILVER, WE WOULD HAVE , WE D HAVE A NITROGEN, 14.01, AND WE D HAVE 3 X 16 FOR 48 FOR THAT, SO.88, 12, 9, 6, 1, GRAMS. SO THAT TAKES CARE OF THAT, AND NOW WE COULD CALCULATE 9.76 DIVIDED BY , AND WE HAVE MOLES OF AGNO3. STEP NUMBER TWO WOULD BE TO SAY HOW MANY MOLES OF CAI2 DO WE HAVE THEN IN THE REACTION, AND WE HAVE 5.75 X 10-2 MOLES OF AGNO3, MULTIPLIED BY 1 MOLE OF CAI2 PER 2 MOLES OF AGNO3. SO THOSE UNITS CANCEL, NOW WE HAVE MOLES OF KNOWN SUBSTANCE,

10 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 10 AND SO WORKING THAT PART OUT TO GET A NUMBER. WE HAVE THEN 2.88 X 10-2 MOLES OF CAI2. NOW THE THIRD STEP TAKES US TO THIS RIGHT HERE. WE WANT TO FIND OUT WHAT VOLUME OF THE SOLUTION WE NEED. WE KNOW HOW MANY MOLES WE NEED, WE NEED 2.88 X WE KNOW HOW MANY MOLES THERE ARE IN A LITER, BASED ON THE MOLARITY, AND OF COURSE WE KNOW THE NUMBER OF MOLES AND WE KNOW THE NUMBER OF MOLES PER LITER. WE SHOULD BE ABLE TO EASILY DETERMINE THE NUMBER OF LITERS OF SOLUTION THAT WE NEED, AND THEN OF COURSE TO CONVERT TO MILLILITERS. SO, MILLILITERS, CAI2, IS EQUAL TO 2.88 X 10-2 MOLES. NOW I M GOING TO DROP THE UNIT OFF HERE JUST BECAUSE IT S CROWDING IT A LITTLE BIT, MULTIPLIED BY 1 LITER OF SOLUTION, WHICH CONTAINS MOLES. THAT S OUR MOLARITY, BACK UP THERE AT THE VERY TOP HERE WE SEE THE MOLARITY OF THE SOLUTION WAS.186. THAT S.186 MOLE OVER 1 LITER, OR WE CAN WRITE IT AS 1 LITER OVER THEN.186 MOLE. NOW NOTICE THAT MOLES WILL HAVE CANCELLED, BUT WE ASKED FOR MILLILITERS, SO OUR FINAL STEP WOULD BE 10 3 MILLILITERS PER LITER, AND SO NOW WE HAVE A FINAL ANSWER THEN EXPRESSED IN TERMS OF MILLILITERS. SO LET S SEE HERE I HAVE DIVIDED BY.186 AND MULTIPLIED BY 1000 AND WE HAVE, WE WOULD NEED 154 MILLILITERS OF THE CALCIUM IODIDE SOLUTION IN ORDER TO REACT THE 9.76 GRAMS OF SILVER NITRATE. THIS IS A TYPE OF PROBLEM ESPECIALLY THAT I M TALKING ABOUT THAT THIS THIRD STEP IS CERTAINLY PROBABLY EASIER IF YOU RE DOING IT AS A SEPARATE STEP. YOU COULD COMBINE THE FIRST TWO STEPS, AS A MATTER OF FACT, THAT S NOT A BAD WAY TO ASK HOW MANY MOLES OF THE UNKNOWN, AND CALCULATE THAT AS ONE STEP AND THEN CONVERT TO WHATEVER IT IS THAT YOU NEED IN THE NEXT ONE. ANY QUESTION ON THIS ONE? ALRIGHT, LET S LOOK AT AN EXAMPLE THEN INVOLVING TWO SOLUTIONS. HERE WE HAVE THE REACTION OF PHOSPHORIC ACID WITH POTASSIUM HYDROXIDE. THIS IS CALLED AN ACID-BASE REACTION, AND IN CHAPTER FIVE, OUR NEXT CHAPTER, WE RE GOING TO TALK EXTENSIVELY ABOUT ACIDS AND BASES. SO HERE WE HAVE THE REACTION OF AN ACID, PHOSPHORIC ACID WITH A BASE, POTASSIUM HYDROXIDE, AND BOTH ARE SOLUTIONS. WE FIND THAT IT TAKES 15.6 MILLILITERS OF.525 MOLAR PHOSPHORIC ACID TO REACT WITH 25 MILLILITERS OF THE SOLUTION OF POTASSIUM HYDROXIDE. WHAT WE WANT TO KNOW IS WHAT IS THE MOLARITY OF THE

11 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 11 POTASSIUM HYDROXIDE? SO, AGAIN, THEN OUR FIRST STEP IS GOING TO BE TO DETERMINE THE NUMBER OF MOLE OF KNOWN, THEN THE NUMBER OF MOLES OF UNKNOWN USING OUR EQUATION, AND THEN FINALLY PUTTING IT INTO WHATEVER UNITS WE WANT. THIS TIME LET S GO AHEAD AND DO IT IN TWO STEPS, SO LET S ASK THE QUESTION HERE HOW MANY MOLES OF KOH DO I HAVE, HAVE I REACTED? ALRIGHT, WE HAVE.0156 LITER X MOLE OF H3PO4 PER LITER. SO WE VE USED UP OUR VOLUME, LITERS, CONCENTRATION, MOLES PER LITER, SO THAT WE COULD DETERMINE THE NUMBER OF MOLES OF KNOWN MATERIAL. NOW WE RE GOING TO BRING IN OUR BALANCED EQUATION, SO MULTIPLIED BY 3 MOLES OF KOH PER 1 MOLE OF H3PO4. NOTICE OUR 3 TO 1 STOICHIOMETRIC RELATIONSHIP. AND SO AT THIS POINT THEN NOTICE THAT WHAT WE HAVE ARE MOLES OF POTASSIUM HYDROXIDE, OUR UNKNOWN SUBSTANCE. SO, WORKING THIS OUT HEN QUICKLY, LET S SEE WE HAVE.0156 MULTIPLIED BY.525, AND MULTIPLIED BY 3, AND WE END UP WITH MOLES OF KOH. STEP 1 AND 2 HAVE BEEN COMPLETED AS ONE. NOW STEP NUMBER THREE IS TO SOLVE FOR WHATEVER UNITS IT IS THAT WE ARE ASKED FOR. THIS TIME IT S ASKING US FOR WHAT? THE MOLARITY OF KOH. LET S STOP AND THINK HERE, MOLARITY OF KOH. WHAT WOULD WE NEED FOR UNITS? MOLES OF KOH OVER LITERS OF KOH SOLUTION. THAT S WHAT WE NEED TO GET OUR ANSWER. DO WE HAVE THAT INFORMATION AVAILABLE? WELL WE JUST CALCULATED THIS RIGHT THERE, AND THE LITERS OF KOH SOLUTION WAS GIVEN TO US IN THE PROBLEM, AND SO IF WE PLUG IN OUR TWO VALUES THAT WE NOW HAVE, MOLE DIVIDED BY THE VOLUME EXPRESSED IN LITERS, LITERS, WE NOW SEE THAT WE HAVE AN ANSWER IN MOLES PER LITER, WHICH OF COURSE IS A CORRECT UNIT FOR OUR MOLARITY. AND WE CAN SEE THAT WITHOUT WORKING THROUGH HERE. WELL I GUESS I CAN PLUG IT IN HERE, BUT IT LOOKS VERY CLOSE TO BEING A 1 MOLAR SOLUTION, JUST SLIGHTLY LESS THAN THAT. SO DIVIDE BY.025 AND WE WOULD HAVE A MOLARITY THEN OF MOLAR SOLUTION OF POTASSIUM HYDROXIDE. THIS IS A VERY USEFUL METHOD OF DETERMINING CONCENTRATION OF ONE SOLUTION BY REACTING IT WITH A SOLUTION OF KNOWN CONCENTRATION, AND IF WE HAVE A BALANCED CHEMICAL EQUATION THEN WE CAN DETERMINE WHAT THE MOLARITY OF THE UNKNOWN SOLUTION IS, UNKNOWN ONLY FROM THE STANDPOINT OF THE QUANTITATIVE INFORMATION. WE KNOW IT S KOH. ANY

12 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 12 QUESTION ON ANY STEP OF THAT ONE? ALRIGHT, BY THE WAY, I VE NOTICED ON SOME HOMEWORK SETS THAT OCCASIONALLY PEOPLE ARE DISREGARDING THE ZEROS. THESE ZEROS, WHEN IT SAYS 25.0 MILLILITERS, THAT.0, THAT ZERO IS A SIGNIFICANT FIGURE, AND SO IF YOU RE GOING TO REWRITE IT IN LITERS YOU STILL NEED TO PUT THAT ZERO THERE AT LEAST TO KEEP IN MIND THAT YOU RE USING THREE SIGNIFICANT FIGURES. IF YOU WRITE THAT AS.025, AND THEN YOU GO AHEAD AND MAKE YOUR CALCULATION AND THEN YOU LOOK BACK AT THE NUMBER YOU WROTE DOWN YOU MIGHT PUT YOUR ANSWER IN TWO. THAT S INCORRECT. THOSE ZEROS AFTER DECIMAL POINTS, PROCEEDED BY NUMBERS ARE SIGNIFICANT AND NEED TO BE INCLUDED IN DETERMINING SIGNIFICANT FIGURES. WELL ONE LAST ONE OF THIS TYPE HERE QUICKLY. AGAIN, THIS IS A SOLUTION WITH A SOLUTION, BUT THIS TIME INSTEAD OF ASKING FOR THE MOLARITY OF THE UNKNOWN WE RE ASKED HOW MANY MILLILITERS OF THE UNKNOWN DO WE NEED TO REACT WITH THIS SOLUTION? SO AGAIN WE COULD GO AHEAD AND SET THIS UP. FIRST OF ALL WE LL DO ONE STEP TO DETERMINE THE NUMBER OF MOLES OF THE UNKNOWN SUBSTANCE, AND THEN WE LL DO THE FINAL STEP TO CONVERT TO WHAT THE UNITS ARE. SO WE RE GOING TO ASK THE QUESTION MOLES OF CUNO3, THAT S EQUAL TO.0320 LITERS, 32 MILLILITERS TIMES MOLE OF KIO3 PER LITER. THAT S ITS MOLARITY, AND SO THOSE UNITS CANCEL. SO NOW WE HAVE MOLES OF THE KNOWN SUBSTANCE, NOW WE NEED TO CONVERT THEN 1 MOLE OF THE COPPER NITRATE, A LITTLE CROWDED THERE, PER 2 MOLES OF THE KIO3 CHEMICAL CALLED POTASSIUM IODATE, AND NOTICE THEN THAT MOLES OF KIO3 CANCEL OUT, AND WE HAVE MOLES OF THE UNKNOWN MATERIAL. SO CALCULATING THAT THEN LET S SEE, WE HAVE.032 X.0544, AND DIVIDED BY 2, AND WE HAVE 8.70 X 10-3 MOLES OF CUNO3. NOW WE GO BACK, WHAT WAS IT THAT WE WERE ACTUALLY TRYING TO DETERMINE? WE WERE ACTUALLY NEEDING TO DETERMINE THE VOLUME OF COPPER NITRATE THAT WAS NEEDED THEN TO GIVE US THIS NUMBER OF MOLES OF COPPER NITRATE SOLID. WELL AGAIN, KEEPING IN MIND THAT THIS GIVES US A RELATIONSHIP OF MOLES TO LITERS, THEN WE CAN USE THAT TO CONVERT TO VOLUME OF COPPER NITRATE IN LITERS AND THE OF COURSE CONVERT TO MILLILITERS. SO FINAL SETUP THEN, MILLILITERS WILL BE EQUAL TO 8.70 X 10-3 MOLES, MULTIPLIED BY 1 LITER OF THE COPPER NITRATE SOLUTION OVER MOLES. SO NOW

13 CHM 105 & 106 UNIT TWO, LECTURE EIGHT 13 MOLES WILL CANCEL AND THEN MULTIPLIED BY 10 3 MILLILITERS. NOW WE SHOULD HAVE TECHNICALLY WROTE DOWN COPPER NITRATE AND COPPER NITRATE IN EACH OF THOSE, AND BUT SPACE CROWDS A LITTLE BIT SO WE LL JUST SET IT UP THIS WAY. NOW WE HAVE AN ANSWER IN THEN MILLILITERS. ALRIGHT, NOW ONE OTHER THING THAT I SHOULD MENTION, IT IS IMPORTANT TO WATCH THE UNITS BECAUSE WE RE DEALING WITH TWO DIFFERENT CONCENTRATIONS. IF YOU DON T PAY VERY CLOSE ATTENTION TO WHICH NUMBER YOU RE WRITING IN AS THE SECOND LITERS OVER SOLUTION, OVER THE MOLES, OF COURSE YOU MIGHT ACCIDENTALLY PUT IN THE WRONG ONE. SO IT S REALLY BEST TO CARRY THOSE UNITS, AND ON A PIECE OF PAPER IT S NOT THAT DIFFICULT TO DO. WE LET S SEE, 10-3 AND 10 3 WOULD CANCEL AND WE HAVE ABOUT 8.7 OVER.886, SO SOMETHING VERY CLOSE TO 10 MOLAR IN CONCENTRATION, 8.7 DIVIDED BY.886 GIVES US THEN 9.82 MOLAR SOLUTION OF COPPER NITRATE. ANY QUESTION ON ANY STEP OF THAT? IN OUR NEXT LECTURE WE RE GOING TO LOOK AT ONE MORE CONCENTRATION UNIT CALLED MOLALITY. WE RE GOING TO LOOK AT SOM E PROPERTIES OF SOLUTIONS THAT WE CALL COLLIGATIVE PROPERTIES AND WE RE GOING TO GO ON AND TAKE A LOOK AT SOME NON-AQUEOUS SOLUTIONS. SO FAR WE VE BEEN DEALING WITH WATER AS THE SOLVENT. WE RE GOING TO LOOK AT SOME OTHER SOLVENTS IN OUR NEXT LECTURE.

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