CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 1 IN OUR PREVIOUS LECTURE WE WERE LOOKING AT EQUILIBRIUM APPLICATIONS TO DISSOLVING

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1 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 1 CHM 105/106 Program 46: Unit 5 Lecture 6 IN OUR PREVIOUS LECTURE WE WERE LOOKING AT EQUILIBRIUM APPLICATIONS TO DISSOLVING AND PRECIPITATION PROCESSES WHICH WE REFERRED TO AS K SP OR SOLUBILITY PRODUCT EQUILIBRIUM, AND TODAY WE RE GOING TO FOCUS PRIMARILY ON EQUILIBRIUMS INVOLVING ACIDS AND BASES. BEFORE WE LOOK AT THE EQUILIBRIUM INVOLVING ACIDS AND BASES WHICH DEAL WITH WHAT WE CALL WEAK ACIDS AND WEAK BASES WE LL DO JUST A LITTLE LOOK FIRST OF ALL AT A STRONG ACID AND SOME OF THE CALCULATIONS THAT WE HAVE DONE PREVIOUSLY RELATIVE TO SUCH THINGS AS HYDRONIUM ION, ph, ET CETERA, AS A LITTLE REVIEW AND THEN GO ON AND APPLY THIS TO WEAK ACID WHERE WE RE GOING TO HAVE AN EQUILIBRIUM INVOLVED. MENTIONED AT THE END OF THE LAST LECTURE IF WE HAD REPRESENTED AN ACID AS HA AND PLACING IT THEN WITH WATER AS A SOLVENT, H 2 O LIQUID, THE ACID DONATES A PROTON TO THE WATER FORMING THE HYDRONIUM ION AND THE ANION THE NEGATIVE ION OF THE ACID MOLECULE. NOW IN THIS PARTICULAR REACTION NOTICE ONE DIFFERENCE BETWEEN THIS AND THE EQUATIONS WE WROTE FOR DISSOLVING THE OTHER DAY IS THAT I VE INDICATED HERE ONLY A SINGLE ARROW. A SINGLE ARROW IMPLIES THAT THE SYSTEM IS NOT GOING TO REACH EQUILIBRIUM. IT IS GOING ONLY ONE DIRECTION, AND SO ALL OF THE REACTANT WILL BE CONVERTED TO PRODUCT. IN OTHER WORDS THE K IS SO LARGE THAT WE DON T EVER CONSIDER IT REACHING AN EQUILIBRIUM. SO WE SAY THAT IN A STRONG ACID OR ALSO IN THE REACTION OF A STRONG BASE THAT THE REACTION IS A HUNDRED PERCENT. IN OTHER WORDS THAT IT HAS TOTALLY GONE TO THE PRODUCT SIDE. IF IT HAS GONE TO THE PRODUCT SIDE 100% THEN WHATEVER AMOUNT OF ACID WE STARTED WITH, IF WE STARTED WITH 10 MOLECULES OF THESE FOR INSTANCE ALL 100 WOULD HAVE BEEN CONVERTED TO THE HYDRONIUM ION. SO IN OTHER WORDS WE CAN THEN MAKE THE STATEMENT THAT THE HYDRONIUM ION CONCENTRATION FOR A STRONG ACID NOW, THE HYDRONIUM ION CONCENTRATION IS EQUAL TO WHATEVER THE MOLARITY OF THE ACID ITSELF WAS BECAUSE ALL OF THE ACID HERE HAS BEEN CONVERTED TO THE HYDRONIUM ION IN THE PROCESS. SO IF WE STARTED WITH A MOLE OF THIS WE WOULD END UP WITH A MOLE OF

2 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 2 THAT. SO THIS IS OUR RELATIONSHIP THEN. SO NOW WE WANT TO APPLY THIS IDEA DEALING WITH A STRONG ACID TO SPECIFICALLY THEN A SOLUTION OF 3.5 X 10-3 MOLAR HYDROCHLORIC ACID, HCL. REMEMBER THE STRONG ACIDS THAT WE VE TALKED ABOUT, NITRIC ACID, HYDROCHLORIC, SULFURIC, THESE ARE THE STRONG ACIDS, HYDROBROMIC, HYDROIODIC, ET CETERA. THE STRONG ACIDS, SO THIS RELATIONSHIP HOLDS TRUE. WELL THIS FIRST PART, CALCULATE THE HYDRONIUM ION CONCENTRATION OF COURSE IS PRETTY STRAIGHTFORWARD BECAUSE WE JUST SAID THAT THE HYDRONIUM IS EQUAL TO THE MOLARITY OF THE ACID. SO WE CAN THEN SAY THAT THE HYDRONIUM ION CONCENTRATION, REMEMBER BRACKETS MEANS MOLES PER LITER, IS EQUAL TO 3.5 X 10-3 MOLAR. SO THAT ANSWERS THE FIRST QUESTION. THE SECOND QUESTION IS WHAT IS THE HYDROXIDE ION CONCENTRATION? THIS TAKES US BACK TO CHAPTER FIVE WHERE WE TALKED ABOUT THE EQUILIBRIUM BETWEEN HYDRONIUM AND HYDROXIDE. AT THAT TIME WE REALLY DIDN T SAY EQUILIBRIUM AS MUCH AS W E JUST SAID THAT IN ANY AQUEOUS SOLUTION THAT THE PRODUCT OF HYDRONIUM TIMES HYDROXIDE WAS ALWAYS EQUAL TO A NUMERIC VALUE OF 1 X SO HYDRONIUM ION CONCENTRATION TIMES THE HYDROXIDE ION CONCENTRATION IS EQUAL TO 1.0 X 10-14, CHAPTER FIVE DEALING WITH OUR FIRST LOOK AT ACIDS AND BASES. NOW THIS PARTICULAR RELATIONSHIP HERE REALLY DOES COME ABOUT BECAUSE OF AN EQUILIBRIUM. THIS IS, WE DIDN T CALL IT THAT AT THAT TIME BUT THIS IS REALLY CALLED K W THE IONIZATION CONSTANT FOR WATER. K W, OF COURSE WE HADN T TALKED ABOUT SYSTEM EQUILIBRIUM IN THAT CHAPTER SO WE JUST TALK ABOUT THE NUMERIC RELATIONSHIP BETWEEN THE TWO. AND WE SEE THAT IF WE KNOW THE HYDRONIUM ION AND WE KNOW THE NUMERIC VALUE FOR THE EQUILIBRIUM CONSTANT FOR THAT SYSTEM WE OBVIOUSLY CAN SOLVE FOR THE HYDROXIDE. SO THE HYDROXIDE ION CONCENTRATION THEN IS EQUAL TO K W DIVIDED BY THE HYDRONIUM ION CONCENTRATION. ALRIGHT, AND WE PLUG IN THEN THE TWO NUMERIC VALUES 1.0 X FOR THE TOP NUMBER AND OUR BOTTOM NUMBER FROM THIS STEP RIGHT HERE, 3.5, I M TRYING TO FIND ONE PEN THAT LL WRITE HALFWAY DECENT HERE. 3.5 X SO IF WE NUMERICALLY SOLVE THIS THEN WE WILL HAVE WHAT THE HYDROXIDE ION CONCENTRATION IS. SO ONE EXPONENTIAL MINUS 14 DIVIDED BY 3.5 EXPONENTIAL MINUS THREE AND WE HAVE A HYDROXIDE ION

3 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 3 CONCENTRATION THE OF 2.9 X SO WE VE NOW ANSWERED PART, THE FIRST PART, THE HYDRONIUM ION. WE VE NOW ANSWERED THE SECOND PART BY USING THE EQUILIBRIUM THAT EXISTS IN ANY WATER SYSTEM OF AN ACID OR A BASE, THAT BEING THIS TIMES THIS IS EQUAL TO K W AND NOW WE RE READY TO GO ON TO THE FINAL STEP OF THE CALCULATION WHICH WAS CALCULATE THE ph. WE RECALL ph IS EQUAL TO MINUS LOG OF THE HYDRONIUM ION CONCENTRATION. AGAIN YOU CAN REFER BACK TO CHAPTER FIVE IF YOU HAVE ANY QUESTION ABOUT ph AND CALCULATION. THERE ARE MANY EXAMPLES OF IT IN CHAPTER FIVE AND SO ph IS EQUAL TO MINUS THE LOG OF 3.5 X 10-3, AND SO IF WE PLUG IN THEN 3.5 EXPONENTIAL MINUS THREE AND TAKE THE LOG OF THAT AND TAKE THE NEGATIVE WE HAVE A VALUE OF ph OF 2.46 FOR A VALUE. NOW NOT THAT IT IS CRITICAL, BUT KEEPING IN MIND SIGNIFICANT FIGURES YOU MIGHT SAY WELL DID I NOT WRITE TOO MANY SIGNIFICANT FIGURES IN THE ANSWER? AGAIN, REFERRING BACK TO CHAPTER FIVE WHEN WE DID THE MATH INVOLVING ph WE INDICATED THAT ONLY THE NUMBERS AFTER THE DECIMAL POINT ARE SIGNIFICANT. THIS IS REALLY A POWER OF TEN INDICATOR AND SO WE ONLY ARE SHOWING TWO SIGNIFICANT FIGURES. WE HAD TWO AND THE NUMBER WE RE WORKING WITH SO WE HAVE SHOWN IT CORRECTLY. ALRIGHT, ANY QUESTIONS ON ANY ONE OF THOSE RELATIONSHIPS? THESE ARE IMPORTANT RELATIONSHIPS RELATIVE TO DEALING WITH EITHER A STRONG ACID OR A WEAK ACID. IN THIS PARTICULAR CASE WE VE ELIMINATED PART OF THE EQUILIBRIUM ASPECT OF THE ACID ITSELF. AS WE GO ON NOW AND LOOK AT SOME WEAK ACID EQUILIBRIUM WE LL SEE THAT WE HAVE TO MAKE SOME ADDITIONAL CALCULATIONS. SO ANY QUESTIONS ON ANY STEP OF THAT PROCESS? ALRIGHT, HEARING NONE THEN WE KNOW THAT EVERYBODY IS NOW AN EXPERT AT CALCULATING ph FROM STRONG ACIDS. S LOOK AT A NEW PROBLEM HERE. CALCULATE THE SAME THINGS, HYDRONIUM, HYDROXIDE, AND ph OF A 0.57 MOLAR SOLUTION OF HYDROACETIC ACID, HN 3. AND THE K A, THE EQUILIBRIUM CONSTANT, NOW NOTICE THAT WHEN WE TALK ABOUT DISSOLVING WE TALKED ABOUT K S. WHEN WE TALKED ABOUT WATER WE USE K W. WHEN WE TALK ABOUT ACIDS, WEAK ACIDS, WE USE K A. ALL OF THESE STILL ARE THE SAME, THEY RE K EQ. THEY RE ALL EQUILIBRIUM EXPRESSIONS, EQUILIBRIUM CONSTANTS. OKAY, WE JUST USE THE LITTLE SUBSCRIPTS TO AGAIN IDENTIFY THE FACT THAT WE RE TALKING ABOUT

4 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 4 AN ACID HERE OR WE RE TALKING ABOUT WATER IN THE CASE OF K W. OR WE RE TALKING ABOUT DISSOLVING PROCESS IF WE USE K SP. THEY DON T MEAN SOMETHING DIFFERENT, THEY JUST ARE TALKING ABOUT A SPECIFIC SYSTEM. ALRIGHT, THIS IS A WEAK ACID HOWEVER WHICH MEANS IN FACT THAT IT IS GOING TO REACH EQUILIBRIUM. SO NOT 100% OF IT IS DISSOCIATED, SO THE FIRST THING WE RE GOING TO HAVE TO DO HERE IS WE RE GOING TO HAVE TO WRITE AN ACID/BASE EQUATION FOR THE REACTION. SO WE HAVE HN 3 REACTS WITH WATER AND THIS TIME WE LL HAVE A DOUBLE ARROW. THIS WILL DONATE A PROTON TO THE WATER SO WE RE GONNA END UP WITH H 3 O + PLUS AN N - 3 ION. NOW, IN THIS PARTICULAR CASE, IN ORDER TO CALCULATE THE HYDRONIUM ION HERE WE SOMEHOW HAVE TO FIGURE OUT HOW MUCH OF THE ACID HAS REACTED. PREVIOUSLY IN THE LAST ONE WITH A STRONG ACID WE SAID 100% REACTED SO WHATEVER WE STARTED WITH WAS WHAT WE ENDED UP WITH BUT THAT S NOT TRUE WITH A WEAK ACID. WE NEED TO FIND OUT HOW MUCH HYDRONIUM IS FORMING, HOW MUCH ACID HAS REACTED. WE DON T KNOW HOW MUCH. WELL AS WE DID YESTERDAY THEN WE COULD SAY LET X EQUAL THE MOLES PER LITER OF ACID REACTING. OKAY. NOW, IF X EQUALS THE MOLES PER LITER OF ACID REACTING THEN HOW MANY OR WHAT WILL THE CONCENTRATION OF THE HYDRONIUM ION BE AT EQUILIBRIUM? X. THANK YOU. SAME THING WE SAID YESTERDAY, X REPRESENTED THE NUMBER OF MOLES OF SOLID THAT DISSOLVE THEN X ALSO REPRESENTS THE CONCENTRATION OF THE ANION IN SOLUTION, OR THE CATION IN SOLUTION IF THEY RE A ONE TO ONE STOICHIOMETRIC RATIO. AND WE SEE THAT ONE ACID GIVES US ONE HYDRONIUM. TEN GIVES US TEN. X GIVES US X. X IS THE AMOUNT THAT IS REACTING SO THEREFORE AT EQUILIBRIUM, AT EQUILIBRIUM THE HYDRONIUM ION CONCENTRATION IS GOING TO BE X. WHAT IS THE N - 3 ION CONCENTRATION GOING TO BE? X. ONE TO ONE STOICHIOMETRIC RATIO THERE ALSO. ALRIGHT, SO WE HAVE X AT EQUILIBRIUM. NOW AT EQUILIBRIUM HOW MUCH HN 3 DO WE HAVE? ANYONE? ALRIGHT, LET S SUPPOSE THAT I HAD A DOLLAR AND I WENT DOWN TO THE VENDING MACHINE AND I SPENT 35 CENTS ON A CANDY BAR PROBABLY CAN T GET ONE FOR 35 CENTS WE LL FIND SOMETHING CHEAP TO BUY FOR 35 CENTS. HOW MUCH MONEY DO I HAVE LEFT? THIS IS NOT REAL HIGH MATH HERE. THIS IS MAYBE I'VE MISSED SOMETHING SOMEPLACE. ALRIGHT. HOW

5 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 5 MUCH MONEY DO I HAVE LEFT? 65!! VERY GOOD, HOW DID YOU DETERMINE THE 65? YOU SUBT RACTED WHAT YOU HAD SPENT FROM WHAT YOU HAD. ALRIGHT, NOW HOW MUCH DIFFERENCE IS IT FROM WHAT WE RE TALKING ABOUT HERE? I STARTED WITH.57 MOLES PER LITER OF ACID. HOW MUCH OF IT DID I USE UP? X! HOW MUCH DO I HAVE LEFT? (STUDENT RESPONSE NOT AUDIBLE) OTHER WAY AROUND..57 MINUS X. OKAY, WE DIDN T TAKE THE THIRTY FIVE CENTS AND SUBTRACT THE DOLLAR. WE TOOK THE DOLLAR WHICH WAS THE AMOUNT WE STARTED WITH AND SUBTRACTED THE AMOUNT THAT WE SPENT. WE TAKE THE AMOUNT OF ACID THAT WE STARTED WITH AND SUBTRACTED THE AMOUNT THAT REACTED, WHICH WAS X, AND SO WHEN WE GET ALL DONE, WHEN THIS DOUBLE ARROW IS ESTABLISHED, WHEN WE VE REACHED EQUILIBRIUM WE RE GOING TO HAVE 0.57 X. THAT S HOW MUCH WE STARTED WITH. THIS IS HOW MUCH REACTED. THIS IS HOW MUCH W E HAVE LEFT. WELL YOU MIGHT SAY HOW IS THAT GOING TO HELP US SOLVE THE PROBLEM? WELL WE KNOW THAT THIS IS EQUILIBRIUM AND THEREFORE WE CAN WRITE AN EQUILIBRIUM EXPRESSION. K IS EQUAL TO THE CONCENTRATION OF THE HYDRONIUM ION TIMES THE CONCENTRATION OF THE N - 3 ION, ALL DIVIDED BY THE CONCENTRATION OF THE HN 3 EQUILIBRIUM CONCENTRATIONS. WE HAVE TO HAVE EQUILIBRIUM REACHED FIRST. NOTICE I DIDN T PUT WATER IN THE BOTTOM TERM BECAUSE REMEMBER WE DON T INCLUDE LIQUID WATER IN THE EQUILIBRIUM EXPRESSION. IT S A CONSTANT IN ITSELF. WELL. DO I HAVE THINGS TO PUT INTO THESE VALUES? WELL YES I DO. WE JUST DEFINED IT AS X FOR THE HYDRONIUM AND WE DEFINED X FOR THE N 3 AND WE DEFINED 0.57 MINUS X FOR THE UNREACTED ACID AND IN THE PROBLEM IT SAYS THAT ALL OF THIS MUST NUMERICALLY BE EQUAL TO 1.5 X THAT S THE K A, THAT S THE EQUILIBRIUM CONSTANT FOR THAT PARTICULAR CHEMICAL PROCESS, AND IT DOESN T MAKE ANY DIFFERENCE HOW MUCH WE START WITH. THIS IS AN EQUILIBRIUM VALUE. IT S ALWAYS GOING TO BE THAT NUMERIC VALUE AT A PARTICULAR TEMPERATURE, AND THE TEMPERATURE WE RE USUALLY TALKING ABOUT IS 25 CELSIUS DEGREES. WELL NOW YOU SEE WE ONLY HAVE ONE UNKNOWN IN THERE. WE JUST HAVE X, AND MATHEMATICALLY WE CAN NOW SOLVE FOR X. WELL IF WE WERE TO CARRY OUT THIS W HOLE PROCESS TO SOLVE FOR X AS YOU MIGHT LOOK. WE VE GOT X TIME X, THAT S X SQUARED AND WE VE GOT THIS NUMBER

6 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 6 TIMES A NUMBER SO WE HAVE A NUMBER AND WE HAVE AN X TIMES A NUMBER SO WE HAVE AN X AND IF WE HAVE A SYSTEM THAT LOOKS LIKE THIS X SQUARED, X AND A NUMBER, WHAT KIND OF EQUATION ARE WE LOOKING AT IF WE RE GOING TO MATHEMATICALLY SOLVE IT? KIND OF EQUATION HAS THOSE THREE FACTORS IN IT/ QUADRATIC EQUATION. THIS IS A QUADRATIC EQUATION, AND MATHEMATICALLY WE HAVE A WAY TO SOLVE A QUADRATIC EQUATION. SO WE COULD USE THAT SETUP AND WE COULD MATHEMATICALLY SOLVE FOR IT, BUT BASICALLY IF WE CAN GET BY WE RE GOING TO USE A SHORTCUT METHOD OR A SIMPLER METHOD TO GET AN ANSWER. WE RE GOING TO MAKE AN ASSUMPTION AND HOPEFULLY THE ASSUMPTION IS CORRECT AND WE DON T NEED TO DO THE QUADRATIC EQUATION, NOT THAT WE CAN T BUT WE CAN FIND A SHORTER METHOD WHICH GIVES US A GOOD APPROXIMATION. I M GOING TO LOOK RIGHT HERE. NOW THIS REACTION DOESN T GO BASED ON THE TEN TO THE MINUS FIFTH. IT TELLS US THAT THERE S NOT A WHOLE LOT OF PRODUCTS WHEN EQUILIBRIUM IS REACHED AND IF THERE S NOT A WHOLE LOT OF PRODUCTS IT MEANS THAT WE DIDN T USE UP A WHOLE LOT OF OUR ACID IN THE FIRST PLACE. AND IF THAT S THE CASE WE MIGHT START BY MAKING THE ASSUMPTION THAT THE AMOUNT THAT DISAPPEARED, THE MINUS X IS SORT OF INSIGNIFICANT. IT S KIND OF LIKE IF WE HAD A DOLLAR AND WE WENT IN TO A STORE AND WE HAPPENED TO FIND A TWO-CENT PIECE OF GUM. CAN T IMAGINE WHAT STORE BUT LET S JUST SUPPOSE THAT THERE WAS SUCH A THING. SO WE SPEND TWO CENTS OF OUR ONE HUNDRED. WELL YOU SEE THAT TWO CENTS MINUS FROM SUBTRACTED FROM A HUNDRED REALLY GIVES US VERY LITTLE DIFFERENCE. 98 CENTS IN OUR POCKET OR A HUNDRED CENTS IN OUR POCKET ISN T TOO MUCH DIFFERENCE AND SO THAT S WHAT WE RE SAYING HERE. WE VE USED UP VERY LITTLE OF THAT ORIGINAL ACID AND IF THAT IS CORRECT THEN WE CAN DISREGARD THE MINUS X. NOW WE WOULDN T SAY THAT WE COULD DISREGARD 35 CENTS THAT WE SPENT. I MEAN THERE S QUITE A DIFFERENCE IN HAVING A HUNDRED CENTS IN OUR POCKET VERSUS 65 CENTS, THAT S AN APPRECIABLE AMOUNT. AND IF IT TURNED OUT THAT WE WERE WRONG THAT THIS X WAS FAIRLY LARGE THEN OF COURSE WE WOULD HAVE TO GO BACK AND SOLVE USING THE QUADRATIC. WELL WHAT WE RE GOING TO DO IS OUR FIRST TIME THROUGH WE RE MERELY GOING TO SAY WE THINK THAT THIS IS SMALL COMPARED TO THIS AND BY SMALL WE

7 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 7 MEAN THAT IT SHOULD BE 10% OR LESS OF THE NUMBER THAT WE RE SUBTRACTING FROM. SO AS LONG AS X IS SMALLER THAN.057 THEN IT S REALLY AN INSIGNIFICANT PART OF OUR CALCULATION. WELL IF WE DISREGARD THAT NOW THEN AND WE THEN GO AHEAD AND DO AN APPROXIMATE SOLUTION WE LL HAVE X TIMES X, X TIMES X X 2 IS EQUAL TO 1.5 X 10-2, 10-5 EXCUSE ME, TIMES TAKING THIS NUMBER MULTIPLYING BY THAT AMOUNT. SO WE VE DONE JUST A LITTLE ALGEBRA REA RRANGEMENT AND THEN WE COULD GET A NUMERIC VALUE FOR THIS SO LET S DO THAT QUICKLY. 1.5 EXPONENTIAL MINUS 5 TIMES X NOW THIS IS SORT OF LIKE WHAT WE HAD THAT WE SOLVED BEFORE WITH THE X AND X FOR THE TWO PARTICLES WHEN WE TALKED ABOUT SOLUBILITY, AND X SQUARED. SO HOW ARE WE GOING TO SOLVE THIS? WE MERELY TAKE THE SQUARE ROOT OF THIS SIDE AND TAKE THE SQUARE ROOT OF THIS SIDE AND X THEN BECOMES EQUAL TO THEN THE SQUARE ROOT OF THAT. SO I NOW HIT THE SQUARE ROOT AND I HAVE AN ANSWER OF 2.9 X 10-3, OR I CAN WRITE THAT IF WE WANT TO AS NOW THE ONLY REASON I M DOING THAT IS SO THAT WE CAN LOOK BACK HERE AND SAY WAS THAT REALLY SMALL COMPARED TO THIS NUMBER HERE AND I THINK YOU CAN SEE THAT IT S CERTAINLY MUCH LESS THAN 10%. IT S EVEN LESS THAN 1% OF THE NUMBER. SO IT S LESS THAN LOSING A PENNY OUT OF OUR HUNDRED THAT WE HAVE TO START WITH, AND THAT S AN INSIGNIFICANT AS FAR AS FOR QUANTITATIVE CALCULATIONS IN EQUILIBRIUM. YES? (STUDENT RESPONSE NOT AUDIBLE) THE SQUARE ROOT? OH, BECAUSE X COULD BE MINUS 2.9 X YES, WE CAN DISREGARD THAT BECAUSE X CAN T BE ANY SMALLER THAN WHAT? X CAN T BE ANY SMALLER THAN ZERO. I MEAN IF NONE OF THE ACID REACTED HOW MUCH HYDRONIUM WOULD WE GET? ZERO. WE CAN T HAVE A DISAPPEARING. IT CAN T BE LESS THAN ZERO. WE EITHER HAVE SOME OR NONE IN THE SYSTEM, SO YES. THE ANSWER TO THE QUESTION, THE QUESTION WAS DO WE NEED TO CONCERN OURSELF WITH THE FACT THAT A SQUARE ROOT COULD GIVE US A NEGATIVE NUMBER AS WELL AS A POSITIVE VALUE? AND THE ANSWER IS NO WE DON T NEED TO WORRY ABOUT THE NEGATIVE VALUE BECAUSE IT IS NOT A POSSIBLE SOLUTION TO THE PROBLEM. MATHEMATICALLY IT S FEASIBLE BUT REALITY IT IS NOT SO WE DON T NEED TO WORRY ABOUT IT. NOW, WE SOLVED FOR X, AND X IF WE GO BACK UP HERE, NOTICE THAT X REPRESENTED THE HYDRONIUM ION. AND THAT WAS

8 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 8 OUR FIRST QUESTION IN THE PROBLEM CALCULATE THE HYDRONIUM ION FOR THIS ACID. SO WE NOW CAN SAY THE HYDRONIUM ION CONCENTRATION IS EQUAL TO 2.9 X AND I DON T KNOW IF WE VE GOTTEN DIFFERENT FILM OR WHAT BUT THEY DON T SEEM TO WANT TO HOLD THE INK AT ALL. NOW, THAT S STEP ONE. ANY QUESTIONS UP TO THIS POINT? YES? (STUDENT RESPONSE NOT AUDIBLE) I DID IT JUST BECAUSE I M LAZY AND DIDN T WAN TO DO A QUADRATIC EQUATION. (STUDENT RESPONSE NOT AUDILE_ WELL, YES IT S A GOOD QUESTION. THE QUESTION WAS WHAT JUSTIFICATION MIGHT I HAVE USED TO SUSPECT THAT X, WE RE TALKING ABOUT THE MINUS X, THAT THE MINUS X ITSELF IS SMALL? THE CLUE IS THE VALUE FOR K. BECAUSE REMEMBER THAT IF K IS MUCH LESS THAN.1 IT MEANS THAT THE REACTANT SIDE IS GREATLY FAVORED AND THE SMALLER K IS EVEN THE FURTHER THE REACTANT SIDE IS FAVORED. IN OTHER WORDS BECAUSE K IS RATHER SMALL 10-5, I DON T EXPECT TO HAVE A WHOLE LOT OF PRODUCT AND BECAUSE THE AMOUNT OF PRODUCT IS REPRESENTED BY X THEN I SUSPECTED THAT AS COMPARED TO WHAT I STARTED WITH THERE IT S RATHER INSIGNIFICANT. SO THAT WAS THE REASONING BEHIND THAT. NOW BACK TO OUR ORIGINAL PROBLEM WHICH WAS CALCULATE THE HYDRONIUM. WE VE DONE THAT. CALCULATE THE HYDROXIDE AND CALCULATE THE ph WERE THE OTHER TWO STEPS. SO LET S GO AHEAD KEEPING IN MIND OUR ANSWER THERE LET S JUST GO AHEAD AND DO ph. ph IS EQUAL TO MINUS LOG THEN OF 2. I DON T NEED THE BRACKETS THERE, REWRITE IT ph IS EQUAL TO MINUS LOG OF2.9 X 10-3 AND SO ALL WE HAVE TO DO NOW IS TAKE THE LOGARITHM OF THE NUMBER THAT WE HAD AND WE HAVE A ph EQUAL TO NOW THIS IS AN ACID SO QUITE OBVIOUSLY AS AND ACID THE ph SHOULD BE LESS THAN 7. REMEMBER THE CUTOFF POINT NEUTRAL IS 7, ACIDS ARE ph LESS THAN 7 AND SO OUR ANSWER MAKES SENSE RELATIVE TO WHAT WE KNOW THE ph FOR AN ACID SHOULD BE. THE FINAL CALCULATION THEN WAS TO SOLVE FOR THE HYDROXIDE ION CONCENTRATION AND AGAIN WE GO BACK TO THE FACT K W IS EQUAL TO HYDRONIUM TIMES HYDROXIDE AND IS ALWAYS EQUAL TO 1.0 X I SAY ALWAYS, I MEAN ALWAYS IF THE TEMPERATURE IS 25 DEGREES. SO WE KNOW THIS ONE WE WANNA SOLVE FOR THAT ONE SO HYDROXIDE ION IS EQUAL TO 1.0 X DIVIDED BY 2.9 X 10-3, AND THAT LOOKS LIKE IT SHOULD BE PRETTY CLOSE TO 3 X OKAY. ANY QUESTIONS ON

9 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 9 ANY STEP OF THAT ONE. ALRIGHT. WELL LET S LOOK AT ANOTHER EXAMPLE HERE. CALCULATE THE HYDROXIDE ION, HYDRONIUM ION, AND ph OF A 0.61 MOLAR SOLUTION OF METHYL AMINE. METHYL AMINE. NOTICE THAT THE EQUILIBRIUM CONSTANT THIS TIME SAYS K B. WELL THAT S A BASE. SO IF WE SEE K A THAT MEANS WE RE TALKING ABOUT AN ACID SYSTEM. WE RE TALKING ABOUT THE COMPOUND DONATING A PROTON. IF WE RE TALKING ABOUT A BASE WE RE TALKING ABOUT A CHEMICAL THAT ACCEPTS A PROTON. ALRIGHT, THE FIRST THING THAT WE NEED TO DO IS AGAIN WRITE A CHEMICAL REACTION FOR THIS OVERALL PROCESS. SO WE HAVE CH 3 NH 2 REACTING WITH WATER, H 2 O. NOW THIS IS A BASE AND SEEING THAT THE NUMBER IS SMALL IT OBVIOUSLY IT S A WEAK BASE SO IF IT S A BASE IT S ACCEPTING A PROTON FROM THE WATER SO WE RE GOING TO END UP WITH CH 3 NH + 3 AND OH -. OKAY, SO WHEN WE HAVE A BASE REACTION WE END UP PRODUCING A PRODUCT SIDE THAT HAS HYDROXIDES. WHEN WE HAVE AN ACID REACTION WE PRODUCE A PRODUCT SIDE THAT HAS HYDRONIUM IONS. AGAIN BACK TO LOOKING AT CHAPTER FIVE ON ACID BASE SOLUTIONS. ALRIGHT, AGAIN NOW HERE S WHAT WE INITIALLY STARTED WITH SO LET S LOOK AT IT THIS WAY THIS TIME. WE LL JUST WRITE IT A LITTLE DIFFERENT. WE INITIALLY HAD 0.61 MOLES OF THIS, MOLES PER LITER. AND WE HAD ZERO OF THIS AND ZERO OF THIS AND NOW WE WANNA KNOW WHAT DO WE HAVE AT EQUILIBRIUM. WELL TO GET TO EQUILIBRIUM SOME OF THIS HAS TO REACT. HOW MUCH REACT WE DON T KNOW. HOW DO WE DEFINE THINGS WE DON T KNOW? X. SO WE RE GONNA SAY LET X EQUAL THE MOLES PER LITER OF BASE REACTING. LIKE WE DID BEFORE. SO ONCE WE VE MADE THAT STATEMENT NOW WHEN EQUILIBRIUM HAS BEEN REACHED HOW MUCH OF THE METHYL AMMONIUM ION WOULD WE HAVE FORMED? X. AND HOW MUCH HYDROXIDE WOULD WE HAVE FORMED? X. AND HOW MUCH OF THE BASE, HOW MUCH OF THE UNREACTED BASE WOULD WE HAVE LEFT?.61. AGAIN OUR DOLLAR IN THE POCKET MINUS THE AMOUNT THAT WE SPENT, MINUS X. NOW, AGAIN WE HAVE ONLY ONE UNKNOWN HERE, ALL X S, AND OUR NEXT THING THEN WOULD BE TO LOOK AT AND SEE IF WE CAN COME UP WITH A WAY FOR SOLVING FOR X AND OF COURSE ONE OF THE WAYS THAT WE HAVE IS A K, AN EQUILIBRIUM EXPRESSION, K B. K B IS EQUAL TO THE CONCENTRATION OF THE CH3NH 3 - DIDN T GIVE A WHOLE LOT OF ROOM THERE TIMES THE HYDROXIDE ION

10 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 10 CONCENTRATION, ALL DIVIDED BY THE CONCENTRATION OF THE CH 3 NH 2, AND WE KNOW NUMERICALLY THAT THIS IS EQUAL TO 4.8 X THAT S GIVEN IN A TABLE. SO PLUGGING IN THE CONCENTRATION AT EQUILIBRIUM FOR THE FIRST ION IS X. THE CONCENTRATION OF THE HYDROXIDE ION AT EQUILIBRIUM IS X. THE CONCENTRATION OF THE UNREACTED BASE 0.61 X, AND THAT S ALL EQUAL TO 4.8 X 10-4, DIDN T LEAVE MUCH ROOM THERE. ALRIGHT, NOW ONCE AGAIN NOTICE THAT WE HAVE ONLY ONE UNKNOWN IN THERE, X, AND WE HAVE A NUMERIC VALUE TO WORK FROM SO WE CAN GO AHEAD AND SOLVE IT. ONCE AGAIN IF WE WERE GOING TO MAKE A FULL SOLUTION WE WOULD HAVE TO DO A QUADRATIC EQUATION, BUT IF WE CAN GET BY WITHOUT DOING A QUADRATIC EQUATION WE LL GO AHEAD AND MAKE AN APPROXIMATE SOLUTION FIRST AND THEN WE LL COME BACK AND MAKE A FINAL CALCULATION. NOW IF OUR APPROXIMATE ANSWER COMES OUT WHERE THIS IS SMALL RELATIVE TO THIS THEN OF COURSE WE'RE DONE WITH THAT STEP OF THE CALCULATION. YES WE HAVE A QUESTION? (STUDENT RESPONSE NOT AUDIBLE) QUESTION WAS IS THIS TOP ONE CH 3 NH 3? YES IT IS. CH 3 NH 3 WITH A PLUS IS THIS PRODUCT. ALRIGHT REMEMBER EQUILIBRIUM EXPRESSIONS ARE ALWAYS PRODUCTS ON TOP DIVIDED BY REACTANTS ON THE BOTTOM. OKAY SOLVING FOR THIS THEN X TIMES X GIVES US X 2, AND WE RE GONNA HAVE THIS NUMBER TIMES THIS NUMBER SO WE RE GOING TO HAVE 4.8 X 10-4 TIMES.61 AND WE LL FIGURE OUT WHAT THAT NUMBER IS HERE QUICKLY SO WE HAVE 4.8 EXPONENTIAL 4 TIMES.61 AND WE END UP WITH AN ANSWER OF 2.9 X TO GET AN ANSWER FOR X NOW OF COURSE WE WOULD TAKE THE SQUARE ROOT OF BOTH SIDES AND SO X IS EQUAL TO THE SQUARE ROOT 1.7 X OR, IF WE WROTE IT IN DECIMAL FORM NOW THE QUESTION IS IS THAT NUMBER THAT WE JUST CALCULATED 0.17, IS THAT SMALLER THAN 10% OF THIS NUMBER HERE AND IT IS BECAUSE 10% OF THAT NUMBER WOULD BE WHAT?.06 AND CERTAINLY.01 IS LESS THAN THAT. SO THEREFORE, THEREFORE WE RE JUSTIFIED IN DISREGARDING THE MINUS X AND WE DON T HAVE TO DOA QUADRATIC. NOW IN THE PROBLEMS THAT YOU RE ASSIGNED AND IN THE PROBLEMS THAT ARE WORKED IN THE BOOK IN THE ACIDS THAT ARE USED AND THE BASES THAT ARE USED THIS MINUS X TERM IS ALWAYS GOING TO BE INSIGNIFICANT. YOU WILL NOT NEED TO DO ANY QUADRATIC EQUATION SOLUTIONS, BUT IN REAL, THE REAL WORLD SOME OF

11 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 11 THESE ACIDS ARE STRONG ENOUGH OR THE SOLUTION IS DILUTE ENOUGH THAT THE MINUS X TERM CAN T BE DISREGARDED AND SO MATHEMATICALLY WE WOULD NEED TO SOLVE A QUADRATIC EQUATION. ALRIGHT NOW WHAT IS X? WELL IF WE GO BACK UP HERE WE SAID THAT X WAS THE HYDROXIDE ION AND THAT ANSWERS OUR FIRST QUESTION OF THE PROBLEM SO THE HYDROXIDE ION IS EQUAL TO 1.7 X THAT WAS OUR FIRST QUESTION. NOW WE GO BACK TO THEN ASKING THE QUESTION WHAT IS THE HYDRONIUM ION? AND SO WE HAVE HYDRONIUM ION TIMES THE HYDROXIDE ION IS EQUAL TO 1.0 X 10-14, LL WATER SOLUTIONS. THIS REMAINS CONSTANT AND SO HYDRONIUM ION IS EQUAL TO 1.0 X DIVIDED BY THE HYDROXIDE ION AND THAT IS THE ION CONCENTRATION WE SOLVE FOR USING OUR EQUILIBRIUM AND THAT TURNED OUT TO BE 1.7 X SO IF WE NOW MAKE OUR CALCULATION HERE WE THEN HAVE, UH, WE END UP WITH AN ANSWER OF 5.8 X SO 5.8 X IS THE HYDRONIUM ION CONCENTRATION. SO WE NOW HAVE THE HYDROXIDE ION CONCENTRATION CALCULATED FROM OUR EQUILIBRIUM, THE HYDRONIUM ION CALCULATED FROM THE EQUILIBRIUM OF THE WATER K W, AND THEN FINALLY OF COURSE WE CAN ANSWER THE LAST QUESTION WHICH IS ph. ph IS EQUAL TO MINUS LOG OF 5.8 X AND WE HAVE A ph OF WE RECALL THAT IF ph IS GREATER THAN 7 THE SYSTEM IS ABASE AND QUITE OBVIOUSLY IS GREATER THAN 7 SO THEREFOR IT IS A BASE. OKAY. WELL ANY QUESTION ON ANY STEP OF THAT CALCULATION? THE SAME TYPE BUT DEALING WITH A WEAK BASE INSTEAD OF WITH A WEAK ACID. ONE OF THE REALLY VALUABLE RELATIONSHIPS INVOLVING ACIDS AND BASES ARE WHAT WE CALL BUFFERS AND BUFFER SOLUTIONS. BUFFERS ARE IMPORTANT IN COMMERCIAL PROCESSES. BUFFERS ARE IMPORTANT IN OUR BIOLOGICAL PROCESSES, AND BUFFERS ARE IMPORTANT IN TERMS OF SOMETIMES MEDICINAL PROCESSES. SO WE HAVE A LOT OF THINGS WHERE BUFFERS BECOME INVOLVED. A BUFFER, OR A BUFFERED SOLUTION IS ONE WHOSE ph DOES NOT CHANGE IF WE ADD SOME STRONG ACID OR STRONG BASE. ph IS A CONSTANT FOR A BUFFER. OKAY, SO THAT IS ONE OF THE CHARACTERISTICS THAT S IMPORTANT. A BUFFER HAS A CONSTANT ph. SO WHETHER WE ADD A LITTLE ACID TO IT OR ADD A LITTLE BASE TO IT DOESN T AFFECT THE ph TO ANY GREAT EXTENT. FOR INSTANCE OUR BLOOD IS BUFFERED IN OUR BODY. OUR BLOOD IS BUFFERED TO A ph OF ABOUT 7.2, I THINK

12 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX SOMEBODY MAYBE THAT S INVOLVED IN THE HEALTHCARE AREA CAN CORRECT ME ON THAT IF I M WRONG IT S SLIGHTLY BASIC. SO WHETHER I DRINK SOMETHING THAT S ACIDIC, SOME SODA POP, EAT SOME ACIDIC FOOD, TOMATOES, PUT VINEGAR AND OIL ON MY SALAD AND EAT THAT WHATEVER THE CASE IS IT DOESN T AFFECT THE ph OF THE BLOOD. THE ONLY THING THAT AFFECTS THE ph OF THE BLOOD USUALLY IS SOM ETHING BIOLOGICALLY WRONG, AND AS A MATTER OF FACT, ONE OF THE FIRST TESTS THEY DO IF YOU RE SICK IS DRAW A LITTLE SAMPLE OF BLOOD WELL OF COURSE OBVIOUSLY THE FIRST IS TO CHECK TEMPERATURE BUT IF YOU GO A STEP FURTHER YOU DRAW A SAMPLE OF BLOOD AND ONE OF THE FIRST TESTS ON THE BLOOD IS ph. IF THAT ph IS OFF VERY FAR FROM NORMAL IT INDICATES THAT THERE IS A SERIOUS BIOLOGICAL PROBLEM GOING ON. A SERIOUS INFECTION, A SERIOUS MALFUNCTION OF AN ORGAN OR SOMETHING OF THAT NATURE BECAUSE THE BLOOD SHOULD NOT CHANGE ITS ph. IT IS BUFFERED. IT S BUFFERED WITH SEVERAL THINGS. ONE IS A HYDROGEN PHOSPHATE AND DIHYDROGEN PHOSPHATE. IT S ALSO BUFFERED WITH HYDROGEN CARBONATE AND CARBONATE. SO THERE S SEVERAL BUFFERS. IT S ALSO BUFFERED WITH SOME PROTEINS THAT ARE AMINO ACIDS. ALRIGHT SO A BUFFER HAS A CONSTANT ph. NOW WHAT COMPOSES A BUFFER? A BUFFER MUST CONTAIN TWO THINGS. COMBINATION OF A WEAK ACID AND THE CONJUGATE BASE THAT IS, THE ANION OF THE ACID. SO IF WE HAVE SOMETHING LIKE A WEAK ACID OF HYDROGEN CYANIDE, CYANIC ACID THEN THAT WOULD BE THE WEAK ACID AND CN - WOULD BE THE CONJUGATE BASE. THAT S WHAT S LEFT AFTER THE PROTON IS DONATED. OR A BUFFER MAY EXIST OF A WEAK BASE AND THE ACID THAT IS FORMED FROM THAT WEAK BASE. FOR INSTANCE IN THAT ONE WE COULD USE WELL WE KNOW THAT BASES TURN LITMUS BLUE SO WE LL MAKE THE BASE BLUE HERE. SO NH 3, AMMONIA IS A WEAK BASE AND NH + 4 IS THE ACID SPECIES THAT WE GET AFTER IT ACCEPT A PROTON. SO A WEAK BASE AND A CONJUGATE ACID AND THAT THEN GIVES US BUFFERS EITHER COMBINATION. WELL LET S LOOK AT A QUANTITATIVE RELATIONSHIP RELATIVE TO A BUFFER. IT ASKS US TO CALCULATE THE CONCENTRATION OF A HYDRONIUM ION AND THE ph FOR A SYSTEM WHICH IS MADE U OF.6 MOLAR HYPOCHLOROUS ACID HClO AND 1.56 MOLAR SODIUM HYPOCHLORITE. THE K A FOR HYPOCHLOROUS ACID IS 3.0 X NOW

13 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 13 THE FIRST THING WE HAVE TO DO IS WE HAVE TO WRITE AN EQUATION FOR THE REACTION OF A WEAK ACID WHICH IN THIS CASE WOULD BE THE HClO. SO HClO REACTS WITH H 2 O LIQUID TO PRODUCE H 3 O + PLUS ClO -. OKAY THAT S JUST OUR REGULAR ACID REACTING WITH WATER DONATING A PROTON FORMING THE HYDRONIUM ION AND THE ANION. NOW LET'S ASK FIRST OF ALL WHAT IS THE INITIAL CONDITION THAT WE HAVE? NOW THAT WE VE DUMPED THESE THINGS TOGETHER WHAT DO WE INITIALLY HAVE IN THE SYSTEM? HOW MUCH ACID DO WE INITIALLY HAVE? OKAY, HOW MUCH HYDRONIUM DO WE INITIALLY HAVE? ZERO. UNTIL SOME OF THE ACID REACTS WE DON T HAVE ANY HYDRONIUM ION. HOW MUCH OF THE ClO - DO WE HAVE INITIALLY? I HEARD ZERO BUT THAT S NOT CORRECT. WE REALLY HAVE, THIS IS A COMPOUND AND REMEMBER ALL 1A AND 2A METAL COMPOUNDS ARE IONIC. THEY RE IONIC WHEN WE DISSOLVE THEM IN THERE AND SO IF WE HAVE THIS WE REALLY HAVE THEN Na + IONS AND ClO - IONS IN SOLUTION. SO WE REALLY HAVE 1.56 OF THE ClO IONS ALREA DY PRESENT BECAUSE WE DUMPED THEM IN. NOT BECAUSE THEY WERE FORMED BY THE ACID-BASE REACTION BUT BECAUSE THEY RE IN THERE INITIALLY. WE DUMPED THESE TWO TOGETHER TO MAKE THIS SOLUTION. NOW WE RE GOING TO LET THE ACID REACT WITH THE WATER AND SO WE LL SA Y HERE LET X EQUAL THE MOLES PER LITER, MAYBE IT S THE PENS. TRY TO HERE. MOLES PER LITER X EQUAL MOLES PER LITER OF HClO REACTING. ALRIGHT, NOW IF WE DO THAT AND THEN WE GO BACK AND ASK NOW WHAT WILL WE HAVE ONCE EQUILIBRIUM IS REACHED? HOW MUCH OF THIS WILL WE HAVE AFTER EQUILIBRIUM HAS BEEN REACHED? SEE THIS IS JUST LIKE THE LAST TWO WE DID. (STUDENT RESPONSE NOT AUDIBLE) YEAH!.76 MINUS X. THAT S WHAT WE STARTED WITH. WE JUST SAID LET X EQUAL THE AMOUNT THAT REACTED SO THE AMOUNT THAT WE RE GOING TO HAVE LEFT IS 0.76 X. HOW MUCH HYDRONIUM ION IS GOING TO BE FORMED? X. BECAUSE THAT S WHAT WE SAID AND IT S A ONE TO ONE. NOW IF WE GOT TWO HYDRONIUMS PER ACID OF COURSE IT WOULD BE 2X. WE LOOKED AT AN EXAMPLE PROBLEM WITH SOLUBILITY OF THAT TYPE. WHAT IS THE EQUILIBRIUM CONCENTRATION OF THE HYPOCHLORITE ION? (STUDENT RESPONSE- NOT AUDIBLE) VERY GOOD X, I HEARD SOMEBODY SAY. YES. WE STARTED WITH THAT AMOUNT AND IF ANYTHING REACTED UP HERE WE RE GOING TO GET SOME ADDITIONAL AMOUNT. SO WE LOST

14 CHM 105 & 106 MO1 UNIT FIVE, LECTURE SIX 14 SOME OF THE ACID BUT WE HAVE ADDED SOME OF THE HYPOCHLORITE ION. NOTICE THAT THE ONLY THING NOW LEFT THAT IS AN X ALL BY ITSELF IS THE HYDRONIUM ION AND WHAT WAS OUR FIRST QUESTION? CALCULATE THE HYDRONIUM ION FOR THE SYSTEM. SO IF WE NOW PUT THIS IN AN EQUILIBRIUM EXPRESSION AND SOLVE FOR X WE WILL THEN BE ABLE TO DETERMINE THE HYDRONIUM ION AND CONVERT IT TO ph. AND IN OUR NEXT LECTURE WE LL GO AHEAD AND COMPLETE THIS CALCULATION ON THE BUFFER AND WE LL TALK ABOUT SOME REAL BUFFER SYSTEMS AS THEY APPLY TO OUR BIOLOGICAL SYSTEMS.