- WELCOME TO ACIDS AND BASES PART 9.
THIS VIDEO TUTORIAL WILL FOCUS ON SERIAL DILUTIONS
AND THE BEAUTIFUL ALIGNMENT THEY HAVE WITH PH
AND LOGARITHM SCALE.
LET'S GET STARTED.
ALL RIGHTY. SO DILUTION, SO WE'VE HEARD THAT WORD BEFORE
BACK IN AN EARLIER CHAPTER WE LEARNED THE DILUTION EQUATION,
TYPICALLY, WE'LL REPRESENT C1 OR C2
AS THE MORE CONCENTRATED SOLUTION,
AND THEN C2 AND V2 WILL BE THE MORE DILUTE SOLUTION.
SO, OF COURSE THE C'S REPRESENT CONCENTRATION
AND THE V'S REPRESENT VOLUME.
ALL RIGHT. AND THERE IS A VIDEO TUTORIAL ON DILUTIONS
AND SOLUTIONS EARLIER IN THE SERIES IF YOU NEED TO,
A REFRESHER.
THE BEAUTY HERE THOUGH IS THAT FOR SERIAL DILUTIONS,
WE ARE GOING TO WANT TO SOLVE FOR C2,
SO THIS IS UNIQUE TO THE SERIAL DILUTION SITUATION.
SO IF WE REARRANGE OUR DILUTION CALCULATION FOR C2,
WE WOULD SEE THAT C2 = (C1V1)/V2.
NOW, THE COOL PART IS WHEN WE'RE DOING A ONE TO TEN DILUTION,
WHAT THAT SAYS IS THAT THE VOLUME--THE INITIAL VOLUME, V1,
IS REFERRED TO AS THE 1,
AND WE'LL JUST SAY 1 ML.
AND THEN THE 10 REFERS TO V2, AND WE'LL JUST USE MILLILITERS.
THE MAIN POINT HERE IS THAT WHATEVER VOLUMES YOU USE,
THE UNITS MUST MATCH.
SO AS LONG AS YOUR VOLUME UNITS MATCH,
THIS TUTORIAL WILL GUIDE YOU TO THE CORRECT ANSWER.
SO NOW THAT WE'VE IDENTIFIED THE VALUES FOR V1 AND V2,
WE CAN SEE THAT DOING A ONE TO TEN DILUTION,
THE C2 WILL C1 DIVIDED BY 10.
SO THERE IS OUR DILUTION FORMULA FOR A ONE TO TEN
SERIAL DILUTION SERIES.
AND THE REASON WHY WE LIKE TO DO ONE TO TENS
IS BECAUSE OF THE LOGARITHMIC NATURE OF PH,
WHICH IN BASE 10, SO WE HAVE A BEAUTIFUL COMPLEMENT.
LET'S GO THROUGH THE SERIAL DILUTION PROCESS.
SO WE BEGIN WITH SOLUTION A,
THIS IS OUR ORIGINAL STOCK SOLUTION.
AND SO TYPICALLY WHEN YOU'RE IN THE LAB,
YOU'LL START WITH A CAREFULLY PREPARED ORIGINAL STOCK SOLUTION
AND THEN WE CAN MAKE A SERIES OF DILUTIONS,
SO WE'LL CALL IT A SERIAL DILUTION.
SO THIS IS SOLUTION A.
SO LET'S SEE WHAT WE KNOW ABOUT SOLUTION A.
SO SOLUTION A, WE CAN READ THE BOTTLE, IT'S 1 MOLAR HCL.
AND WE UNDERSTAND ACID/BASE CHEMISTRY.
SO WE KNOW IF WE HAVE HYDROCHLORIC ACID,
THAT'S IT'S A STRONG ACID AND IT FULLY IONIZES TO H+.
SO IF WE SEE WE HAVE 1 MOLAR HCL,
WE KNOW THAT WE ALSO HAVE 1 MOLAR H+
BECAUSE IT FULLY IONIZES INTO THE HYDRONIUM OR THE HYDROGEN
IONS AND THE CHLORIDE IONS.
SO NOW WE CAN CALCULATE THE PH WHICH IS THE NEGATIVE LOG
OF THE HYDROGEN ION CONCENTRATION.
SO THE PH IS THE NEGATIVE LOG OF 1 WHICH IS 0.
SO THIS FIRST SOLUTION, PH = 0.
NOW, THAT WE KNOW ABOUT SOLUTION A,
LET'S LOOK AT SOLUTION B.
SO HERE'S SOLUTION B. WE'LL WORK ON SOLUTION B
RIGHT HERE IN THIS SPACE.
SO, NOW WE NEED TO USE OUR SERIAL DILUTION CALCULATION.
SO THE CONCENTRATION OF SOLUTION B
WILL BE THE CONCENTRATION OF SOLUTION A DIVIDED BY 10.
SO THE CONCENTRATION OF SOLUTION A,
AND WE'RE JUST GOING TO FOCUS ON THE HYDROGEN IONS HERE
BECAUSE WE'RE LINKING TO PH.
SO THAT WOULD BE 1 MOLAR DIVIDED BY 10 OR 0.1 MOLAR H+.
SO IF YOU'RE WONDERING WHERE THE 10 MILLILITERS CAME FROM,
THE FINAL VOLUME IS 10 MILLILITERS BECUASE...
WHERE DID THIS 10 COME FROM?
WE HAVE THE 1 MILLILITERS OF SOLUTION A
AND THEN 9 MILLILITERS OF WATER,
SO WE HAVE 10 MILLILITERS AS OUR FINAL VOLUME.
SO FROM THERE, THEN THE PH IS THE NEGATIVE LOG OF H+
WHICH WE HAVE CALCULATED TO BE .1 MOLAR,
SO THE PH = 1.
SO NOW WE CAN PUT THAT INFORMATION ON OUR DIAGRAM.
PH = 1.
SO WE'RE SEEING THE LOVELY SYNCHRONICITY
BETWEEN 1 TO 10 SERIAL DILUTIONS AND THE PH SCALE.
LET'S DO ONE LAST SOLUTION
JUST TO MAKE SURE EVERYBODY UNDERSTANDS.
LET'S LOOK AT SOLUTION C.
SO FOR SOLUTION C, THE CONCENTRATION OF SOLUTION C
WILL BE THE CONCENTRATION OF SOLUTION B DIVIDED BY 10.
SO THAT'LL BE .1 MOLAR DIVIDED BY 10, AND THAT'S FOR H+,
SQUEEZE THAT IN.
SO THE CONCENTRATION OF SOLUTION C WILL BE .01 MOLAR H+.
AND TO TAKE THE PH, NEGATIVE LOG OF .01 MOLAR,
AND SO THE PH WOULD EQUAL 2.
SO WE'LL WRITE THAT INFORMATION RIGHT HERE.
PH = 2.
SO I KNOW THAT THIS IS AN IMPORTANT SET OF CALCULATIONS
ESPECIALLY FOR THE MICROBIOLOGY CLASS, IF THAT'S IN YOUR FUTURE.
SO LET'S--NOW THAT WE HAVE THE BASIC UNDERSTANDING,
LET'S PRACTICE A LITTLE BIT MORE ON THE NEXT PAGE.
OKAY. SO THIS TIME, WE WILL START WITH A STOCK SOLUTION
THAT HAS AN ORIGINAL CONCENTRATION OF 10
TO THE -3 MOLAR.
SO THIS WILL BE OUR ORIGINAL SOLUTION A.
OKAY. AND SO LET'S MAKE SURE EVERYONE UNDERSTANDS.
WE DID THIS ON THE PREVIOUS PAGE, BUT LET'S MAKE SURE,
RIGHT, THERE'S THE ACID DISSOCIATION REACTION.
SO THIS WILL BECOME H+ + CO-.
SO WE KNOW THAT THE CONCENTRATION OF HCL
IS CONVERTED, COMPLETELY INTO H+ AND CO-
BECAUSE HAVE A STRONG ACID.
SO A STRONG ACID GIVES US A HUNDRED PERCENT IONIZATION.
SO, THE CALCULATIONS I'M SHOWING YOU ONLY WORK FOR STRONG ACIDS.
WITH WEAK ACIDS IT BECOMES MUCH MORE COMPLICATED.
NOW, LET'S DRAW A DIAGRAM SO WE CAN CONCEPTUALIZE
EXACTLY WHAT'S HAPPENING.
TYPICALLY, OUR ORIGINAL SOLUTION COMES IN A BOTTLE,
THAT WOULD BE SOLUTION A AND THIS IS 10 TO THE -3 MOLAR H+
BASED OFF OUR REACTION.
AND THEN WE'RE GOING TO TAKE 1 MILLILITER OF SOLUTION A,
AND WE'RE GOING TO COMBINE THAT WITH 9 MLS OF WATER
INTO SOLUTION B.
WE'LL MIX THAT WELL.
THEN WE'LL TAKE 1 ML OF B AND 9 MLS OF WATER,
AND THAT WILL PRODUCE SOLUTION C.
AND THEN FINALLY WE WILL TAKE 1 MILLILITER OF C
AND WE'LL ADD 9 MLS OF WATER AND THAT'LL PRODUCE SOLUTION D.
SO NOW, THAT WE UNDERSTAND THE DILUTION STEPS,
WE WILL PERFORM THE CALCULATIONS.
WELL, THE ORIGINAL SOLUTION WE ALREADY KNOW THE CONCENTRATION,
IT'S 1.0 X 10 TO THE -3 MOLAR.
AND SO OF COURSE FOR THE PH,
WE JUST SIMPLY TAKE THE NEGATIVE LOG OF H3O+,
SO THE PH WOULD BE 3.00.
AND IF YOU'RE WONDERING ABOUT THE SIG FIGS,
PLEASE REFER TO THE PH TUTORIAL.
SO NOW, FOR THE FIRST DILUTION,
THE CONCENTRATION OF 'A' DIVIDED BY 10,
SO THAT WOULD BE 1.0 X 10 TO THE -3 DIVIDED BY 10 MOLAR,
SO THAT WOULD BE 1.0 X 10 TO THE -4 IS THE RESULT.
YOU SHOULD BE ABLE TO DO THIS IN YOUR HEAD,
BUT IF YOU NEED TO USE A CALCULATOR,
TRY TO TRANSITION AWAY BECAUSE IN YOUR BIO CLASSES,
THEY WON'T LET YOU USE A CALCULATOR.
SO FOR C, WE'LL HAVE THE CONCENTRATION OF B
DIVIDED BY 10.
LET'S PUT THAT. C. OKAY.
SO OUR CONCENTRATION FOR B WAS 1.0 X 10 TO THE -4 MOLAR.
AND THEN WE'RE GOING TO DIVIDE THAT BY 10.
AND SO SOLUTION C, 1.0 X 10 TO THE - 5 MOLAR,
WHICH WOULD GIVE US A PH OF 5.00.
AND THEN LASTLY TO DETERMINE THE CONCENTRATION OF SOLUTION D,
WE WOULD TAKE THE CONCENTRATION OF C AND DIVIDE BY 10.
SO THE CONCENTRATION OF C, WE DETERMINE IT TO BE
1.0 X 10 TO THE -5 MOLAR DIVIDED BY 10. 1.0 X 10 TO THE -6 MOLAR,
WHICH OF COURSE, WOULD BE A PH OF 6.
AND THEN JUST TO KIND OF TAKE THIS OPPORTUNITY,
THERE'S A LITTLE SPACE OFF DOWN THAT PAPER,
MIGHT AS WELL WRAP THIS AROUND.
FOR SOLUTION D, HOW WOULD WE CALCULATE
THE HYDROXIDE CONCENTRATION?
YOU WOULD USE THE KW.
SO WE ALWAYS KNOW THE KW. THE KW IS 10 TO THE -14.
AND WE KNOW THE HYDRONIUM CONCENTRATION, IT'S RIGHT HERE.
SO WE CAN SOLVE FOR HYDROXIDE.
SO THE HYDROXIDE CONCENTRATION WILL BE 1 X 10 TO THE -14
DIVIDED BY 1.01 X 10 TO THE -6,
THEN SO OUR HYDROXIDE CONCENTRATION,
1.0 X 10 TO THE -8 MOLAR HYDROXIDE.
THAT CONCLUDES OUR VIDEO TUTORIAL ON SERIAL DILUTIONS
AND THEIR RELATIONSHIP TO THE PH SCALE.
PLEASE TAKE SOME TIME NOW TO WORK A FEW HOMEWORK PROBLEMS
TO REINFORCE YOUR UNDERSTANDING.