Titration introduction | Chemistry | Khan Academy
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Titration introduction | Chemistry | Khan Academy


Titration is a procedure for determining the concentration of a solution. And so let’s say we’re starting
with an acidic solution. So in here let’s say we
have some hydrochloric acid. So we have come HCl. And we know the volume of HCL, let’s say we’re starting
with 20.0 milliliters of HCl. But we don’t know the concentration right? So question mark here for
the concentration of HCl. We can find out that concentration
by doing a titration. Next we need to add a few drops
of an acid base indicator. So to this flask we’re also going to add a few drops of an acid base indicator. We’re gonna use phenolphthalein. And phenolphthalein is colorless in acid but turns pink in the presence of base. And since we have our
phenolphthalein in acid right now we have a clear solution. There’s no color to it. Up here we’re gonna have
our standard solution right? We’re gonna have a known
concentration of sodium hydroxide. So let’s say we have a
solution of sodium hydroxide and the concentration is zero
point one zero zero molar. And we’re ready to start our titration. So we allow the sodium hydroxide to drip into our flask containing
our HCl and our indicator. And the acid in the
base will react, right? So we get an acid base
neutralization reaction. HCl plus NaOH right? If we think about the products,
this would be OH minus, this would be H plus, H plus and OH minus give us H2O. And our other product we would
have Na plus and Cl minus, which give us NaCl, or sodium chloride. So let’s say we add a
certain volume of base right? So now this would be higher, and we see our solution turn light pink. Alright so let’s say we see
our solution turn light pink and it stays light pink. That means that all of the acid has been neutralized by the base. And we have a tiny amount
of excess base present, and that’s causing the acid
base indicator to remain pink. So a tiny excess of base means we’ve neutralized
all of the acid present. And where the indicator changes color, this is called the end point
of a titration, alright? So when our solution changes color, that’s the end point of our titration. And here we stop and we
check and see the volume of base that we used in our titration. So if we started right here, if we started with that much base, let’s say we ended down here, alright? So we still have a
little bit of base left. And this would be the volume of base that we used in the titration. Alright so we have a
change in volume here, and let’s say that it’s 48.6 milliliters. So it took 48.6 milliliters of our base to completely neutralize the
acid that we had present. And so we can now calculate
the concentration of the HCl. Alright so let’s go ahead and do that, and let’s start with the
concentration of sodium hydroxide. Alright we know that we started with point one zero zero molar
solution of sodium hydroxide. So point one zero zero molar. And molarity is equal to mols over liters. Alright so this is equal
to mols over liters. And our goal is to figure
out how many mols of base that we used to neutralize
the acid that was present. Alright so we can take our
volume here, 48.6 mililiters and we can convert that into liters. Alright so just move your decimal place three places to the left. So one, two, three. So that’s point zero
four eight six liters. So this is equal to mols over zero point zero four eight six liters. And so let’s get some more space. Alright let me just rewrite
this really quickly. Zero point one zero zero is equal to X over zero point zero four eight six. So we’re just solving
for X, and X represents the mols of sodium hydroxide
that were necessary to neutralize the acid
that we had present. Alright so when you solve for X, you get zero point zero
zero four eight six mols of sodium hydroxide used in our titration. Next you look at the balanced
equation for what happened . So if I look at my balanced equation alright there’s a one here and there’s a one here. So we have a one to one mol ratio. And the equivalence point is where just enough of your standard
solution has been added to completely react with the
solution that’s being titrated. And at the equivalence point, all of the acid has been neutralized. Right? So it’s completely reacted. And since we have a one to one mol ratio, if I used this many mols
of sodium hydroxide, that must be how many mols
of HCl that we had present in our original solution. So therefore, I can go ahead
and write that I must have had zero point zero zero four eight six mols of HCl present in the flask
before we started our titration. Right and I knew that because
of the one to one mol ratio. Remember our goal was to find
the concentration of HCl. The original concentration. And concentration, molarity is equal to mols over liters. So now I know how many mols of HCl I had, and my original volume of HCl was 20 milliliters right? So right up here we had 20 milliliters. So I need to convert that into liters. So I move my decimal place one two three. So I get point zero two liters. So now our final step here to calculate the concentration of HCl, right so the concentration
of HCl is equal to how many mols of HCl we have, which is zero point zero
zero four eight six mols, over liters of solution. And we had 20 milliliters
which is equal to zero point zero two zero zero liters. Alright so now we can
take out our calculator and do this calculation to find the concentration of HCl
that we started with. Point zero zero four eight six, all right and we’re gonna divide that by point zero two zero zero. And we get zero point two
four three for our answer. So the concentration of HCl is equal to zero point two four three molar. So we’ve solved for the
original concentration of HCl. There’s a shortcut way to do this problem, and the shortcut way would
be to do the molarity times the volume of the acid is equal to the molarity times
the volume of the base used. So MV is equal to MV. So let’s say we have the
acid over here on the left, and the base over here on the right. So the molarity of the acid
is what we’re trying to find. So I’ll just make that X. The volume of the acid
that we started with, you can just leave this in
milliliters if you want, 20 point zero milliliters is how much of the acid we started with. And for the base, we
knew the concentration of the base that we used
in our titration right? It was zero point one zero zero molar. And we also knew the volume of base that we used to completely
neutralize the acid. We used 48.6 milliliters. And notice how the mLs
would cancel out here. Right and we can just
go ahead and do the math and solve for X. So we get out the calculator,
and we need to multiply 48.6 times point one zero zero. Alright and so we get four
point eight six obviously. And then if we divide by 20 we will get our answer of
zero point two four three. So X is equal to zero
point two four three molar. And this shortcut way works pretty well when you’re dealing with a
strong acid and a strong base and a one to one molar relationship. Alright in the next video
we’ll do a problem where the mol ratio is no longer one to one.

About James Carlton

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90 thoughts on “Titration introduction | Chemistry | Khan Academy

  1. What would be really helpful would be a video only showing examples which are visual with all three acid-base titrations.

  2. Thanks for the video, just one question: is it possible to find the opposite to this, NaOH concentration? using the same method seen in this video? For instance, If I know the concentration of HCl and I need to find the concentration of NaOH?

  3. When you went to find the concentration of HCL, why didn't you use the formula concentration = mass/volume? Wouldn't that have worked? Concentration's unit is in grams per cubic decimeter or in mols per dm^3

  4. Hi, I'm trying to perform a titration to find the average amount of lactic acid concentration in yogurt and how much is in a certain amount of bacteria(such as streptococcus thermopilus). I am confused on what my standard solution would be, pls help?

  5. 3 weeks until the end of chemistry! I am so freaking done with school this year. I am loosing my sanity!

  6. Why can't my teacher actually explain stuff to us like this is stead of giving us packets with reading

  7. Chemistry's born to mess with minds…….:(……………titration calculations were easy though…..:P

  8. Real life question. What good is it to know the molarity of an acid if it's already been neutralized?

    I'm guessing it was only a sample of acid. Idk. Anyone?

  9. With the help of khan academy I will survive exams.
    Needless to say that my teachers are so unqualified😑

  10. I study in one of the best schools of my country but I develop my understandings only through online vids…

  11. I mean… Why did my chem teacher give the class a titration concentration table the day before a test for review without explaining it? Why does this look a little complicated, what am I doing here, why wont they let me sleep, why am I pulling an all nighter for this test, why did he give all the review the day before the test? Im going to be okay 😀

  12. thank you for this video, I've just begun IB HL chemistry and it was already getting stressful, this video helped a lot!! thanks!!!

  13. ASDFHFJGKGKGL I just need to accept im getting a C in Chem. But it’s fine. I’m fine. EVERYTHING IS FINE GODDAMMIT

  14. why the fuck people don't subscribe this channel it's very important channel at least more important than those Paul's Chanel

  15. I didn't understand what a standard solution, equivalence point, and that stuff was about – even though I got taught that in chemistry but I still don't understand it 🙁

  16. I was so happy with this as a link to show my students until you got to 7:30 and told them it's acceptable to use M1V1 = M2V2. You don't START with the context that it works only for 1:1 reactions and that it is not an acceptable general case for titrations. Most of the time, students stop watching after they figure out the way to do the problem! I suffered through EIGHT YEARS of titrations being set up as dilution problems regardless of the mole ratio, and this just perpetuates that error. You have to present the context FIRST, or, better yet, don't present the method that way at all!

  17. You assumed early on that I knew how to "Solve for X" now I can't move forth with the rest of the video #whygodwhy #nowIknowwhyoneshowtheirwork

  18. can you not do this in two steps using ratios?
    20ml : 48.6ml
    0.1 moles : x moles

    20/0.1= 200 so 48.6/200= 0.243 M

    or will this not work every time?

  19. What's different from [HCl] and X ?لماذا لم نبقي تركيز الحامض[HCl] نفسه لتركيز (X) ?
    Please choose me…
    Thanks for explaining 😊😊

  20. Only if teachers would bloody earn more, so that you'd get competent teachers. Students have review systems but teachers dont. Fuck all teachers who suck.

  21. Stupid teacher couldn't explain this for a week yet I get a clear understanding in a few minutes. Screw paying for school. I want a refund from my University. FML.

  22. I prefer the short cut method of MV=MV .The problem is that it only applies wen the mole ratio for both is 1:1.anyway thanks❤

  23. I love how I already graduated w/ a B.S. in biology and I understand these issues so much more AFTER I am done with school. Now that I'm studying for the MCAT and actually watching videos all of these topics make so much more sense. Hate that I didn't apply myself in school. Fantastic vid.

  24. Please dear : if we have water tank has avolume of water 25000 liter ,and pH=4 ,And i want to add sodium hydroxide to the tank until reach pH=7…determine concentration of sodium hydroxide that you added to keep pH=7??

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