In a titration, a solution of known concentration (the titrant) is added to a solution of the substance being studied (the analyte). In an acid-base titration, the titrant is a strong base or a strong acid, and the analyte is an acid or a base, respectively. The point in a titration when the titrant and analyte are present in stoichiometric amounts is called the equivalence point. This point coincides closely to the endpoint of the titration, which can be identified using an indicator. Created by Jay.
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- That MCAT does not allow for a calculator... how would you do this then?(21 votes)
- I find that writing everything in scientific notation is really helpful! For example, at6:39you can do it like this:
.00486 mol / .02 L = 4.86 x 10^-3 / 2 x 10^-2
I like to separate it out like this (4.8 / 2) x 10^-3 - (-2)
2.4 x 10-1 = .24 M(8 votes)
Please correct me if I'm wrong
I've read that phenolphthalein changes color when the pH is from above the range of 8.3 to 10.0
1: Using it we have the change in color when we have more OH- than those used to neutralize the strong acid.
2: As the range is from 8.3 to 10.0, we would have an error margin of about 2 decimal points in the total concentration of H+(17 votes)
- Clarification to conclusion 1: The solution will NOT change color at the exact instant that [OH-] first exceeds [H+], because the pH at this point would only be very slightly above 7 (which is not in the pH range of phenol). A color change will happen when [OH-] ions is about (2*10^-6 M), because this is when pH is at 8.3. The color change marks the end point of the titration. Hope this helps!(16 votes)
- What is the difference between end point2:21and equivalence point?(8 votes)
- Great question! The equivalence point is almost more of a mathematical concept, it is when the moles of H+ in solution equals the moles of OH- in solution, and all the acid (or base) in the original solution is neutralized.
But how do we know when the equivalence point has been reached? We know we have reached the equivalence point by adding an acid-base indicator such as phenolphthalein, we look for when the solution turns pink. But as Jay mentions at0:43, phenolphthalein turns pink when there is base present. If there is base present, that means we are at least a teensy bit past the equivalence point, since there isn't any base at the equivalence point. When the solution turns pink and we stop the titration, that is called the end point. But the endpoint is usually just a little bit past the equivalence point, because that is how the indicators work.(17 votes)
- does M (mol) means also N (nomality) ?(2 votes)
- M is Molarity (Not mol)
N=M*n factor (For redox reactions)
N=M*acidity (For bases)
N=M*basicity (For acids)(11 votes)
- For the short cut equation does the molar ratio between acid and base matter. Would you need to need to do the calculation differently if for every 2 mol of compound A you need 3 mol of compound B to make the product?(5 votes)
- the equation we use here is CaVa/CbVb=na/nb
where Ca is concentration of acid and Va is volume of acid
Cb is concentration of base and Vb is volume of base
na/nb is the mole ratio of the acid and base(3 votes)
- Which is correct Ca(OCl)Cl or CaOCl(subscript)2? And what is the chemical name of the compound?(3 votes)
- the correct formula is CaOCl(subscript)2
it is an inorganic compound called calcium hypochlorite or u may commonly know it as bleaching powder, used as a disinfectant. as it is a mixture of lime and calcium chloride, it is often called as chlorine powder.
hope this helps :)(5 votes)
- When calculating the [HCl], wouldn't the final volume be the initial 20 mL (from HCl) plus the additional 48.6 mL (NaOH)? And thus, the concentration would ~0.071 M instead of 0.243 M?(3 votes)
- You are calculating the concentration of HCl at the beginning of the titration, that is, when the HCl was in the original 20 mL.(4 votes)
- Why should an acid base titration use a single indicator rather than a mixed indicator?(3 votes)
- A single indicator should be used so that you can correctly interpret the result (the color). If more than one indicator is used then it may be difficult to tell which indicator is producing the color you are seeing.(3 votes)
- What do I use Titration for?(2 votes)
- You use titration to find the molarity of a substance. You may have the mL or L, but no other information, which is why you would need to use a known molarity base to help you determine the unknown molarity of your acid.(4 votes)
- why didn't he just use m1v1=m2v2 in the first place?(2 votes)
- You can use the dilution formula for a strong base/acid titration where the acid and base neutralize each other in an equal mole-to-mole ratio (i.e. 1:1, 2:2, etc.), but this is the only case where you can do so. A titration involving say a weak acid and a strong base would not work if you did the dilution formula there because buffer solutions arise.
Remember, the point of the video is to teach people about titrations. Can't really do that if you just calculate the unknown concentration without even doing the titration. Really the dilution formula is there to confirm the results from the titration and that they agree with each other.
Hope that helps.(4 votes)
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.