Strong Titrate-Strong Titrant Curves - Video Tutorials & Practice Problems

Strong Acid-Strong Base Titration Curve

You were going to say that in these two types of titration curves, both the titrate and tran represent a strong acid or base. Here, we're gonna pay attention to the left side first and in the following video, we'll take a look at the right side. All right. So here, let's just focus on the left side. In this one, it represents a strong acid, strong base titration curve. We can see here that our T trait, which is it. Well, we're starting out at a Ph that's much lower than seven. So in this case, our tight trait would have to be a strong acid as we begin to add strong base to it. We see that the ph is increasing over time. So our tight trend would have to be a strong base. Now, here, the PH starts below seven and we see there's a sharp increase um as we add base and we can see around 60 MLS of our strong base being added. That's why we have the steepest climb in terms of our ph. Now here, the equivalent point is this red spot right here, which is the dead center portion of the steepest climb, we can see that at that point, the PH is equal to seven. Now, here we're going to say after the equivalent points or after that red dot the strong acid is neutralized and excess strong base remains. So at this point, when we get to the equivalence point, all of the strong acid has been completely neutralized by the strong base. But since we keep adding more and more base, we're gonna have some excess base remaining. That's why it continues to increase. But eventually it'll level off. OK. So here we see a gradual leveling off around just under ph of 13. So here when we're dealing with a strong acid, strong base titration curve, the first thing discussed would be our titrate. It's being titrated by this titrate of the strong base and doing this gives us the sigmoidal shape of our titration curve. And we see that it starts out at a low ph and then eventually gets to a higher ph. So now that we've seen this type of sigmoidal curve, click on the next video and let's look at the right side or flipping rolls and seeing how that curve will look.

Strong Base-Strong Acid Titration Curve

Now, let's take a look at a strong base, strong acid titration curve. We can see here that this also has a sigmoidal shape, but it's going the opposite direction. We can see that we're starting out at a Ph. That's well above seven. Since we're starting out a Ph, that's well above seven, that must mean our tit rate is a strong base. And here we're adding our tight Trent over time and that's causing a decrease in our ph. So our tight trend here would be a strong acid. So here we can see that our ph starts above seven since we have a strong base to begin with and it decreases sharply with added acid. Now, here we look at the steepest drop, the steepest um drop in terms of our Ph. It happens here. Depth center of that gives us this red dot which is our equivalence point. It too would give us a Ph equal to seven at the equivalence point. So keep this in mind when both the titrate and T Trent are strong, the PH will be equal to seven at the equivalence point like regardless of who's the tight trait and who's the tight trait because again, it's based on strength, since they're both strong, it's kind of like a draw. So their ph is equal to seven at the equivalence point. Now, after the equivalence point, we're talking about this portion here, we see that the strong base has been completely neutralized and excess strong acid remains. So when we get to the equivalence point, all of the strong acid has destroyed all of the strong base, but we continue to add more and more, more strong acid. So that's why our ph continues to drop. Eventually it'll level off. And we see here that it's leveling off around a ph of one. So just remember in both cases, we have a sigmoidal shape because both the a the tran and titrate are strong species because they're both strong. The PH is equal to seven at the equivalence point, right. So just keep in mind the similarities between these two types of titration curves.

Strong Titrate-Strong Titrant Curves Example

Here, it says, consider the titration of 100 mls of 0.500 molar hydro brom acid solution with 100 and 20 mls of 0.450 molar potassium hydroxide solution, which species would be in excess? All right. So there's two ways we can approach this question. One way is we could figure out what the equivalent volume of our tight trait if our tight trait will be. So let's say I made this the tran, then we can say that M acid times V acid equals M base times V base. And we could calculate the volume of our base to reach the equivalence point. If the volume we calculated is less than this. And that means koh is in excess because we've gone beyond our equivalent volume. So let's just do it that way. Then we're gonna look at another way we can approach the same question. So here we say we have 0.500 molar of hydro Bromma acid and then 100 MLS, then we have 0.450 molar of potassium hydroxide. And we're looking for its equivalence volume, divide both sides by 0.450 molar and we get our equivalent volume for our strong base as being 100 and 11.1 mL. So that's how much strong base we need to add to get to the equivalence point. We have a vine that's even greater than that, which would mean that K weight would be in excess. So Koh is our species that will be in essence, another way we could at attempt this question is to remember that moles equals leader times molarity. If we divide both these volumes by 1000 we get liters of each. All means multiply. So I've multiplied them by their molars. Doing. This would give me 0.0500 moles of H pr and then multiplying the the liters time similarity of this would give me 0.0540 moles of Koh. This is also telling me that hey, we have more Koh than H pr more moles. So Koh would again, would be an excess. So again, two different ways we can approach this type of question and they're asking you to figure out which one is an excess, but they're giving you the volumes of both and the molarity of both make either one, the tight print and then try to figure out its equivalent volume. If the equivalent volume calculated is lower than what they're giving to you, then you're in excess. You could also calculate the moles of each, whichever one has greater moles would be the one in excess, right. So take those two approaches, you'll find the answer. In this particular case, the answer is option B.

Which combination would give a pH = 7.0 at the equivalence point?

a) HNO₃ and NH₃                      b) HCl and NH₄⁺                        c) HC₂H₃O₂ and NaOH               d) HBr and NaH

• A. HNO₃ and NH₃
• B. HCl and NH₄⁺
• C. HC₂H₃O₂ and NaOH
• D. HBr and NaH