Weak Titrate-Strong Titrant Curves - Online Tutor, Practice Problems & Exam Prep

When titrating a weak acid or base with a strong titrant, a buffer region forms where the pH changes gradually. This region indicates the presence of a buffer system that resists changes in pH. The half-equivalence point occurs within this buffer region, where the concentration of the weak acid equals its conjugate base, creating an ideal buffer situation. At the equivalence point, the pH change is steepest; for a weak acid with a strong base, the pH will be greater than 7, while for a weak base with a strong acid, it will be less than 7. Beyond the equivalence point, excess strong titrant is present, leading to a further gradual change in pH until it plateaus. The strong component in the reaction ultimately determines the pH at the equivalence point, with strong acids driving it below 7 and strong bases above 7.

Acid-Base Titration Curve

In this video, we now take a look at a weak titrate being titrated by a strong titrate. We're going to say under this type of titration between a weak titrate and a strong titrant, a buffer region can exist. So if we take a look, we have a weak titrate, strong titrate curve. And here there are a few key features that we need to keep in mind, all right?

So our pure titrant is just again the beginning of the titration before any Titan has been added. So we're starting off here. The buffer region is just a region where the pH is most resistant to change. So what we need to realize here is that we're starting out with pure titrate and RPH is going to increase pretty quickly initially, but then it's going to start leveling off and gradually increasing. So in this region here that I've highlighted, that's why we have the slow gradual grind up in pH. That's because at that point our buffer has been created.

Now we'll see mathematically how this works out, but for right now, just realize that we start out with pure titrate in the beginning, but near the beginning there's a sharp increase in pH. And then there's a moment where a buffer is created, and because a buffer is created, there's a slowdown in the gradual increase of the pH. Now of course, eventually you add too much titrate and the buffer is destroyed, and then you start increasing your pH drastically again.

So here we have our buffer region. Here we're going to say our half equivalence point. This is just a midpoint within the buffer region where the amount of weak acid is equal to the amount of conjugate base. This is where we have an ideal buffer. So midway here around here, so now the equivalence point. This is just the middle region of the curve itself that has the steepest incline. Here this would be our equivalence point. So this is our equivalence point and this is our half equivalence point. They're related to each other.

So it takes about 50 mL to get to these equivalence point. So it will take about half of that 25 mL to get to the half equivalence point. After the equivalence point, which is this region here, this is where there is an excess of titrate still being added. So when it comes to weak titrate and strong titrate, these are the key features you need to keep in mind when looking at a typical curve.

Weak Titrate-Strong Titrant Curves Example

Instead of the titration of 55 mills of 0.120 molar hydrocyanic acid with 0.160 molar lithium hydroxide, I'll give you the volume needed to reach the half equivalence point. All right, so let's look at the following steps that we can utilize to help us find this answer.

So step one, we determine the volume of the titrant to reach the equivalence point. All right, so who's our titrate? We're not given the volume of looking hydroxide. That's because it's behaving as a titrant here. We'd say that M acid times V acid equals M base times V base. When it comes to asset based hydration, there are formulas that you can keep in mind. So our equivalent volume can be calculated using this equation. Acid times V acid equals M base times V base.

Now before the equivalence point, we have a buffer region that can exist and because of that we utilize the Henderson Hasselbeck equation. Here I'm just showing the variant where we have PKA involved. pH=pKa+log(conjugate baseweak acid) and then our half equivalence point is where the amount of weak acid and conjugate base are equal to one another, meaning that this would be equal to 1 and so we drop out because log(1)=0. The Henderson Hasselbeck equation would simplify to just pH=pKa, and if we were to isolate Ka, it would be restructured as Ka=10-pH.

Now let's go back. We're using the equivalence volume formula so that we can figure out the volume of the titrin to reach the equivalence point. So here we're going to say we have 0.120 molar of our acid, 55 MLS of our acid. Our base is 0.160 molar and we don't know the volume of our base. Divide both sides by 0.160 molar so cancels cancels volume of our base. To get to the equivalence point would be 41.25 milliliters.

Step 2 Utilize the correct formula based on the region of the titration curve. Utilize this step only if necessary. All right, so we just determined the volume to get to the equivalence point. Well, the half equivalence point would just mean we need half of that volume. So all we do here is we divide this by two and that will give you my half equivalence volume. What y'all just put as B sub half. So dividing 41.25 MLS by two gives me 20.625 milliliters. So this would represent the volume to get to the half equivalence point for this particular example question.

Consider the titration of 30.0 mL of 0.100 M HC₂H₃O₂ with 0.100 M NaNH₂. Which volume of NaNH₂ would take place within the buffer region? 

a) 30.0 mL                    b) 50.0 mL                    c) 10.0 mL                     d) 100.0 mL                   e) 1.5 L

• A. 30.0 mL
• B. 50.0 mL
• C. 10.0 mL
• D. 100.0 mL
• E. 1.5 L

Weak Titrate-Strong Titrant Curve

In these two types of titration curves, the pH changes gradually before the equivalence point, because we're going to have a buffer region that can be formed. So here we're going to take a look at the left side. In the following video we'll take a look at the other side, but let's focus on this left side first.

Here we have a weak acid strong base titration curve. We're starting off with a pH less than 7. So because the pH is less than 7 here, it would have to be the weak acid because if it was a strong base let's it would be above 7. Here we're slowly adding our strong base to it. So our strong base is the tight trend. Because we're starting out with an acid, we see that the pH is less than 7 and it increases gradually with added base.

That's because initially there's a jump in pH, but we get to a point where a buffer has been created and so there's a slow grind up in terms of our pH. This highlighted part is where our buffer region has been created. Now here, because we have something weak versus something strong, the strong species will dominate. Therefore the pH of the equivalence point would be greater than 7. So here we can see the pH is around 8:00 or so.

Now, after the equivalence point. So we're talking about. After this red dot, we're going to say the weak acid is neutralized. An excess strong base remains. We just keep adding more and more strong base even after all that weak acid has been neutralized. So This is why our pH continues to grow and and increase over time. But eventually it's going to plateau because it can't go up forever, right?

So when it comes to this weak asset, strong base titration curve, these are the key features to keep in mind when looking at a typical curve. When we have a weak acid versus a strong base, whatever, a strong will dominate since it's a base that's the strong species. It wants the pH to be above 7 at the equivalence point, right. So that's what we see, our equivalence pH greater than 7. It's around 8:00 or so here, right? So now that we've taken a look at this titration curve, click on the next video and let's look at the opposite trend.

Buffer Region

Now let's take a look at the opposite print. Here, our weak base will be our titrate. We're starting at a pH above 7, and we're adding strong acid to it. Strong acid will be our titrant. Now because we're starting out with a base, we'd expect our pH to be greater than 7 initially and it decreases gradually with acid. That's because initially there's going to be a quick drop in pH, but then it's going to kind of level off because in this highlighted region, a buffer region has been created.

So this is going to resist additional drop, steady clear drop in pH. It's going to gradually slow down. It's still going to be decreasing, just not at the same fast rate. Now here, what's the pH at the equivalence point? Remember, we have a weak species and a strong species. The strong species will dictate where the pH will be at the equivalence point. Because it's a strong acid, it wants an acidic pH, so it wants the pH to be less than seven at the equivalence point. Here we can see that the equivalence point is around a pH of five.

Now after the equivalence point, so after the red dot, we can see that the weak base is neutralized, but we still keep adding strong acid. So there's going to be some excess strong acid remaining. Eventually it's going to level off and plateau. And when again, when it comes to weak and strong, whatever is strong will dictate what kind of pH we will have at the equivalence point. In the first one, since we had a strong base, the pH was above 7 at the equivalence point. In this one, the acid is what's strong. So we're going to have a pH less than seven at the equivalence point.

So keep this in mind when looking at these two types of titration curves.

Weak Titrate-Strong Titrant Curves Example

At the half equivalence point, the concentration of weak acid is equal to the concentration of conjugate base. Consider the titration of 100 MLS of 200 molar of a weak monoprotic acid, meaning it only has one acidic H plus with 50M LS of .200 molar sodium hydroxide. Determine RK value of the weak acid if the pH is 4.18.

Now here they're giving us amounts of our weak monoprotic acid and our strong base titrate. But that's not the key to the question. I tell us that we're talking about the path equivalence point. So we would know that these amounts would somehow get us to this half equivalence point. Here we're going to say if we're at the half equivalence point, pH equals PKA.

And remember I talked about earlier how we can rearrange this formula to help us isolate ka. How do we do this? Well, we're going to say that PKA really needs negative log of ka. So if we take, if we divide both sides by -1 by both sides by -1, that's going to be negative pH equals log of ka and then take the inverse log of both sides. So when we take the inverse log of both sides, it's going to be

10 - pH =

When I take the inverse of log of the right side, log just cancels out or disappears so it becomes this formula.

K A = 10 - pH

Again. This happens when worth 1/2 equivalence point. That's because weak acid and conjugate base are equal to each other in amount. So all we do is we do

10 - 4.18

So when we do

10 - 4.18

that's going to give me

6.61 * 10 - 5

as the KA for this weak monoprotic acid. So this would be our final answer.

In titration of NH₃ and HCl, what is the predominant species beyond the equivalence point?

a) NH₃                                      b) HCl                                       c) CH₃NH­₂                                d) Cl^–

• A. NH₃
• B. HCl
• C. CH₃NH₂
• D. Cl^–

Here’s what students ask on this topic:

Which point on the titration curve is the point where the pH equals the pKa for BH⁺?

The point on a titration curve where the pH equals the pKa for BH⁺ (a conjugate acid of a weak base B) is known as the half-equivalence point or the midpoint of the buffering region. This is the stage in the titration where exactly half of the BH⁺ has been deprotonated to form B. At this point, the concentrations of the weak base (B) and its conjugate acid (BH⁺) are equal. Since the Henderson-Hasselbalch equation states that

pH = pKa + log(^[B]/_[BH⁺_])

when [B] = [BH⁺], the log term becomes log(1), which is 0. Therefore, at the half-equivalence point, pH = pKa. This is a key concept in understanding buffer systems and acid-base titrations.

What species is/are present in the solution at point 2, given the following titration curve for a weak acid/strong base titration?

At point 2 of a titration curve for a weak acid with a strong base, you are likely observing the part of the curve where the weak acid has started to react with the strong base but has not yet reached the equivalence point. At this stage, the species present in the solution are primarily the unreacted weak acid (HA) and its conjugate base (A^-), which is formed as the acid donates protons (H⁺) to the hydroxide ions (OH^-) from the strong base.

The chemical reaction occurring is:

HA + OH^- → A^- + H₂O

This means that in the solution, you have a mixture of the original weak acid and the conjugate base that it forms upon reacting with the base. This combination of HA and A^- can establish a buffer system, which resists changes in pH. The exact ratio of HA to A^- depends on how much base has been added at this point and can be calculated using the Henderson-Hasselbalch equation if the pKa of the acid and the pH of the solution are known.

A weak unknown monoprotic acid is titrated with a strong base. The titration curve is shown below. What is the Ka for the unknown acid?

To determine the Ka (acid dissociation constant) for a weak monoprotic acid from a titration curve with a strong base, you need to identify the half-equivalence point on the curve. At the half-equivalence point, the amount of acid is equal to the amount of its conjugate base, and the pH at this point is equal to the pKa of the acid.

1. Locate the half-equivalence point on the titration curve. This is the point where exactly half of the acid has been neutralized by the base. It is also the point where the curve begins to rise the most steeply.
2. Read the pH value at the half-equivalence point directly from the curve.
3. Convert the pKa to Ka using the formula 10^-pKa.

For example, if the pH at the half-equivalence point is 4.75, then the pKa is 4.75, and the Ka is 10