Diprotic Acid Titrations - Online Tutor, Practice Problems & Exam Prep

In the titration of a diprotic acid, two equivalence points are crucial. The first equivalence volume can be calculated using the equation: $C^{1}V=C^{2}V$. The Henderson-Hasselbalch equation helps determine pH before reaching the equivalence point, while after, the pH increases significantly due to the formation of conjugate bases. Understanding these concepts is essential for mastering acid-base reactions and their calculations.

Diprotic Acid Titrations

With the titration of a diprotic acid, we now have to keep in mind the existence of 2 equivalence points as well as all the equations and calculations involved between those. If we take a look here, we’re going to say, as always, we should determine what the equivalence volume will be in terms of our titrant. Here, our diprotic acid will serve as the analyte and our strong base will serve as the titrant. Because we have the existence of 2 equivalence points, we’re going to have to find 2 equivalence volumes. We’re going to say first that the molarity of my acid times the volume of my acid equals the molarity of the base times the first equivalence point.

When we input the values, we have 0.100 molar times 100 mLs equals 0.050 molar times the first equivalence point. Divide both sides here by 0.050 molar. So our first equivalence volume would be 200 milliliters. That’s how much it takes to get to the first equivalence point. To get to the second equivalence point, we would just need another 200 milliliters.

We would need 400 milliliters in total to get to the second equivalence point. All right. Now, before we've added any strong base, we essentially just have a weak diprotic acid. With a weak diprotic acid, we have to employ the use of an ice chart. From that information, we can set up our equilibrium expression.

With this first point in our titration, we’re dealing with the removal of the first acidic hydrogen from our sulfuric acid solution which will be donated to water. We’ll create bisulfite as a product, as well as H₃O⁺. Because we’re talking about removing the first H⁺ ion, that means we’re dealing with K_a1. K_a1 for sulfuric acid is equal to 1.6 × 10^-2. We’d use that K value here within our equilibrium. Remember, we can always do the 5 percent approximation method to see if we can ignore the minus x within our equilibrium expression. You would take the initial concentration of your weak diprotic acid, which is 0.100 molar, and you divide it by the K that we’re using, which is 1.6 × 10^-2.

If you do that, you'll find that this ratio is not greater than 500. Therefore, you would not be able to ignore this minus x. We would keep the minus x, and as a result, have to use the quadratic formula to solve for x itself. Once you use the quadratic formula, you would find that x equals 0.032792 molar. Once you have that value for H⁺, you could just take the negative log, and from that, you can find the pH, which is 1.48.

Just remember, at this point, the titration technically hasn't begun yet. Once we start adding our strong base to our sulfuric acid solution, then our titration can commence. We’re going to have to be aware of where exactly within the titration are we dealing with calculations. Are we dealing with at the equivalence point, before the equivalence point, or after the equivalence point? Since there are 2 of them, it becomes even more complex.

Click on to the next video and see what happens when we start to add KOH to our sulfuric acid solution.

Diprotic Acid Titrations

Recall we need 200 milliliters to get to our first equivalence point. Here, we're dealing with calculations before the equivalence point. Here, we have the titration of 100 ml of 0.100 molar sulfuric acid with 75 milliliters of 0.050 molar potassium hydroxide. At this point, before we've gotten to our first equivalence point, we'll have the production of a buffer. Therefore, we'll have to rely on the Henderson-Hasselbalch equation to find our pH.

Remember, within this setup where we're dealing with our weak acid and our strong base, our units will be either millimoles or moles. Remember, moles is just liters times molarity, and millimoles is milliliters times molarity. Here, if we divide these milliliters by 1000 and then multiply by the molarities, we'll find the moles of each one. Here, from that, we input them into our chart as our initial values. Water, we could ignore.

We don't have any initial amount for bisulfite here, so it's 0 initially. Now, on the reactant side, the smaller moles will subtract from the larger moles. We subtract 0.00375 moles from both the reactants. Remember, the law of conservation of mass, whatever we lose as reactants, is really just getting changed or reformatted into products. We have the gaining of the same amount of moles as products.

At the end of this, we'll have left weak acid and conjugate base, which is why we have a buffer. Here, using the Henderson-Hasselbalch equation, we plug in pH=pKa 1+log conjugate baseweak acid because we're still dealing with the loss of the first acidic hydrogen from our diprotic acid. So that'd be pH=−log (1.6×10−2)+log (0.00375 moles of bisulfite ion0.00625 moles of sulfuric acid). Doing that will give me a pH of 1.57.

Realize that as we're adding more and more potassium hydroxide, we should see an increase in our pH. The jump in pH is not very large here because we have the production of a buffer. Buffers resist large changes in pH, which is why we've only seen a small uptick in our pH at this point. Once we get to the first equivalence point, the buffer will be destroyed, and we'll see a large increase in our pH. Click on the next video and see what happens in terms of the first equivalence point and the calculations involved.

Diprotic Acid Titrations

At the first equivalence point, we've now reached a portion of our titration where we'll have equal moles of our weak acid and strong base. Both of them, because they have the same amount of moles, will completely neutralize one another. In the end, both will be 0. But remember according to the law of conservation of mass, we don't actually lose the weak acid. It gets reformatted to form this conjugate base.

At the end, this is how many moles we have of our conjugate base. Because of that, we can use this information to determine our pH. We first need to figure out the formal concentration of our conjugate base form. That would just be the initial concentration of our weak acid, which is 0.100 molar, times the volume of the acid used, which is 100 mL, over the volume of the solution, which is 100 mL plus 200 mL. In the end, that gives me a formal concentration of 0.0333 molar.

With that formal concentration, we now can use the following equation to figure out the concentration of H⁺. Remember, this is an equation that we have seen before when dealing with the intermediate form of a diprotic acid. When we talked about diprotic acids, we discussed finding the concentration of the intermediate form. K1 is just the Ka1 for the intermediate form of the diprotic acid, which is 1.6×10-2. K2 for sulfuric acid is 6.4×10-8. Kw is just our standard 1.0×10-14. We plug all that information in to find the concentration of H⁺ which comes out to 2.63×10-5 molar.

Because you know the concentration of H⁺, you know what the pH is because all you'd have to do is take the negative log of that number and you'd have your pH. So here, we'd have 4.58 as our pH up to the first equivalence point. Notice that there is a jump in terms of the pH. That's because once we've reached our first equivalence point, the buffer, as we know it, has been destroyed so there's nothing preventing the jumping up of our pH. Remember, we’re dealing with a diprotic acid so we still have additional titration points within the whole process.

We still have to work our way up to the second equivalence point and then the calculations after that second equivalence point. We'll continue on with our journey in terms of the titration of this diprotic acid.

Diprotic Acid Titrations

At this point, we've passed our first equivalence point. Now, here we have the titration of 100 ml of 0.100 molar sulfuric acid with 250 ml of 0.050 molar potassium hydroxide. Remember, we only needed 200 ml of our strong base to get to the first equivalence point. We are 50 ml over that. So we've passed the first equivalence point.

When it comes to your titrant, you have to input the excess moles involved. We're 50 ml over our 200 ml needed to get to the first equivalence point. The excess volume is 50 ml and it's still of 0.050 molar KOH. Dividing this by 1,000 then multiplying it by the molarity gave us these excess moles here of our strong base. Now remember, that's the technique we use in terms of figuring out the excess moles of our titrant, the strong species.

Now when it comes to this compound here, remember at our first equivalence point, we had H₂SO₃ reacting with KOH to produce our conjugate base. Remember at the equivalence point, all of this had been destroyed and all of this had been destroyed. What we had left was just this. We had 0.0100 moles of it left. Those moles that we had left at the first equivalence point are the moles that we place here.

For the titrant, again, we look for the excess moles and for the species it's reacting with, we use the moles that we would find at the first equivalence point. Then we're going to say here that this represents our new weak acid and this is the new conjugate base. Here we have none of the conjugate base has been formed just yet. We're focusing on the reactant moles. The smaller moles will subtract from the larger moles.

That's why we're subtracting out 0.00250 moles from both of these. At the end, there are still some weak acid left. Then remember, we're going to form that same amount of moles here as products, so we're going to have some conjugate base left. We've gone past the first equivalence point so we're no longer talking about Ka₁. We're now dealing with removing the second and last H+ ion to produce this conjugate base here.

Because we're talking about removing the second H+ ion, we're now dealing with Ka₂. From this information, we can plug in what we know. We say pH here equals negative log of 6.4 × 10^-8 plus we're going to do the log of the conjugate base plus log of conjugate base, 0.00250 moles of the sulfide ion divided by 0.00750 moles of our bisulfite ion. Remember, we've gone past the first equivalence point and now we're dealing with our journey towards the second equivalence point.

That's why we're dealing with Ka₂ now. When we punch these in, we get 6.72 as our pH. Again, continuing onward from where we first started with the titration up to this point, you can see that the pH is continuously increasing. This is what's happening when we're adding a strong base to our weak diprotic solution. Now that we've seen what's happening after the first equivalence point, click on the next video to see what happens when we reach the second equivalence point.

Diprotic Acid Titrations

So we've reached our second equivalence point. Realize that we have the continuation of our reaction. So we have our bisulfide ion continuing to react with KOH to get to this second equivalence point. In this process, we have the production of our sulfide ion plus water, which we don't care about. So at the second equivalence point, we're going to have equal moles of both of these compounds.

In the process, they both get completely neutralized, so they're both gone. What we have remaining is our sulfite ion, which represents our weak conjugate base. Now realize we're talking about removing the second H⁺ to produce this final conjugate base. Since we're talking about the second H⁺, that means we're dealing with Ka₂.

Now, here we're going to find the formal concentration of our sulfite ion because it's what's left. So we have the initial concentration of our weak acid, which is 0.100 molar, times the volume of our acid, which is 100 mL, divided by the volume of our solution, which is 100 mL for the acid plus 400 mL for our strong base. Here when we do this, that's going to give me 0.020 molar as the formal concentration for my sulfide ion, my weak conjugate base. Now, that would go in here. Since we're dealing with \( \text{SO}_3^{2-} \), we can't use Ka₂.

We're going to have to convert that into Kb. We have to recall from earlier sections that the relationship between Ka₂ and Kb is that \( \text{Ka}_2 \times \text{Kb} = \text{k}_\text{w} \). Divide both sides here by Ka₂, and we'll have what Kb equals. We'd have \( 1.0 \times 10^{-14} \) divided by Ka₂ which is \( 6.4 \times 10^{-8} \). Here, that would give me a Kb of \( 1.56 \times 10^{-7} \).

That would be the Kb that we use. Again, as always, we would also do the 5 percent approximation method. We do the initial concentration divided by now our Kb, and see if that gives us a ratio greater than 500. Our initial concentration is the formal concentration we discovered, divided by the Kb that we just calculated. Here, you would see that that gives us a value greater than 500, which would mean that we can ignore this minus x here.

We'd plug all that in and solve for x. Once you find x, realize that that would represent OH^- concentration. Here, that x would equal \( 5.59 \times 10^{-5} \). When you take the negative log of it, you would find your pOH. That will come out to being 4.25.

Then if you subtract it from 14, you'd have your pH, which is 9.75 for our pH. As we've been seeing, the more strong base we add, the higher the pH will get. Again, we've reached another equivalence point, so there should be another jump in our pH. Remember, the largest increases in pH happen around the equivalence point. The largest jump really happens around the first equivalence point and then around the second equivalence point, there's still another jump.

It may not be as large as the first equivalence point. Now that we've seen what happens at the second equivalence point, click on to the final video to see what happens after we've passed the second equivalence point within our titration.

Diprotic Acid Titrations

We finally reached the point where we are beyond the second equivalence point. Here we have the titration of 100 ml of 0.100 molar sulfuric acid with 420 ml of 0.050 molar potassium hydroxide. Remember, to reach the second equivalence point, we need 400 milliliters of potassium hydroxide. Here we have 420 ml's. We have an excess of 20 milliliters of potassium hydroxide.

At this point, we have still the reaction of our bisulfite plus KOH to produce our sulfide ion. All of this has been destroyed because we've gone beyond the equivalence point, so we're going to have an excess of our strong base and, of course, we'll have some conjugate base as well. But the strong base has a greater impact on the overall pH, so we're going to focus on finding its excess. Here, to find the concentration of the excess base, we say it's equal to the initial concentration of the strong base which is 0.050 molar times the volume of excess base. We only needed 400 ml's to get to the second equivalence point.

We are 20 ml's above that so that's 20 ml's over the total volume which is 100 ml's plus 420 ml's. That gives me a concentration of 0.001923 molar. When you take the negative log of that, you'll get 2.72 as your pOH. Therefore, when you subtract it from 14, pH will give you 11.28.

That will represent our final pH up to this point after the second equivalence point. Remember, with monoprotic acids, we have to follow steps along the way to determine where in our titration our reaction is occurring. The same thing can be applied for this diprotic acid. You just have to determine, are you dealing with calculations, before, after, or at the first equivalence point or second equivalence point? That guides you in the method needed to find your pH.

So hopefully, you guys are able to follow along and make sure you go back if necessary to look at different portions of the titrations of a diprotic acid.

Diprotic Acid Titrations Calculations 1

Here it states, calculate the pH of 100 mL's of 0.25 molar carbonic acid when 70 mL's of 0.25 molar of sodium hydroxide are added. Here we're given the \( Ka_1 \) and \( Ka_2 \) of our diprotic acid. With any type of titrations, we need to first determine what the equivalence volume of our titrant will be. We're going to say our titrant is our strong base. We're going to say molarity of our acid times volume of our acid equals molarity of our base times the equivalence volume of our strong base.

Plug in the values that we have. And then here, we'll find out what our equivalence volume will be. So divide both sides now by 0.25 molar, we'll have our first equivalence volume is 100 mL's. We need a 100 mL's of our NaOH to get to the first equivalence point. Since we're dealing with the diprotic acid, we would need double that to get to the second equivalence point.

200 mL's would be needed to get to the second equivalence point. Here, we're only dealing with 70 mL's of NaOH. We haven't reached our first equivalence point yet. The calculations involved deal with getting before the first equivalence point. Remember, before the equivalence point is reached, we have to show the stoichiometric relationship between our diprotic acid and our strong base.

Here we have our diprotic acid plus NaOH helps to produce sodium bicarbonate as our conjugate base plus water. We have our initial, our change, and our final amount. When it comes to this stoichiometric relationship, our units can be either millimoles or moles. Here we ignore the water. We'll divide these mL's by 1,000 and multiply them by their molarity to get our moles.

Here we're going to have 0.0250 moles. Here we're going to have 0.0175 moles. And here, this is 0. Focus on the reactants. The smaller moles will subtract from the larger moles.

So subtract 0.0175, and then we're going to add 0.0175 here. Bring down everything. So, we'll have 0.0075 moles of my weak acid remaining. And we'll have 0.0175 moles of the conjugate base remaining. Again, before we reach our first equivalence point or second equivalence point, we have the production of a buffer.

Here since we're dealing with the first acidic hydrogen being removed to create this conjugate base, that means we're dealing with \( Ka_1 \). It'd be pH equals \( pKa_1 \) plus \( \log \) of my conjugate base over my weak acid. That'd be \( -\log(4.3 \times 10^{-7}) + \log \left(\frac{0.0175 \mathrm{ moles} \text{ of sodium bicarbonate}}{0.0075 \mathrm{ moles} \text{ of carbonic acid}}\right) \). When we punch all that in, we'll get a pH of 6.73. As always, in our titration, even if it's with a diprotic acid, you should always determine what the equivalence volume of our titrant is first, then see where within your titration our calculations will take us.

We know that it's occurring before the first equivalence point, so a buffer will be created. Therefore, we'll have to rely on the Henderson-Hasselbalch equation. Since we're dealing with the diprotic acid, we must be aware that are we dealing with the first acidic hydrogen coming off and therefore use \( Ka_1 \)? Or are we talking about removing the second hydrogen and therefore \( Ka_2 \) will be needed? As long as you keep these things in mind, you'll be able to find the pH for the solution.

Now that you've seen this example, move on to example 2 and see if you can find the pH for this next question. Once you do, come back and see if your answer matches up with mine.

Diprotic Acid Titrations Calculations 1

So here we have to calculate the pH of 75 ml of 0.10 molar phosphorous acid with 80 ml of 0.15 molar sodium hydroxide added. We're told that the Ka1 and Ka2 are 5.0×10-2 and 2.0×10-7 respectively. We have to calculate our equivalence volumes. We declare our titrant as the sodium hydroxide. The equation is macid∙Vacid=mbase∙Veq.

Plugging in what we know: 0.10 molar ∙ 75 ml = 0.15 molar ∙ Veq. Dividing both sides by 0.15 molar, we get our first equivalence volume as being 50 milliliters. To reach our second equivalence volume, we would need 100 milliliters. Here, we have 80 ml being used.

This means we have passed the first equivalence point and have not quite reached the second equivalence point yet. We have our phosphorous acid here, and we're considering the formation of H₂PO₃^- with our first Ka. However, now we're discussing this conjugate reacting with another mole of NaOH to produce HPO₃^2- + water.

We're dealing with Ka2. At the first equivalence point, the moles of acid would have been equal to the moles of the strong base, completely neutralizing each other. We would have had dihydrogen phosphite, H₂PO₃^-, left. Its moles would be equal to the moles of our initial diprotic acid, calculated by dividing by 1,000 and multiplying by the molarity, giving 0.0075 moles.

After having an excess of 30 ml of 0.15 molar NaOH, we find 0.0045 moles of excess NaOH (volume converted to liters and multiplied by molarity). Focusing on the reactants, the smaller moles subtract from the larger ones, resulting in 0.0030 moles of HPO₃^2- formed. Now dealing with the Henderson-Hasselbalch equation for the pH, since we're before the second equivalence point:

pH=-log(Ka2)+log(conjugate baseweak acid) gives a pH of 6.88.

Remember, the ability to calculate the equivalence volumes for our titrant is crucial in determining our exact location in the titration. Once we know our position, we can utilize either equations or methods accordingly to find the pH or pOH.

Diprotic Acid Titrations Calculations 2

Here it states, find the pH when 100 ml of 0.1 molar dibasic compound B was titrated with 11 ml of a 1 molar hydrochloric acid solution. Alright. So here we're dealing with a dibasic compound instead of a diprotic compound, but the methods are basically the same. We need to determine what the equivalence volumes are for our titrant. Our titrant is our strong species.

In this case, it would be the hydrochloric acid. We're going to say here that molarity of my acid times the equivalence volume of my strong acid equals molarity of the base times the volume of the base. The molarity of HCl is 1 molar. We don't know what its equivalence volume is. The molarity of my dibasic compound is 0.1 molar and its volume is 100 ml.

Divide both sides here by 1 molar, which is not really going to change anything. K. So then here, molarities cancel out. So the equivalence volume for the first one is 10 ml. To get to the second equivalence point of my dibasic compound, it would take double that.

It would be 20 milliliters. We can see here that we have 11 milliliters of hydrochloric acid. We've gone beyond the first equivalence point, but we haven't quite reached the second equivalence point yet. We're going to say we're dealing with after the first equivalence point. We have to think about what's happening here stoichiometrically.

Here, we're going to say that we have our dibasic compound B reacts with HCl originally. It's going to donate an H⁺ to the dibasic compound to give me BH⁺ plus Cl^-. This dealt with accepting the first H⁺ from HCl. So this means we're dealing with Kb 1. And by association, pKb 1.

Then this compound could have reacted with a second mole of HCl to accept another H⁺. And accepting a second H⁺ means we're dealing with Kb 2 and therefore pKb 2. Again, we're past the first equivalence point. That means that we've already passed this first equation and now we're dealing with this second equation. We're trying to accept a second H⁺ ion.

So the equation is BH⁺ + HCl gives us BH₂²⁺ + Cl^-. Here we have initial, change and final. We ignore the spectator ion here. Now we have to determine what the initial moles would be here for my reactants. We're going to say once we reached the first equivalence point, we would have had equal moles of these two reactants.

How do we determine what those moles would be? We just divide here by 1000 and multiply by the molarity. So we'd have here or we could just keep it as millimoles. It doesn't really matter. We just multiply the milliliters times the molarity.

So we'd have 10 millimoles and 10 millimoles at the first equivalence point. Remember, they're both destroyed in the process, but we'd have this being created. So we'd have 10 millimoles of this initially. Then remember, for the strong titrant species, we need the excess moles after the first equivalence point has been passed. We needed 10 milliliters of HCl to get to the first equivalence point.

We have an excess of 1 milliliter left. We'd say 1 milli of 1 molar HCl. When we multiply those two together, they give me 1 millimole. We wouldn't have any of this created just yet. Now the smaller millimoles would subtract from the larger millimoles.

So we'd have 0 and then this would be 9.0. Here, this would increase by 1 millimole. And we'd have at the end, we'd have this weak acid and its conjugate base, so we'd have a buffer. Now realize here, we're dealing with again Kb 2 in the process and therefore by extension, we're dealing with pKa on pKb 2. Remember, there is a relationship between acid constants and base constants.

Here, we'd say that Kb 2 is connected to Ka 1, and they equal Kw. We would say here, if we took the negative log of both sides, that would give us a new equation. Pkb 2 plus pka 1 equals 14. Because remember, when we take the negative log of Kw which is \(1.0 \times 10^{-14}\), that would give me 14 as a value. Alright.

I'm going to use the pkb 2 value given to us which was 8, and use that to find pka 1. Subtract both sides by 8, so we'd see that pKa 1 equals 6. So pH equals pKa 1 plus log of conjugate base over weak acid. Again, we had to convert pKb to pKa because we're using the Henderson Hasselbalch equation because we're dealing with weak acid and conjugate base at the end. So that'd be 6.0 + log of our conjugate base would be 9 millimoles divided by 1 millimole of our weak acid.

That would give me at the end 6.95 as my final pH. Just remember, although we're dealing with a dibasic compound, we can still apply some of the methods that we've learned in terms of diprotic species. Knowing that and using that guides us to the correct answer. So just remember the fundamentals we learned in terms of finding out where exactly in the titration process our question is directing us to. Are we dealing with calculations? Remember, there are two equivalence points because we're dealing with diprotic or dibasic species.

Now that you've seen this, attempt to do the example question that's left at the bottom of the page.

Diprotic Acid Titrations Calculations 2

Carbonic acid is a diprotic acid that dissociates, losing its 2 protons to create bicarbonate and carbonate ions according to the following reactions given below. Here, we have carbonic acid which breaks up into bicarbonate ion and hydronium ion. We know that in reality, it's really just carbonic acid donating an H⁺ to water to form H₃O⁺. But here, we just have a simplified view of it. The bicarbonate that forms can further dissociate to form our carbonate ion and another hydronium ion.

We say here as a diprotic acid system, it has 2 dissociation constants: pKa 1 and pKa 2 for the 2 steps. In the reaction, you titrate 50 ml solution of 0.50 molar carbonic acid with 1 molar solution of sodium hydroxide. Now here, it says, what would be the expected pH after the addition of 35 ml of the NaOH titrant? We know that the titrant is the strong species within our titration. Here, we're going to say, m_acid × v_acid = m_base × v_equivalence of our titrant.

So here we're going to plug in the numbers that we know. Molarity equals 1 molar. And we're looking for the equivalence volume. Divide both sides by 1 molar. So, here my equivalence volume equals 50 milliliters.

Then we'd say here actually, my equivalence volume 1 is equal to 25 milliliters and my equivalence volume 2 is equal to 50 milliliters. We need 25 ml to get to our first equivalence point and we're given 35 ml. We're dealing with calculations after the first equivalence point. That means that we're dealing with bicarbonate ion essentially helping to create carbonate ion. Here, we're going to have HCO₃^- + NaOH.

It's going to create carbonate ion + water. We don't care about the sodium. It's just going to act as a spectator ion. Again, remember, after the first equivalence point, we need the excess moles of our titrant.

So we needed only 25 ml, but we have 35 ml. So, we have an excess of 10 ml of NaOH multiplied by its molarity to find our millimoles. So here, that would give us 10 millimoles. Then here, remember the amount of bicarbonate that we have is equal to the moles, the total moles of our carbonic acid. So here, that would be 25 millimoles.

Here, we ignore water. We'd have 0 of this. The smaller millimoles would subtract from the larger millimoles. So we'd have 15 millimoles, 0, and then we'd have 10 millimoles here. So at the end, we'd have weak acid and some conjugate base.

We'd have a buffer. Remember, we're talking about removing the second acidic hydrogen to form carbonate ion. That means we're dealing with Ka 2 and therefore pKa 2. pH equals pKa 2 + log of conjugate base over weak acid. That's 10.30 + log of my conjugate base which is 10 millimoles divided by 15 millimoles of my weak acid.

Weak acid form. Here that would give me a pH equal to 10.12. Like all the examples we've seen thus far, we know that we have to determine where exactly within our titration our calculations are leading us. Here, we're dealing with calculations after the first equivalence point. Knowing that, we have to set up a stoichiometric relationship between the reacting compounds.

Right now we realize that if we're before the first equivalence point, then we're going to form a buffer. All we have to do is figure out the amount of conjugate base and weak acid remaining and from there determine the pH using the Henderson-Hasselbalch equation.

Here’s what students ask on this topic:

What are the two equivalence points in a diprotic acid titration?

In a diprotic acid titration, there are two equivalence points. The first equivalence point occurs when the first proton (H⁺) is completely neutralized by the titrant, resulting in the formation of the intermediate conjugate base. The second equivalence point is reached when the second proton is neutralized, leading to the formation of the final conjugate base. These points are crucial for calculating the titration curve and understanding the pH changes during the titration process.

How do you calculate the first equivalence volume in a diprotic acid titration?

The first equivalence volume in a diprotic acid titration can be calculated using the equation:

$C_{1}V=C_{2}V$

Here, $C_1$ and $C_2$ are the molarities of the acid and base, respectively, and $V$ represents the volumes. By rearranging the equation and solving for the volume of the titrant, you can determine the first equivalence volume.

How is the Henderson-Hasselbalch equation used in diprotic acid titrations?

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution before reaching the equivalence point in a diprotic acid titration. The equation is:

$\mathrm{pH}=\mathrm{pK}{a}_1+\frac{\log (\frac{\mathrm{[A}-]}{\mathrm{[HA]}}}{}$

Here, $\mathrm{pK}{a}_1$ is the negative logarithm of the first dissociation constant, $\mathrm{[A}-$ is the concentration of the conjugate base, and $\mathrm{[HA]}$ is the concentration of the weak acid. This equation helps in calculating the pH of the solution as the titration progresses.

What happens to the pH at the first equivalence point in a diprotic acid titration?

At the first equivalence point in a diprotic acid titration, the pH experiences a significant jump. This is because the weak acid and the strong base have completely neutralized each other, forming the intermediate conjugate base. The buffer system is destroyed, leading to a noticeable increase in pH. The pH at this point can be calculated using the concentration of the conjugate base formed and the relevant equilibrium expressions.

How do you calculate the pH after the second equivalence point in a diprotic acid titration?

After the second equivalence point in a diprotic acid titration, the pH is determined by the excess strong base. To calculate the pH, first find the concentration of the excess base using:

$\mathrm{Concentration}=\frac{\mathrm{Initial}\mathrm{Concentration}\mathrm{of}\mathrm{Base}\times \mathrm{Excess}\mathrm{Volume}}{\mathrm{Total}\mathrm{Volume}}$

Then, calculate the pOH by taking the negative logarithm of the hydroxide ion concentration and convert it to pH by subtracting from 14:

$\mathrm{pH}=14-\mathrm{pOH}$