Titrations: Diprotic & Polyprotic Buffers - Online Tutor, Practice Problems & Exam Prep

Diprotic and triprotic buffers are crucial in maintaining pH levels in solutions with multiple acidic hydrogens. Diprotic acids, represented as H₂A, can dissociate twice, each dissociation having a unique acid ionization constant (K_a1 and K_a2). The Henderson-Hasselbalch equation,

$\mathrm{pH}=\mathrm{pK}{a}_{\mathrm{}}+\log \left(\frac{\mathrm{base}}{\mathrm{acid}}\right)$

is used twice for diprotic buffers, once for each dissociation step. The conjugate base pairs (K_a1 with K_b2, and K_a2 with K_b1) relate through the water ion product (K_w). Triprotic buffers, like H₃A, have three dissociation steps and three corresponding Henderson-Hasselbalch equations, each representing the relationship between different forms of the acid (acidic, intermediate, and basic). The conjugate pairs (K_a1 with K_b3, K_a2 with K_b2, and K_a3 with K_b1) also relate to K_w. Understanding these relationships and equations is essential for calculating pH changes in polyprotic buffer systems.

Diprotic Buffers

Now when it comes to dichotic buffers, we're going to say that they deal with the presence of two equivalence points and two Henderson Hasselbeck equations as a result of 2K values. Remember, diprotic means you have two acidic hydrogens, so you can lose 2H pluses, each with its own unique K value. Here we're going to say the relationships between the equivalence point and equations are shown as.

So here we have our dissociation steps. First, remember a diprotic acid is in the common form of H₂A here. This is its form where it has both acidic hydrogens. So we're going to say that this is the acidic form. K_A1 deals with losing that first acidic hydrogen. When it does, it's now HA^-. This is called the intermediate form. Now it still has another acidic hydrogen, so if it decides to lose that, that means we're dealing with K₂ and we finally get to this last phase A^2- where it's lost both of acidic hydrogens. So it would be a basic form.

Now we could also think in the reverse. We could start out with the basic form and we can gain our first H to give us our intermediate form. Gaining that first H plus means we're dealing with AB₁. This intermediate form could then accept a second H plus to reform its acidic form. Accepting that second H plus would give us KB₂. Now we'd see that K₁ and KB₂ are paired and K₂ and KB₁ are paired. That takes us to the K_aKB equations. So here we say K_a1 times KB₂ gives us k_W, K_A2 times KB₁ gives us k_W.

Now the Henderson Hasselbeck equation. There's two of them because we have two acidic hydrogen. Now remember the general form, the regular form of our Henderson Hasselbeck equation is

pH = pK a 1 + log ( base acid )

So if we take a look at this first Henderson Hasselbeck equation, it connects together the acidic form and the intermediate form. So here we're talking about losing the first acidic hydrogen. So it would be pK_A1 plus log of base over acid here. Remember the base has one last H plus, so the base here would be the intermediate form and the acid would be the acidic form.

Now the second Henderson Hasselbeck equation deals with the intermediate form of the basic form. Here we're talking about losing the 2nd acidic hydrogen to create the basic form. So we're dealing with K₂ and therefore pK_A2 plus log of base over acid. Here again, the base is the one with one less H plus ion, so that will be the basic form, and the asset form here would actually be the intermediate. Remember, an intermediate can act as an asteroid base by donating an H or accepting an H. So these will be the two Henderson Hasselbeck equations associated with a typical diprotic buffer solution.

Titrations: Diprotic & Polyprotic Buffers Example

Here it says calculate the pH of 100 MLS of 250 zero molar carbonic acid with 120 MLS of 0.250 molar sodium hydroxide are added. Now here we're giving K1 and K2 of our carbonic acid solution. What we do first is we need to calculate the equivalence volume needed to by the strong species to reach the first equivalence point. This will tell us how much we need to get to the first equivalence point and all successive ones afterwards.

Now here we're going to do M acid times V acid equals M base times V base. Our strong species is the sodium hydroxide. So here we do of the acid, 250 molar times 100 MLS equals 0.250 molar and volume of the base. We're looking for that. We divide both sides by 0.250 molar and the volume of the base needed to reach the first equivalence point is 100 MPs. This will help us determine how much tightrin was actually reacting in terms of this titration.

So we're going to say volume of strong speech to reacting equals your actual volume. So the volume given to us in the question minus your equivalence volume. So if we did this, it'd be 120 MLS. -100 MLS. There'll be 20 amounts. Now why is that? Well, here, when we calculated the first equivalence volume, this tells me how much volume of NaOH is needed to get to the first equivalence point, which is 100. But in reality, we've gone beyond the first equivalence point because we have a volume greater than 100. We have 120 MLS, so we've gone beyond the first equivalence point.

The 2nd equivalence point would just be another 100 on top of that. So we need 200 MLS to get to the second equivalent point. We're not quite there. So here we're going to use steps one to three to set up the ICF chart. We're going to say before the first equivalence point, volume. You start with the acidic form and PK one after the first equivalence point volume, you're going to start with the intermediate form and PKA two. We are beyond the first equivalence point volume at 120 miles. So we're dealing with the intermediate form of carbonic acid, which is bicarbonate.

So HCL₃ minus here, it's still reacting with base. And over here we get our sodium bicarbonate. Actually we get our this would donate in H so we get water here and then we have NA₂CO₃ here. We don't really worry too much about balancing of this equation. We have this equation here given to us, the sodium's or spectator ions, so they don't really matter. We don't need to actually include them if we desire. We're going to say that this is an ICF chart since it's weak and strong mixing together. So we have initial change. Final remember with an ICF chart we use moles as the units.

So the moles of my intermediate will come from the moles of the original die protic species. So remember moles is liters times molarity. So divide the MLS by 1000 and multiply by the molarity, and we'll have the moles of our intermediate form, so .025 moles. Then we would take the reacting volume of our strong species, which is 20 ML divided by 1000, and then multiply that by the original molarity of the strong base. When we do that, we get 0.005 moles, our conjugate base. Here we don't have anything of it originally, so and then here water the liquid so we ignore it.

Now look at the reactant side. The smaller moles will subtract from the larger moles, so we'd have 0.020 moles left of my intermediate form, 0 left of my strong species. Now based on the law of conservation of mass, whatever we lose as reactants, we gain as products. So this would be plus .005 moles. Now what do we have left? At the end? We have the intermediate form and then we have the final basic form. And here just to simplify, we'll just keep it as CL₃2 minus. So what we fundamentally have here is we have basin acid remaining, so we have a buffer.

So Step 4 says the Henderson Hasselbeck equation is used for a buffer to find the pH of a solution. Using the final row, use the moles of your weak acid and its conjugate base to find the pH. Here we are after the first equivalence point, so we're using PK2. So we're going to say PK2 plus log of base over acid. So here our PKA two is negative log of 5.6 * 10^-11 wizard base, the one with less hydrogen. So that would be the carbonate ion, so 0.005 moles. And then who is our acid form? The one with one extra H plus, so that's our bicarbonate form. So that's 0.020 moles. Plug this into our calculator and we get 9.65 as the pH for this titration.

Polyprotic Buffers

Now when it comes to polyportic buffers, we typically are dealing with Tri protic species. Because of this, we're going to say it deals with the presence of three equivalence points and Henderson Hasselbeck equations as a result of three K_A values. Here we're going to say the relationship between the equivalence points and Henderson Hasselbeck equations are shown as.

So first we're going to talk about the dissociation steps here. H₃A is the generic form of a triprotic acid. So here it has all of its H⁺ ions in its possession. So this is the acidic form. So now here we're talking about giving away the first H, so that B K_A1. Doing that gives us H₂A^-. This would represent our intermediate form 1 that can continue to donate in H or donates a second H⁺ so that B K_A2. That gives us HA^2-, which is our intermediate form, 2. This one has an H⁺, so it too can donate an H, which will deal with K_A3. Doing that gives us A^3-, which is the basic form.

Now we could also go the opposite way, where the basic form gained its first H⁺ to create the intermediate form 2. Gaining your first H is K_B1. The intermediate form 2 can gain another H⁺, so that would be K_B2 to give us intermediate form one. And then finally the intermediate form one can gain that last H⁺ it needs to recreate the acidic form. So that'd be K_B3. We'd see here that these two are grouped together, these two are grouped together, these two are grouped together, so that segues into our K_AK_B equations.

Here we'd say K_A1 times K_B3 equals K_W, K_A2 times K_B2 equals K_W, and K_A3 times K_B1 equals K_W. From there, we could talk about the different types of Henderson Hasselbeck equation. So the first one deals with the acidic form and the intermediate form 1. We're talking about the first K_A here. So this would be pK_A1 plus. Now all of these are based on this simpler idea of Henderson Hasselbeck equation is pH equals pK_A plus log of base over acid. The base is the one with one less hydrogen. The acid is the one that has one extra hydrogen.

So here it D be plus log of base over assets, so that would be H₂A^- over H₃A. The next one deals with the two intermediate forms, so they're connected by K_A2. So this would be pK_A2 plus log of base over assets. So that would be HA^2-, / H₂A^-. And then finally here intermediate form 2 and basic form give us the last Henderson Hasselbeck equation. We're dealing with K_A3 in this regard. So this would be pH equals pK_A3 plus A^3-, / HA^2-. So these would be the three Henderson Hasselbeck equation associated with a typical tripodic buffer.

Titrations: Diprotic & Polyprotic Buffers Example

Here it says to calculate the pH of 30 mills of 0.10 molar of this particular triprotic acid when 20 MLS of 20 molar sodium hydroxide are active. Here we're given the 3K's of our triprotic acid. So what we start with in doing is we calculate the equivalence volume needed by the strong species to reach the first equivalence point. If we know that, then we can determine divine needed for each successive equivalence point.

So here we're going to say M acid times V acid equals M base times V base. Here our acid is 0.10 molar. Its volume IS30ML. Here we have .20 molar and we're looking for the volume of the base. Divide both sides by .20 molar, so volume of our base equals 15 MLS. So that would mean that the equivalence volume one would take 15 MLS. If we wanted to get to the second equivalence volume, we'd add another 15, so that would be 30 amounts. And if we're trying to get to the third equivalence point, it would take another 15 M LS, so we need 45 M LS.

All right, so we're at 20 M LS. So we've gone beyond the first equivalence point, but we haven't quite reached the second because it takes 30 to get to the second equivalence point. So here the volume of strong species reacting would be our actual volume given to us within the question minus your equivalence volume actual is 20 MLS. We only need 15 MLS to get to the first equivalence point, so we're at 5 MLS beyond the first equivalence point. Now here we set U steps one to three to use the ICF chart O we're going to say before the 1st equivalence point. We're going to say we start with the acidic form and PK one.

After the first equivalence point, volume, you start with intermediate one and PKA 2. This is where we are. And then we're going to say after the second equivalence point, volume. You start with the intermediate two form and K3. So here again, we're just after the first equivalence point. We haven't reached the 2nd equivalence point, all right, So we're going to use the intermediate form one and PKA two. So the second the first intermediate form would lose 1H plus, so it would be H₂C₃ or C₆H₅O₇^-. Let's go back up here to make sure. Yep, so that would be the first intermediate. And it's still reacting with an Oh. So the acid would donate an H plus to the base to give us water and then we get our second intermediate form being created.

So the H it's going to be C₆H₅O₇^2- equals here. I don't include the sodium, it's just a spectator ion. This is an ICF chart since we're mixing weak and strong, so we have initial change final. Now here the amount of my intermediate form is equal to the amount of my diprotic acid because that's where the dot the intermediate form originates from. And so we're going to do 101 thirty ML divided by 1000 to get liters and multiply by the molarity. That will give me the moles of the intermediate form. One comes up to .003 moles. Remember moles equals liter times molarity.

Here for the base we would do the reacting volume which is five ML. So you divide it by 1000 and then multiply it by the molarity of the strong base. So when we do that we get 0.001 moles. We don't have any of the second intermediate form, so that's zero. Water is a liquid sword. Couldn't ignore it. Now look at the reactant side. The smaller moles will subtract from the larger moles. OK. So we have that and then we're going to say here whatever we lose on the reactant side, we gain on the product side. So it's going to be plus 001, bring down everything. So this is zero and then this is here .001.

Now coming down here, we have to fill this out. So what we have at the end is we have the intermediate form, one which can serve as our weak asset form and here we have the intermediate form 2 which has one less hydrogen. So this can serve as our conjugate base form. So we're going to use the final rule, we're going to use the moles of the weak acid and it's conjugate base define pH. So again we're after the first equivalence points we're using PK2 and that be log of our base over our asset. So our base is the conjugate base, our acid is the weak acid form. Alright, so we're going to plug these numbers in here and when we do that, that's going to give us our new pH.

So then plus log of that's what we're going to do, 1.7 * 10^-5 here, this is .001, this is .002. So we're going to do that and that gives me 4.46 as my pH. So all we're doing here is we're adding what we got from the ICF chart into our Henderson Hasselbeck equation. Doing that helps us to isolate the pH of our solution.

Calculate the pH of 75.0 mL of a 0.10 M of phosphorous acid, H₃PO₃, when 80.0 mL of 0.15 M NaOH are added. K_a1 = 5.0 × 10⁻², K_a2 = 2.0 × 10⁻⁷.

• A. 7.12
• B. 1.35
• C. 6.88
• D. 12.65

Find the pH when 100.0 mL of a 0.1 M dibasic compound B (pK_b1 = 4.00; pK_b2 = 8.00) was titrated with 11 mL of a 1.00 M HCl.

• A. 8.00
• B. 6.95
• C. 3.00
• D. 7.05

Suppose you have 50.1 mL of a H₃PO₄ solution that you titrate with 15.4 mL of 0.10 M KOH solution to reach the endpoint. What is the concentration of H₃PO₄ of the original H₃PO₄ solution?

• A. 0.00200 M
• B. 0.00154 M
• C. 0.00649 M
• D. 0.0307 M

Here’s what students ask on this topic:

How does the titration curve of a diprotic acid differ from that of a monoprotic acid?

A titration curve represents the pH change of a solution as an acid or base is added. For a monoprotic acid, which has one proton to donate, the titration curve shows a single, distinct equivalence point where the amount of acid equals the amount of base added. This is typically characterized by a sharp increase in pH at the equivalence point.

In contrast, a diprotic acid, which can donate two protons, has two distinct stages in its titration curve, each corresponding to the deprotonation of one of the two acidic protons. This results in two equivalence points on the curve. The first equivalence point occurs when the first proton is completely neutralized, and the second when the second proton is neutralized. The pH will rise significantly at each equivalence point, but the increase is often less sharp for the second proton because it is less acidic than the first. Additionally, there is a buffer region between the two stages where the pH changes more gradually, corresponding to the presence of both the acid and its anion.

How many equivalence points would you expect to observe in a diprotic acid titration?

In a titration involving a diprotic acid, you would typically expect to observe two equivalence points. This is because a diprotic acid has two acidic protons that can be donated to a base. The first equivalence point occurs when the first proton is completely neutralized by the base, and the second equivalence point is reached when the second proton has been neutralized. Each equivalence point corresponds to the stoichiometric neutralization of one of the acidic protons, and they can often be identified by a sudden change in the pH of the solution, which is usually monitored with a pH meter or indicated by a color change if a pH indicator is used. The exact pH values at which these points occur depend on the particular diprotic acid being titrated and the strength of the base used.

How does diprotic acid affect titration?

Diprotic acids are unique in titration because they can donate two protons ($H^{+}$ ions) per molecule, unlike monoprotic acids which donate only one. This results in a titration curve with two distinct stages and equivalence points. Each equivalence point corresponds to the neutralization of one of the acidic protons.

During a titration with a strong base, the first equivalence point occurs when the first proton is completely neutralized. The titration curve shows a first pH jump here. As the titration continues, the second proton begins to dissociate and react with the added base. The second equivalence point is reached when the second proton is neutralized. This is typically at a higher pH than the first equivalence point because the second proton is usually less acidic (has a higher pKa) than the first.

The presence of two equivalence points allows for the analysis of each dissociation step separately, which can be useful in determining the acid's dissociation constants (Ka values) and the strength of each acidic proton.