We've taken a look at monoprotic acids, diprotic acids, and now we're taking a look at polyprotic acids. This means the inclusion of additional equivalence points and even more calculations involved with titrations. As always, we look for the equivalent volume of our titrant being used. Here, our titrant is a strong base. We're going to say here, molarity of my acid times volume of my acid equals molarity of my base times the equivalence volume of my strong base. Here, we'll plug in the values. When we divide both sides by the 0.100 molar of the strong base, we'll see that the first equivalence volume here will be 50 mLs. Since we're dealing with phosphoric acid, that means we have 3 equivalence points involved. So we're going to have 3 equivalence volumes needed. To get to the second equivalence point, we need an additional 50 mL. So that'd be 100 mL. To get to the final and third equivalence point, we would need another 50 mL. That would be a 150 mL. Those are the volumes needed for each equivalence point. Before any strong base has been added, we essentially just have a weak acid. Therefore, we can set up an ICE chart in order to determine our equilibrium expression. Since we're dealing with removing the first acidic hydrogen from phosphoric acid, that means we're dealing with Ka1. So the acid donates an H+ to water to produce H_{2}PO_{4}^{−}, which is dihydrogen phosphate, plus a hydronium ion. Remember, we'd have the initial concentration of our acid, but no initial concentrations of our products because they haven't yet formed. We're losing reactants in order to make products. Bringing down everything helps us to come up with our expression. Remember, we could do our 5% approximation method to help us determine, we have k1, k2, and Ka3. So Ka1 is 7.5 × 10^{−3}. K2 is 6.2 × 10^{−8}, and then we have 4.8 × 10^{−13}. If we use the initial concentration of 0.100 molar and divide it by the Ka that we're using in this example, which is 7.5 × 10^{−3}, we would not get a value greater than 500. Therefore, we have to keep the minus x in our expression and perform the quadratic formula. Using the quadratic formula, we'd find out that x, which equals our H^{+} concentration, was going to be equal to approximately 0.0239 molar. By taking the negative log of that concentration, we'd find out that our pH is approximately 1.62. At this point, we haven't even commenced titrations yet. We haven't added any strong base. This is our initial pH based on just the concentration of phosphoric acid. Once we start adding our strong base to this solution, we add NaOH and OH^{−} to our weak acid solution.

- 1. Chemical Measurements1h 50m
- 2. Tools of the Trade1h 17m
- 3. Experimental Error1h 52m
- 4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
- 6. Chemical Equilibrium3h 41m
- 7. Activity and the Systematic Treatment of Equilibrium1h 0m
- 8. Monoprotic Acid-Base Equilibria1h 53m
- 9. Polyprotic Acid-Base Equilibria2h 17m
- 10. Acid-Base Titrations2h 37m
- 11. EDTA Titrations1h 34m
- 12. Advanced Topics in Equilibrium1h 16m
- 13. Fundamentals of Electrochemistry2h 19m
- 14. Electrodes and Potentiometry41m
- 15. Redox Titrations1h 14m
- 16. Electroanalytical Techniques57m
- 17. Fundamentals of Spectrophotometry50m

# Polyprotic Titrations - Online Tutor, Practice Problems & Exam Prep

In titrations involving polyprotic acids like phosphoric acid, multiple equivalence points are crucial for understanding acid-base reactions. The first equivalence point occurs when equal moles of weak acid and strong base react, forming a conjugate base. The Henderson-Hasselbalch equation is used to calculate pH at various stages, reflecting the acid ionization constants (K_{a}) and concentrations of the acid and its conjugate base. As strong base is added, pH increases, demonstrating the transition from weak acid to its conjugate base, ultimately leading to a strong base excess beyond the second equivalence point.

These types of titrations revolve around the three or more Ka values of the polyprotic acid.

## Polyprotic Acid Titrations

### Polyprotic Acid Titrations

#### Video transcript

### Polyprotic Acid Titrations

#### Video transcript

Before we reach the first equivalence point between our weak acid and our strong base, we will have the formation of a buffer. At some point in our calculations, we're going to have to use the Henderson-Hasselbalch equation. Remember, to get to the first equivalence point, we need 50 milliliters of our titrant, NaOH. Here, we only have 30 mL. We would show the stoichiometric relationship between my weak acid and my strong base by writing out a balanced equation. Within this, when we're filling out our chart, we will be dealing with either millimoles or moles. Here, I decided to just divide the milliliters by 1,000 to get liters and multiply them by their molarities. So we'd have the initial moles of my weak acid and my strong base. Our conjugate base, we would have nothing initially, so it'd be 0. Now remember, we look at the reactant side, the smaller mole total would subtract from the larger mole total. At the end, we'd have nothing left of the strong base. We'd have some of our weak acid remaining. Through the law of conservation of mass, we would produce some conjugate base. We'd have weak acid, we'd have conjugate base, and therefore, we would have a buffer. Because of this, we can employ the Henderson-Hasselbalch equation. Now realize since we haven't reached the first equivalence point, we're basically talking about how Ka_{1} is attached to the removal of the first acidic hydrogen from phosphoric acid. So Henderson-Hasselbalch's equation will be:

That would just be negative log of our Ka_{1} which is 7.5 × 10^{-3} plus log of my conjugate base, 0.00300 moles, over my weak acid. Then when we plug that in, that would give me 2.30 as my pH up to this point. So realize that we've begun our titrations by adding some strong base, and we see that our pH has increased as a result. Click over to the next video and see what happens when we get to our first equivalence point.

### Polyprotic Acid Titrations

#### Video transcript

We've now reached the 1st equivalence point within our titration. At this point, we have equal moles of our weak acid and strong base. Because they're equal in terms of moles, they will completely destroy one another. At the end, we have 0 of each. But remember, we have the conservation of mass. We're going to form some conjugate base. At the end, we have some of it remaining. In terms of calculating pH at this point, we would discover what the formal concentration is of our conjugate base. We'd say formal concentration, which is f, equals the initial concentration of my acid which is 0.100 molar times the volume of the acid which is 50 mL divided by the volume of the solution. So 50 mL of the acid and 50 mL of the strong base. That would give me a concentration of 0.050 molar.

What we would then do is we'd plug in all the values that we need to figure out the concentration of H+ ions. Remember, this here is the conjugate base, but it represents the first intermediate form of my polyprotic acid. This equation, we should be familiar with because we used it when talking about calculations dealing with polyprotic acids. So Ka1, we said was 7.5×10-3, and Ka2, we said was 6.2×10-8. Those will be the values we plug in for K1 and K2 respectively. Kw is 1.0×10-14. When you plug all those values in, we'd find a concentration of H+ equal to 2.01×10-5 molar.

Since we know what the concentration of H+ is, we could find out what pH is, which is just simply the negative log of H+. That will give us 4.70 for the pH at the first equivalence point. Remember, we're dealing with a polyprotic species, so we still have equivalent point number 2 and number 3 to go through as well as all the titration points in between and after them. At this point, we've seen that adding strong base is showing us an increase in our pH over time. We'll continue onward with our discussion of adding even more strong base and seeing what else the pH can do as we're adding more of it.

### Polyprotic Acid Titrations

#### Video transcript

At this point, we've passed the first equivalence point. Now, we have the titration of 50 ml of 0.100 molar phosphoric acid with 70 ml of 0.100 molar NaOH. Remember, we needed 50 ml of NaOH to get to our first equivalence point. We are 20 milliliters beyond that point. Here, at the first equivalence point, we had only left dihydrogen phosphate and the moles of it that we had left were these moles here. We have an excess of 20 ml, so it'll be 20 ml times the concentration of NaOH. Converting that into moles gives us this value here. Hydrogen phosphate, which is our conjugate base, hasn't been formed at all. That's why initially it's 0 moles. As usual, we subtract the smaller moles from the larger moles. As a result of this, we'll have a portion of weak acid remaining and we have the generation of some conjugate base. Because we're talking about the removal of the second H_{+} from our polyprotic acid, we're dealing with K_{a2}. As a result of this, we're dealing with pK_{a2} within the Henderson Hasselbalch equation. By inputting the values, we have pH here equals negative log of K_{a2} which is 6.02×10-8 plus log of my conjugate base in moles over weak acid. That gives me a pH of 7.03. Now that we've seen this, we can take a look at what happens at the second equivalence point. Click over to the next video and see what happens once we get to the second equivalence point.

### Polyprotic Acid Titrations

#### Video transcript

At the second equivalence point, all of our dihydrogen phosphate has been totally neutralized by our NaOH. Both of these will be gone. It's okay though because remember, we have the conservation of mass. So whatever we lost here in the form of dihydrogen phosphate gets formed here in the form of hydrogen phosphate. This is the conjugate base that we're looking at. Here we'd find out its formal concentration. We'd use the initial concentration of my weak acid which is 0.100 molar, multiply it by the volume of the acid divided by the volume of the solution. So that's 50 ml plus 100 ml which is 150 ml's. So that would give me 0.0333 molar. Then by inputting values for k2, k3, the formal concentration as well as kw, we can find out the concentration of H^{+}. Remember, this formula we've seen before when dealing with polyprotic acids. This was the formula to find the concentration of my intermediate 2 from a polyprotic species. Here when you plug in the values that we need, we will get an H^{+} concentration of 2.20×10-10 molar. By having that, we can determine what our pH is. We just take the negative log of that value and that'll give me a pH of 9.66. We can see that our pH of course is increasing as we're adding more and more strong base.

Click over to the next video and see what happens after we've passed the second equivalence point.

### Polyprotic Acid Titrations

#### Video transcript

We've passed the second equivalence point. Realize here that we had dihydrogen phosphate reacting with NaOH to give us hydrogen phosphate. We are beyond the second equivalence point, which means we'll have an excess of NaOH. This has been totally destroyed, so it's no longer around. We would form some of this conjugate base here, but remember, strong species have a bigger impact on the overall pH than conjugates would. We'll focus on determining the concentration of our strong base.

Here, we'd say the concentration of our strong base is equal to the initial concentration of the strong base which is 0.100 molar times the volume of excess. So we needed 100 mL to get to the second equivalence point. We are 15 mL over that. So that is our excess divided by our total volume which is 50 mL plus 115 mL, which is 165 mL. This gives us a concentration of 0.009091 molar.

All we do now is determine what the pOH is by taking the negative log of that number. That would give me 2.04 as the pOH. pH would just be 14 minus that value. At this point, we'd have a pH of 11.96 once we've passed the second equivalence point. We'll quickly approach the third equivalence point and see what our pH would register at that point.

pH = 14 - pOH Concentration = initial concentration × volume of excess total volume pOH = - log concentration### Here’s what students ask on this topic:

What are polyprotic acids and how do they differ from monoprotic acids in titrations?

Polyprotic acids are acids that can donate more than one proton (H^{+}) per molecule, such as phosphoric acid (H_{3}PO_{4}). In titrations, they differ from monoprotic acids, which donate only one proton, by having multiple equivalence points. Each equivalence point corresponds to the donation of a proton. For example, phosphoric acid has three equivalence points, each representing the removal of one proton. This results in more complex calculations and the need to consider multiple acid dissociation constants (K_{a1}, K_{a2}, K_{a3}).

How do you calculate the pH at the first equivalence point in a polyprotic titration?

At the first equivalence point, the moles of the weak acid equal the moles of the strong base, forming the conjugate base. To calculate the pH, first determine the formal concentration of the conjugate base. Use the formula:

$F=\frac{C}{}$_{0} _{0}_{0} + _{t}

where C_{0} is the initial concentration of the acid, V_{0} is the volume of the acid, and V_{t} is the volume of the titrant. Then, use the equilibrium expression involving K_{a1} and K_{a2} to find [H^{+}]. Finally, calculate pH as:

$\mathrm{pH}=\; -\mathrm{log}([H+])$

What is the Henderson-Hasselbalch equation and how is it used in polyprotic titrations?

The Henderson-Hasselbalch equation is used to estimate the pH of a buffer solution. It is given by:

$\mathrm{pH}=\mathrm{pK}$_{a} + ^{-}]

where pK_{a} is the negative logarithm of the acid dissociation constant (K_{a}), [A^{-}] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. In polyprotic titrations, this equation is used at various stages to calculate the pH, particularly before reaching each equivalence point, where a buffer system is present.

How do you determine the pH after the second equivalence point in a polyprotic titration?

After the second equivalence point, the solution contains an excess of the strong base (e.g., NaOH). To determine the pH, first calculate the concentration of the excess strong base using:

$\mathrm{[OH-]}=\frac{C}{}$_{base} _{excess}_{total}

where C_{base} is the concentration of the strong base, V_{excess} is the volume of excess base, and V_{total} is the total volume of the solution. Then, find the pOH:

$\mathrm{pOH}=\; -\mathrm{log}([OH-])$

Finally, convert pOH to pH:

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

What are the key steps in performing a polyprotic titration involving phosphoric acid?

Key steps in performing a polyprotic titration involving phosphoric acid (H_{3}PO_{4}) include:

- Determine the initial pH of the weak acid solution using its initial concentration and K
_{a1}. - Add the strong base (e.g., NaOH) incrementally and calculate the pH at various points using the Henderson-Hasselbalch equation before reaching the first equivalence point.
- At the first equivalence point, calculate the pH using the formal concentration of the conjugate base and the equilibrium expression involving K
_{a1}and K_{a2}. - Continue adding the strong base and use the Henderson-Hasselbalch equation for the second buffer region involving K
_{a2}. - At the second equivalence point, calculate the pH using the formal concentration of the second conjugate base and the equilibrium expression involving K
_{a2}and K_{a3}. - After the second equivalence point, calculate the pH based on the excess strong base concentration.