Polyprotic Acids and Bases: Study with Video Lessons, Practice Problems & Examples

In polyprotic acids, the removal of acidic hydrogens involves distinct acid ionization constants: K1, K2, and K3. The fully protonated form behaves like a monoprotic acid, while the fully deprotonated form acts as a base. Intermediate forms require specific equations to calculate [H⁺] concentrations, utilizing overlapping constants such as K_a and K_b.

Polyprotic Acids

So at this point, we've mastered diprotic acids and it's onwards towards polyprotic acids. Now because of the knowledge we've gained from looking at diprotic acids, we can now apply them to our understanding of polyprotic acids. Here it says, for polyprotic acids which we'll represent by H₃A, their equations can be illustrated by the examples given below. With H₃A, we're dealing with the fully acidic form of our polyprotic or in this case triprotic acid. Here it's going to give away its first acidic hydrogen to water, thereby creating H₃O⁺. Because we're talking about removing the first acidic hydrogen, we're dealing with K_a1.

Here it's equal to products over reactants. So it's H₂A^- times H₃O⁺ divided by H₃A^-. This compound here still has acidic hydrogens it can still donate so it continues onwards and gives another H⁺ to a second water molecule creating HA₂^2- with H₃O⁺. Again, we're dealing with K_a2 now because we're removing the second acidic hydrogen. This is HA₂^2- times H₃O⁺ divided by H₂A^-.

Then finally this species that's created can give away another H⁺ to a third water molecule to produce A₃^3- plus H₃O⁺. Our equation at the end will be A₃^3- times H₃O⁺ over H₂A^2-. As we said earlier with diprotic species, remember removing the first acidic hydrogen is the easiest. With each sequential hydrogen afterward, it becomes harder and harder to remove. This weak acid will make a small amount of these products because that's the easiest H⁺ to remove but then making these products will have very little being formed, and with the third phase, even less so. Less and less products are being formed because it becomes harder and harder to remove the next acidic H⁺. Just keep in mind some of the fundamentals when looking at a triprotic or a polyprotic acid form here.

Now click on to the next video and see how we tackle polyprotic bases.

Polyprotic Bases

We've looked at the polyprotic acid form. Now, let's look at the polyprotic base form. Here, we're dealing with a 3-. We have the potential of absorbing or accepting 3 H⁺ ions. Here, we're dealing with the basic form where it has no acidic hydrogens on it.

Here, it's going to just accept an H⁺ from water thereby creating HA_2-. By donating an H⁺ away, water itself becomes OH^-. Now, because we're talking about accepting the first acidic hydrogen, we're dealing with K_b1. So here would be 2 minus times OH^- divided by A_3-. Here, we're now going to deal with 2-. With 2-, water is now going to donate a second H⁺ over to it to make it into H₂A_- and OH^-. Here we have, H2AOH-HA2-.

Finally, this one that we created can react with a third water molecule to produce these new products times OH^- divided by H₂A_-. Realize here by examining the polyprotic acid form and the polyprotic base form, we've gone through a lot of different variations of our polyprotic species. We've gone through the fully protonated version in this acid form which could have lost an H⁺ to become H₂A_-. That could have continued to lose another H⁺ to become HA_2-. And then finally, we have A_3- after it's lost all of its H⁺.

This understanding will help us realize the connections between the different forms of these compounds and the different K_a and K_b values that are associated with them. Click on to the next video to look at the final portion when talking about polyprotic species.

Polyprotic Dissociation Constants

Now that we've looked at the polyprotic acid form as well as its base form, let's connect them all together. Now, here as we said that we have the fully protonated version. Here we have our intermediate one after it's lost its 1st acidic hydrogen. Then it's lost its 2nd to give us intermediate 2 and then we have our basic form here. Realize to remove the first acidic hydrogen means that we have Ka1 involved.

To remove the second is Ka2 and finally to remove the 3rd and last hydrogen is Ka3. Going the opposite way, adding the first acidic hydrogen would be Kb1. Adding the second would be Kb2 and adding the third would be Kb3. Just like diprotic species, we have overlapping of equilibrium constants.

We can see that Ka1 and Kb3 overlap so that's why Ka1⁢Kb3=kw. Then we have ka2⁢kb2=kw because they overlap here. And then finally, we have ka3⁢kb1=kw because they overlap finally here. Now when dealing with polyprotic acids, we're going to say here that when we have the fully protonated version, H3A, where it has all 3 of its acidic hydrogens, we're going to treat it as though we would a monoprotic acid and use Ka1. When we're dealing with A3− where it's lost all of its hydrogens and all it can do is accept the first one, we're going to treat it as though it's a base and use Kb1.

Then finally if we're dealing with intermediate 1 where it's lost its 1st acidic hydrogen, we use this equation here to determine the concentration of H+. Here it's ka1⁢ka2⁢ the initial concentration or formal concentration plus ka1⁢kw∕ ka1+ your formal or initial concentration. Recall that this is the same equation we used for the intermediate of a diprotic species. But now we're talking about it in terms of the first intermediate of a polyprotic or triprotic species. Finally, here we have our 2nd intermediate form.

Since we're talking about removing the second hydrogen to get to this form, we deal with Ka2⁢Ka3⁢ the formal or initial concentration of the species plus Ka2⁢kw∕ Ka2+ initial concentration or formal concentration. Remember, when we're looking at a polyprotic or a triprotic species, we have 4 possible outcomes. When we're dealing with the fully acid form, the fully basic form and then its 2 intermediates. The intermediate forms are easier because we just have to recall the formulas, plug in the values and find our concentrations of H+. With the acid form, we'd have to set up an ICE chart and use Ka1.

With the base form, we'd have to set up an ICE chart and use Kb1. Just recall all of these different avenues we can take in order to determine our pH or H+ concentration for each one of these compounds.

Polyprotic Acid Calculations

Here it says, calculate the equilibrium concentrations of phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate, and the hydronium ion for 0.35 molar of phosphoric acid. All right. We're dealing with a triprotic or polyprotic acid, highlighted by the fact that we have 3 Ka values given to us. In this triprotic or polyprotic acid case, we have all 3 acidic hydrogens. Therefore, we know that it's going to start off as an acid.

It will react with water. Since it's in its acidic form, we know that it's going to donate an H plus to water. By doing this, it becomes H₂PO₄^- and water becomes H₃O⁺. Here we're going to say initial change equilibrium. We know that we're going to ignore solids and liquids.

The initial concentration of this weak acid is 0.35 molar. These initially are 0. We're losing the reactant so that we can build up our products. Bring down everything. Since we're talking about removing the first acidic hydrogen or donating it to water, we know that we're dealing with Ka₁.

Here, that's equal to our products over our reactants times H₃O⁺ divided by our reactant. Now we plug in the values we know for each term. So Ka1 is \(7.2 \times 10^{-3}\). Both of these are x so that's \(x^2\) on top divided by \(0.35 - x\) on the bottom. We do the 5 percent approximation method here, so initial concentration of 0.35 divided by the Ka1 value we're using.

When we do that, you should see that you don't get a value greater than 500. Because of that, you'll have to keep the minus x here and perform the quadratic formula. We're just multiplying both sides by \(0.35x\). So distribute, distribute. When we do that, we're gonna get \(0.00252 - 7.2 \times 10^{-3}x = x^2\).

Our lead term is the \(x^2\) since it has the highest power, so everything has to be moved over to its side. When we do that, we're gonna get \(x^2 + 7.2 \times 10^{-3}x - 0.00252 = 0\). This represents my a, my b, and my c. So using the quadratic formula, we're gonna have \(-7.2 \times 10^{-3} \pm \sqrt{(7.2 \times 10^{-3})^2 - 4 \times 1 \times -0.00252} \div 2 \times 1\). Here, that's going to give us \(-7.2 \times 10^{-3} \pm 0.100657 \div 2\).

Now, remember, because it's plus or minus, that means we're going to get 2 different variables, answers for x. So x can equal 0.0467 molar. Or if you do the negative approach, x equals negative 0.0539 molar. Realize here that at equilibrium, you cannot have a negative value. It's not possible.

So that means that this negative x gets dropped out. So that x represents our correct answer. Now we're gonna say at equilibrium, H₃PO₄ equals \(0.35 - x\). So plug in the value we just found for x. So that equals 0.3033 molar.

At equilibrium, H₂PO₄^- equals H₃O⁺ because they both equal x. Their concentrations are 0.0467 molar. Alright. So far, we've found the concentration for our phosphoric acid, our dihydrogen phosphate as well as our hydronium ion. We still have to figure out for our other compounds.

We still have to figure out what our hydrogen phosphate as well as our phosphate ions will be. Now, we need to realize at this point, I've created this first intermediate. Because it still possesses acidic hydrogens, it can continue to donate H⁺ to other water molecules and that's what's going to happen. We're going to write a new equation. Now that dihydrogen phosphate will react with another water molecule.

It'll donate an H⁺ to that new water molecule and therefore become HPO₄^2- + H₃O⁺ being created. We have again initial change equilibrium. Again, we ignore water because it's a liquid. Initially, HPO₄^2- this is the first time we're seeing this so initially it's 0. At equilibrium in the previous ICE chart we found out that both your dihydrogen phosphate and your hydronium ion equaled x which equaled this number.

So they still equal that number and now we're creating a new ICE chart so their initial amounts are that number. Now we're going to say again we're still losing reactants here in order to make these products. Bring down everything. Now we're going to say here at this point since we're dealing with the second acidic hydrogen being removed to create our second intermediate, we're dealing with K₂ equals products over reactants. Coming back up here, let's look at what Ka2 is.

K₁ is \(7.2 \times 10^{-3}\). Ka₂ is \(6.3 \times 10^{-8}\). Look at the differences in the powers. Negative 3 to negative 8. Remember I've said before in the past that the first acidic hydrogen is the easiest and each hydrogen afterward becomes harder and harder to remove which is why we see a big decrease in our Ka values.

They're different by a magnitude of 5. What does that mean? That means that H₂PO₄^-, although it acts as an acid and donates H⁺ because its Ka₂ is so small, very very little product is formed which means it's gonna lose very little bit of x which means that this x is so small that it's not gonna really affect the final amount for H₂PO₄^- and therefore it can be ignored. Same thing can be said for H₃O⁺. This plus x, because x is such a small number, it's not gonna really increase the concentration of H₃O⁺ so it can be ignored.

This x we have to keep around because we're looking for the concentration of HPO₄^2- and that's all we have for there. We need that x there. Going back again, ka₂ is \(6.3 \times 10^{-8}\). Plug in what we know. Hydrogen phosphate equals x.

Hydronium ion is 0.0467. And then same thing with dihydrogen phosphate. It's also equal to 0.0467. What do we see here? We see that these two numbers are the same so they cancel out.

That means that at equilibrium, the concentration of our second intermediate hydrogen phosphate equals x which equals Ka₂. So it equals \(6.3 \times 10^{-8}\) molar. Alright. So we've found basically all of our species except for the last remaining one which is our phosphate ion. So this compound still has on it another H⁺.

So it could react with a third mole of water. So we're going to write yet another equation. So we have now HPO₄^2- It's going to react with our 3rd and final water because we don't have any more H plusses to donate after this. So it donates an H⁺ to become PO₄^3-. Water becomes H₃O⁺.

Since we're running out of room guys, let me take myself out so we can finish this last portion. Alright. So how do we work this out? We have initial, change, equilibrium. Water is a liquid so it's ignored.

This is the first time we're seeing the phosphate ion here. Right. So it's the first time we've seen it. So we know that initially it should be 0 because this is the first time it's appearing. Now, we found out what, our hydrogen phosphate is in the second ICE chart.

We found out it was equal to x so this number is what it's starting off with in this new ice chart. Then hydronium ion really hasn't changed because this is what we found out it was equal to in the first ice chart. And again, we said that x is such a small number. It doesn't really affect its, final amount. So it's still this number of 0.0467.

We're still losing reactants to try to make products. So we have that. Alright. Now we're gonna say we're dealing with Ka₃ which equals PO₄^3- times H₃O⁺ divided by HPO₄^2-. Now, this Ka₃ let's go up here and see what Ka₃ is.

So Ka₃ here is \(4.2 \times 10^{-13}\). It is even smaller than Ka₁ and Ka₂. Much smaller. Which means that very little product at this point is being formed. Which means that the minus x and plus x are even that much smaller and had even less of an impact on my final equilibrium amounts.

So what does that mean? So like in the 2nd ICE chart that means we can ignore this, minus x and we can ignore this plus x. Because they're so small, they're not gonna affect those equilibrium amounts at the end for those 2 compounds. So, we're just gonna plug in what we know. So we saw it was \(4.2 \times 10^{-13}\) for Ka₃.

This is equal to x. H₃O⁺ is 0.0467, and hydrogen phosphate is equal to \(6.3 \times 10^{-8}\). All we have to do at this point is solve for x and that'll represent the equilibrium amount of phosphate ion. Alright. So, multiply both sides by \(6.3 \times 10^{-8}\).

So that's gonna give me \(2.646 \times 10^{-20} = x \times 0.0467\). Divide both sides now by 0.0467. And we'll have our x variable at the end. So x here, which represents the concentration of phosphate ion because at equilibrium it equals x, equals \(5.67 \times 10^{-19}\) molar. And what you should realize is what ... The rest of the text is missing. If you can provide it, I'll continue the formatting and review as needed!

Polyprotic Acid Calculations 2

Here it says determine the pH of 0.250 molar sodium hydrogen phosphate. They tell us that phosphoric acid contains \(ka_1\), \(ka_2\) and \(ka_3\) with those respective values. Realize here that they're giving us information on phosphoric acid and we're only doing that because we're going to have to utilize one or more of those \(Ka\) values. Here we're dealing with sodium hydrogen phosphate. Remember, for phosphoric acid it's a triprotic or polyprotic acid.

So its acidic form is \(H_3PO_4\). When it loses that first acidic hydrogen, it becomes \(H_2PO_4^{-}\). When it loses that next one, it becomes \(HPO_4^{2-}\) and then finally becomes \(PO_4^{3-}\). In this particular question, we're dealing with sodium hydrogen phosphate which is \(HPO_4^{2-}\) that's involved. We're dealing with intermediate 2. Remember, with the intermediates, we don't have to use any types of ice charts or anything. All we have to recall is the formulas to help us determine the concentration of hydronium ion.

For this intermediate 2, we're saying hydronium ion concentration is equal to the square root of \(ka_2 \times ka_3 \times\) the formal or initial concentration plus \(ka_2 \times kw\), the dissociation constant of water, divided by \(ka_2 +\) formal concentration or initial concentration. All we have to do now is plug these numbers in and we'll find our hydronium ion concentration. So let's plug these values in: \(6.2 \times 10^{-8}\), \(4.2 \times 10^{-13}\), \(0.250\) molar plus \(6.2 \times 10^{-8}\), \(1.0 \times 10^{-14}\) divided by \(6.2 \times 10^{-8} + 0.250\) molar. When we work out everything within there, it's going to be the square root of \(2.852 \times 10^{-20}\).

When we take the square root of that, \(H^+\) equals \(1.69 \times 10^{-10}\), which means that pH is just simply the negative log of that concentration of \(H^+\). So negative log of \(1.69 \times 10^{-10}\), which gives me \(9.77\) as the pH for sodium hydrogen phosphate. Just remember, when looking at these polyprotic species, you have to identify what particular form are you dealing with. Are you dealing with the fully protonated acidic form? Are you dealing with the deprotonated basic form?

Are you dealing with one of the intermediate forms? Knowing which form you're dealing with helps you determine which path to take to figure out what your pH of the solution will be at the end. Now that you've seen this example, go on to example 2 and see if you can identify what form we're dealing with and as a result, what path to take in order to determine the pH.

Polyprotic Acid Calculations 1

Here it states, "Determine the pH of 0.150 molar citric acid." Here we're dealing with the fully protonated form of citric acid. It has all 3 of its acidic hydrogens. We know it has 3 acidic hydrogens because it possesses 3 Ka values. Because we're dealing with its first acidic form, that means we're going to deal with Ka1.

The acid itself will react with water because it is the acid, that means that water will be the base. So, it will donate an H⁺ to the water, becoming H₂C₆H₅O₇^-. Water gains that H⁺ to become H₃O⁺. We know we have initial change equilibrium. Remember, we ignore solids and liquids.

The initial concentration is 0.150 molar. These products initially haven't formed yet so they're 0. We lose reactants in order to make products. So we bring down everything. Now at this point, we'd say that our Ka1 equals products over reactants.

So x2/0.150-x. We can use our 5% approximation method to see if we can ignore that minus x, and therefore, ignore the quadratic formula. So we do initial concentration divided by the constant that we're using, which in this case is Ka1. As long as the ratio is greater than 500, we can ignore the minus x. Now if we took that initial concentration and divided it by our Ka value, we would get approximately 202.7 which is not greater than 500.

We have to keep the minus x here in our expression and do the quadratic formula. Since we're going to need room guys, let me take myself out of the image. Alright. So, we're going to plug in our Ka1 which is 7.4×10-4, x2/0.150-x. So we're going to multiply both sides by 0.150 minus x.

This gets distributed. So we'll have 1.11×10-4-7.4×10-4x=x2. Because this x here has the largest power, it's the lead term, so everything has to be moved over to its side. So add 7.4×10-4x to both sides. Subtract 1.11×10-4 from both sides.

Now we have x2+7.4×10-4x-1.11×10-4. Remember, your quadratic formula is -b±b2-4ac2a. So let's input those values that we know. So negative 7.4 times 10 to the negative 4 plus or minus 7.4 times 10 to the negative 4 squared, minus 4 times 1 times negative 1.11 times 10 to the negative 4 divided by 2 times 1.

Remember, it's plus or minus, so we're going to get 2 values. So x will equal negative 7.4 times 10 to the negative 4 plus so when I take the square root of everything within there, it gives me 0.021084, or x equals negative 7.4 times 10 to the negative 4 minus 0.021084, divided by 2. So my x will equal 0.010172 molar, or negative 0.010912 molar. Remember, these x values that we find represent these x values here in our ice chart. Remember, at equilibrium, you cannot have a negative concentration.

It's not possible. It's like having negative time. We cannot use this x. This is the x that we would use. That x is equal to H₃O⁺. pH, remember, is negative log of H₃O⁺ or H⁺. Plug in that concentration we just found.

That equals 1.99 as my pH. Remember, when it comes to polyprotic acids, it's always imperative to know which form you're dealing with. Knowing the form will dictate which path we take in order to find our pH.

Here’s what students ask on this topic:

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

Polyprotic acids are acids that can donate more than one proton (H⁺) per molecule. They have multiple acidic hydrogens, each with its own dissociation constant (K_a1, K_a2, K_a3, etc.). In contrast, monoprotic acids can donate only one proton per molecule and have a single dissociation constant (K_a). The dissociation of polyprotic acids occurs in steps, with each step having a distinct equilibrium constant. For example, H₃A dissociates to H₂A^-, then to HA^2-, and finally to A^3-, each step involving a different K_a value.

How do you calculate the pH of a solution containing a polyprotic acid?

To calculate the pH of a solution containing a polyprotic acid, you need to consider the stepwise dissociation of the acid. For the fully protonated form (e.g., H₃A), use the first dissociation constant (K_a1) and set up an ICE (Initial, Change, Equilibrium) chart to find [H⁺]. For intermediate forms, use specific equations that involve overlapping constants. For example, for the first intermediate (H₂A^-), use the equation involving K_a1 and K_a2. For the fully deprotonated form (A^3-), treat it as a base and use K_b1 to find [OH^-], then convert to pH.

What are the equilibrium constants K_a and K_b in the context of polyprotic acids?

In the context of polyprotic acids, K_a (acid dissociation constant) represents the equilibrium constant for the dissociation of each acidic hydrogen. For example, K_a1 is for the first dissociation, K_a2 for the second, and K_a3 for the third. K_b (base dissociation constant) represents the equilibrium constant for the acceptance of protons by the deprotonated forms. For instance, K_b1 is for the first proton acceptance by the fully deprotonated form (A^3-), K_b2 for the second, and K_b3 for the third. These constants are related by the equation K_a * K_b = K_w, where K_w is the ion-product constant of water.

How do you determine the concentration of H⁺ for intermediate forms of polyprotic acids?

To determine the concentration of H⁺ for intermediate forms of polyprotic acids, use specific equations that involve the dissociation constants and the initial concentration of the acid. For the first intermediate (H₂A^-), the equation is:

Ka1Ka2C + Ka1KwKa1 + C

where C is the initial concentration. For the second intermediate (HA^2-), the equation is:

Ka2Ka3C + Ka2KwKa2 + C

These equations help calculate [H⁺] for the intermediate forms.

What is the significance of overlapping equilibrium constants in polyprotic acids?

Overlapping equilibrium constants in polyprotic acids are significant because they simplify the calculations for certain forms of the acid. For example, K_a1 and K_b3 overlap, meaning K_a1 * K_b3 = K_w. This relationship helps in understanding the behavior of the acid and its conjugate base forms. When dealing with the fully protonated form (H₃A), you use K_a1 as if it were a monoprotic acid. For the fully deprotonated form (A^3-), you use K_b1 to treat it as a base. These overlaps allow for more straightforward calculations of pH and [H⁺] for different forms of the polyprotic acid.