Diprotic Acids and Bases - Online Tutor, Practice Problems & Exam Prep

Diprotic acids, such as carbonic acid, can donate two hydrogen ions (H⁺), leading to two distinct dissociation constants, K_a1 and K_a2. K_a1 is always greater than K_a2, with K_a1 representing the loss of the first hydrogen ion and K_a2 the second. The intermediate form of the acid, after donating the first hydrogen ion, is an anion (e.g., HCO₃⁻), and after donating the second, it becomes a fully realized base (e.g., CO₃²⁻). The relationship between K_a and the corresponding K_b is such that K_a1 is paired with K_b2 and K_a2 with K_b1, with both products equaling the ion product constant for water (K_w). Equilibrium expressions for these reactions are determined by the ratio of the concentrations of the products over reactants. Understanding these concepts is crucial for analyzing the behavior of diprotic acids in various chemical reactions and their equilibrium states.

Ka values of Diprotic Acids

For diprotic acids. We're going to use the general formula of H2A here. This diprotic acid has 2H+ ions and as a result it can donate theoretically 2 acidic hydrogen's. This would give it K1 values.

Now in terms of K magnitude, we're going to say that K1 is always larger than KA2, and we're going to say here that KA1 deals with donating the first acidic hydrogen or acidic proton. Remember, acidic hydrogen or acidic proton is represented by H+. K2 deals with donating the 2nd acidic proton or hydrogen.

So understanding these basic ideas of diprotic acids will lead us into discussing how it relates to KA and KB now, and also how it relates to general equilibrium expressions of a typical diprotic acid.

3 forms of Diprotic Acids

Now the relationship between the K values and the respective K_B values are shown as. So in an association step we have the different steps of the diprotic acid donating its H ion to some unknown base and the first one. In the first version we have H₂A. This represents the fully realized diprotic acid. It has both of its acidic protons still attached to it. So since it's it's fully realized diprotic acid form, this would be the acidic form.

We know that K_A1 deals with donating the first acidic proton, so K₁ here is what allows us to go from H₂A to HA^-. It's donated its first acidic H⁺ ion to some unknown base here. This is the midpoint in terms of the diprotic acid in its journey of donating both H⁺ ions. Since it's the midpoint, we call this the intermediate form somewhere in the middle. Now K₂ deals with donating the second H ion. So donating the second H⁺ ion changes HA^- to A^2-. Here it's lost both of its acidic H⁺ ions. It's no longer an acid, it's a fully realized base. So we're going to say that this is the basic form.

Now we have to think about this in terms of K_A and K_B. We went from H₂A to A^2- talking about K_A. But what happens if we go the opposite way? Let's say we started with the basic form. We're going to say here, going from A^2- to HA^-. That means we're accepting our first H⁺ ion. Accepting your first H⁺ ion means that you're dealing with K_B1, and let's say we're going from HA^- to H₂A. We have to accept that second H⁺ ion to get there. So this would deal with K_B2.

Now you can see that these two and these two are on the same die arrows or dual arrows connecting them together. So because of this we're going to say our K_A-K_B equations are K_A1 is connected to K_B2 and together their product equals K_W. And over here we can say that K_A2 is connected to K_B1 and that equals K_W. So it's important because we have multiple acidic H ions. You have to remember which K_A lines up with which K_B.

Now here we have a good example of a typical diprotic acid in the form of carbonic acid. So here it's a weak acid so we'd have it reacting with water in its liquid form. It is the acid, so it donates an H⁺ to water. It would become the carbonate ion, which is HCO₃^-. Water would gain an H and become H₃O⁺. Here we're talking about donating the first H ion away so we're dealing with K_A1. The equilibrium expression would be products over reactants. So it would be the carbonate ion times the hydronium ion divided by the concentration of carbonic acid.

Now we've lost our first H ion. Now we have only one left, so we're here in the bicarbonate form. It still has an H⁺, so it technically could still donate it away to another water molecule. So let's say we have another water molecule involved. This is still an acid. It could donate that second H⁺ and become carbonate ions, CO₃^2-, and water gains an H⁺ and becomes H₃O⁺. We're talking about giving away the second and last H⁺ ion, so this would be K_A2. It also equals products over reactants, so it'd be carbonate ion times hydronium ion divided by bicarbonate ion.

So these two would represent the equilibrium expression for basically the diprotic acid form and the intermediate form in the form of bicarbonate, right. So keep in mind diprotic species are a little bit more complicated. We're dealing with multiple H⁺ ions, so you got to pay attention to what kind of K_A and K_B is involved.

Diprotic Acids and Bases Example

Carbonic acid, which is H₂CO₃, represents a weak diprotic acid with AK₁ and AKA₂. Here we need to determine the base association constant associated with the carbonate ion. All right, so the carbonate ion represents our diprotic species, where it has lost all of its H⁺ ions. So it represents the basic form.

Since we're dealing with the basic form, it would still react with water, and since it's a base, water would act as the acid. Water would donate an H to it to create the bicarbonate ion plus the hydroxide ion. We're talking about accepting our first H ion, and we're talking about accepting our first H ion. That means we're dealing with KB₁.

Now how does this relate to our different Kas? Recall that KA₂ is what's connected to KB₁, and their product is equal to kW. Now we're going to say we know what KA₂ is. We know what kW is. We can find out what KB₁ is. KB₁ equals kW divided by KA₂, so that's 1.0×10-14/5.6×10-11.

When you punch that into your calculator, you'll get back 1.8×10-4. So this will represent the value for KB₁ of the carbonate ion.

Determine the equilibrium expression for the K_a2 of hydrosulfuric acid, H₂S?

• A. K_a2 = [HS⁻][H₃O⁺]/[H₂S][H₂O]
• B. K_a2 = [S²⁻][H₃O⁺]/[HS⁻][H₂O]
• C. K_a2 = [HS⁻][H₃O⁺]/[H₂S]
• D. K_a2 = [S²⁻][H₃O⁺]/[HS⁻]

Here’s what students ask on this topic:

What is a diprotic acid?

A diprotic acid is a type of acid that has the capacity to donate two protons or hydrogen ions (H⁺) per molecule in an aqueous solution. This donation of protons occurs in a stepwise manner, with the first proton typically being released more easily than the second. The dissociation of a diprotic acid happens in two distinct stages, each characterized by its own dissociation constant (K_a). The first dissociation constant (K_a1) corresponds to the release of the first proton, resulting in the formation of a monoprotic anion. The second dissociation constant (K_a2) is for the release of the second proton, leading to the formation of a diprotic anion. Examples of diprotic acids include sulfuric acid (H₂SO₄), carbonic acid (H₂CO₃), and oxalic acid (C₂H₂O₄). Each of these acids undergoes two separate ionization reactions in water, reflecting their diprotic nature.

What is an example of a diprotic acid?

A diprotic acid is an acid that can donate two protons or hydrogen ions (H⁺) per molecule in an aqueous solution. An example of a diprotic acid is sulfuric acid (H₂SO₄). In the first stage of ionization, sulfuric acid donates a single proton to form the hydrogen sulfate ion (HSO₄^-), which makes it a monoprotic acid at this point. However, the hydrogen sulfate ion can further donate a second proton to form the sulfate ion (SO₄^2-), which demonstrates the diprotic nature of sulfuric acid. This two-step dissociation process is a characteristic of diprotic acids, where each proton is released in a separate reaction, often with different dissociation constants. The first dissociation constant (K_a1) is typically larger than the second (K_a2), indicating that the first proton is more easily lost than the second.