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

Triprotic acids, such as phosphoric acid, have the ability to donate three protons (H⁺ ions), resulting in three distinct acid dissociation constants (K_a values). The K_a values decrease with each proton lost, as it becomes increasingly difficult to lose subsequent protons. The first dissociation constant (K_a1) is associated with the loss of the first proton, K_a2 with the second, and K_a3 with the third. The basic form of a triprotic acid, after losing all three protons, can accept protons in reverse, leading to three corresponding base dissociation constants (K_b values). These K_b values are inversely related to the K_a values, with K_a1 related to K_b3, K_a2 to K_b2, and K_a3 to K_b1, each pair multiplying to equal the ion product constant for water (K_W). Understanding these relationships and the equilibrium expressions for each dissociation step is crucial in analyzing the behavior of triprotic acids in various chemical contexts.

Ka values of Triprotic Acids

For triprotic acid. We're going to use the general formula of H₃A. Because of this formula, we say that it can donate 3 acidic hydrogens, and as a result of these three acidic hydrogens we have three possible K_A values.

Now we're going to say here, in terms of K magnitude, K₁ is always larger than K₂, which is always larger than K₃. And that's because with each H⁺ lost, it becomes harder to lose that second or third H⁺ ion. The harder it is to lose the smaller K_A is.

Here we're going to say that K_A1 deals with donating the first acidic proton, K₂ deals with donating the 2nd acidic proton, and K₃ deals with donating the 3rd and final acidic proton for our triprotic acid, right? So keep this in mind when we're dealing with triprotic acids, we have to worry about what kind of K is involved. Are we talking about donating the first, second or third proton for a given species?

4 forms of Triprotic Acids

For a triprotic acid, we're going to see the relationships between the K values and their respective KB values are shown as all right. So when it comes to a triprotic acid, sometimes you'll hear the term polyprotic acid. That just means it has more than two. OK, but for this level of chemistry, triprotic is a sealing.

Now with a triprotic acid, we actually have four different forms that it can take as it slowly gives up an H⁺ at each step. In the first one we have H₃A. It has full possession of all of its acidic H ions, so this would represent the acidic form Here. Giving away the first H would create H₂A^-, Giving away that first H⁺ is tied to K_a1 here, this is somewhere in the middle, so we're going to call this the intermediate form 1.

H₂A^- can donate the second H⁺ and become HA₂^-, giving away the second H⁺ is connected to K_a2. This is somewhat in the middle, so this is the intermediate form 2. Then finally HA₂^- can donate away its final H⁺ ion to create A^3-. So this would give us K_a3. This final form has lost all of its acidic H⁺ ions. This is the basic form.

Now, we talked about the relationship between K_a and K_b. So how is K_b related to this? Well, let's say we're starting out with the basic form and we decided we're going this way. Accepting the first H ion would change A^3- into HA₂^-. Accepting the first H⁺ means we're connected to K_b1. This form here could accept an H and become this intermediate form 1. Here accepting the second H⁺ would be K_b2, and then finally H₂A^- could accept the final H to become the acidic form again. So accepting that third H⁺ would be connected to K_b3.

This line of thinking is what connects these sets of K and K_b values together. So K_a1 is connected to K_b3 and then multiplying together gives us K_w. You would say K_a2 times K_b2 is connected to K_w and then finally K_a3 times K_b1 gives us K_w. Here. If we take a look at equilibrium expressions, we have phosphoric acid, which is a pretty common type of triprotic acid here. So here we're going to say it's a weak acid, so it would react with water in its liquid form.

Since it's an acid, it gives away an H⁺ and it becomes H₂PO₄^- and water becomes H₃O⁺. We're talking about giving away the first K The first H⁺ we're dealing with K_a1. K_a1 is products over reactants. So this would be dihydrogen phosphate times the hydronium ion divided by phosphoric acid. We're continuing. We're now starting out with the intermediate form 1. It can react with the second water molecule donating its H⁺ to become HPO₄^2-, which is hydrogen phosphate and the hydronium ion.

We're talking about losing the second H⁺, so this is K_a2. Products over reactants will give me hydrogen phosphate times the hydronium ion. And then we'd say here we have dihydrogen phosphate on the bottom. And then finally we have our intermediate form, one that was created in the top in the equation above. It can react with yet another water molecule, giving away an H⁺ to create itself into phosphate ion and then hydronium ion again. We're talking about giving away the 3rd and final H⁺, So this would be K_a3.

Products over reactions would be phosphate times hydronium ion divided by hydrogen phosphate ion. So these would represent the equilibrium expressions based on the different stages of a triprotic acid donating an H⁺ away to various water molecules, right? So you can see that triprotic acid, there's a lot more involved. Make sure you're tracking correctly what kind of K_a or K_b is involved, depending on the question asked, right? So just remember, triprotic acids have 3 acidic hydrogens, which makes things a little bit more tricky.

Triprotic Acids and Bases Example

Provided association equation associated with the K₂ value for the tripodic acid of pyrophosphosphoric acid. Here we're going to say K₂ means we're dealing with losing its second H⁺ ion, meaning that we've already lost the first H⁺ ion.

If we've already lost the 1st, that means we're starting out now as H₃P₂O₇^-, 1. Here it would react with water. Since it's acting as an acid, it would donate an H⁺ to the water molecule. Doing so would change it into H₂P₂O₇^2- aqueous and then water.

Accepting an H would become H₃O⁺. This would represent the equation associated with the K₂ of this particular triprotic acid.

Determine the equilibrium expression for the K_a3 value of citric acid, H₃C₆H₅O₇?

• A. K_a3 = [H₂C₆H₅O₇⁻][H₃O⁺]/[H₃C₆H₅O₇]
• B. K_a3 = [HC₆H₅O₇²⁻][H₃O⁺]/[H₂C₆H₅O₇⁻]
• C. K_a3 = [C₆H₅O₇³⁻][H₃O⁺]/[HC₆H₅O₇²⁻]
• D. K_a3 = [C₆H₅O₇³⁻][H₃O⁺]/[HC₆H₅O₇²⁻][H₂O]

Here’s what students ask on this topic:

What is triprotic acid?

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Triprotic acid is a hypothetical term and does not refer to a specific chemical compound in standard scientific nomenclature. However, the term "triprotic" suggests an acid with three proton-donating groups, meaning it can donate three hydrogen ions (<sup>H+</sup>) when dissolved in water. Acids with multiple ionizable hydrogen atoms are known as polyprotic acids, and a triprotic acid would be a type of polyprotic acid.

In a broader sense, if an acid can donate three protons, it undergoes dissociation in stages, with each hydrogen ion being released sequentially. With each proton donation, the acid becomes a different species: first, the intact acid (<sub>H3</sub>A), then the dihydrogen ion (<sub>H2</sub>A<sup>-</sup>), followed by the hydrogen ion (HA<sup>2-</sup>), and finally the triply-deprotonated ion (A<sup>3-</sup>). Each of these species would exist in equilibrium in a solution, with the position of the equilibrium depending on the pH of the solution and the acid's dissociation constants (Ka values).

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