Triprotic Acids and Bases - Video Tutorials & Practice Problems
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concept
Ka values of Triprotic Acids
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For trip protic acid, we're gonna use the general formula of H three A because of this formula, we say that it can donate three acidic hydrogens. And as a result of these three acidic hydrogens, we have three possible K A values. Now, we're gonna say here in terms of K A magnitude K A one is always larger than K A two, which is always larger than K three. And that's because with each H plus lost, it becomes harder to lose that second or third H plus ion, the harder it is to lose, the smaller K A is here, we're going to say that K one deals with donating the first acidic proton K two deals with donating the second acidic proton and K three deals with donating the third and final acidic proton for our trip protic acid, right? So, keep this in mind when we're dealing with trip protic acids, we have to worry about what kind of K is involved are we talking about donating the 1st, 2nd or third proton for our given species?
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concept
4 forms of Triprotic Acids
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For atriotic acid, we're gonna see the relationships between the K values and their respective KB values are shown as, all right. So when it comes to a trip protic acid, sometimes you'll hear the term polypro acid, that just means it has more than two. OK. But for this level of chemistry, trip protic is the ceiling. Now with a tr protic acid, we actually have four different forms that it can take as it slowly gives up an H plus at each step. In the first one, we have H three A, it has full possession of all of its acidic H plus ions. So this would represent the acidic form here giving away the first H plus would create H to A minus giving away that first H plus is tied to K one here. This is somewhere in the middle. So we're gonna call this the intermediate form one H two A minus can donate the second H plus and become H A two minus. Giving away the second H plus is connected to K A two. This is somewhat in the middle. So this is the intermediate form too. Then finally H A two minus. Can donate, donate away its final H plus ion to create a three minus. So this would give us K A three. This final form has lost all of its acidic H plus ions. This is the basic form. Now, we talked about the relationship between K A and KB. So how is KB related to this? Well, let's say we're starting out with the basic form and we decide we're going this way, accepting the first H plus ion would change a three minus into H A two minus. Accepting the first H plus means we're connected to KB one. This form here could accept an H plus and become this intermediate form. One here, accepting the second H plus would be KB two. And then finally H two A minus could accept the final H plus to become the acidic form again. So accepting that third H plus would be connected to KB three. This line of thinking is what connects these sets of K A and KB values together. So K A one is connected to KB three and then multiplying together gives us KW. Here, we'd say K A two times KB two. It is connected to KW. And then finally K A three times KB one gives us KW. Here, if we take a look at equilibrium expressions, we have phosphoric acid, which is a pretty common type of trip protic acid here. So here we're gonna 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 plus and it becomes H two po four minus and water becomes H +30 plus. We're talking about giving away the first K, the first H plus we're dealing with K A one, K, one is products over reactants. So this would be di hydrogen phosphate times the hydro ion divided by phosphoric acid. We're continuing. We're now starting out with the intermediate form. One, it can react with a second water molecule donating its H plus to become HPO +42 minus which is hydrogen phosphate. And the hydro, we're talking about losing the second H plus. So this is K two products over reactants will give me hydrogen phosphate times the hydro ion. And then we'd say here we have di hydrogen 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 a yet another water molecule giving away an H plus to create itself into phosphate ion. And then hydro ion again, we're talking about giving away the third and final H plus. So this would be K A three products over reactants would be phosphate time hydro divided by hydrogen phosphate ion. So these would represent the equilibrium expressions based on the different stages of Atriotic acid donating an H plus away to various water molecules, right. So you can see that trip protic acids, there's a lot more involved, make sure you're tracking correctly. What kind of K A or KB is involved, depending on the question asked. Right. So, just remember, trip protic acids have three acidic hydrogens, which makes things a little bit more tricky.
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example
Triprotic Acids and Bases Example
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Provided the association equation associated with the K two value for the trip protic acid of pyrophosphate acid. Here, we're going to say K two means we're dealing with losing its second H plus ion, meaning that we've already lost the first H plus ion. If we've already lost the first, that means we're starting out now as H three P +207 minus one. Here, it would react with water. Since it's acting as an acid, it would donate an H plus to the water molecule. Doing so would change it into H two P 207 to minus aqueous and then water accepting an H plus would become H 30 plus. This would represent the equation associated with the K two of this particular trip protic acid.
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Problem
Problem
Determine the equilibrium expression for the Ka3 value of citric acid, H3C6H5O7?
A
Ka3 = [H2C6H5O7−][H3O+]/[H3C6H5O7]
B
Ka3 = [HC6H5O72−][H3O+]/[H2C6H5O7−]
C
Ka3 = [C6H5O73−][H3O+]/[HC6H5O72−]
D
Ka3 = [C6H5O73−][H3O+]/[HC6H5O72−][H2O]
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