Diprotic Acids and Bases - Video Tutorials & Practice Problems
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1
concept
Ka values of Diprotic Acids
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59s
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For Dipro acids, we're gonna use the general formula of H two A. Here, this Dipro acid has two H plus ions. And as a result, it can donate theoretically to acidic hydrogens. This would give it two K A values. Now, in terms of K A magnitude, we're gonna say that K A one is always larger than K A two. And we're gonna say here that K A one deals with donating the first acidic hydrogen or acidic proton, remember acidic hydrogen or acidic proton is represented by H plus K A two deals with donating the second acidic proton or hydrogen. So, understanding these basic ideas of Dipro acids will lead us into discussing how it relates to K A and KB now and also how it relates to general equilibrium expressions of a typical Dipro acid.
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concept
3 forms of Diprotic Acids
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4m
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Now, the relationship between the K values and the respective KB values are shown as. So in an association step, we have the different steps of the Dipro acid donating its H plus ion to some unknown base. And the first one, in the first version, we have H two A, this represents the fully realized dicrotic acid. It has both of its acidic protons still attached to it. So since it's, it's fully realized di protic acid form, this would be the acidic form. We know that K A one deals with donating the first acidic proton. So K A one here is what allows us to go from H two A to H A minus. It's donated its first acidic H plus ion to some unknown base here. This is the midpoint in terms of the Dipro acid in its journey of donating both H plus ions. Since it's the midpoint, we call this the intermediate form somewhere in the middle. Now K two deals with donating the second H plus ion. So donating the second H plus ion changes H A minus to A two minus here. It's lost both of its acidic H plus ions. It's no longer an acid. It's a fully realized base. So we're gonna say that this is the basic four. Now, we have to think about this in terms of K A and KB, we went from H two A to A two minus talking about K A. But what happens if we go the opposite way? Let's say we started with the basic four. We're gonna say here going from H A two minus to H A minus. That means we're accepting our first H plus ion, accepting your first H plus ion means that you're dealing with KB one. And let's say we're going from H A minus to H two A, we have to accept that second H plus ion to get there. So this would deal with KB two. Now, you can see that these two and these two are on the same dye arrows or um dual arrows connecting them together. So because of this, we're gonna say our Kakb equations are K A one is connected to KB two and together their product equals KW. And over here, we can say that RK A two is connected to our KB one and that equals KW. So it's important because we have multiple acidic H plus ions. You have to remember which K A lines up with what, which KB. Now, here we have a good example of a typical dip protic 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 donated H plus to water. It would become the carbonate ion which is HCO three minus water would gain an H plus and become H 30 plus. Here, we're talking about donating the first H plus I in a way. So we're dealing with K A one, the equilibrium expression would be products over reactants. So it would be the carbonate ion times the hydro ion divided by the concentration of carbonic acid. Now, we've lost our first H plus ion. Now we have only one left. So we're here in the bicarbonate form. It still has an H plus. 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 plus and become carbonate ion co +32 minus and water gains in H plus and becomes H +30 plus. We're talking about giving away the second and last H plus ion. So this would be K A two, it also equals products overreacted. So it'd be carbonate ion times hydro ion divided by bicarbonate ion. So these two would represent the equilibrium expressions for the, basically the Dipro acid form and the intermediate form in the form of bicarbonate, right. So keep in mind, Diro species are a little bit more complicated. We're dealing with multiple H plus ions So you gotta pay attention to what kind of K A and KB is involved.
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example
Diprotic Acids and Bases Example
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1m
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Carbonic acid, which is H two co three represents a weak Dipro acid with a K one and A K A two. Here, we need to determine the based association constant associated with the carbonate ion, right? So the carbonate ion represents our Diro species where it has lost all of its H plus 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 plus to it to create the bicarbonate ion plus the hydroxide. I, we're talking about accepting our first H plus ion. If we're talking about accepting our first H plus ion, that means we're dealing with KB one. Now, how does this relate to our different Kas recall that K A two is what's connected to KB one and their product is equal to KW. Now we're going to say we know what K A two is, we know what KW is. We can find out what KB one is KB one equals KW divided by K A two. So that's 1.0 times 10 to the negative 14 divided by 5.6 times 10 to the negative 11. When you punch that into your calculator, you'll get back 1.8 times 10 to the minus four. So this will represent the value for KB one of the carbonate.
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Problem
Problem
Determine the equilibrium expression for the Ka2 of hydrosulfuric acid, H2S?
A
Ka2 = [HS−][H3O+]/[H2S][H2O]
B
Ka2 = [S2−][H3O+]/[HS−][H2O]
C
Ka2 = [HS−][H3O+]/[H2S]
D
Ka2 = [S2−][H3O+]/[HS−]
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