18. Aqueous Equilibrium
Titrations: Diprotic & Polyprotic Buffers
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Now when it comes to polyprotic buffers, we typically are dealing with triprotic species. Because of this, we're going to say it deals with the presence of 3 equivalence points in Henderson Hasselbalch equations as a result of 3 Ka values. Here, we're gonna say the relationship between the equivalence points and Henderson Hasselbalch equations are shown as So first, we're gonna talk about the dissociation steps. Here, H three a is the generic form of a triprotic acid. So here it has all of its H+ ions in its possession, so this is the acidic form. Now here, we're talking about giving away the first H plus, so that'd be K a 1. Doing that gives us h two a minus. This would represent our intermediate form 1. That can continue to donate an h plus, so it donates a second h plus, so that'd be k a 2. That gives us h a 2 minus, which is our intermediate form 2. This one has an H plus, so it too can donate an H plus, which would deal with k a 3. Doing that gives us a 3 minus, which is the basic form. Now we could also go the opposite way, where the basic form gains its first h plus to create the intermediate form 2, Gaining your first h plus is k b 1. The intermediate form 2 can gain another h plus, so that'd be k b 2, to give us intermediate form 1. And then finally, the intermediate form 1 can gain that last H plus it needs to recreate the acidic form, so that'd be k b 3. We'd see here that these 2 are grouped together, these 2 are grouped together, these 2 are grouped together. So that segues into our KaKB equations. Here we'd say Ka 1 times k b 3 equal k w, ka2 times kv2 equals kw, and ka3 times kv1 equals kw. From there, we could talk about the different types of Henderson Hasselbalch equation. So the first one deals with the acidic form and the intermediate form 1. We're talking about the first Ka here, so this would be p k 1 plus now all of these are based on this simpler idea of Henderson Hasselbalch equation is pH equals pKa plus log of base over acid. The base is the one with one less hydrogen, the acid is the one that has one extra hydrogen. So here, it'd be plus log of base over acid, so that'd be h 2 a minus over h 3 a. The next one deals with the 2 intermediate forms, so they're connected by Ka 2, so this would be pKA 2, plus log of base over acid, so that'd be h a two minus over h two a minus. And then finally here, intermediate form 2 and basic form give us the last Henderson Hasselbalch equation. We're dealing with Ka 3 in this regard, so this would be pH equals p k a 3 plus a 3 minus over h a 2 minus. So these would be the 3 Henderson Hasselbalch equation associated with a typical triprotic buffer.
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