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Ch.17 - Applications of Aqueous Equilibria

Chapter 17, Problem 152a

In qualitative analysis, Ca2+ and Ba2+ are seperated from Na+, K+, Mg2+ by adding aqueous (NH4)2CO3 to a solution that also contains aqueous NH3 (Figure 17.18). Assume that the concentrations after mixing are 0.080 M (NH4)2CO3 and 0.16 M NH3. (a) List all the Bronsted-Lowry acids and bases present initially, and identify the principal reaction.

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Hey everyone, we're told an Aquarius solution contains ammonium sulfide list, all the bronze said Lowry acid and base present initially and write the principal acid base reaction first. Let's go ahead and determine our bronze said Lowry acid. Now, as we've learned, our Bronson, Lowry acid is going to be our proton donor and in this case it's going to be our ammonium ion. Next let's go ahead and determine our bronze, said larry bass. And as we've learned, our bronze said Lowry base is going to be our proton except er and in this case it's going to be our sulfide ion. Now let's go ahead and determine our principal acid base reaction. In order to do so we need to look at our KBS and R. K. A's. So first let's go ahead and look at ammonia. When we look at the KB of Ammonia, we find that it is 1.7 times 10 to the -5. Now let's go ahead and determine our K. A. For our ammonium ion to determine our K. A. We're going to take our K. W and divide this by our KB. So in this case it's going to be one times 10 to the negative 14, divided by 1.7 times 10 to the negative five. Now this is going to get us to a K. A. Of 5.8 times 10 to the negative 10. Now let's go ahead and look at our hydro sulfide. When we look up the K. A of our hydra sulfide, we find that it is 1.3 times 10 to the -13. Now let's go ahead and look at our sulfide ion. We want to determine our KB by taking our KW and dividing this by our K. A. So again we're going to take one times 10 to the negative 14 and divide this by our K. A. Of 1.3 times 10 to the negative 13. This will get us to a K. b. 0.0769. So why is this important? While this is important? Because in acid based reactions, proton transfer will occur between the stronger acid and the stronger base. So when we compare these we can see that our ammonium ion is the stronger acid. While our sulfide ion is the stronger base. Now that we have this information. Let's go ahead and write out our principal acid based reaction. So taking our ammonium ion, we're going to react this with our sulfide ion. And when we do that we're going to end up with ammonia plus our hydro sulfide. And this is going to be our final answers. Now. I hope that made sense. And let us know if you have any questions
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