The amino acid glycine 1H2N¬CH2¬COOH2 can participate in the foll...

Question

The amino acid glycine 1H2N¬CH2¬COOH2 can participate in the following equilibria in water:
H2N¬CH2¬COOH + H2OΔ H2N¬CH2¬COO- + H3O+ Ka = 4.3 * 10-3

H2N¬CH2¬COOH + H2OΔ+H3N¬CH2¬COOH + OH- Kb = 6.0 * 10-5

(a) Use the values of Ka and Kb to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: H2N¬CH2¬COOH Δ +H3N¬CH2¬COO-

Accepted Answer

Hello. Everyone in this video, we're determining for the equilibrium constant of the formation of ours order ion. So we know that alan is an amino acid that can both have equilibrium in water. So we can go ahead and first realize that the formation of his order ion is requiring a whole bunch of steps so we can go ahead and apply hess's law to this. What has law states is that regardless of how many steps are in the reaction, the total entropy change is the sum of all the changes. So we see here that we have the starting regions of Alan nine and we will get this equation here to be our ideal equation. Or they find no equation and we see here on the right side of our first reaction that we have in O. H. And of course that relates it with a K. B. Value. Same thing for the second reaction with H 30 plus and then we have our K. Value. All right, so let's go ahead and route the reactions. Then again we're starting with our ally nine. That is our H two N. C H C H three C. O. H. Reacting with our H 20. Which is our water. To go ahead and give us a charged a means that we have positive H three N. The rest of the molecules the same. We have C H C H three C. O. H. And now we spit out a hydride R. O. H minus again of course we relate that with our KB value. I won't put our numerical values just yet because we don't need it just yet. Then our second reaction we have our H. Two N. Again C H C. H. Three C. 00. H. Reacting with our water. But this time instead of pro nation we're gonna go ahead. D. Pro Nate. So we will get a H. Two N. C. H C. H three C. O minus. So we just took out this proton here and then our second product will be our H. 30. Plus. Again. We have our HBO plus that we relate this with a K. Value. The third reaction that's not included in our given is the reaction of our two um ion products. So that is just our H. 30. Plus reacting with R. O. H minus two. Give us two moles of H 20. Sarah waters and that constant is going to be one over K. W. And of course we have the net reaction of H two N. C. H three nope. I meant to put C. H. C. H. Three C. O. H. And then our product is of course are charged. I mean then ch CH three and our depot donated C. 00. Group. Alright so now we can go ahead and take the three concepts that we have and multiply them. So maybe we can do this in a different color on the right side. So if we multiply these let's just put um not the numerical values first we know that we have our K. A. Multiplied by R K B. Over R k w R k w R Over KW is because we have one over K W. So we have these two on the top. So now plugging in our numerical values, then We have 4.5 times 10 to the -3, Multiplied with 7.4 times 10 to the -5. And it's going to all be over our Kw, which is of course one times 10 to the negative 14. So I'm gonna put these values into my calculator once I do, so I can get the value of 3.33 times 10 to the positive seven. So then this value right here is going to be my final answer for this problem. That's going to be the equilibrium constant for the reaction.

Watch next