Bronsted-Lowry Acids and Bases

So we started with the simplest definitions of acids and bases on the radius model where the acid was just increasing the amount of H plus ions in solution and the base was increasing the amount of O. H minus concentration and solution. After that, we move on to the Bronston Laurie definitions of acids and bases. Now in 1923 Jonas Bronson and thomas Laury developed their new set of definitions for what constitutes an acid and a base. Now, according to them, acids are considered proton, which is the same thing as your hydrogen ion or hydro name ion donors and bases are considered to be proton. Except ear's. So unlike uranus acids and bases, they are not limited to aqueous environments. So we're expanding out from solvent of water and moving into other solvents, other situations in which this definition still holds true. Now here every Arrhenius acid is a Bronston Laurie acid. So think about it. Our example of hcl for Arrhenius acid. Here under the Arrhenius acid, we said it broke up into ions to give us a plus. Now the boston Laurie model, what's really going on is hcl is donating its H plus to water to give us a 30 plus, which is the same thing as H plus, that's why everyone is acid is a boston Laurie acid. Now every Iranians base is also a boston Laurie base if we had any O. H. That'll which minus which is the base since it's negative it could easily accept an H plus ion from hcl if it had to and because it can accept an H plus from hcl, it constitutes a boston Laurie base. Now our boston lori, acids and bases are just called our conjugate acid base pairs. When we're talking about a reaction. Hopefully this is just quick review for all of you. Um, this was a predominant portion of general chemistry or we talked about acid based reactions. So we're just getting the definitions out of the way before we head on to more advanced calculations. Now here remember Arrhenius was the simplest definition of an acid and base Bronston lori is just taking a few steps further into talking about more complex city that don't involve water as a solvent. We still have another form of acids and bases that still remain after this one. But right now we're focused on our assets being proton donors and our base is being proton except ear's now that we've got the fundamentals down. Try to do example one. Hopefully you guys remember these types of questions from jen cam in order to get the correct answer. For example, one if you don't quite remember, don't worry, just come back and we'll quickly go over this example

So here we have to write the formula of the conjugate base for the following compound. So here conjugate base means that this is acting as an acid. So if we want to write its conjugate base, that means we have to remove NH plus from the compound itself. Remember if they're asking for the conjugate base? Just removing H Plus from the compound and you'll have your conjugate base. So here removing an H. Plus from it gives us S. +03. Remember we're not only losing a hydrogen but we're losing a positive charge here, it's already negative one. Losing a positive charge means it becomes more negative. So becomes S. +032 minus. In actuality what's going on here is we have a chess. So three minus reacting with water, it is acting as the acid, water is acting as the base. So it's going to donate an H plus away and that's how it becomes S. +032 minus. In the process, water accepts an H. Plus to become a church +30. Plus here, sulfate ion would represent the conjugate base of hydrogen sulfate sulfate or by sulfate which is H. S. +03 minus now that you've seen that example. I think of it the opposite way in which we have to find the contra acid of the example given below. Um Once you do that, click on the next video and see if your answer matches up with mine

So here we have to write the conjugate acid for the compound given. So if we want its conjugate acid, that must mean that this is it acts as a base and accepts an H. Plus. So here by accepting an H. Plus it becomes H two P. 03. Here, you're not only accepting a hydrogen, but you're accepting a plus one charge as well, its charges already two minus. By accepting a plus one, it becomes more positive. So now it's H two P. +03. What's happening is that it's reacting with water? It is acting as the base now, so water is acting as the acid. Water is gonna donate an H. Plus to it. That's how it becomes a church to P. +03 minus. And in the process by losing an H. Plus, water becomes O. H minus. So we'd say here that this is called di hydrogen, phosphate. H two P 03 minus would represent the conjugate acid of our starting material of HP oh 32 minus. Now that we've seen those two examples, put it all together and classify each one of the following compounds under the words provided. Once you do that, come back and see how your answer matches up with mine

So this question is pretty straightforward in terms of recognizing Bronston Laurie, acid and Bronston lori base and its conjugate. So here we just simply have to identify the acid base, conjugate acid and conjugate bases in the following reaction. So here we have H. F. Here we have water in the process of donating and accepting an H. Plus, we produce fluoride ion and the hydro nia My on we can see that HF transitions into F minus. It lost an H plus. So it has to represent the acid. Water accepts an H plus to go from water to H +30 plus. So it must be the base. Remember whatever you are as a reactant, you're the opposite as a conjugate. If HF is the acid F minus would be its conjugate base. And if water is the base H30 plus is its conjugate acid. Now realize that H F and F minus, they're different by one hydrogen. So they would represent the conjugate acid base pair to one another. Water and H 30 plus are different by one hydrogen. One H plus. So they would represent another conjugate acid base pair. So the idea of Bronston Laurie is pretty simple and straightforward. This is important when it comes to calculating the ph of weak acids and weak based solutions because you have to think of the reactions in this format, you have your acid and bases as reactant and your conjugate as products. This will help later on with the formation of ice charts in order to calculate the concentration of H Plus or oh h minus two at the end, determine the ph or P o. H of solutions. Now that you've got that down, move on to the last classification of acids and bases.