 Now, as with some of the previous videos that we've already prepared for this module, this is another area where you really need to be making sure that you address this section practically. So you will have the opportunity in class to carry out an experiment to calculate the enthalpy of neutralization. So this video is really just to give you a bit of background, a little bit of theory behind that, and also a theoretical value for you to have a look at when you do your final calculations. We've already looked at the fact that there are two main types of neutralization reactions that we've looked at, and in terms of ionic equations, the H plus ions combining with OH minus ions, both of these in solution, in aqueous solution, will form H2O. Now, this is obviously part of a wider equation or a general equation that relates to the combination of an acid and a base forming a salt and water. But we know for a lot of these reactions, the salt that's formed is a soluble salt, and therefore the ions remain in the solution. They don't actually come out. So we refer to them as spectator ions. We don't include them in our equation. Of course, we also have an example of a neutralization reaction between, say, an acid and a carbonate, both of which also are in solution, although the carbonate can be a solid. And in this case, we will produce H2O liquid plus CO2 gas. So these are the two equations that represent certainly the net ionic equations that represent types of neutralization. In this particular video, we're going to be concentrating on this one for the time being, but it would be quite useful in class for you to carry out an experiment with both types of neutralization and compare the values for the enthalpy that you calculate. So I guess the most important thing that we need to be aware of is that all neutralization reactions are exothermic. So if I rewrite that first equation as H plus aqueous plus OH minus aqueous forms H2O liquid, then I can now identify the change in enthalpy as a negative value. In fact, the value of that particular delta H value is minus 57 kilojoules per mile of water that is produced. So I could update my little calculation here or my little formula here to this one. Minus 57 kilojoules per mile. Now, this is the theoretical value that we want to sort of give you as the value of enthalpy of neutralization. It's important when you're carrying out experiments that you think about how you design those experiments, what sort of substances you're going to put together in order to create a neutralization reaction and how you're going to calculate that value. Of course, we do have an equation that allows us to do this, which is the Q equals minus MC delta T equation, which is the specific heat equation. And from this, we can calculate the delta H value.