 Now the final question that we want to look at is the pH of resultant solutions when you mix two solutions together. So we'll do exactly the same thing that we did before but this time we just need to be aware of whether or not there's a solution in excess or and or one that's a limiting agent. So in this case I've got lithium hydroxide and I'm reacting it with nitric acid. This is a nicer one because our ratios are a little easier. We have lithium nitrate as our salt in solution and we form water. And this equation is already balanced so our mole ratios are 1 to 1 to 1 to 1 so that's nice. Then what I'm going to do is I need to work out the number of moles of each of these. So let's put the values that we have in and see where we go from here. Now for the lithium first it's a 1 molar solution of lithium hydroxide and we have 20 ml so let me make that straight away 0.02 litres. For the nitric acid I'll just do the lighter blue and it is a 0.5 molar solution, 0.5 molar and we have 30 ml so it's 0.03 litres. Now as I did before I'm going to go up for each of these and I'm going to have a look at my total number of moles. So in the first case I've got my name on my CV so I'm going to multiply my concentration by volume and I will get 0.02 times 1 so that's not too difficult, 0.02 moles. But for the nitric acid I have 0.5 times 0.03 so this one is going to be 0.015 moles. So when this reaction occurs because my ratio is 1 to 1 I have a limiting agent. I have one of my species which cannot fully react with the other so this is limiting and this is in excess. What I want to do then is I want to look at the one that's in excess because that's going to be the critical one. So as a result of this I have an excess of lithium hydroxide and the number of moles I have is 0.02 minus 0.015 which is going to be 0.005 moles. Because I have 0.05 moles my final volume is equal to 20 plus 30 which is 50 ml because I've added both of these solutions together so now I have a final volume of 50 ml so therefore my final concentration is going to be number of moles over volume which is 0.005 divided by and 50 ml is going to be 0.05 litres and when I do that I'm going to end up with a final concentration of 0.01 moles per litre of my lithium hydroxide. Now the problem that I have to solve now is that I need the final pH and in order to get the final pH I'm going to have to actually do a two step calculation because I'm going to have to start with the POH. So again just in red here let's put the POH is equal to minus the log base 10 concentration of OH minus ions and because this is the value that's going in there I'm going to have minus the log base 10 running at a little space here but let's just do it anyway base 10 of 0.1 and that's going to be a POH equal to 1. Now the problem is that's not the pH but we do know that 14 minus the POH is equal to the pH and therefore the pH of this solution is going to be 13. Now this is a very basic solution so even though we've tried to neutralize these solutions we find that because one of these was in excess it remained in the solution afterwards and still ended up therefore with a very high pH solution. This has been a lot this big video lots of worked examples here for you and I'm sure you do plenty more in class good luck with them keep practicing and thanks for watching.