So here are example states, consider the tight rations of 50 mls of 500.150 moller of methyl amine. It has a K B a 4.4 times 10 to the negative four with 75 ml of 750.200 molar hydrochloric acid here. They tell us to calculate the ph so we have a weak base here since it's KB value is less than one and it's being tight traded by this hcl. So we're gonna write this down. So CH three NH two plus hcea the acid will protein eight or give over an H plus to the CH three NH two group to make it CH three. NH three positive what's left behind? A cl minus since we have an acid and a base high trading one another its initial change final. We don't care about this cl minus which is a neutral ionic salt. We're gonna remember if we're dealing with an I. C. F. We need the units and moles. So divide the mls by 1000. To get leaders Multiply them by their polarities, will give us the moles of both. The methyl amine and the hydrochloric acid. So those numbers would come out to be .0075 moles And .015 moles. We don't have anything initially of this compound. So it's zero. Now looking at the reactant line, we're gonna say the smaller moles will subtract from the larger moles. So at the end we'll have zero left of our weak base and we'll have some strong acid remaining. Now we're gonna say, over here to B plus that amount of moles at the end. What we have remaining is strong acid and we have weak acid. Now remember the strong acid is gonna have a bigger effect on the ph so we're gonna focus on that instead we need its concentration so we take the moles of it left divided by the total volume. So 0.50 liters of methyl amine plus 0.75 liters of our hydrochloric acid. That gives me 0.60 moller of hcl. And because I have the polarity of a strong acid, I can just take the negative log of that concentration and that will give me ph So PH equals negative log of the .060 concentration, Which comes out to 1.22. We can see here since we're left with some strong acid at the end. This must be a question dealing with after the equivalence point. Because remember after the equivalence point, we'll have some excess strong acid remaining. Now that we've seen this example, continue to the next question. And let's see if you can apply the concept that we've gone over in terms of a weak base strong acid filtration
