So here it states, find the ph when 100 mls of 1000.1 Moeller die basic compound B was tight treated with 11 mls of a one molar hydrochloric acid solution. Alright, so here we're dealing with a dye basic compound instead of a byproduct compound. But the methods are basically the same. We need to determine what the equivalent volumes are. For our tie Trent are tight. Trent is our strong species. In this case it would be the hydrochloric acid. So we're going to say here that polarity of my acid times the equivalent volume of my strong acid equals more clarity of the base times the volume of the base. The polarity of Hcl is one molar. We don't know what it's equivalent volume is the polarity of my di basic compound is .1 molar and its volume is 100 mls. The Vibe both sides here by one Molar, which is not really gonna change anything. Okay, so then hear melodies cancel out. So the equivalent volume for the first one is 10 mls. And to get to the second equivalence point of my die basic compound, it would take double that. So it would be 20 mL. We can see here that we have 11 ml of hydrochloric acid. So we've gone beyond the first equivalence point but we haven't quite reached the second equivalence point yet. We're gonna say we're dealing with after the first equivalence point. So we have to think about what's happening here. Stark geometrically. So here we're gonna say that we have our diet basic compound B reacts with hcl. Originally it's gonna donate an H plus to the dye basic compound to give me B H positive plus cl minus. This dealt with accepting the first H plus from HcL. So this means we're dealing with KB one. And by association PKB one, then this compound could have reacted with a second mole of Hcl to accept another H. Positive. And accepting a second H positive means we're dealing with KB two and therefore K. B two. So again, we're past the first equivalence point. So that means that we've already passed this first equation and now we're dealing with this second equation. We're trying to accept a second H plus ion. So the equation is B H positive plus Hcl gives us B H +22 plus plus cl minus. So here we have initial change and final we ignore this spectator ion here. Now we have to determine what the initial moles would be here for my reactant, we're gonna say once we reached the first equivalence point, we would have had equal moles of these to react ints. And how do we determine what those moles would be? Well, we just divide here by 1000 multiplied by the polarity. So we have here. Or we could just keep it as millie moles. It doesn't really matter. We just multiply the milliliters times of polarity. So we'd have 10 million moles and 10 million moles at the first equivalence point. Remember they're both destroyed in the process. But we'd have this being created. So we'd have 10 million moles of this initially. Then remember for the strong titrate, the Titan species, we need the excess moles after the first equivalence point has been passed. So we needed 10 middle leaders of hcl to get to the first equivalence point. We have an excess of one millimeter left. So we'd say one million of one molar hcl. So when we multiply those two together, they give me one million more. We wouldn't have any of this created just yet. Now the smaller millie moles would subtract from the larger millie moles. So we have zero and then this would be 9.0 Here. This would increase by one million more. And we'd have at the end we'd have this weak acid and its conjugate base. So we'd have a buffer. Now realize here we're dealing with Again, two in the process. And therefore by extension, we're dealing with PKA on PKB two. Remember there is a relationship between acid constance and based constance here, we'd say that KB two is connected to K one and they equal KW and we would say here, if we took the negative log of both sides, that would give us a new equation. P K B two plus p K a one equals 14. Because remember when we take the negative log of KW which is 1.0 times 10 to the -14. That would give me 14 as a value. Alright, so I'm gonna use the P K. B. two value given to us which was eight. And use that to find p. K. one. So subtract both sides by eight. So we see that PK- one equals 6. So P H equals P K. A. One plus log of conjugate base over weak acid. Again, we have to convert P K. B to P K. A. Because we're using the Henderson Hasselbach equation because we're dealing with weak acid and conjugate base at the end. So that'd be 6.0 plus log of Are conjugate base would be nine million moles Divided by one million mole of our weak acid. So that would give me at the end 6.95 as my final ph Okay, so just remember although we're dealing with a dye basic compound, we can still apply some of the methods that we've learned in terms of di protic species, knowing that and using that guides us to the correct answer. So just remember the fundamentals we learned in terms of the in terms of finding out where exactly in the titillation process our question is directing us to are we dealing with calculations before the equivalence point? After the equivalence point? At the equivalence point. And remember there are two equivalent points because we're dealing with dye product or die basic species Now that you've seen this attempt to do the example question that's left at the bottom of the page.