in these next set of videos, we're gonna take a look at week based strong ascetic rations. Now in a weak base, strong ascetic trey shin, we're going to say that our weak base will represent are an elite. So our beginning material and our strong acid will just be our tight Trent which we're adding to it. So we should expect as we're adding strong acid to our weak base, we should expect our ph to drop because we're adding strong acid to it. Now we're gonna say whenever you titrate a weak species, which in this case is a weak base and with a strong species which in this case is a strong acid. The fact that we're doing an asana based mixture means that this titrate a shin will require an I. C. F chart. Now when we say I C F I C F stands for initial change final. And whereas ice chart requires polarity as the units and I C. F chart requires moles as the units. Now here we're gonna follow a road map for the different points within a week based strong acid penetration. Now, before we start things off with this situation, um sometimes it's important to know what the equivalent volume is. We're gonna say, we calculate the equivalent volume which will abbreviate as b in order to determine the vine of titrate required to reach the equivalence point. Remember at the equivalence point, your moles of acid will equal your moles of base and remember moles itself equals leaders, times more clarity. So if we take a look here, it says the temptation of 100 mls of 1000.100 moller ammonia with 0.20 molar hydrochloric acid. So at the equivalent corner, moles are equal to each other, moles are leaders times polarity or volume times polarity. So we can say here am acid times v acid equals and base times V base here. I'm not specifying if I want my volume and leaders or milliliters so we can keep these millimeters here, Plug in the polarity of our acid, which is 0.20 Moeller of hydrochloric acid. We don't know its volume. To get to the equivalence point, We know here that we have 0.100 moller of our base ammonia and we have 100 amounts of it, Divide both sides here by 0.20 Moller Polarities cancel out and we have the volume of our acid which equals 50 amounts. Now this is important to know because as we're slowly adding the amount of strong acid to our week, based based on the volume added, we'll know if we're dealing with tight rations at the equivalence point, before the equivalence point or after the equivalence point. Now, before any of this strong base on a strong acid is added. We essentially just have a weak base by itself. We're gonna say here that a weak base or weak acid requires an ice chart in order to determine its ph So here they're telling us, we have the hydration of 100 mls of 1000.100 moller ammonia with zero mls of 00.20 molar hydrochloric acid. We have 20 mls here, I mean zero mls here of hcl, which means it's not a factor in our calculations, we only have the weak base because the acid hasn't been added yet. We don't need this volume just yet. So all we bring down into the ice chart is polarity. Water is a liquid. And remember in a nice chart, we ignore liquids and solids. Remember that if this is the base, water will behave as the acid acids act as proton donors. So water would donate an H plus two NH three to give us NH four positive and water would become O H minus. Now, initially, we don't have any of those products. So initially there's zero polarity. Now looking at the change line, remember we lose reactant in order to create products. So we have minus X here plus X plus X. Bringing down everything, we have 0.100 minus X plus X and plus X. Now with a weak base. We use KB. Our base dissociation constant. We'd say here, KB equals products overreacting. So it equals X squared divided by the initial concentration minus X. Because we're dealing with a base here when we find X. That gives us O H minus. If I know the concentration of O H minus, I can determine P O H. Because it equals negative log of O h minus. And then if I know p O H, then I know ph because ph equals 14 minus P. O. H. We're going to say here that this minus acts that is part of my equilibrium expression. It can be ignored if it is not significant. And the way I determine if it's significant or not, I use what's called the 5% approximation method. All that is is I take the initial concentration of our weak base in this case and divided by its KB value. If this ratio here is greater than 500 then I can ignore the minus X within my um equilibrium expression, thereby going away from the quadratic formula and making the solution faster to get. So if we took the initial concentration here and we did 0.100 moller of ammonia and divided by its KB. In your book, the kB of ammonia is approximately 1.76 times 10 to the negative five. When we do that we get 5,681 a ratio that's definitely greater than 500. That means I can ignore the minus X here and avoid the quadratic formula. So I would take this Kb plug it in equals X squared over the initial concentration of my weak base. Multiply these together. So 1.76 times 10 to the negative six equals x squared take the square root of both sides. So x here will equal 1.33 times 10 to the -3 Mueller. And because X here gives me O h minus, that means I know what concentration of minuses by taking the negative log of that. That gives me p o H. So that comes out to 2.877. And if I know p o h, I know ph which is 14 minus p o h. So that equals 11 That equals 11 123. So we haven't started to add any of our strong acid just yet. So right now, we're dealing with a simple ice chart. Click on the next video and let's see what starts happening to our ph as I slowly begin to add my strong acid tie Trent.

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Weak Base-Strong Acid Titrations

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So here we're finally going to start adding our strong acid titrate. Remember we're gonna require 50 mls of hydrochloric acid in order to reach our equivalence point, we're gonna say once our acid and base begin to mix, we use an I. C. F. Chart to determine the ph here we only have 20 ml of hydrochloric acid being added. Not quite enough to reach the equivalence point. That's why we're at this point where it's before the equivalence point. Now, here's the attrition of 100 mls of 1000.100 moller ammonia with 20 miles of 200.20 molar hydrochloric acid. Now remember in an I. C. F chart, the units have to be in moles and moles equals leaders times more clarity. We're going to divide these mls by 1000 and multiplied by the polarity. That's gonna give me the moles of my weak base which can be seen as conjugate base and the moles of my strong acid. We don't have any of our weak acid or conjugate acid. So initially it's zero. Remember look on the react inside, the smaller moles will subtract from the larger moles. So what we have left is .006 moles of this conjugate base which can be seen as a weak base will have zero left of our strong acid based on the law of conservation of mass matter can't be created or destroyed. It just changes form. So this increases by that same amount. Okay, so here at the end we have weak acid which can be seen as conjugate acid, whichever way you look at it and realize at the end of this process, what do we have? Well, we have a weak acid, we have a conjugate base. And remember with those two that helps to give us a buffer. If we have a buffer then we utilize the Henderson Hasselbach equation here it will be P H equals negative log of K. A. Because that's what P K. Is At the moment. All we have is the KB of NH three As 1.76 times 10 to the -3. Remember K A equals K W divided by K B. So that'll be 1.0 times 10 to the negative 14 divided by 1.76 times 10 to the negative five That comes out and gives us 5.68 times 10 to the -10. We plug that in Plus log of our conjugate base or weak base which is an h. three divided by our weak acid or conjugate acid depending on how you look at it. Of ammonia. My. Huh? So that gives me at the end a ph of about 10 746. So we can see that we started out with a ph of just over 11. But as we slowly start adding our strong acid tight trend, our Ph begins to decrease and we can expect as we have more and more strong acid that's going to continue now that we've seen the calculations before the equivalence point let's move on to at the equivalence point.

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Weak Base-Strong Acid Titrations

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So now we're dealing with a tin tray shin at the equivalence point. So we're gonna say at the equivalence point of a weak base, strong ascetic trey shin, the solution is acidic And because it's acidic, that means it's ph will be less than seven When we're dealing with the temperature of 25°C. So here we have 100 mls of 1000.100 moller ammonia being tight traded with 50 ml of 500.200 moller hcl. Remember when I divide these mls by 1000 And multiply them by their polarities. That gives me the moles of both compounds. Doing that, we see that we have 0.100 of both. They will completely neutralize one another. So at the end we have zero of both. But we're gonna have plus .100 of ammonium ion. Now many mayan represents our conjugate acid or weak acid. And remember when you have a weak acid, you have to utilize an ice chart in order to determine ph And remember with an ice chart, the units have to be in malaria T. So what we're gonna do here is we're going to say to calculate the new polarity of our weak acid. We take the moles left of it, which is 0.100 moles of ammonium ion and divided by the total volume and that would be the volume of NH three which was 30.100 liters plus the volume of Hcl which was 0.50 liters. Now together when we do that, that's gonna give me a polarity of .667 moller of ammonium ion. So we take that malaria and we plug it into our ice trucked so we can see at the equivalence point there's a lot more work involved because we first have to do an I. C. F. Because we have a tie tray shin. What's left as a product will be a weak species and therefore requiring ice chart in order to solve for P. H. In a nice chart, water which is a liquid is ignored. These products here initially are zero in a nice chart. We say that we lose react ints in order to make product bring down everything. Now what we're gonna say here is we have to isolate what X. Is because X here will give me a 30. Plus. We're gonna utilize this K. Is our asset association constant of our weak acid. We calculated that as one, not as five. On the previous page, we calculate that as five point 68 times 10 to the negative 10 equals X squared divided by the concentration of my weak acid. So when we plug that in, that gives me X squared equals 3.78 nine times 10 to the -10. Take the square root of both sides. So X equals 1.95 times 10 to the -5. So that will be equal to my H. 30. Plus. And because of that we can just take the negative log of that number and that will give me my ph, which comes out to 4.71. So we can see here that our ph continues to drop as we add more and more strong acid titrate here at the equivalence point, both of our weak base and strong acid have been completely neutralized. But as a result of their neutralization, we have the creation of a conjugate acid or weak acid. Because it's an acid, it's gonna give us a ph less than seven. Now that we've seen how to tackle this type of hydration at the equivalence point. Move on to the next video where we're talking about going beyond the equivalence point.

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Weak Base-Strong Acid Titrations

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we're now dealing with tight rations. After the equivalence point. So we're gonna say after the equivalence point of a week based strong acid tray shin, we will have excess strong acid remaining. Remember it required 50 mls 50 mls of hydrochloric acid to reach the equivalence point. And now we're at 60 mls as before. Since we're having an acid and a base titrate one another. We have to use an I. C. F. Divide the mls by 1000 to get leaders and multiply them by the polarities. Gives us these moles Again, look at the reactant side. The smaller mold total will subtract from the larger one. So what we have left at the end is zero of my conjugate base or weak base but I'll still have some strong acid remaining. Also have some weak acid remaining. Now the strong acid has a much bigger presence in changing the ph So we're gonna go with the strong acid amount. We're gonna say here we need to find its concentration. So the concentration of hcl equals the molds of it left divided by the total leaders. So that would be 0.100 liters plus 0.60 liters. So that comes out to .125 moller of Hcl. Now that I have the polarity of a strong acid. Remember then it's simply the negative log of that concentration to find our ph So it'll be negative log a 0.125 Moeller. And that gives me 0.903 as my final ph So notice from the very beginning of our titrate, 10 of this week base we first initially started with a p. H, just above 11 when it was the weak base by itself as we gradually added more strong acid. We've seen that our ph decreases to this point right here, after the equivalence point where it's just below one. Remember the steps, remember the spots within this tight tray shin and whether we're using an ice chart or an I. C. F chart or a combination of the two, and you'll be able to determine the ph of this type of filtration.

Weak Base-Strong Acid Titrations Calculations

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Weak Base-Strong Acid Titrations Calculations

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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

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example

Weak Base-Strong Acid Titrations Calculations

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So here in this practice question, it says calculate the ph of the solution resulting from the mixing of 75 mls of 750.100 moller sodium acetate and 75 ml of 750.150 molar acetic acid with 0.40 moles of per caloric acid. Here we're told the K value of acetic acid is 1.8 times 10 to the negative five. Alright, so we're coming into a situation where there are actually three species all mixing together. But remember, the point of these types of filtration is we're adding a strong tight rent to a weak species are strong tight. Right here would be the strong acid that strong acid. We know what's going to be a reactant. An acid will react with a base. So this strong acid would have to react with this conjugate base. So this conjugate base would have to also be our reactant. The acid will produce. Will protein eight or give an H. Plus to the acetate ion, creating acetic acid Over here as a product. Plus we'd have what's left as N A C L +04. That would represent a neutral salt. So we don't really care about it because we have an acid and base involved in this situation. We do an I C. F. Chart. Remember in an I C. F chart, we use moles as our units. So we divide these mls by 1000 Multiplied by their polarities gives us moles of each. So for this sodium acetate, I'd have .0075 moles. They tell us the moles already of this per chlorate gasses. So that's .0040 moles. And then these leaders multiplied by this polarity gives me .01125 moles. Now remember we look at the reacting side, I know there's stuff on the product side but we've always looked at the reacting side because that's what's involved in the titillation on the react inside. The smaller moles will subtract for the large from the larger moles. So at the end we'll have .0035 moles of this conjugate base. We'll have zero left of our strong acid Here on this side, it would increase by this many moles. So then we have .015-5 moles at the end. What do we have? We have conjugate base remaining, we have weak acid remaining. So we have a buffer. And because we have a buffer we use the Henderson Hasselbach equation. So P H equals P K A plus log of conjugate base over weak acid. Remember P K. Is just the negative log of K. A. So bring down this K value plus log of our conjugate base over our weak acid. So as a result of this, when we plug it in, That's gonna give me a value of 4.11 as the ph So in this question because we still have a buffer at the end of this tray. Shen we must be dealing with a tin tray. Shen before the equivalence point. Remember before the equivalence point, we still have the presence of a buffer, so utilize the henderson hassle back to isolate your ph at the end.