Strong Acid-Strong Base Titrations - Video Tutorials & Practice Problems

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Strong Base-Strong Acid Titration

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Strong Base-Strong Acid Titration

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So in this video we're going to take a look at strong base, strong ascetic rations. Now we're going to say here that in a strong base, strong ascetic tray shin, we're going to say that the initial compound represents are an elite. And we're gonna make this second structure our titrate. Okay, now we're gonna say whenever we have a strong acid or strong base, we never used an ice chart. Remember ice charts are used predominantly for weak acids and weak faces because they don't disassociate 100%. We're using a nice chart to determine what their concentrations are at equilibrium. And from there we can figure out P. H. Or P. O. H. Now, whenever we titrate to species like in this case a strong acid and a strong base. Then we use what's called an I. C. F. Chart and it uses the units of moles. Now remember an ice chart stands for initial change equilibrium. And when we have two strong species tight trading one another, we're gonna use an I. C. F. And this stands for initial change final. Now the following roadmap is what we're going to use whenever we have a strong acid, strong based filtration. So first we start out with the equivalent volume, the volume that's necessary to get to the equivalence point between these two substances. Now we're gonna say, we calculate the equivalent volume V. In order to determine the amount of tight trend or volume titrate required to reach the equivalence point here, we have the tray shin of 150 mls of 1500.100 moller N. A. O H. Which is a strong base with 00.50 moller nitric acid here, nitric acid is acting as our titrate. We're adding it to the N A. O. H. Now remember at the equivalence point we have equal moles of our acid and our base at the equivalence point we could say that m acid. So the polarity of the acid times the volume of the acid equals polarity of the base times volume of the base. Remember we said at the equivalence point our moles are equal well remember moles equals more clarity times volume. So by multiplying these two, that's giving me the molds of acid and I'm multiplying these two. That's giving me the moles of base. This equation is saying that our molds of acid equals our molds a base which is true. At the equivalence point we're gonna plug in these values. So the polarity of our base and A O. H. Is 00.100 100. The polarity of our volume of our base is 150. The polarity of our acid is 0.50 moller and we don't know what its volume is, Divide both sides by .050 Moller. See here that our polarities will cancel out and we'll be left with mls which are the units for our volume. So when we plug that in. That gives us an answer of 300 ml. That means it will take 300 ml of our nitric acid in order to reach the equivalence point. In terms of this situation now we dis term in the equivalent volume. Now let's talk about before we added any strong acid to our mixture. We're gonna say here before any of the strong acid titrate is added. We only have a strong base initially. Remember strong acid, strong base associate completely. So their concentrations can help us determine the amount of H plus or oh minus present. So here we have the titillation of 150 mls of 1500.100 moller, sodium hydroxide with zero mls a 00.50 molar nitric acid. Notice here we have no strong acid volume being added. So all we have within our jar within our container is strong base. Now remember because it's a strong base it associates completely. So it's gonna break up into an A. Plus and O. H minus completely because it's a strong electrolyte. The concentration of N. A. O. H. Is equal to the concentration of each ion. So now we have the concentration of O. H minus our hydroxide ion. If you know the concentration of your hydroxide ion you can calculate P. O. H. Because remember P O. H. Negative log of P. O. H. P. O. H equals negative blog o H minus. We plug in this concentration of 0.100 that gives me a P. O. H. Of one. If we know what P. O. H. Is. We know what ph is because P. H. Equals 14 minus P. O. H. So we plug in this one that was for P. O. H. And what we'll see is that R P. H equals 13. So here we're just going through the different steps of art. It trey shin between the strong base and the strong acid. At this point all we have a strong base so we can easily find the P. O. H. Of that solution. Now, as we move on to the next video, we'll talk about what starts to happen as I slowly add my strong acid to my strong base.

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Strong Base-Strong Acid Titration

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So remember, it's gonna take us 300 MS of nitric acid to reach the equivalence point. So until that point we're gonna be dealing with tight rations before the equivalence point. Now, once our acid and base begin to mix, we use an I. C. F. To determine the ph now this is true, no matter which one is the analytic and which one is the thai trade. As soon as an acid and base mix together a tin tray shin is occurring and you have to utilize the I. C. F. Chart. Now, if we take a look at this example, it says the tray shin of 150 mls appoint 100 moller sodium hydroxide with 100 and 20 mls a 200.50 molar nitric acid. We can see that we're at 100 and 20 mls. So we haven't reached the 300 mls necessary to reach the equivalence point. Remember also that when we're dealing with an I. C. F. Chart, the units will be in moles moles itself equals leaders, times more clarity. Remember this is just a rearrangement of the polarity equation. If we're trying to isolate moles, we multiply both sides by leaders and that's why moles equals leaders. Now dividing these by a 1000 will give us our leaders of sodium hydroxide and nitric acid. So that's 0.150 liters of means multiply so times 0.100 moller this is 0.120 liters times 0.50 moller multiplying those gives me the moles of N. A. O. H which I can plug into my I. C. F. Chart and moles of nitric acid. Now in our I. C. F. Chart, the two compounds that are involved in the tie trey shin are reacted. In this case it's a strong acid and a strong base. Remember an acid acts as a proton donor. So the H. N. 03 will donate an H. Plus to the O. H. Minus H. Plus an O. H minus combining together gives us water. Then the N. A. Plus and the N. +03 minus combined together to give us N. A. N. +03. This represents our conjugate base. Now that we have our initial moles, we're gonna go to change now in an I. C. F. Chart where we look at the moles of the reactant and the smaller amount of moles represents our limiting amount. It will subtract from the larger moles At the end, we'll have zero left of the strong acid. And when we subtract these two numbers we'll have .009 moles of our strong base. But remember the law of conservation matter is neither created nor destroyed. We're not really losing that amount. What's happening is being reformed into our conjugate base here as a product. So whereas we lose .006 moles as reactant. It's just being reformed. So we're gaining .006 moles over here as product. Bring this down at the end of this, what we have remaining is strong base and conjugate base. Now the conjugate base is not gonna affect the ph as far as the strong base. Well so again if we have a strong species remaining, that's what we care about. So we're gonna ignore this conjugate base amount. So before we've reached the equivalence point we don't have quite enough strong asset to completely neutralize our strong base. Because we have strong base remaining. We're gonna have to determine its new concentration. And from that we can determine our P. O. H. So we take a look here. We're gonna say the new concentration of my strong base, abbreviated as S. B. Equals moles left after that hydration divided by total leaders used. So we have here 0.6 moles of N. A. O. H. Remaining divided by our total volume. Our total volume would be this point 150 L plus This .120 l together. That will give me a concentration of 0.33 molar of N. A. O. H. Now because I have the polarity of a strong base, I can just take the negative log of that concentration to give me P. O. H. So p. o. h. equals negative log of .033. So that gives me one point 48 as my P. O. H. If I know P. O. H. I know ph because ph equals 14 minus P. O. H. So that comes out to 12.52. So at this point before we've reached the equivalence point, we still have XS strong base remaining. So we should expect a ph that's above seven here. We've worked it out. And we see that it's 12.52. Now, what we're gonna see next is what happens when we get to the equivalence point in this tray, shin of a strong base with a strong acid. So click on the next video and let's see what that entails.

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Strong Base-Strong Acid Titration

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We've now reached our equivalence point. So we've used the 300 ml of nitric acid necessary to reach this step. Now we're gonna stay here at the equivalence point of a strong base, strong ascetic trey shin the solution is neutral and that's because we have equal moles of our strong acid and strong base which completely neutralize one another. We're gonna say since our solution is neutral, our ph would automatically be equal to seven. Remember this is true at 25 degrees Celsius. Remember that if our temperature changes that affects our kW constant. And therefore the definition of neutral would change to a number other than seven. So again a neutral solution. Its ph is equal to seven at exactly 25 degrees Celsius. Now here we have the attrition of 150 mls of 1500.100 moller N A O H. With our equivalent volume of 300 mls of 3000.50 molar nitric acid. So if we divide these by 1000 to get leaders and we multiply them by their polarities. We see that we have equal moles of our strong base and our strong acid which makes sense because at the equivalence point we have equal moles of both. Now remember we look on the react inside the smaller moles will subtract from the larger moles since they're both the same, They completely eliminate one another. So at the end we have zero moles of both. Through the use of law of conservation of mass. We know that will have .015 moles of this conjugate base. But here's the thing when a strong acid and strong base completely neutralize each other, the conjugate base that's formed represents a neutral salt. This is further justification to explain why our pH is seven. Ny our solution overall is neutral. So again, remember when we have a strong acid and a strong base with one another and they're at the equivalence point, our ph overall will equal seven. Now that we've seen what happens at the equivalence point, let's look and see what happens when we go after the equivalence point.

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Strong Base-Strong Acid Titration

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We've now gone beyond the equivalence point. Remember we needed 300 ml of nitric acid and at this point we've used 310 mls. Now we're gonna stay here after the equivalence point of a strong base strong acid filtration. We will have excess strong acid remaining as before we divide these mls by 1000 to get leaders multiplying them by their polarities, gives us the moles of N A O. H and H N 03. So here, as always we look at the react inside and remember the smaller moles which are our limiting moles will subtract from the larger ones when I do that. I'll have no base left. But I'll have just a little bit of my strong acid. And that's all that's really necessary here again, following the law of conservation of mass would actually be increasing on this side. But that doesn't matter because we talked about how that's a neutral salt. So it doesn't influence the ph But here we're going to say that we're gonna take these this amount of moles of our strong acid and find its concentration. So we're gonna say the new concentration of our strong acid will be the moles of it left divided by the total leaders. So we have here .005 moles of nitric acid divided by The total volume, which is this .15 zero leaders plus This .310 L. When we plug that in. That will give you a polarity of about .001087 moller a channel three. Since it's a strong acid When I take the negative log of that, that'll give me ph so P H equals negative log of H plus which is the negative log of that concentration I just wrote Gives me approximately a ph of 2.96 as my final answer. So as we've been slowly adding strong acid to our strong base, we can see that our ph has decreased. We started at a ph much higher than seven. And now at this part since we have access strong acid remaining, we're gonna have a ph much lower than seven. Now correction in terms of this neutral salt. If we don't take into consideration things such as activity coefficients, then we can say here that this neutral salt won't influence our ph at all. But here we're not talking about that type of topic. So we're gonna say at this point we care more about the fact that we have excess strong acid than we do that neutral salt that conjugate base. But these are all the steps that we need to take when we're doing a strong base, strong acid titrate in. Now that we've seen every step along the way, let's start doing some additional example in practice questions when it comes to these types of situations

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Strong Base-Strong Acid Calculations

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So here it states calculate the ph of the solution resulting from the tight rations of 100 and 50 mls of 500.20 molar sodium hydroxide with any mls of 0.15 molar hydrochloric acid. All right. So, we have here are strong base and it's being tight traded by this strong acid. So remember when we have an acid and base undergoing penetration, we should set up an I. C. F. Chart. So we start out with our N A. O. H plus hBr high trading one another. Now, we're going to say here that remember your acid is a proton donor. So it would donate an H plus B O. H minus. And that would create water. The N A plus and the B R minus left behind were combined together. And that would produce our conjugate base. Remember we have I. C. F. Which is initial change. Final water is a liquid. Remember liquid and solids are not involved in ice charts or I. C. F. Charts. Remember an I. C. F. Chart uses moles as the units. So we'll divide these mls by 1000 to change them into leaders and then multiply by their polarities. When we do that, we'll get 10000.30 moles of n A O H 0.1 to zero moles of H B. R. We're not giving any information on N A B A B R. So initially it's zero. Now looking at the reactive side, the smaller mold total, which is our limiting amount will subtract from the larger moles. So we're gonna subtract .01-0 from both. So at the end we'll have no strong acid remaining and we'll have a little bit of strong base remaining that the law of conservation of mass this side would increase. But here we're only concerned with the strong species because it will have a greater impact on the overall ph So we have excess strong base remaining within this hydration. That means that we are before the equivalence point. We haven't added enough strong acid to completely neutralize and overtake the strong base. So now we need the concentration of N. A. O. H. So we're gonna take the molds left And divided by the total volume used. So that would be .150 L plus .080 L. So when we work that out we should get .078 moller in your calculator. Now that we have a concentration of our strong base, we take the negative log of that and that will give us P. O. H. It would be negative log of .078 which will give me 1.11 as my P O H. If I know my p O H, then I know my ph because ph equals 14 minus P O H. So that comes out to as 12.89. So that's the setup for this first example. Remember we're just implementing the changes that we know in terms of our strong base strong acid filtration. Follow the steps to determine where you are. In terms of arbitration. In reference to the equivalence point, you'll know what you'll have at the end and then take the necessary steps in order to find your p O. H. Or P. H at the end. Now that we've seen this problem, move on to the next one.

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Strong Base-Strong Acid Calculations Practice

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So we now have this practice question. It states calculate the ph of the solution resulting from the tight rations of 100 mls of 1000.30 moller lithium hydride with 150 mls of 1500.40 moller hydro biotic acid. So we have our strong base here initially and we're tight trading it with this strong acid. So again, those two are tight trading one another. So there there are initial react ints. We're gonna say here that we know L. I plus an L. I. Plus and I minus. We'll combine together to give us our conjugate base here. Here, we're not going to get our typical water as a product. What's gonna happen here is this H minus will combine with this H plus and we actually get H two gas, which would mean if this reaction took place. We have the evolution of gas. Now we're going to say that this is I C. F. Because it's a acid and base high trading one another. Remember that initial change final, We know that the units in an I. C. F. Chart have to be in moles. So divide your mls by 1000 to get leaders and multiplied by the polarities to get our moles. So that's gonna give me 10000.30 moles. And then this is gonna give me .060 moles H two. There's a gas that's going to escape this solution. So we don't have to worry about it here. Um We're gonna say now we have lithium iodide initially zero. Looking on the react inside the limiting moles will subtract from the larger moles. That'll be minus 0.30 minus 0.30 at the end. We have none of the strong base remaining and we still have some of our strong acid through the law of conservation of mass. We have the increase here, so we'll have this many moles of our conscious base. But again, we're focusing on the strong acid remaining since we have a strong acid remaining, we have to find its concentration now. So we're gonna stay here. The concentration of our strong acid equals the molds left behind. After the tight rations complete, divided by the total volume. The total volume would be 0.100 mls. On that point point 100 liters plus .15 L. When we plug that into our calculators, you should get .12 moller H I since it's a strong acid when I take the negative log of that concentration, that will give me ph So that comes out to .92. So pretty acidic, very acidic. So again, in this case, because we have excess strong acid left at the end, that means we are after the equivalence point. So once we find the concentration of our strong acid, just take the negative log and you'll find your ph

Strong Acid-Strong Base Titration

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Strong Acid-Strong Base Titration

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We now will take a look at strong acid, strong base tight rations. In this case when I use the term strong acid, strong base filtration. I'm making our strong acid are an elite and it's going to be tight traded by our strong base, our strong base will be our tightrope. So here we provided the road map to help guide us on the methods to calculate ph at different points of the tray shin. So first of all, we're going to say that it's important to know the equivalent volume when dealing with um this type of filtration. So we're gonna stay here, calculate the equivalent volume V. E. In order to determine the volume of titrate required to reach the equivalence point. So here we have 120 mls of 1200.250 molar hydrochloric acid, which is our strong acid with 0.200 Mohler K O H, which is potassium hydroxide are strong base. So we say it's gonna be m acid times v acid equals m base times v bits. We plug in the polarity of our strong assets, that's .250 moller The volume of our strong acid, which is 1:20 MLS equals the polarity of my strong base. And then we don't know the volume of the base here. So what we do here is we're gonna divide both sides by .200. So that's gonna cancel out the polarities here and we'll be left with volume of the base in milliliters. When we work this out. The volume of the base, which will represent my equivalent volume equals 150 mls. So it'll take 150 MS of this potassium hydroxide for us to reach the equivalence point. Now, before any of this strong base is added. So before the Thai tradition has even started, what we will have initially is just strong acid. And remember if we have a strong acid, we're gonna say it's concentration is equal to the concentration of H plus ions. So if we take a look here, we have 0.250 molar hcl, we don't have to involve the volume yet that volume comes into play. Once we start the titrate in process. So once we start adding the strong base, that's when we utilize the volume. Because we only have strong acid here by itself, just focus on this polarity. So since it's a strong acid, it associates 100% creating 100% of H plus 100% of cl minus Now, we're going to say here, the concentration of the strong acid is equal to the concentration of each of those ions. So we have exactly 1000.2 50 moller H plus. And if we know the polarity or concentration of H plus ions, then we can calculate ph because remember P H equals the negative log of H plus. So we take this and we plug it in When we take the negative blog that that gives me APOA ph of .602. So before we added any strong base, this is our initial ph. Now, as we start to slowly add the strong base to our strong acid, we should expect the ph to gradually increase. And that's what we're going to see here. In terms of the steps involved in this hydration. Alright, So now that we've seen what happens before any strong base is added, let's move on to the next video and see what process do we take as this potassium hydroxide is slowly being added.

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Strong Acid-Strong Base Titration

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So remember we're gonna need 100 and 50 mls of potassium I drop side in order to reach the equivalence point. So as we begin to add potassium hydroxide until we get to this number will be dealing with calculations before the equivalence point. So if we take a look here, it says um once our acid and base begin to mix, we use an I. C. F. Chart to determine the ph So when I see f chart remember is stands for initial change final. If we take a look here, we have the determination of 100 and 20 mls of 200.250 molar hcl with 100 mls of 1000.200 Mueller K O H. So we haven't quite reached our equivalent volume yet. We're only at 100 in an I. C. F. Chart, the units have to be in moles. Remember moles equals leaders, times more clarity. That's just a rearrangement of the polarity equation. When we multiply both sides here by leaders, We can isolate our moles so we're gonna have to divide these mls by 1000. That'll give us leaders of each so that give us a 0.120 liters of means multiply so that it would multiply with the polarity of each compound. This would give me 0.100 liters multiplied by the 0.200 moller. So doing that gives me the moles of the moles of hcl which came out as 0.30 moles plugging this and gave me 0.0 to zero moles of K. O. H. Now we're going to say that in this reaction. Remember acid is a proton donor. So the hcl donated an H plus two O H minus together they formed this water here as a product and then we're going to say next that the K positive left behind and the cl minus left behind combined together to give me this conjugate base. Now, since we are dealing with a situation before the equivalence point will have some strong acid remaining at the end. That's because there's not enough strong base to completely neutralize and overtake the strong acid. So if we go back to the I. C. F chart, we're gonna look at the react inside, We have these two mol amounts. We're going to say that the smaller moles which is our limiting amount will deduct from the larger moles. So subtract .0-0 to attract .0-0 at the end, we'll have no strong based remaining and then we'll have some strong acid left. The law of conservation of mass. Remember matter cannot be created or destroyed. It just changes form. So in reality, we're not exactly losing these these compounds. They're not really they're not disappearing from existence. They're just being reformed into this conjugate base. So as on this side on the reactant side we have a loss of .02 On this side here will have an addition of .020. So bring that down. Now in the tight rations of a strong acid and a strong base. We have the creation of a neutral salt. So this neutral salt, if we don't take into account things such as activity coefficients, we can ignore it. Okay, activity coefficients. We'll talk about later on when we try to connect it to P. H. And p O. H. But for now this neutral salt will not influence our ph so we can ignore it. So what's important here is we have a strong acid left at the end. Now this strong acid because we have it, we can determine its concentration. So we're gonna say here that the strong acid, we're gonna take the molds left, we're gonna divided by the total leaders. So the molds left of our strong acid are 0.10 moles. The total volume would be this point 120 l plus This 0.100 l Altogether. That gives me .045 moller hcl. And since I have the concentration of a strong acid when I take the negative log of it, that'll give me my ph because remember we set up above that, the concentration of a strong acid is equal to the concentration of H Plus so that's gonna equal 1.35. So we can see that we've added some strong base and it's caused my ph to increase. We started out around 0.602. Initially when we just have the strong acid by itself adding the base is going to cause the ph to increase over time. At this point now, we're at 1.35 now that we've seen the calculation setup that we use when we're dealing with before the equivalence point, move on to the next video and see what happens when we get to the equivalence point between this strong acid and strong base.

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Strong Acid-Strong Base Titration

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So we're calm. It's gonna take 150 MLS of potassium hydroxide in order to reach the equivalence point. Now in this question, we've finally reached that toll. So we're gonna say at the equivalence point of a strong acid, strong based filtration. The solution is neutral And the P. H. will equal seven at 25 degrees Celsius. So remember when we reach the equivalence point between a strong acid and a strong base, no matter which one is the analytics and which one is the tie Trent? Our ph should be equal to seven at 25 degrees Celsius. They completely neutralize each other and we'll be left with a conjugate base. Uh that is a neutral salt. That neutral salt. If we don't take into account things such as activity coefficients that it will have no impact on the ph So if we take a look here, we divide these by 1000. That's gonna give us leaders of each. Multiplying them by their polarities will give us moles. So doing that, we see that we have identical molds of K. O. H. And H C. L. Which makes sense because at the equivalence point we have equal moles of our acid and our base when they subtract from one another, we'll see that we have zero left of both. We have the addition of our conjugate base here on this side which again, since it's a neutral salt, it doesn't have an impact on the ph this further justifies why our solution is neutral at the end. So again, whether we're doing a strong acid, strong based hydration or strong base strong acid hydration. As long as both species are strong, when we get to the equivalence point, we're gonna have a ph that will be neutral. Okay, so we're gonna have a ph equal to seven. If the if the temperature is also at 25°C. Now that we've seen this, let's move on to the last video and see what happens when we go beyond the equivalence point.

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Strong Acid-Strong Base Titration

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So here we've gone beyond the equivalent volume. So remember we needed 150 mls of K. O. H. So after the equivalence point of a strong acid, strong based hydration, we will have excess strong base remaining. So we have excess strong base. So if we take a look here, we divide these mls by 1000 To get leaders and then multiply them by their polarities. That will give us the moles of each compound that gives us .036 moles of potassium hydroxide and .030 moles of hydrochloric acid. Looking at the reacting side, we subtract the smaller mold total From the larger one As a result of this hydrochloric acid is zero at the end And then we'll have .006 moles of K. O. H. What we're gonna do is realize that we're gonna add this same amount to the other side since it's a neutral salt, we will ignore it now because we're after the equivalence point, we're going to have strong base present. That's our excess base left at the end. If I can find its new concentration that can give me the concentration of O H minus. So we're gonna take the moles that we have left of KO H And divided by the total volume which is .120 L plus .180 L. So that gives me zero 20 for my coach. The majority of my coach since I know the polarity of my strong base, taking the negative log of that will give me my P. O. H. So plug that in That gives me 1.70. And if I know P O. H then I know ph because ph equals 14 minus P. O. H. So we have 12.3 as our final ph. So we can see that as we've added strong base to our initial strong acid. We've seen that the ph has jumped up, starting at a lowly 0.602. All the way up to this final ph of 12.3 and that's what's going on here are strong bases. Driving up our ph now that we've seen all the steps at different points of this tray, shin move onto the next video and let's start doing some practice and example questions.

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example

Strong Acid-Strong Base Calculations 1

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So here the example says calculate the ph of the solution resulting from the thai tradition of 50 mls of 500.10 molar hydrochloric acid with 20 ml of 200.30 moller sodium hydroxide. So we have our strong acid here. That is being tight traded by a strong base. We're gonna write down our compound. So H. I. Plus an A. O. H. We know that H. I will protein it the O. H. Minus to give us a church to O. So it gives an H. Plus to the O. H. Minus. Then we're gonna have N ai positive remaining and I minus remaining. And they combined together to give us our conjugate base because we have an acid and a base thai trading one another. We set up our I. C. F. Chart. So initial change. Final water is a liquid liquid and solid are not found in either ice charts or I. C. F. Charts. Remember in an I. C. F. Chart, we need the units to be in moles. So divide the mls by 1000 to get leaders, then multiply them by their polarities. That will give us the moles of H. I and N A. O. H. When we do that, we're gonna get 00.50 moles And .0060 moles. Initially our conjugate base is zero. Since we're not giving any information on it on the react inside the smaller molds, which is our limiting amount will deduct or subtract from the larger moles. So subtract 0.50 point 0050 at the end. We're gonna have zero left of our strong acid. That means that we've gone beyond the equivalence point where there's going to be an excess of strong based remaining. So this is .0010 moles remaining. Here we have our conjugate base, it's a neutral salt. So we don't care about it. We care about the strong species because it has a greater impact on our final ph since we have a strong base remaining, we just need to find its concentration. So we're gonna take the molds left of it and divided by the total volume, Which would be zero. So point 050 liters plus 500.20 liters. Alright, so then when we plug all that into our calculator, we'll get a polarity of 0.14 Moeller of N A. O. H. Since we know the concentration of our base, we can use that to help us find P. O. H. So negative log of O H minus. So plug that concentration in. So that equals 1.85. Since we know the P O. H, we can find the ph ph would equal 14 minus that P. O. H total. So that comes out to be 12.15. So in this case we had excess strong based titrate left. So we're dealing with after the equivalence point. So find the concentration of that strong base and use it to find P. O. H. And then from there ph now that we've seen this example, question click onto the next video and let's continue our discussion of this type of filtration.

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Strong Acid-Strong Base Calculations 1

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So here are practice question states, calculate the ph of the solution resulting from the penetration of 90 mls of 900.40 moller coric acid with 50 ml of 500.50 moller potassium hydroxide. So again we have a strong acid being titrate it by a strong base. So we're gonna write those down as our reactant plus K. O. H. The H plus from the acid will be given to the O. H minus of the base giving us water. The K positive and cielo three minus left behind will combined together to give us our conjugate base. Since we have an acid and base thai trading one another. We have an I. C. F. Chart so we have initial change. Final water is ignored. Now remember we need moles as our units within this I. C. F. Chart. So we're gonna divide the mls by 1000 to get leaders And then multiply them by the polarities to get the moles of our cleric acid and are potassium hydroxide. When we do that, we should get .036 moles of our strong acid and .025 moles of our strong base and zero of our conjugate base on the react inside the smaller mold which is our limiting amount will subtract from the larger molds. So at the end we'll have zero left of my strong base but 00.11 moles of our strong acid. Since we still have some strong acid left, that means there wasn't enough strong base to completely neutralize it. This is dealing with tight rations before the equivalence point here we're adding this again as as we normally do, but it's a neutral salt, the strong acids. What we care about here, we're gonna say that the concentration of our strong acid we take the moles left divided by the total volume. So that would be 0.90 liters from the strong acid and 0.50 liters of the strong base together. That gives me 0.79 moller of the strong acid. Now because I have a strong acid, its concentration is equal to the concentration of H. Plus. And with that I can determine the ph so plug that in and when I do that I get 1.10 as my final ph So we can see that the titrate in between two strong species is pretty quite simple as long as you adhere to the I. C. F. Truck principles. Okay, so just make sure you see how much moles are left of either one to help us determine if we'll figure out P. O. H. Or P. H. At the end