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Acid-Base Titrations

These types of titrations revolve around the two Ka values of the diprotic acid.

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Diprotic Acid Titrations

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with the titillation of a dia protic acid. We now have to keep in mind the existence of two equivalents points as well as all the equations as well as calculations involved between those two. So, if we take a look here, we're gonna say, we as always we should determine what the equivalent volume will be. In terms of our titrate here are die protic acid will serve as the analytics and our strong base will serve as the tightrope. And because we have the existence of two equivalents points, we're gonna have to find two equivalent volumes. So we're gonna stay here first. That molar itty of my acid times volume of my acid equals polarity of the base times the first equivalence point when we input the values we have 0.0.100 moller Times 100 MLS Equals .050 moller times the first equivalence point Divide both sides here by .050 Moller. So our first equivalent volume would be 200 ml. So that's how much it takes to get to the first equivalence point. And to get to the second equivalence point, we would just need another 200 ml. So we would need 400 ml total to get to the second equivalence point. All right now, before we added any strong base, we essentially just have a week dia protic acid with the weak die protic acid. We have to employ the use of an ice chart from that information, we can set up our equilibrium expression now with these first point in our before we get to our real hydration. We're dealing with the removal of the first acidic hydrogen. So this first acidic hydrogen from our sulfurous acid solution will be donated to water will create by isil fight as a product as well as H 30 plus. Because we're talking about removing the first H plus ion, that means we're dealing with K. A. One K. one for sulfuric acid is equal to 1.6 times 10 to the -2. We'd use that K. A. Value here within our equilibrium expression. Now remember we can always do the 5% approximation method to see can I ignore the minus X. Within my equilibrium expression? You would take the initial concentration of your week die protic acid which is 50.100 moller. And you divide it by the K. That we're using which is 1.6 times 10 to the negative two. If you do that, you'll find that this ratio is not greater than 500. Therefore you would not be able to ignore this minus X. We would keep the minus X. And as a result have to use the quadratic formula to solve for X itself, Once you use the quadratic formula, you would find that X equals .032792 Molar. Once you have that value for H plus you can just take the negative log and from that you can find the ph which is 1.48. So just remember at this point the thai tradition technically hasn't begun yet. Once we start adding our strong base to our sulfurous acid solution, then our titrate, asian is can commence. And we're gonna have to be aware of where Exactly within the tight rations are we dealing with, calculate, are we dealing with at the equivalence point before the equivalence point or after the equivalence point? And since there's two of them, it becomes even more complex, click onto the next video and see what happens when we start to add k O H two R sulfurous acid solution.

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Diprotic Acid Titrations

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so recall. We need 200 mL to get to our first equivalence point here, we're dealing with calculations before the equivalence point here we have the titillation of 100 mls of 1000.100 moller sulfurous acid with 75 mL of 750.50 molar potassium hydroxide. At this point before we've gotten to our first equivalence point, we'll have the production of a buffer and therefore we'll have to rely on the Henderson Hasselbach equation to find our ph Remember within this setup where we're dealing with our weak acid and our strong base our units will be either milly moles or moles. Remember moles is just leaders times molar itty and millie moles is milliliters times more clarity here. If we divide these mls by 1000 and then multiplied by the polarities will find the moles of each one. Hear from that. We input them into our chart As our initial values. Water we could ignore, we don't have any initial amount for by sulfide here. So it's zero initially. Now on the react inside the smaller moles will subtract from the larger moles. So we subtract .00375 moles from both the reactant. And remember the law of conservation of mass whatever we lose as reactant is really just getting changed or reformatted into product. So we have the gaining of the same amount of moles as products. At the end of this will have left weak acid and conjugate base, which is why we have a buffer here. Using the Henderson Hasselbach equation, we plug in P. H equals P. K. One because we're still dealing with the loss of the first acidic hydrogen from our di protic acid plus log of conjugate base over weak acid so that the negative log of 1.6 times 10 to the negative two Plus log of my conjugate base, which is .00375 moles of by Isil fight ion Divided by .00625 moles of sulfurous acid. Doing that would give me a ph of 1.57 realize that as we're adding more and more potassium hydroxide, we should see an increase in r P H. Now the jump in ph is not very large here because we have the production of a buffer. A buffers resist large changes in ph which is why we've only seen a small uptick in our ph at this point. Once we get to the first equivalence point, the buffer will be destroyed and we'll see a large increase in our ph So click on the next video and see what happens in terms of the first equivalence point And the calculations involved

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Diprotic Acid Titrations

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at the first equivalence point, we've now reached a portion of our tight rations where we'll have equal moles of our weak acid and strong base. So both of them because they're both the same amount of moles will completely neutralize one another. So at the end they're both will be zero. But remember that the law of conservation of mass, we really don't lose the weak acid, It gets reformatted to form this conjugate base. So at the end, this is how many moles we have of our conjugate base. Because of that, we can use this information to determine our ph what we first do is we have to figure out the formal concentration Of our conjugate base form. That would just be the initial concentration of our weak acid, which is .100 moller times the volume of the acid used, which is 100 mls over the volume of the solution, which is 100 mls plus 200 mls At the end. That gives me a formal concentration of .0333 molar. With that formal concentration, we now can plug it into this equation to figure out the concentration of H. Plus remember this is an equation that we've seen before when dealing with the intermediate form of a di protic acid. So, when we talked about dia product acids, we talked about finding the concentration of the intermediate form K. One is just the K. One for the dia protic acid, which is 1.6 times 10 to the negative two K. A two for sulfurous acid is 6.4 times 10 to the -8 K. W. is just our standard 1.0 times 10 to the -14. We plug all that information in to find the concentration of H plus which comes out to 2.63 times 10 to the negative five molar. Because you know the concentration of H plus, you know what the ph is because all you have to do is take the negative log of that number and you'd have your ph So here we have 4.58. As our ph up to the first equivalence point, notice that there's a jump in terms of the ph that's because once we've reached our first equivalence point, the buffer as we know it has been destroyed. So there's nothing preventing the jumping up of our ph. Now remember we're dealing with the dia protic acid. So we still have additional titrate in points. Within the whole process, we still have to work our way up to the second equivalence point and then the calculations after that second equivalence point. So we'll continue on with our journey in terms of the attrition of this dia protic acid

4

Diprotic Acid Titrations

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So at this point we've passed our first equivalence point. Now here we have the titillation of 100 mls of 1000.100 moller sulfurous acid with 250 mls of 2500.50 Mueller potassium hydroxide. Remember we only needed 200 mls of our strong base to get to the first equivalence point. We are 50 mls over that. So we've passed the first equivalence point. When it comes to your tie trend, you have to input the excess moles involved. So we're 50 mls over our 200 mls needed to get to the first equivalence point. So the excess volume is 50 mls and it'll still be of 500.50 moller K O H, dividing this by 1000. Then multiplying it by the polarity gave us these excess moles here of our strong base. Now, remember that's the technique we use in terms of figuring out the excess moles of our titrate, the strong species. Now, when it comes to this compound here, remember at our first equivalence point we had H two S. 03 reacting with K. O. H. To produce Are conjugate base. Remember at the equivalence point, all of this had been destroyed and all of this had been destroyed. So what we had left was just this, we had .0100 moles of it left those moles that we had left at the first equivalence point are the moles that we place here. So for the tightrope again, we look for the excess moles. And for the species it's reacting with, we use the moles that we would find at the first equivalence point. Then we're going to say here that this represents our new weak acid and this is the new conjugate base here. We have none of the conjugate base has been formed just yet. We're focusing on the reactant moles. The smaller mold will subtract from the larger moles. So that's why we're subtracting out .00-50 moles from both of these. At the end, we're still gonna have some weak acid left. And then remember, we're gonna form that same amount of moles here as products. So we're gonna have some conjugate base left. We've gone past the first equivalence point. So we're no longer talking about K. A. One. We're now talking about removing the second and last H plus ion to produce this conjugate base here because we're talking about removing the second H plus ion, we're now dealing with K A two. So, from this information, we can plug in myself out the image guys. So from this information we can plug in what we know. So we say ph here equals negative log of 6.4 times 10 to the negative eight plus. We're gonna do the log of the conjugate base Plus log of conjugate base zero 250 moles of the sulfide ion Divided by .00750 moles of our by sulfide ion. So remember we've gone past the first equivalence point. And now we're dealing with the dealing with our journey towards the second equivalence point. That's what we're dealing with K two. Now, when we punch these in, we get 6.72 as our ph So again continuing onward from where we first started with the titrate up to this point, you can see that the ph is continuously increasing and this is what's happening when we're adding strong base to our weak die product solution. Now that we've seen what's happening after the first equivalence point, click on the next video to see what happens when we reach the second equivalence point.

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Diprotic Acid Titrations

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So we've reached our second equivalence point, realize that we have the continuation of our reaction. So we have our by sulfide ion continuing to react with K. O. H. To get to the second equivalence point. And in this process we have the production of our sulfide ion plus water, water which we don't care about. So at the second equivalence point we're gonna have equal moles of both of these compounds. In the process they both get completely neutralized, so they're both gone. So what we have remaining is our sulfide ion, which represents our weak conjugate base. Now realize we're talking about removing the second age plus to produce this final conjugate base. And since we're talking about the second H plus, that means we're dealing with K two. Now here we're gonna find the formal concentration of our sulfide ion because it's what's left. So we have the initial concentration of our weak acid Which is .100 molder times the constant, the volume of our acid which is 100 mls divided by the volume of our solution, which is 100 mls for the acid plus 400 mls For our strong base here when we do this, that's gonna give me .0-0 moller as the formal concentration for my sulfide I on my weak conjugate base. Alright, so now that would go in here, since we're dealing with S. 032 minus. We can't use K two. We're gonna have to convert that into KB. So we have to recall from earlier sections that the relationship between K. A two and K. B. Is that K two times KB one equals kw Divide both sides here by K two and we'll have what KB one equals. So we have 1.0 times 10 to the negative 14 divided by K two. Which is 6.4 times 10 to the negative eight. Here, that would give me a KB one of 1.56 times 10 to the negative seven. So that would be the KB one that we use again. As always. We would also do the 5% approximation method. So we do the initial concentration divided by now our KB one and see if that gives us a ratio greater than 500. So our initial concentration is the formal concentration we discovered divided by the KB. One that we just calculated here, you would see that that gives us a value greater than 500 which would mean that we could ignore this minus X. Here we plug all that in and solve for X. Once you find X realize that that would represent O H minus concentration. So here that X would equal 5.59 times 10 to the negative five. When you take the negative log of it, you will find your P. O. H. That will come out to being 4.25. And then if you subtract it from 14, you'd have your ph which is 9.754 p So as we've been seeing the more strong base, we add, the higher the ph will get again, we've reached another equivalence point. So there should be another jump in our ph Remember the largest increases in ph happened around the equivalence point, the largest jump really happens around the first equivalence point and then around the second equivalence point. There's still another jump. It may not just, it may not be as large as the first equivalence point. So now that we've seen what happens at the second equivalence point, click on to the final video to see what happens after we've passed the second equivalence point within our titillation.

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Diprotic Acid Titrations

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we finally reached the point where we are beyond the second equivalence point. So here we have the titrate, 10 of 100 mls of 1000.100 moller sulfurous acid with 420 ml of 4200.50 molar potassium hydroxide. Remember to reach the second equivalence point, we needed 400 mL of potassium hydroxide. Here we have 420 mls. So we have an excess of 20 mL of potassium hydroxide. So at this point we have still the reaction of our by sulfide plus K. O. H. To produce our sulfide ion. All of this has been destroyed because we've gone beyond the equivalence point. So we're gonna have an excess of our strong base and of course we'll have some conjugate base as well. But the strong base has a greater impact on the overall ph so we're going to focus on finding its excess. So here to find the concentration of the excess base, we say it's equal to the initial concentration of the strong base which is 0.0.50 moller times the volume of excess space. We all needed needed 400 mls to get the second equivalence point, we are 20 mls above that. So that's 20 mls over the total volume which is 100 mls plus 420 mls. So that gives me a concentration of .001923 moller. When you take the negative log of that, you'll get 2.72 as your P. O. H. And therefore when you subtracted from 14 ph will give you 11.28. So that will represent our final ph up to this point after the second equivalence point. So remember, with mono product acids, we have to follow steps along the way to determine where in our tray shin. Our reaction is occurring. The same thing can be applied for this dye protic acid, you just have to determine. Are you dealing with calculations before after or at the first equivalence point or second equivalence point that guides you? And the method needed to find your ph so hopefully you guys are able to follow along and make sure you go back if necessary to look at different portions of the tight rations of a dia protic acid.

7

Diprotic Acid Titrations Calculations 1

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So here it states, calculate the ph of 100 mls of 1000.25 moller carbonic acid when 70 ml of 700.25 molar of sodium hydroxide are added here were given the K. One and K. Two of our diet protic acid. So with any type of tight rations, we need to first determine what the equivalent volume of our tie Trent will be. So we're gonna say our toy train is our strong base. So we're gonna say polarity of our acid times volume of our acid equals polarity of our base times the equivalent volume of our strong base plug in the values that we have. And then here we'll find out what our equivalents volume will be. So divide both sides. Now by .25 moller and we'll have our first equivalent volume is 100 mls. So we need 100 mls of RNA O. H. To get to the first equivalence point. Since we're dealing with the dia protic acid, we would need double that. To get to the second equivalence point. So 200 mls would be needed to get to the second equivalence point here, we're only dealing with 70 mls of N A. O. H. So we haven't reached our first equivalence point yet. So the calculations involved deal with getting before the equivalent the first equivalence point. And remember before the equivalence point is reached, we have to show the stark geometric relationship between our di protic acid and our strong base. So here we have our di protic acid plus an a O. H. Helps to produce sodium bicarbonate as our conjugate base plus water. We have our initial our change in our final amount When it comes to this. Tokyo metric relationship, our units can be either millie moles or moles. Here we ignore the water. We'll divide these mls by 1000 and multiply them by their molar itty to get our moles. So here we're gonna have 10000.0 to 50 moles. And here we're gonna have .0175 moles. And here this is zero Focus on the reactant, the smaller moles or subtract from the larger moles. So subtract .0175.0175. And then we're gonna add .0175 here. Bring down everything. So we have .0075 moles of my weak acid remaining. And we'll have .0175 moles of the conjugate base remaining again before we reach our first equivalence point or second equivalence point, we have the production of a buffer here. Since we're dealing with the first acidic hydrogen being removed to create this conjugate base, that means we're dealing with a K. A. One. So it would be PH equals PK. one plus log of my conjugate base over my weak acid. So that would be negative log of 4.3 times 10 to the negative seven Plus log of my conjugate base which is .0175 moles of sodium bicarbonate Divided by zero some 5 moles of carbonic acid. When we punch all that in, we'll get a ph of 6.73. So as always in our tight rations, even if it's with a dia protic acid, you should always determine what the equivalent volume of our tight trend is. First to see where within your titrate. In. Our calculations will take us. We know that it's occurring before the first equivalents point. So a buffer will be created. Therefore, we'll have to rely on the Henderson Hasselbach equation. Since we're dealing with the dye product acid, we must be aware that are we dealing with the first acidic hydrogen coming off and therefore use K one or are we talking about removing the second hydrogen and therefore K two will be needed? As long as you keep these things in mind, you'll be able to find the ph for the solution now that you've seen this example, move on to example two and see if you can find the ph for this next question. Once you do come back and see if your answer matches up with mine

8

Diprotic Acid Titrations Calculations 1

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So here we have to calculate the ph of 75 ml, of 750.10 moller phosphorous acid with 80 mls of 800.15 molar sodium hydroxide are added. Now we're told that the K one and K two are 5.0 times 10 to the negative two and 2.0 times 10 to the negative seven respectively. What we have to do is calculate what are equivalent volumes are. So, we're gonna say here are tight train is the sodium hydroxide. So, we're gonna have em acid times V acid equals m base times the equivalent volume of the strong base plug in what we know. So this is 0.10 moller Times 75 MLS Equals .15 Moller times the equivalent volume, Divide both sides by 0.15 Moller And we're gonna get our first equivalent volume as being 50 ml. And to get to our second equivalent volume, We would need 100 ml here. We have 80 mls being used. So that means we have passed the first equivalence point. And we haven't just we haven't quite reached the second equivalence point just yet. So, realize that we have our phosphorous acid here and we're talking about creating H two p. O three minus here. So that would involve my K. A. one. But remember we've gone beyond this point. So now we're talking about this particular one reacting with another mole of N. A. O. H. To produce HP oh 32 minus plus water. Yet again, We're dealing with K two. So it is the second equation in which we have to deal with finding the ph All right, so, we're gonna say here that we have a church to P. 03 minus plus an A. O. H. We're gonna produce H P. 032 minus plus water. We don't really care about the sodium. We're just gonna say that's a spectator ion. So we have initial change and final. Now at the first equivalence point, we would have had equal moles of my acid as well as my strong base. They would have totally destroyed each other. And what we would have had left at the first equivalence point is di hydrogen phosphate. So H two P. 03 minus. That's what we would have had. And it's moles. How do we determine its moles? Well, it's moles would be equal to the moles of our initial die protic acid. So divide this by 1000 and multiplied by the polarity. So the initial moles here will be 10000.75 moles. Now, we needed 50 mls of N A. O. H. To get to the first equivalence point here, we would have an excess of 30 mls left. Remember when we passed the first equivalence point we're looking for the excess moles remaining. So the excess volume is 30 MLS Of 0.15 Moller. So you would divide this by 1000 to get leaders and multiplied by the polarity. So we have 10000.45 moles of N. A. O. H excess this would be zero initially focus on the reactant. The smaller moles would subtract from the larger ones. So this would be .0030 moles. This would be zero. We would gain this number of moles. So we have now our new weak acid are new conjugate base amounts so we can deal with the Henderson hassle back. So because we're dealing with the second equipment well, since we're dealing with before the second equivalence point, that means we're dealing with PK two. So that would be negative log of K2 plus log of conjugate base over weak acid. So that would give me at the end a ph of 6.88. So again guys, it's important that we first are able to calculate the equivalent volumes for our titrate. Once we do that, we can tell exactly where we are within our given titrate in question. Once we know the location of our tight rations, we can employ either equations or set up some I. C. F. Chart in order to determine our concentration of H plus or oh H minus on our way to determining ph So just remember the fundamental steps needed. And you'll always be able to find your P. H. Or P O H.

9

Diprotic Acid Titrations Calculations 2

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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.

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Diprotic Acid Titrations Calculations 2

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So, carbonic acid is a die protic acid that associates losing its two protons to create bicarbonate and carbonate ion according to the following reactions given below. So here we have carbonic acid, which breaks up into bicarbonate ion and hydro nia. My on. We know that in reality it's really just the carbonic acid donating an H plus to water to form H +30 plus. But here we just have a simplified view of it that bicarbonate that forms can further um dissociate to form or carbonate ion and another hydro knee um, ion. Now we say here as a dye protic acid system, it has to dissociation constants that of P K one and K two. For the two steps in the reaction, you titrate 50 ml solution of 500.50 moller carbonic acid with one molar solution of sodium hydroxide. Now here it says, what would be the expected ph after the addition of 35 mls of the N A O h tightened so we know that the title is the strong species within our tight rations. Hey, we're gonna say, em acid times v acid equals m base times the equivalent volume of our tight trend. So here we're gonna plug in the numbers that we know ml equals one molar and we're looking for the equivalent volume, Divide both sides by one Moeller. So here my equivalent volume equals 50 ml. And then we'd say here, actually my equivalent volume one is equal to 25 mL. And my equivalent volume two is equal to 50 mL. So we need 25 mls to get to our first equivalence point and were given 35 mls. So we're dealing with calculations after the first equivalence point. So after The 1st equivalence point, so that means that we're dealing with bicarbonate ion, basically helping to create carbonate ion. So here we're going to have H C 03 minus plus an A. O. H. It's gonna create carbonate ion plus water. We don't care about the sodium, it's just gonna act as a spectator ion. Again, remember after the first equivalence point, we need the excess moles of our tie Trent. Right? So we needed only 25 mls but we have 35 mls. So we have an excess of 20 of 10 mls of N A. O H multiplied times its polarity to find our milly moles. So here, that would give us 10 million moles. Then here, remember, the amount of bicarbonate that we have is equal to the moles. The total moles of our carbonic acid. So here, that would be 25 million moles. Here, we ignore water. We have zero of this. The smaller millie moles would subtract from the larger millie moles. So we have 15 million moles zero and then we'd have 10 million moles here. So at the end we'd have weak acid and some conjugate base. So we'd have a buffer. Remember we're talking about removing the second acidic hydrogen form carbonate ion. That means we're dealing with K A two and therefore P K two. So PH equals PKA two plus log off conjugate base over weak acid. So that's 10.30 plus log of my conjugate base, which is 10 million moles, Divided by 15 million moles of my weak acid. We casted form. So here that would give me a ph equal to 10.12. So like all the examples we've seen thus far, we know that we have to determine where exactly within our tie Trish in our calculations are leading us. So here we're dealing with calculations after the first equivalence point, knowing that we have to start up stoking metric relationship between the reacting compounds. Right now, we realize that if we're before the first equivalence point, then we're gonna form a buffer. All we have to do is figure out the amount of conjugate base and weak acid remaining. And from there determine the ph by using the Henderson Hasselbach equation

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