Titrations: Diprotic & Polyprotic Buffers - Video Tutorials & Practice Problems
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1
concept
Diprotic Buffers
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Now, when it comes to dip protic buffers, we're going to say that they deal with the presence of two equivalent points and two Henderson Hasselback equation as a result of two K A values. Remember, Dipro means you have two acidic hydrogens. So you can lose two H pluses each with its own unique K A value. Here, we're going to say the relationships between the equivalent points and equations are shown as. So here we have our dissociation steps first. Remember a dip protic acid is in the common form of H two A here, this is its form where it has both acidic hydrogens. So we're gonna say that this is the acidic form K A one deals with losing that first acidic hydrogen when it does it's now H A minus, this is called the intermediate form. Now it still has another acidic hydrogen. So if it decides to lose that, that means we're dealing with K two and we finally get to this last phase A two minus where it's lost both of acidic hydrogens. So it would be a basic form. Now, we could also think in the reverse, we could start out with the basic form and we can gain our first H plus to give us our intermediate form. Gaining that first H plus means we're dealing with a B one. This intermediate form could then accept a second H plus to reform its acidic form, accepting that second H plus would give us KB two. Now we see that K A one and KB two are paired and K A two and KB one are paired that takes us to the KAKB equations. So here would say K A one times KB two gives us KW K A two times KB one gives us KW. Now the Henderson Hasselback equations, there's two of them because we have two acidic hydrogens. Now remember the general form, the regular form of our Henderson Hasselback equation is PH equals PK A plus log of base over acid. So if we take a look at this first Henderson Hasselback equation, it connects together the acidic form and the intermediate form. So here we're talking about losing the first acidic hydrogen. So it'd be PK A one plus log of base over acid here, remember the base has one less H plus. So the base here would be the intermediate form and the acid would be the acidic form. Now, the second Henderson Hasselback equation deals with the intermediate form of the basic form. Here, we're talking about losing the second acidic hydrogen to create the basic form. So we're dealing with K A two and therefore PK A two plus log of base over acid. Here. Again, the base is the one with one less H plus ion. So that will be the basic form. And the acid form here would actually be the intermediate. Remember, an intermediate can act as an A or base by donating an H plus or accepting an H plus. So these will be the two Henderson Hasselback equations associated with a typical Dipro buffer solution.
2
example
Titrations: Diprotic & Polyprotic Buffers Example
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Here it says, calculate the ph of 100 mls of 0.250 molar carbonic acid with 100 and 20 mls of 0.250 molar sodium hydroxide are added. Now, here we're giving K one and K two of our carbonic acid solution. What we do first is we need to calculate the equivalent volume needed to by the strong species to reach the first equivalence point. This will tell us how much we need to get to the first equivalent point in all successive ones afterwards. Now, here we're gonna do m acid times V acid equals m base times V base. Our strong species is the sodium hydroxide. So here we do of the acid 0.250 molar times 100 MLS equals 0.250 molar and V of the base. We're looking for that, we divide both sides by 0.250 molar and the line of the base needed to reach the first equivalent point is 100 g mos. This will help us determine how much tran was actually reacting in terms of this titration. So we're gonna say volume of strong speech to reacting equals your actual volume. So the volume given to us in the question minus your equivalent volume. So if we did this, it'd be 100 and 20 mls minus 100 mls. So there'll be 20 amounts. Now, why is that? Well, here, when we calculated the first equivalent volume, this tells me how much volume of NAOH is needed to get to the first equivalence point, which is 100. But in reality, we've gone beyond the first equivalence point because we have a fine, greater than 100 we have 100 and 20 MLS. So we've gone beyond the first equivalence point, the second equivalence point would just be another 100 on top of that. So we need 200 MLS to get to the second equivalence point, but we're not quite there. So here we're gonna use steps 1 to 3 to set up the IC F chart. We're gonna say before the first equivalents point volume, you start with the acidic form and PK one after the first equivalent point volume, you're gonna start with the intermediate form. And PK A two, we are beyond the first equivalence point volume at 100 and 20 os. So we're dealing with the intermediate form of carbonic acid which is bicarbonate. So HCO three minus here, it's still reacting with base. And over here we get our sodium bicarbonate, aren't, you know, we get our, this would donate in H plus. So we get water here and then we'd have N A two co three here. We don't really worry too much about balancing of this equation. We have this equation here given to us the sodiums or spectator ions. So they don't, they don't really matter. We don't need to actually include them. If we desire. We're gonna say that this is an IC F chart since it's weak and strong mixing together. So we have initial change. Fine. Remember with an IC F chart, we use moles as the units. So the moles of my intermediate will come from the moles of the original Dipro species. So remember, moles is liters times molarity. So divide the MLS by 1000 and multiply by the molarity and we'll have the moles of our intermediate form. So 0.0 to five moles, then we would take the reacting volume of our strong species, which is 20 MLS divided by 1000. And then multiply that by the original molarity of the strong base. When we do that, we get 0.005 moles, our, our conjugate base. Here, we don't have anything of it originally. And then here water is a liquid. So we ignore. Now look at the reactant side, the smaller moles will subtract from the larger moles. So we'd have 0.020 moles left of my intermediate form zero, left of my strong species. Now based on the law of conservation of mass, whatever we lose as reactants, we gain as products. So this would be plus 0.005 moles. Now, what do we have left? At the end? We have the intermediate form and then we have the final basic form in here. Just a simple fight. We'll just keep it as co 32 minus. So what we fundamentally have here is we have basin acid remaining. So we have a buffer. So step four says the Henderson Hasselback equation is used for a buffer to find the page of a solution using the fire ro use the moles of your weak acid and its conjugate base to find the Ph. Here we are after the first equivalence point. So we're using PK A two. So we're gonna say PK two plus log of base over acid. So here our PK A two is negative log of 5.6 times 10 to the negative 11. Who is our base, the one with less hydrogen. So that would be the carbonate ion. So 0.005 moles and then who is our acid form? The one with one extra H plus? So that's our bicarbonate form. So that's 0.020 moles plug this into our calculator. And we get 9.65 as the Ph for this titration.
3
concept
Polyprotic Buffers
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Now, when it comes to poly protic buffers, we typically are dealing with tri protic species because of this, we're gonna say it deals with the presence of three equivalence points. And Henderson Hasselback equations as a result of three K A values here, we're gonna say the relationship between the equivalence points and Henderson Hasel equations are shown as. So first, we're gonna talk about the dissociation steps. Here H three A is the generic form of atriotic acid. So here it has all of its H plus ions in its possession. So this is the acidic form. Now, here we're talking about giving away the first H plus. So that BK A one doing that gives us H two A minus. This would represent our intermediate form one that can continue to donate an H plus or donates a second H plus. So that'd be K A two that gives us H A two minus, which is our intermediate form two, this one has an H plus. So it too can donate an H plus, which would deal with K A three doing that gives us a three minus which is the basic form. Now, we could also go the opposite way where the basic form gained its first H plus to create the intermediate form too. Gaining your first H plus is KB one, the intermediate form two can gain another H plus. So that'd be KB two to give us intermediate form one. And then finally, the intermediate form one can gain the last H plus. It needs to recreate the acidic form. So that'd be KB three, we'd see here that these two are grouped together. These two are grouped together, these two are grouped together. So that segues into our KAKB equations here, we'd say K A one times KB three equal KW K A two times KB two equals KW and K A three times KB one equals KW. From there, we could talk about the different types of Henderson Hasselback equation. So the first one deals with the acidic form and the intermediate form one, we're talking about the first K A here. So this would be PK A one plus. Now, all of these are based on this simpler idea of Henderson Hasselback equation is PH equals PK A plus log of base over acid. The base is the one with one less hydrogen. The acid is the one that has one extra. He so here it'd be plus a lot of the base over assets. So that'd be H two A minus over H three A. The next one deals with the two intermediate forms. So they're connected by K A two. So this would be PK A two plus log of base over acid. So that'd be H A two minus over H two A minus. And then finally, here, intermediate form two and basic form give us the last Henderson Hasselback equation. We're dealing with K A three in this regard. So this would be PH equals PK A three plus A three minus over H A two minus. So these would be the three Henderson hasselback equation associated with a tr a typical tri protic buffer.
4
example
Titrations: Diprotic & Polyprotic Buffers Example
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Here, it says to calculate the ph of 30 MLS a 0.10 molar of this particular tr protic acid. When 20 MLS, a 0.20 molar sodium hydroxide. At here, we're given the three kas of our trip protic acid. So what we start with in doing is we calculate the equivalent volume needed by the strong species to reach the first equivalence point. If we know that, then we can determine the divine needed for each successive equivalence point. So here we're going to say M acid times V acid equals M base times V base. Here, our acid is 0.10 more. Its volume is 30 amounts. Here, we have 0.20 molar and we're looking for the vine of the base, divide both sides by 0.20 molar. So V of our base equals 15 amounts. So that would mean that the equivalent volume one would take 15 MLS. If we wanted to get to the second equivalent volume, we'd add another 15. So there'd be 30 MLS. And if we're trying to get to the third equivalence point, it would take another 15 MLS. So we need 45 MLS. All right. So we're at 20 MLS. So we've gone beyond the first equivalence point, but we haven't quite reached the second because it takes 30 to get to the second equivalence point. So here, the volume of strong species reacting would be our actual volume given to us within the question minus your equivalence volume actual is 20 MLS. We only need 15 MLS to get to the first equivalence point. So we're five MLS beyond the first equivalence point. Now, here we set up steps 1 to 3 to use the ICF chart. So we're gonna say before the first equivalent point, we're gonna say we start with the acidic form and PK one, after the first equivalent 0.1 you start with intermediate one and PK A two, this is where we are. And then we're gonna say after the second equivalent point volume, you start with the intermediate two form and PK A three. So here again, we're just after the first equivalence point, we have it reached the second equivalence point. All right. So we're gonna use the intermediate form one and PK A two. So the second in the, the, the first intermediate form would lose one H plus. So it'd be H two C three or C six H 507 minus. Let's go back up here. Make sure. So that would be the first intermediate and it's still reacting with an A oh. So the acid would donate an H plus to the base to give us water and then we get our second intermediate form being created. So the H it's gonna be C six H 507 to minus aqueous. Here, I don't include the sodium. It's just a spectator ion. This is an IC F chart since we're mixing weak and strong. So we have initial change fine. Now here, the amount of my intermediate form is equal to the amount of my Dipro acid because that's where the dot the intermediate form originates from. And so we're gonna do one hun 130 ml divided by 1000 to get liters and multiply by the molarity that will give me the moles of the intermediate form. One that comes out to 0.003 moles. Remember moles equals liter times molarity here for the base, we would do the reacting volume which is five ML. So you divide it by 1000 and then multiply it by the molarity of the strong base. OK. So when we do that, we get 0.001 moles, we don't have any of the second intermediate form. So that's zero. Water is a liquid sword can ignore it. Now look at the reactant side, the smaller moles will subtract from the larger moles. OK. So we have that and then we're gonna say here, whatever we lose on the reactant side, we gain on the product side. So it's gonna be plus 1001, bring down everything. So this is zero. And then this is here point 001. Now coming down here, we have to fill this out. So what we have at the end is we have the intermediate form one which can serve as our weak acid form. And here we have the intermediate form two which has one less hydrogen. So this can serve as our base form. So we're gonna use the fun, we're gonna use the moles of the weak acid and its conjugate base to find ph. So again, we're after the first equivalence points, we're using PK A two ne be log of our base over our acid. So our base is the conjugate base or acid is the weak acid form. All right. So we're gonna plug these numbers in here and when we do that, that's gonna give us our new Ph. So then plus log of it's so we're gonna do 1.7 times 10 to negative five here. This is 0.001 this is 0.002. So we're gonna do that and that gives me 4.46 as my ph. So all we're doing here is we're adding uh what we got from the IC F chart into our Henderson House Eva Equation. Doing that helps us to isolate the Ph of our solution.
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Problem
Problem
Calculate the pH of 75.0 mL of a 0.10 M of phosphorous acid, H3PO3, when 80.0 mL of 0.15 M NaOH are added. Ka1 = 5.0 × 10−2, Ka2 = 2.0 × 10−7.
A
7.12
B
1.35
C
6.88
D
12.65
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Problem
Problem
Find the pH when 100.0 mL of a 0.1 M dibasic compound B (pKb1 = 4.00; pKb2 = 8.00) was titrated with 11 mL of a 1.00 M HCl.
A
8.00
B
6.95
C
3.00
D
7.05
7
Problem
Problem
Suppose you have 50.1 mL of a H3PO4 solution that you titrate with 15.4 mL of 0.10 M KOH solution to reach the endpoint. What is the concentration of H3PO4 of the original H3PO4 solution?
A
0.00200 M
B
0.00154 M
C
0.00649 M
D
0.0307 M
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