Diprotic Acids and Bases Calculations - Video Tutorials & Practice Problems
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1
concept
Sulfuric Acid
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When it comes to dip protic acids. Just remember that sulfuric acid, which is H two. So four represents the only strong Dipro acid, all others will be considered weak. Here, we're going to say in terms of acidity, the first acidic proton associates completely and the second acidic proton only partially taking into account, both acidic protons will help give us a complete picture of the concentration of H plus ions. With that information, we can calculate ph. So just remember how would this look, we have sulfuric acid here reacting with water. Since it's strong, we'd have a solid arrow going forward. This would create HSO four minus aqueous plus H +30 plus. We've lost the first acidic hydrogen. So that's K one. But now we're trying to lose the second acidic hydrogen that's HSO four minus aqueous plus H2O liquid at this point, it's considered weak. So instead of having a solid arrow going forward, we have reversible arrows, meaning only a portion of the reactants break down to give us products here. This would deal with K two where we're losing the second acidic hydrogen, we create our sulfate ion plus additional H +30 plus ion. So this is the picture you have to take when it comes to sulfuric acid. The first acidic proton is lost completely, but only the second one, partially.
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example
Diprotic Acids and Bases Calculations Example
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Calculate the ph of a 0.0550 molar solution of sulfuric acid. Here, we're given the K one and the K two for sulfuric acid. Now notice that K one is a pretty large number, a number much greater than one. That's because the first proton of sulfuric acid is about strong Dipro acid K two though is a number smaller than one. That's because removing the second H plus ion is much more difficult. It's a weaker version of the original strong Dipro acid. All right, with this information, let's get to it. Step one. We're gonna use the Bronson definition to predict the predict the products formed when H two sl four reacts with water. So here we have H two. So four aqueous plus h2o liquid. Here, we're going to say that it's a strong acid. So it reacts completely with the water donating an entire H plus ion to the water molecule to create HSL four minus aqueous plus H +30 plus aqueous. Now, the initial concentration here of the sulfuric acid is 0.0550 molar. We're gonna say sulfuric acid completely dissociates ini initially, meaning the concentration of sulfuric acid equals the concentration of hydro ion and the intermediate form. So these would also have the same initial concentration. With this information, we can go on step two. In step two, we set up an ice chart for the intermediate form created in step one and reacted also with water. So the intermediate form um formed in step one was hydrogen sulfate. So hso four minus this is also called bisulfate. So here it reacts with water as an acid donating an H plus to the water molecule, it becomes the sulfate ion and water gaining in H plus becomes the hydro ion. Here we have is the initial concentration 0.0550. But also for the hydro waters of liquid liquids and solids are ignored in an ice chart. We don't have any information on the basic form. Now here using the initial row, we're gonna place the amounts from step one for the intermediate form and the H 30 plus which we did, we place a zero for the basic form. We now for step four, we're gonna say we lose reactants to make products. And we're gonna say here that using the change row, you're gonna place a minus X for the reactants and A plus X for the products. With this information, we just bring down everything that we have and that will give us our equilibrium rub OK. So bringing down everything this would be 0.0550. Minus X plus X plus X. Now our ice chart is filled in, we can say using the equilibrium row, set up the equilibrium constant expression with K A two and solve for X. Now we're using K A two because now we're talking about losing this last and final H plus of the dip protic species to create our basic form. At the end, here, we're going to say K A two equals products over reactants. So sulfate ion time hydro ion divided by the bisulfate ion K A two. When we plug that number in K A two, here would equal 1.2 times 10 to the negative two at equilibrium. Both of these products are X. So that's X squared divided by 0.0 And actually it wouldn't just be that it be 0.0550 plus X. So that actually be 0.0550 plus X times X divided by 0.0550 minus six. All right. So as we can see, this is gonna be a little bit tricky in terms of solving um this expression, we're gonna have to set up a quadratic formula in order to isolate X here. Now, here uh once we do that, we're going to solve the for ph by adding the hydro ion concentrations that we found in parts one and five to determine the complete H plus concentration or total H plus concentration. All right. So let's get to it, we're going to cross multiply these two together. So basically, this distributes here and this distributes here. So when we do that, what are we gonna get initially when we do that, we're gonna get initially 6.6 times 10 to the minus four, minus 1.2 times 10 to the minus two X equals this X can distribute here and distribute here. So initially, that's gonna give me 0.0550 X plus X squared, we have all these X variables. But if this X that has the highest exponent of two, so it's our lead term. That means everything has to be moved over to the right side. We're going to add this to both sides here. So it adds together with this X and they were gonna subtract this from both sides. So when we do that, we're gonna get X squared plus 0.067 X minus 6.6 times 10 to the minus four. This will be my A, my B and my C and with this, we can set up the quadratic formula. So still not done. So remember the quadratic formula itself is negative B plus or minus square root of B squared minus four AC over two A. Let's start plugging in the values that we have, right. So here we said that B is going to be negative 0.067 plus or minus 0.067 squared minus four times one don't forget the negative sign of C. So negative 6.6 times 10 to the minus four divided by two times one, we're gonna solve for everything in here and then take the square root of it. So when we do that, we're gonna get 0.067 plus or minus 0.08443 divided by two because this is plus or minus. That means there's two possible values for X one value is if we added 0.08443 which would give me initially 0.008715 or we could do minus 0.08443 which would give me negative 0.075 715. Remember the correct AX is the one that where you can place it anywhere on the equilibrium row and it gives you back a positive value. The negative one will, will not work because if you place it here, it would give you a negative concentration at equilibrium, which is not possible. That means that we cannot utilize this one. So now we're going to say what is our total H plus concentration? Remember H plus is the same thing as H +30 plus. It would be equal to this. This is the amount that we had initially from the first set up. Step one. And then we just found out what X is the additional H 30 plus that was created. So we're gonna say 0.0550 plus X. So that's 0.0550 plus the X that we found of 0.008715. When we add those together, we're gonna get 0.063715 that is my total amount of H 30 plus. With that, we can find PH which is negative log of H 30 plus. So plug that in and with this information, we can finally find RPH, which comes out to be 1.20. So this would be the final Ph of mito fur acid solution.
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example
Diprotic Acids and Bases Calculations Example
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4m
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When it comes to calculating the ph of weak Dipro acids, it's important to remember to utilize only K A one to calculate the ph of the acidic form of a weak dip protic acid. Here in this question, it says we have to calculate the ph of 0.115 molar carbonic acid solution. Here, we're given its K A one and it's K A two values. So step one is, we're gonna set up an ice chart for the weak Dipro acid that hasn't reacting with water. We're gonna use the Bronston or definition to predict the products formed. Here. This carbonic acid is gonna react with water which is in its liquid phase. Since it is the acid, it's going to donate an H plus to the water which acts as a base. This would create the bicarbonate ion or hydrogen carbonate ion which is HCO three minus one plus water. Gaining an H plus becomes H +30 plus. This is a weak Dipro acid. So we're utilizing an ice chart which stands for initial change equilibrium. Here we're going to follow step two. Next, using the initial row place the amount given for the weak Dipro acid. So 0.115 molar place a zero for any substance not given an initial amount. Well, here water is a liquid. Liquids and solids within ice charts are ignored. The products initially are zero. We're gonna say here we lose all step three, we're gonna lose reactants to make products. So here we're gonna say using the change row place A minus X for the reactants and A plus X for the products minus accents were losing reactants and plus A and were making products bring down everything. So 0.115 minus X plus X plus X with our ice chart filled in, we go to step four using the equilibrium row, set up the equilibrium constant expression with K A one and solve for X. So here K A one equals products over reactants. So it equal bicarbonate ion times the hydro ion divided by carbonic acid. With this equilibrium expression set up, we can place numbers within the spaces for K one and the different concentrations. And what's important here is we can check to see if a shorter can be utilized to avoid the quadratic formula to do this. We utilize what's called the 500 approximation method where we take the initial concentration of our weak dip protic acid and we divided by K A one. If we get a value greater than 500 we could ignore the minus X within our equilibrium expression. So here we have 0.115 molar divided by K A one. When we punch that into our calculators, we get 267,000, 441.9. So we got a number much greater than 500 which means within our equilibrium expression, we can ignore the minus X. And by doing that, we can avoid the quadratic formula altogether. So we come back up here, K one is 4.3 times 10 to the minus seven. Both products are X. So that's X squared divided by 0.115 cross multiply X squared equals 4.945 times 10 to the negative eight, take the square root of both sides, X equals 2.2 to 4 times 10 to the minus four. So we just found out what X is, this takes us to step five. Since we're looking at the acidic form of a weak dicrotic acid, the X variable will equal the hydro ion concentration. And with that, we can figure out ph, so we're gonna say here P equals negative log of H three plus concentration. So the negative log of 2.224 times 10 to the negative four punch that into our calculators. And we'll get 3.65 as the Ph for the acidic form of this weak Dipro acid in the form of carbonic acid.
4
example
Diprotic Acids and Bases Calculations Example
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So in this video, we're gonna learn how to calculate the concentration of the basic form of a weak Dipro acid. Now this is a special case. So pay close attention. We're going to say when given the initial concentration of our weak Dipro acid, the concentration of its basic form equals the K A two value. Now what does a question like this look like here in this example, it says determine the concentration of the carbonate ion when given 0.115 molar of carbonic acid solution. So they're giving us the initial concentration of our weak Dipro acid. So it's as acidic form. And they're asking you to determine the concentration of its basic form. Remember for a weak Dipro acid, we have the acidic form where it has both H plus ions. When it loses its first H plus ion, it becomes its intermediate form. And then when it loosens that second H plus, it becomes its basic form. Like we set up above if they give you the initial concentration of the weak dichroic acid, then K A two equals the concentration of the basic form. So here we just simply say that the concentration of the car carbonate ion equals K A two. So it equals 5.6 times 10 to the negative 11 molar. That would be our answer. Now, if you don't wanna continue on with this video, you can turn it off because what we're about to do is we're about to show proof of why that is OK. But again, remember if they give you the initial concentration of the wheat Dipro acid, the concentration of the basic form is equal to K A two. For those of you who are continue on with this video. Let's get started. So how do we prove this? Well, step one, we set up a nice chart to calculate the concentration of the intermediate form and the hydro ion. And we're doing this by using K A one. So the carbonic acid will react with water in its liquid face. It donates in H plus to water to become the carbonate ion or bicarbonate ion. Sorry plus the hydro ion. This is a nice chart. So it's initial change equilibrium. Remember with this, we ignore solids and liquids. The initial concentration of my weak acid is 0.115 molar, these are zero, we lose reactants to make product, we bring down everything. Using the 500% approximation method. We would see that we can ignore the minus X. If you're not familiar with this approach, make sure you go back and take a look at my video or I talk about the 500 approximation method. So here K A one equals products over reactants. So we plug in the numbers here. So K A one is 4.3 times 10 to the minus seven equals X squared over 0.115. When you cross multiply, you see that X squared equals 4.945 times 10 to the negative eight. And when you take the square root of both sides, you get X equals 2.224 times 10 to the minus four. Now that we've found X, we found the concentrations of both the intermediate form. So your bicarbonate ion and the hydroxide ion. With that information, we move on to step two, set up another ice chart and using the initial roll place the calculated amounts of the intermediate form and the hydro ion concentration. So here uh we would place 2.224 times 10 to the negative four molar. Here, this is reacting with a second molecule of water creating the carbonate ion, which is our basic form. And we create more hydro whose initial concentration would also be 2.2 to 4 times 10 to the minus four. Again, this number is coming from the first ice chart where we calculated the equilibrium concentrations still a liquid place a zero for the concentration of the basic form. In this case, carbonate ion set all this up again. So this is initial change equilibrium we're gonna say here we lose reactants to make products. So using the change row, again, we do minus X for reactants plus X for products bring down everything in 2.224 times 10 to minus four plus X. So now here we're going to say using the equilibrium ro set up the equilibrium constant expression. But now we're doing with K A two and we're gonna need it to solve for X. Now, we're gonna say we're gonna check our shortcut the 500 approximation method to see if we can avoid the quadratic formula. So here, the initial concentration is not the very first initial concentration. It's the initial concentration of our intermediate floor, which is the bicarbonate. And we're gonna divide it by K A two. When we do this, we're gonna get 3964285.7. So by doing the initial concentration divided by our K two. In this, in this case, we found a value greater than 500. But now we can ignore not only the minus X that we're used to seeing, but also plus X because as we can see here, we had initial concentration of H +30 plus which gave us a plus X here at the end. So we'd ignore those two things. So we ignore this one and we ignore this one. What else happens by ignoring the plus X and the minus X, what cancels out both of these two numbers because they're the same. And what do you have left at the end? You see that X equals the K A two value. Coming back up here in our eye chart, we can see X equals the concentration of the basic form of my dip protic acid. So again, it just simply equals K A two. So this will be our answer. So all of its work was just proved to show that if they give us the initial concentration of the acidic form of the weak dip protic acid, then K A two is equal to the concentration of the basic form. You don't need to do all this work. This was just to prove that OK, just remember you see a question like this. In the beginning, you can say, oh this is a special case, the concentration of the basic form is just simply equal to K A two. All right. So keep that in mind when dealing with this special circumstance, special case.
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Problem
Problem
Which of the following diprotic acid would produce the most acidic solution when dissolved?
A
0.200 M H2SO3 (Ka1 = 1.6 x 10-2 ; Ka2 = 4.6 x 10-5)
B
0.200 M H3PO4 (Ka1 = 7.5 x 10-3 ; Ka2 = 6.2 x 10-8 ; Ka3 = 4.2 x 10-13)
C
0.200 M H2CO3 (Ka1 = 4.3 x 10-7 ; Ka2 = 5.6 x 10-11)
D
0.200 M H2S (Ka1 = 8.9 x 10-8 ; Ka2 = 1.0 x 10-19)
E
0.200 M HC9H7O4 (Ka1 = 3.3 x 10-4)
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Problem
Problem
Determine the pH of 0.115 M Na2S. Hydrosulfuric acid, H2S, possesses Ka1 = 1.0 × 10−7 and Ka2 = 9.1 × 10−8.
A
7.04
B
10.05
C
7.90
D
13.06
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