A LEVEL CHEMISTRY - BUFFER SOLUTIONS CRAM

Added: 2026-05-05

[00:00:00]Hey everyone. So this is my first attempt at a topic cramm video. So basically what that means is I've chosen a topic. So buffer solutions is the first one I'm doing and I've taken 10 past exam questions, put them all together in one file. So you've got 10 questions. The PDF for that file will be attached to the video. Um, and the video itself is me walking through those 10 questions. Uh, I'll also put timestamps into the description so you can go straight to each one if if that works for you. I'd love some feedback on this video if you don't mind because it's the first one I've done so I'm kind of learning as I go. Uh, so any constructive criticism would be greatly appreciated. All right. So, enjoy hopefully and uh I'll do another one.

[00:00:57]Cheers. Bye.

[00:01:01]Okay. So, go through the answers. So, the first thing I'm going to say is the key component of a buffer is a weak acid and a conjugate base. So, from the glycolic acid, we've got a weak acid.

[00:01:13]And from the glycolate ion, we've got the conjugate base. So if we've got these two species in solution, we've got this equilibrium established. So how does it act as a buffer solution? Well, if you add some acid, a small amount of acid, you're obviously increasing the H+ ion concentration. So this reservoir of weak base, so in the case of this buffer, it would be the glycolate ions.

[00:01:39]They can accept the extra H+ ions and send the equilibrium over to the left hand side. if you added a small amount of alkali. So if you add a small amount of hydroxide ions, that's going to lower the H+ concentration and so the reservoir of weak acid will dissociate more and replace the lost H+ ions. So the upshot of both of those processes is that the pH remains roughly the same.

[00:02:13]So if we move on to the second bullet point now. So we got to explain how this buffer can be prepared from ammonia and glycolic acid. So remember we're after the glycolic acid and the glycolate ions being present in the same solution. So the way you do that is you would take an excess of the acid and neutralize it obviously partially with the ammonia.

[00:02:38]So at the end of that process, you would still have some glycolic acid left, but you would also have some glycolate ions.

[00:02:48]So you can see from the equation, if you've got an excess of the glycolic acid, obviously at the end of that reaction, you would still have the glycolic acid left. So there's one of your key components, but you'd also have the glycolate ion, and that's your other key component.

[00:03:09]Okay. So, we'll go through the answers then. So, part A, how does the carbonic acid hydrogen carbonate mixture act as a buffer? So, we've got these two species in solution in the blood. So, we've got a weak acid and we've got the salt ion of a weak acid. So, the conjugate base of the weak acid. Both of these species together act as a buffer solution because if acid is added that means you're increasing the H+ concentration this reservoir of hydrogen carbonate ions can accept the extra H+ ions and send the equilibrium over to the left.

[00:03:49]Conversely, if hydroxide ions enter the blood that they're going to react with the H+ ions and lower the concentration of H+. So the reservoir of carbonic acid dissociates more and puts the H+ ions back into the system and that will maintain the pH at around about that healthy level of of 7.4.

[00:04:16]So moving on to calculation now.

[00:04:18]Obviously the first thing we need to establish is the expression to calculate the H+ concentration of a buffer. I've got a silly way to remember it. If you watch my videos, my teaching videos on this, I call it the cassid over salt expression. Ka of the weak acid times the acid concentration divided by the salt concentration. Cassid over salt. So from the information given, we can calculate the H+ concentration from the pH. We can feed in the um salt to acid concentration and from that we can calculate the Ka for healthy blood. So there's that there. I'll just explain it. So healthy blood H+ concentration is 10 the minus 7.4 that comes out at that.

[00:05:02]Then if we rearrange the um over salt expression for Ka, it rearranges to H+ concentration. Take the salt concentration up there. So times the salt concentration divided by the acid concentration. Feed the numbers in. So the Ka for the healthy blood is that. So now we know that we can feed that back into the calculation for the patient's blood. Remember the patient's blood pH7.2 and we can get the salt to acid concentration from that.

[00:05:37]So there's that calculation there. I'll just quickly explain that. So we want the salt to acid concentration.

[00:05:44]So basically we need to put that up there, that down there. Obviously that comes underneath Ka. So we get that expression there. Put in the Ka we've just calculated. H+ concentration of the patient's blood is 10us 7.2 and that gives us a ratio of 6.63 to 1.

[00:06:15]Okay. So make a start. So we're going to explain why a buffer solution is formed when an excess of proponinoic acid is mixed with aquous sodium hydroxide. So we'll just start by reminding ourselves what the key components of a buffer are.

[00:06:27]So it's a weak acid. So in the case of this question that's going to be your proponinoic acid and conjugate base. So again in the case of this question it's going to be sodium propinoate because we're using sodium hydroxide in this reaction with the proponinoic acid. So there's the equation for the reaction between proponinoic acid and sodium hydroxide. So it's making the salt sodium propinoate and water.

[00:06:54]So if you use an excess of the weak acid, that means at the end of this reaction, you're still going to have some weak acid left, but you're also obviously you've made this salt here. And within that mixture, you've got the key components. You've got your weak acid and you've got your conjugate base.

[00:07:24]Okay, so moving on the second bullet point. Now we've established why we've got a buffer solution. Got these two key components. How does it work or how does it control pH when acid or alkali is added? So if we start with um the effect of acid. So if you add acid, a small amount of acid to this buffer system, you're effectively increasing the H+ concentration. So you've got this reservoir of conjugate base ion from the salt, which can react with the H+ ions or the extra H+ ions put in from the additional acid. It's going to send the equilibrium over to the left. It's going to bring the H+ concentration back down roughly to where it was before and therefore maintain the pH level.

[00:08:12]And then if we look at the flip side of that. So if you add a small amount of alkali, you're adding O minus ions, hydroxide ions, they're going to remove the H+ ions. So the concentration of H+ ions will decrease. So your reservoir of the weak acid now dissociates more to replace the lost H+ ions. And so the equilibrium shifts to the right. And again the H+ concentration will get back to roughly where it was before and therefore the pH is maintained.

[00:08:54]So we'll make a start. You'll see I've already written up the cassid over salt expression be your starting point for any buffers um calculation. So the H+ concentration of a buffer is equal to the Ka multiplied by the acid concentration divided by the salt concentration. So once you've waited through all the information about the sweet, its magic tang, and all that sort of stuff, basically all we want to find is the acid to salt concentration ratio.

[00:09:19]So we make that the subject of this equation. How do we do that? We take the Ka over to the other side and it goes underneath the H+ concentration.

[00:09:30]So we just need to sub in now the H+ concentration and the Ka. And the sort of twist in this question is they haven't given us those. They've actually given us the pH and they've given us the pKa of the lactic acid. So to calculate the H+ concentration from the pH, it's 10 the minus pH. So 10 the minus 3.55 divided by. So to turn pKa into Ka, it's 10 the minus pKa. So this would be 10 the minus 3.86.

[00:10:03]If you put those numbers in your calculator, you'll get an answer of two, which means the ratio needs to be 2:1.

[00:10:09]So in other words, your acid needs to be double the concentration of the salt.

[00:10:16]And this second bullet point, comment on the validity of the prediction that the pH of the suite would give the sweets their magic tank. Well, that's not really valid because there might be other chemicals in the sweet that are contributing to this tangy flavor that they've got. So, it's not essentially linked to the pH of the sweet.

[00:10:44]Okay. Okay. So like I always do when we have questions like this, we'll talk about the key components of a buffer being a weak acid and the conjugate base. So the weak acid in this case is butyricininoic acid and the conjugate base will be the butinate ion.

[00:11:00]So if you think about the reaction that takes place between butyricininoic acid and sodium hydroxide, it's this one here. So we're making sodium butininoate and water. So within that we have got the bututininoate ion and we have got a weak acid. So all we need to do is establish whether at the end of this reaction we've got both present. So the first thing I've done is worked out the initial moles of the two chemicals. So the butininoic acid and the sodium hydroxide. So just concentration times volume in decimeters cubed. So we've got this number of moles of butyricininoic acid, this number of moles of sodium hydroxide. So the next thing we need to do is work out the final moles. So the moles left at the end of the reaction. So this is the limiting reagent. There's the fewest moles of sodium hydroxide. There'll be none of that left. They will make the same number of moles of the salt, the conjugate base. So we're going to have 0.0025 moles of salt made.

[00:12:05]the acids in excess. See that number's bigger than that. It's going to react in a 1:1 ratio with the sodium hydroxide.

[00:12:13]So 0.025 moles of acid will react. So the difference between that and that is what's left. So the important thing is that we have got both key components present at the end of the reaction.

[00:12:26]We've got the weak acid and we've got the conjugate base in the form of this um sodium butininoate. So yes, a buffer will form.

[00:12:41]Okay. So moving on to part B. Now we've got to calculate the pH of the buffer.

[00:12:45]So you can see I've written up my acid over salt expression. So the H+ concentration of the buffer is equal to the Ka for the weak acid multiplied by the acid concentration divided by the salt concentration.

[00:12:58]So we were given the Ka for the acid. It is 1.51 * 10 minus5.

[00:13:06]So we need to multiply that by the acid concentration. Well, we've got the moles of acid that are left at the end of the reaction. The total volume that the everything's in is going to be the combined volume of the two solutions. So remember it was 50 cm cubed of acid was reacted with 50 cm cubed of sodium hydroxide. And so therefore the total volume is going to be 100 cm cubed not.1 decimeters cubed.

[00:13:31]So the concentration is therefore the moles over the volume. So that's why that's there.

[00:13:36]So we're going to divide that by the salt concentration.

[00:13:40]So there's the moles of salt. Remember they're also in that 0.1 decime cubed volume. So that gives us an H+ concentration in the buffer of 6.04 * 10 - 5 moles per decime cubed. So all we need to do now is minus log that to get the pH.

[00:13:59]we get a calculator value for the pH of that and it wants it to two decimal places which is 4.22.

[00:14:09]Now just quickly going back to this stage here. Now technically you could use the mole ratio 0.01 over 0.0025 0025 because they're both in the same volume.

[00:14:21]But I always get my students to work in concentrations because that's what's in the casova salt expression. So that's why I did that.

[00:14:40]Okay. So make a start. And I'm sure you recognize this type of question from previous exam papers. But we'll start by saying the key components of a buffer are a weak acid and the conjugate base of the weak acid. So in this case we've got methaninoic acid is our weak acid and the conjugate base would be the methaninoate ion. So if we think about the reaction that's taking place it's between methaninoic acid and sodium hydroxide. That's going to produce sodium methanoate. So obviously that contains that methaninoate ion and of course you get water. So, we've got to explain why a buffer is formed. So, we're going to look at the the moles that we've got at the start and then the moles that we've got at the end of the reaction. And we just need to prove that we've got both of these key components present in the solution.

[00:15:29]So, first thing we're doing is calculating the initial moles of the acid and the sodium hydroxide. So, concentration times volume. Remember, volume needs to be in decimeters cubed.

[00:15:39]So we've got this many moles of acid and this many moles of sodium hydroxide. So at the end of the reaction, you'll notice that the methaninoic acid is in excess. These react in a 1:1 ratio. So we've got excess moles of acid. So all of the sodium hydroxide will be wiped out, but they will make or they will limit the number of moles of salt formed to 0.4 because of the 1:1 ratio between them. The difference between these two numbers is how much acids left. So at the end of the reaction, we have got some moles of acid left, 0.24 to be exact, and we have got obviously some moles of salt formed, 0.4. So that's why we've got a buffer because we've got both of the key components present.

[00:16:31]Okay, so moving on to the calculation now. So we've got to calculate the pH of the buffer formed. So there's my cassid over salt expression. The H+ concentration of a buffer is equal to K a time the concentration of the acid divided by the concentration of the salt.

[00:16:48]So the Ka was 1.7 * 10 -4.

[00:16:53]So the acid concentration will be the moles left divided by the volume they're in. So it's a nice easy volume this one decimeter cubed. But I'm going to put the the full sort of calculation in just to make the point that uh concentration is moles divided by volume. And we're dividing that by the concentration of the salt 0.4 / 1. Comes out at an H+ concentration of 1.02 * 10us 4 moles per decimal cubed. So all we need to do now is minus log that to get the pH. But remember we need to give it to two decimal places. So we're getting a pH calculator value of that obviously to two decimal places 3.99.

[00:17:46]Okay. So we'll make a start. So if you watched a few of my videos, you'll know I like my little diagrams where I can visualize what's going on in the question. So we've got to create a buffer solution with a pH of 5. The total volume of the buffer is 400 cm cubed and that's initially made up from this 0.2 moles per decime cubed ethaninoic acid. All we're doing is dissolving x g of the sodium ethaninoate salt into there to generate the buffer.

[00:18:15]So obviously we've got to calculate how many grams um of sodium is needed to generate this pH of five. So, as always with any buffer calculation, the first thing we start off with is the cassid over salt expression. It's what I call it anyway. So, the H+ concentration of a buffer is equal to Ka multiplied by the acid concentration divided by the salt concentration.

[00:18:42]So what we need to do in this question is calculate the salt concentration and then from the volume we can work out how many moles of salts needed and then we can turn that into grams. So the acid over salt expression rearranges to that Ka * the acid concentration over the H+ concentration and now we just need to put the numbers in. So the Ka for the acid is 1.75 * 10us 5. The acid concentration was given in the question 0.2 moles per decimeter cubed and the H+ concentration is going to be 10 minus pH. So that's 10 -5.

[00:19:23]So the salt concentration is 0.350 moles per decime cubed. So now we need to work out how many moles of sodium methanoids needed. So moles of CH3 CO N A is concentration time volume. So it needs to be 0.35 concentration. The volume is 400 cm cubed. That's 0.4 of a decime cubed. So that's 0.14 moles of sodium. So the mass of that is going to be the moles times the MR. So that's 0.14.

[00:20:04]The MR is 82 and that comes out at 11.48 g.

[00:20:21]Okay. So we'll make a start. So part A, we've got to calculate the pH of the buffer. You can see I've already written up my casa salt expression. So the H+ concentration of a buffer is Ka time the acid concentration divided by the salt concentration. Now if you notice in the information here it says the concentrations of the proponinoic acid and the proponinoic ions are both 1 m per decimeter cubed. Basically they're going to cancel each other out which will it'll leave H+ concentration equal to K a.

[00:20:52]So we've got the Ka value for the proponinoic acid 1.35 * 10us 5. So that's the H+ concentration as well. So all we need to do now is minus log that to get the pH.

[00:21:05]So to two decimal places that comes out at a pH of 4.87.

[00:21:11]So, moving on to part B, I've written up the equilibrium again for the buffer, and we've got to use this to explain um the effect or how the buffer would respond if a small amount of ammonia solution is added. So, ammonia is a base. It's going to accept H+ ions and form ammonium ions. So, the upshot of that is the H+ concentration is going to drop. So the equilibrium responds by the proponinoic acid dissociates more and puts the H+ ions back. So in other words, the equilibrium shifts to the right.

[00:21:50]Okay, so we'll move on to part C. It's not easy this. Um so if you get this bit right, you're doing really really well.

[00:21:57]Um probably one of the hardest buffer questions I've seen. Right. So, we're told the student adds this many grams of magnesium to one decimal cube to the buffer. So, the magnesium is going to react with the propinoic acid and it's going to take some of the proponinoic acid out of the buffer. And you'll see from the equation, it's actually going to generate some of the propinoate ions.

[00:22:21]So, it's going to put some propenoid ions into the buffer. So, the first thing I've done is worked out the moles of magnesium. Mass over mr. 0.25 25 moles.

[00:22:32]So then if we feed that into the buffer equilibrium, so the mole ratio is telling us that if 0.25 moles of magnesium is reacting, 0.5 moles of acid is going to react because of the ratio. So this is going to change. It's going to go down by 0.5 from the initial one mole per decime cubed. So the new moles or the new concentration because it's all in one decimeter cubed told there is going to be 0.5 moles per decime cubed. And then if we look at what's going to happen to the prop8 ions they're going to increase by 0.5.

[00:23:17]Just go back to this equation. So for every mole of magnesium that reacts we get twice as many moles of proponinoid ions formed. So we're going to get 0.5 moles of proponent ions forming. So that's going to go up from 1 to 1.5 moles per decimeter cubed.

[00:23:36]So if we feed that into the over salt expression to calculate the new H+ concentration, Ka is still the same 1.35 * 10us 5, but the new acid concentration is 0.5. The new salt concentration is 1.5. So that's given us an H+ concentration now of 4.5 * 10 -6 moles per decimeter cubed.

[00:24:04]So the new pH of the buffer is going to be minus log of this new H+ ion concentration. So it's minus log of 4.5 * 10us 6 and to two decimal places it's comes out at 5.35.

[00:24:21]So like I said before, very very well done if you got that right because that is not straightforward.

[00:24:33]Okay. So make a start. So I'll use this diagram to explain what's happening in the question. So the chemists taken um a weak acid and the conjugate base. So the salt, sodium nitrite is a conjugate base. They're obviously the key ingredients for a buffer. So we know the volume and concentration of each of these two components. So the first thing we're going to do is work out how many moles of the nitrous acid and sodium nitrite we've actually put into this buffer solution. So just two concentration times volume calculations.

[00:25:09]Concentration of the acid 0.2 volume in decime cubed 0.2. So we've got that many moles of the acid. And again concentration times volume that many moles of the NO2 minus ion.

[00:25:24]So we'll bring in the acid over salt expression now. So the H+ concentration of the buffer is the Ka of the weak acid multiplied by the acid concentration divided by the salt concentration. So there's the numbers there. Just remember that these moles are in one decimeter cubed anyway. So they're the concentrations as well.

[00:25:44]So that gives us an H+ concentration in the buffer of 3.752 * 10 - 4 mo/ cubed.

[00:25:50]So all we need to do now is minus log this and give our answer to two decimal places.

[00:25:57]And that comes out at 3.43.

[00:26:08]Okay. So make a start. So you'll notice I've written up the equation again, but I've highlighted the key components of a buffer solution. So we've got the weak acid um proponinoic acid in this case and the propinoate ion from the berium propino8. So basically what we've got to do is establish whether or not we have got both of these components present at the end of this reaction. And if the answer's yes, we've made a buffer. So the first thing we got to do is work out from the concentration and volumes of these two chemicals how many moles of each have been used in the reaction.

[00:26:42]So concentration times volume in decimeters cubed remember gives us moles of acid at that moles of berium hydroxide at that. So now we need to work out what our final moles are going to be. So once the reaction's happened what's left in the beaker if you like.

[00:27:00]So we'll start with the moles of berium hydroxide. We've got 0.0.125 moles of those. So if we look at the ratio, we're going to need twice as many moles of proponinoic acid. So 2 times that is 0.0.25 moles. We've got 0.0648. So yeah, we've got more than enough moles of propinoic acid. So at the end of the reaction, our final moles of the acid will be this number here.

[00:27:27]All of the berium hydroxide's going to react because the acid was in excess, but we're also going to make 0.0.125 moles of berium propino8. So what that means is we've got both of the key components present at the end of the reaction and so therefore we have made a buffer solution.