We are ready for our next learning objective and as it says here in the on screen we are going to be talking buffering capacity and how to prepare a buffer for a specific pH. To do this you first have to choice the right substances to be apart of that buffer. You want to choice an acid whose pK value is close to the desired pH you are looking for. Now why would that be? We think of the Henderson-Hasselbalch equation which applies to buffers. The pK_a value of the acid in that buffer is going to be your starting point. This term how much base and how much acid you put in there, that ratio base to acid will tweak it up or down depending upon whether you need it a little bit higher then that pK_a value of pH or a little bit lower of that pK_a value. But you want to start with an acid whose pK_a value is close to the desired pH. If you are looking for a buffer that is in the acid's range, meaning If you are looking for a buffer that is in the acid's range, meaning less than 7. You are going to be going to tables of acids, you are going to be looking at those acids and finding one whose pK_a is close to what you want. But if you wanted to make a buffer and finding one whose pK_a is close to what you want. But if you wanted to make a buffer that has a higher pH lets say great then 7 in the base range, then you want to be choosing a starting material which is a base. So that you have got a basic buffer. But again you are going to be looking at the K_a value of its conjugate acid. So if you were looking at base like ammonia you need to consider acid ammonium and say what is the K_a value of ammonium and gets its pK_a value. So you are looking for something in that range. Once you have decided the substances that will make up your buffer then you are going to take that Henderson-Hasselbalch equation and you are going to solve for the ratio then you are going to take that Henderson-Hasselbalch equation and you are going to solve for the ratio of the base over the acid. And that way you will know how much of what concentration of those two items you are going to need. In order to make that pH. You might want also to consider the moles of base over moles of acid, instead of concentrations Then once you know how many moles of each that you are going to need or the ratios of moles you are going to need then you could say well, how many grams will i need to weight out to get that many moles or I want to do a dilution and I am going to figure out to get that many moles or I want to do a dilution and I am going to figure out molarity of the volume I need to get that number of moles. Now in this problem, I have already told you what the substances are that are going to be making that buffer. I am going to show you the idea of solving for the ratio. Consider the Henderson-Hasselbalch equation pH equals pK_a Consider the Henderson-Hasselbalch equation pH equals pK_a plus the log pH equals pK_a plus the log and I am going to do it in term of moles. plus the log and I am going to do it in term of moles. Number of base over the number of moles of acid. and I am going to do it in term of moles. Number of base over the number of moles of acid. I will plug in that I am trying to Number of base over the number of moles of acid. I will plug in that I am trying to obtain a 4.80 pH and I am going to use the negative log of the acid which is 2.0 x 10 ^ -4 My goal will be to solve for the log of the number moles of base, over the number moles of acid. The negative log of 2.0 x 10 ^-4 is 3.70. So I will subtract 3.70 from both side and have 1.10. Now we have to get rid of the log log function inverse log is 10 to that value. So those will cancel and we will be left with 12.6 equals the number moles of base over the number of moles of acid. Another way to write that would be 12.6 times the number of moles of acid would give you the number of moles of base. So if at this point you have all sort of value that would work and give you a pH of that amount. Lets say we will choose a .10 moles of the acid. If we choose that amount then we will know that if we multiply that by 12.6 that would give me the number of mole in the base I would need. In other words I would need 1.26 moles of base. So if I made a solution that had those two amounts .1 mole of acid 1.26 moles of base then I would know I would have created a buffer with a pH of 4.80 which is what I was trying to achieve. It does not have to be .1 moles. It does not have to be .1 moles. I choose any amount here .1 moles. I choose any amount here lets say I want to have more acid than more base. Maybe I want to have I want 1.0 moles of acid. Well that would tell me that I would need to have a solution with 12.6 moles of base in it. But whatever, that ratio has got to be 12.6 to 1. Now buffers will quit working after a while. You add an acid it will will quit working after a while. You add an acid it will drop the base portion of that buffer down. If you keep on adding acid drop the base portion of that buffer down. If you keep on adding acid you keep dropping it down. Buffering capacity is the amount of acid of base that you can add to that buffer without causing a large change of pH. Eventually you wipe out that buffering capacity. You will have used up the substances and you no longer have a good ratio. Of conjugate acid and conjugate base and you no longer have a good ratio. Of conjugate acid and conjugate base pairs in there, in your buffer. If you look at this diagram down here in the bottom right. We see that they both have the same ratio of base to acid. So the number of moles if we let each ratio of base to acid. So the number of moles if we let each one of those So the number of moles if we let each one of those figures rep[resent a mole the number of moles of base to the number moles of acid the number of moles of base to the number moles of acid is 4 to 4 or 1 over here. The number moles of base to the number of moles of acid is 2 to 2 it would have the same pH. to the number of moles of acid is 2 to 2 it would have the same pH. And it would have, but one would have better buffering capacity then the other. So anytime you increase the concentration of the acid or the base in your buffer, then you increase So anytime you increase the concentration of the acid or the base in your buffer, then you increase the capacity. the base in your buffer, then you increase the capacity. Mess up what is popping up next. We have a diluted buffer and we have a concentrated buffer and certainly the concentrated buffer has a better buffering capacity. You could also do a ratio of acid to base and i will maybe draw it in like this. Lets say I have got four of the acid and I only have two of its conjugate base. If I have a buffer like this it has a good buffering capacity to neutralize it has a good buffering capacity to neutralize an added base. Because there is quite a bit of acid in that buffer but it would not have as good a of acid in that buffer but it would not have as good a buffering capacity again but it would not have as good a buffering capacity again added acid because the added acid would be neutralize by the base form of that buffer. So you could have a buffer in this case because you needed a certain pH range. That would be more effective in this case because you needed a certain pH range. That would be more effective against the addition of an acid verse the addition of a base and vise versa. This is just a diagram we have seen earlier and we can imagine in this one if we had less acid to start with then we had base. And so we are start off with this down here and not all of this up there. A less amount of this acid. that this buffer would be a good buffer against added acid. Because there was quite a bit of the conjugate base in there. but it would not be as affective again added base because I am starting with a lesser amount. So that is the end of our learning objective for four you know what buffer capacity is and you know how to create a buffer with a specific pH range.
