Buffer solution pH calculations
Example of calculating the pH of a buffer solution using the Henderson-Hasselbalch equation, including the pH of the buffer solution after adding some NaOH.
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- At 5.38--> NH4+ reacts with OH- to form more NH3. Since it is an equilibrium reaction, why wont it then move backwards to decrease conc of NH3 and increase conc of NH4+?(29 votes)
- The additional OH- is caused by the addition of the strong base. So these additional OH- molecules are the "shock" to the system. The system counteracts this shock by moving to the right of the equation, thus returning the system to back to equilibrium.
So when the reaction moves to the right, equilibrium is restored. The system is then at equilibrium and there is no cause for the system to move "backwards".
Think of it as a single shock and a single response by the system. Otherwise the logic gets a bit circular and by that logic the system is never able to settle.(16 votes)
- This may seem trivial, but at3:40, why is the hydroxide ion written with the charge on the left-hand side, instead of the right? I've seen it that way consistently in these chemistry videos, but never anywhere else.(6 votes)
- It is preferable to put the charge on the atom that has the charge, so we should write ⁻OH or HO⁻.
However, many people still write the formula as OH⁻. It's OK, as long as you remember that the O atom has the charge.(28 votes)
- At2:06NH4Cl is called an acid, but isn't it a salt?(5 votes)
- It is a salt, but NH4+ is ammonium, which is the conjugate acid of ammonia (NH3).(24 votes)
- How would I be able to calculate the pH of a buffer that includes a polyprotic acid and its conjugate base?
For example, if I were to add a certain amount of the polyprotic acid H2A and a certain amount of its conj base A^2- (obviously in the form of a salt, say Na2A), how would I find the pH? Assume I know the ka1 and ka2 values. Now what? Do I use good ole Henderson Hasselbach? Which pKa do I use? Is there another equation to use in these instances?(5 votes)
- You can still use the Henderson Hasselbach equation for a polyprotic (can give more than two hydrogens, hence needs to have two pKa) but might need to do this twice for depending on the concentration of your different constituents. It is a bit more tedious, but otherwise works the same way.
Btw, amino acids (proteins) are polyprotic and depending which solution they are in, they too will either be acidic or basic buffers.(3 votes)
- how can i identify that solution is buffer solution ? And at4:35how does he know that whole of the NH4Cl is going to dissociate into 0.20M of NH4+(3 votes)
- You need to identify the conjugate acids and bases, and I presume that comes with practice. The same way you know that HCl dissolves to form H+ and Cl-, or H2SO4 form 2H+ and (SO4)2-. In this example with NH4Cl, the conjugate acids and bases are NH4+ and Cl-. Notice how also the way the formula is written will help you identify the conjugate acids and bases (acids come first on the left, bases on the right)(4 votes)
- Very basic question here, but what would be a good way to calculate the logarithm without the use of a calculator?(3 votes)
- There are some tricks for special cases, but in the days before everyone had a calculator, students would have looked up the value of a logarithm in a "log book" (a book the lists a bunch of logarithm values). There isn't a good, simple way to accurately calculate logarithms by hand.(3 votes)
- what happens if you add more acid than base and whipe out all the base(2 votes)
- This question deals with the concepts of buffer capacity and buffer range. A buffer will only be able to soak up so much before being overwhelmed. This is known as its capacity. It's the reason why, in order to get the best buffer possible, you want to have roughly equal amounts of the weak acid [HA] and it's conjugate base [A-]. If you have roughly equal amounts of both and relatively large amounts of both, your buffer can handle a lot of extra acid [H+] or base [A-] being added to it before being overwhelmed.
However, you have to be aware of your buffer's pH. If you took a reaction of a weak acid that has a small Ka value, it will only produce some conjugate base and it's pH might be very low, like a 2. You have to add a lot of conjugate base in order to make this be a buffer (remember, we need roughly equal amounts of the acid and its conjugate base, but in this example, the Ka is low, so we a lot of the weak acid and not much of the conjugate base). So, now that we're adding the conjugate base to make sure we have roughly equal amounts, our pH is no longer 2. It might be somewhere like a 5. So now we have a strong buffer with a lot of capacity, but our pH of this buffer solution is very different from the pH of just the weak acid/conjugate base we started with.(4 votes)
- At the end of the video where you are going to find the pH, you plug in values for the NH3 and NH4+, but then you use the values for pKa and pH. Wouldn't you want to use the pKb to find the pOH and then use that value to find the pH? Thank you.(3 votes)
- I think he specifically wrote the equation with NH4+ on the left side because flipping it this way makes it an acid related question with a weak acid (NH4+) and its conjugate base (NH3). Hasselbach's equation works from the perspective of an acid (note that you can see this if you look at the second part of the equation, where you are calculating log[A-][H+]/[HA].(1 vote)
- I did the exercise without using the Henderson-Hasselbach equation, like it was showed in the last videos. But my thought was like this: the NH4+ would be a conjugate acid, because I was assuming NH3 is a base. So, I would find the concentration of OH- (considering NH3 in an aqueous solution <---> NH4+ + OH- would be formed) and by this, the value of pOH, that should be subtracted by 14 (as pH + pOH = 14).
What did I think wrong? I really got confused... :/(3 votes)
- The 0 isn't the final concentration of OH⁻. The 0 just shows that the OH⁻ provided by NaOH was all used up. In order to find the final concentration, you would need to write down the equilibrium reaction and calculate the final concentrations through Kb.(0 votes)
- Commercial"concentrated hydrochloric acid"is a37%(w/w)solution of HCl in water.(density of HCl is1.017g/mol)calculate the amount of water needed to be added in order to prepare 6.00M of HCl from 2dm3 of the concentrated HCl.(2 votes)
- We know that 37% w/w means that 37g of HCl dissolved in water to make the solution so now using mass and density we will calculate the volume of it. Then by using dilution formula we will calculate the answer.(2 votes)
- [Voiceover] Let's do some buffer solution calculations using the Henderson-Hasselbalch equation. So in the last video I showed you how to derive the Henderson-Hasselbalch equation, and it is pH is equal to the pKa plus the log of the concentration of A minus over the concentration of HA. So we're talking about a conjugate acid-base pair here. HA and A minus. And for our problem HA, the acid, would be NH four plus and the base, A minus, would be NH three or ammonia. So the first thing we need to do, if we're gonna calculate the pH of our buffer solution, is to find the pKa, all right, and our acid is NH four plus. So let's say we already know the Ka value for NH four plus and that's 5.6 times 10 to the negative 10. To find the pKa, all we have to do is take the negative log of that. So the pKa is the negative log of 5.6 times 10 to the negative 10. So let's get out the calculator and let's do that math. So the negative log of 5.6 times 10 to the negative 10. Is going to give us a pKa value of 9.25 when we round. So pKa is equal to 9.25. So we're gonna plug that into our Henderson-Hasselbalch equation right here. So the pH of our buffer solution is equal to 9.25 plus the log of the concentration of A minus, our base. Our base is ammonia, NH three, and our concentration in our buffer solution is .24 molars. We're gonna write .24 here. And that's over the concentration of our acid, that's NH four plus, and our concentration is .20. So this is over .20 here and we can do the math. So let's find the log, the log of .24 divided by .20. And so that is .080. So 9.25 plus .08 is 9.33. So the final pH, or the pH of our buffer solution, I should say, is equal to 9.33. So remember this number for the pH, because we're going to compare what happens to the pH when you add some acid and when you add some base. And so our next problem is adding base to our buffer solution. And we're gonna see what that does to the pH. So now we've added .005 moles of a strong base to our buffer solution. Let's say the total volume is .50 liters. So what is the resulting pH? So we're adding .005 moles of sodium hydroxide, and our total volume is .50. So if we divide moles by liters, that will give us the concentration of sodium hydroxide. .005 divided by .50 is 0.01 molar. So that's our concentration of sodium hydroxide. And since sodium hydroxide is a strong base, that's also our concentration of hydroxide ions in solution. So this is our concentration of hydroxide ions, .01 molar. So we're adding a base and think about what that's going to react with in our buffer solution. So our buffer solution has NH three and NH four plus. The base is going to react with the acids. So hydroxide is going to react with NH four plus. Let's go ahead and write out the buffer reaction here. So NH four plus, ammonium is going to react with hydroxide and this is going to go to completion here. So if NH four plus donates a proton to OH minus, OH minus turns into H 2 O. So we're gonna make water here. And if NH four plus donates a proton, we're left with NH three, so ammonia. Alright, let's think about our concentrations. So we just calculated that we have now .01 molar concentration of sodium hydroxide. For ammonium, that would be .20 molars. So 0.20 molar for our concentration. And for ammonia it was .24. So let's go ahead and write 0.24 over here. So if .01, if we have a concentration of hydroxide ions of .01 molar, all of that is going to react with the ammonium. So we're gonna lose all of this concentration here for hydroxide. And that's going to neutralize the same amount of ammonium over here. So we're gonna be left with, this would give us 0.19 molar for our final concentration of ammonium. Hydroxide we would have zero after it all reacts, And then the ammonium, since the ammonium turns into the ammonia, if we lose this much, we're going to gain the same concentration of ammonia. So over here we put plus 0.01. So the final concentration of ammonia would be 0.25 molar. And now we can use our Henderson-Hasselbalch equation. So let's go ahead and plug everything in. So ph is equal to the pKa. We already calculated the pKa to be 9.25. And then plus, plus the log of the concentration of base, all right, that would be NH three. So the concentration of .25. So this is .25 molar for our concentration, over the concentration of our acid and that's ammonium. So that's over .19. So this is all over .19 here. So if we do that math, let's go ahead and get out the calculator here and let's do this calculation. Log of .25 divided by .19, and we get .12. So 9.25 plus .12 is equal to 9.37. So let's get a little bit more room down here and we're done. The pH is equal to 9.25 plus .12 which is equal to 9.37. So let's compare that to the pH we got in the previous problem. For the buffer solution just starting out it was 9.33. So we added a base and the pH went up a little bit, but a very, very small amount. So this shows you mathematically how a buffer solution resists drastic changes in the pH. Next we're gonna look at what happens when you add some acid. So we're still dealing with our same buffer solution with ammonia and ammonium, NH four plus. But this time, instead of adding base, we're gonna add acid. So we add .03 moles of HCl and let's just pretend like the total volume is .50 liters. And our goal is to calculate the pH of the final solution here. So the first thing we could do is calculate the concentration of HCl. So that would be moles over liters. So that's 0.03 moles divided by our total volume of .50 liters. And .03 divided by .5 gives us 0.06 molar. That's our concentration of HCl. And HCl is a strong acid, so you could think about it as being H plus and Cl minus. And since this is all in water, H plus and H two O would give you H three O plus, or hydronium. So .06 molar is really the concentration of hydronium ions in solution. And so the acid that we add is going to react with the base that's present in our buffer solution. So this time our base is going to react and our base is, of course, ammonia. So let's write out the reaction between ammonia, NH3, and then we have hydronium ions in solution, H 3 O plus. So this reaction goes to completion. And if ammonia picks up a proton, it turns into ammonium, NH4 plus. And if H 3 O plus donates a proton, we're left with H 2 O. So we write H 2 O over here. For our concentrations, we're gonna have .06 molar for our concentration of hydronium ions, so 0.06 molar. And the concentration of ammonia is .24 to start out with. So we have .24. And for ammonium, it's .20. So we write 0.20 here. So all of the hydronium ion is going to react. So we're gonna lose all of it. So we're left with nothing after it all reacts. So it's the same thing for ammonia. So that we're gonna lose the exact same concentration of ammonia here. So we're gonna lose 0.06 molar of ammonia, 'cause this is reacting with H 3 O plus. And so after neutralization, we're left with 0.18 molar for the concentration of ammonia. And whatever we lose for ammonia, we gain for ammonium since ammonia turns into ammonium. So we're going to gain 0.06 molar for our concentration of ammonium after neutralization. So we get 0.26 for our concentration. And now we're ready to use the Henderson-Hasselbalch equation to calculate the final pH. So let's do that. Get some more space down here. So the pH is equal to the pKa, which again we've already calculated in the first problem is 9.25 plus the log of the concentration of the base and that's .18 so we put 0.18 here. Divided by the concentration of the acid, which is NH four plus. So that's 0.26, so 0.26. And we go ahead and take out the calculator and we plug that in. So log of .18 divided by .26 is equal to, is equal to negative .16. So let's go ahead and write that out here. So we have our pH is equal to 9.25 minus 0.16. And so that comes out to 9.09. So the pH is equal to 9.09. So remember for our original buffer solution we had a pH of 9.33. So we added a lot of acid, the pH went down a little bit, but not an extremely large amount. So once again, our buffer solution is able to resist drastic changes in pH.