welcome back folks this is mr johnson with you this is a lecture on the chemistry of buffers it's the first lecture in chapter six of our book it's an extension of our understanding of acids and bases there's a lot coming your way in this lecture it will be a long one if you're not in a mental place to process a bunch of tricky information you should come back a little bit when you can great stuff coming your way sit back relax put your thinking caps on and enjoy this wild ride first thing i'm going to do is define for you what a buffer system is and we're looking at this statement here in green it says an acid-base buffer is a solution that resists changes in ph following the addition of a relatively small amount of a strong acid or a strong base shortly we will learn what the components of a buffer are and how they work but for now know that it's a solution that resists ph changes when a small amount of a strong acid or strong base is added in contradiction to water which if you add even a small amount of strong acid or strong base to water the ph changes dramatically and we'll see mathematical proof of that there's lots of buffer systems that exist in nature our blood for example is very well buffered such that the ph of our blood is held very near 7.35 and that's a crucial range for our ph to remain in so that hemoglobin can work well can release oxygen and gain oxygen in our lungs our blood has in it the components of a buffer system to again keep the ph at a very constant level of 7.35 even if we were to ingest something that would cause the ph to go up or down it is resisted by the components of this buffer system all right so the components of a buffer system are a weak acid in solution with that weak acids conjugate base in appreciable meaning fairly high and approximately equal concentration and we'll we'll come back to the math around that again a buffer system is a weak acid and its conjugate base in the same solution together in appreciable fairly high and in fairly equal concentrations this is the buffer system we'll be looking at for quite a bit of this lecture this is acetic acid or ethanoic acid and it's conjugate base acetate or ethanol but we're going to call this acetic acid and acetate their fonts are written large to represent that the concentration of each of these is fairly high that's the appreciable part and in this case they are equal to each other they're both set at 0.1 molar so this would be a buffer system because it's a weak acid and its conjugate base in appreciable fairly high 0.1 molars fairly high and in this case in equal concentration this buffer system would have a ph lower than seven this is an acidic buffer and the reason why is this weak acid acetic acid has a ka that is greater than the this weak bases kb that's why it's called an acidic buffer but if the base were to have a kb higher than the weak acid it would be considered a basic buffer all buffer systems though have a weak acid in solution with that conjugate base in again appreciable and fairly similar concentration or we could say it the other way we could say a buffer system is a weak base in solution with its conjugate acid in appreciable and equal concentration and whether that buffer system will have a ph above or below seven depends on the ka versus the kb which we'll talk about more in a bit the reason why we want the concentration and we're looking at this section down here in green slash orange of the acid and base to be fairly high is that it will resist ph changes better the higher the concentration of the weak acid and the weak base the more strong acid or base can be added before the ph changes a lot you'll you'll see why mathematically in a bit the reason why we want the concentration of weak acid and weak base to be fairly similar is then that buffer system will work equally well against a strong acid or a strong base if the weak acid in the buffer were to be at a higher concentration than weak base it would buffer better against bases than acids because there's more acid in this buffer than base so again we can't want the concentrations to be high so that enough strong acid or base can be added without the ph changing too much and we want the concentrations to be similar so the buffer system buffers equally well against the addition of a strong base or a strong acid all right we're now going to derive a really important equation for you which later will be turned into another even more important equation called the henderson hasselbach all right so we're looking at the section in pink hair and it reminds us that the ka of a weak acid is equal to that weak acid's conjugate base times hydronium over the undissociated weak acid that's something we know if we isolate hydronium get hydronium by itself algebraically then we see that hydronium equals ka times the concentration of weak acid over that weak acid's conjugate base this is true for all systems and for a buffer system the hydronium concentration in a buffer system will equal the ka of the weak acid in that buffer times the ratio of weak acid concentration to conjugate base again hydronium in a buffer system equals ka times the concentration of that weak acid over the concentration of that weak acid's conjugate base what that means then and we're looking at this section in yellow is that the hydronium concentration and therefore the ph because they're interchangeable is based on two factors the ka of that weak acid and the ratio of the concentration of weak acid to conjugate base the higher the ka of that weak acid the greater the hydronium concentration the lower the ph the greater the ratio of weak acid to conjugate base the greater the hydronium concentration and as well the lower the ph so those are the two factors that will influence the hydronium and therefore the ph of a buffer the ka of the weak acid and the ratio of the concentration of weak acid to conjugate base all right hopefully you're enjoying this so far because there's so many cool concepts and math involved in buffers more is coming your way so we're looking at the section in green here and it's telling us which you may have figured out that if the ratio of weak acid to conjugate base is is one if the concentrations are equal to each other then the hydronium concentration of the buffer system will equal the ka in the example that we're using acetic acid in equal concentration with its conjugate base acetate the ka for acetic acid is 1.8 times 10 to the minus 5 molar which means the hydronium concentration in a buffer with acetic acid and acetate in equal concentration would be that value as well and if we took the negative log of that we'd get the ph of that buffer system which by the way would be the pka for acetic acid also now looking at number two this section in pink once the weak acid and conjugate base have been decided upon for a buffer the ka for that buffer system is fixed that's a constant therefore the only thing that can change the hydronium concentration and therefore the ph of that buffer system would be the ratio of acid to conjugate base which can be altered or manipulated and later we will look at a math problem that asks us to design a buffer system of a particular ph and we'll talk about how that can happen in a little bit now looking at statement three in purple it is telling us that if you add more water to a buffer system you're diluting both the weak acid and conjugate base equally therefore the ratio won't change the ka is still a constant and therefore the hydronium concentration and ph will not change when a buffer system either becomes more dilute or more concentrated just something to remember all right i've just scrolled down i'm on this quick check here on the same page i realize i have not oriented you well to pages in our textbook or on 350 of the physical textbook or 358 of the ebook try this quick check on your own write your answers in your book come on back when you're done and see how your answers compare to mine all right i'm going to give you these answers verbally i can't find my stylus why do we normally attempt to make the concentrations of the weak acid and conjugate base approximately equal that's so that the system will buffer equally well against the addition of a strong base or a strong bass acid and when i say buffer equally well i mean it will resist ph changes equally well against the addition of either a strong base again or a strong acid in a bit we'll see the math to prove number two but if a strong acid is added some of the conjugate base will react with the acid so conjugate base goes down the value of this fraction then goes up because the denominator is going down so the ratio will increase and if that ratio increases hydronium will increase also because they're directly proportional in that equation and as hydronium increases the ph will decrease the solution becomes more acidic as might seem intuitive when you add a strong acid but only a little bit because again it buffers well we'll see the math in a moment all right in three we're looking for good buffer pairs we're looking for weak acids in solution with the weak acids conjugate base hno3 nitric acid is a strong acid so that's not going to work strong acids do not create good buffer systems so go to the right hf hydrofluoric acid is also one of our six strong acids that's not going to work hno2 nitrous acid is a weak acid that would work but it's in a solution with hno3 nitric acid a strong acid so this is a weak acid in solution with a strong acid not going to work all right hcooh is methanoic acid so that's a weak acid that would work maybe and then it's in a solution with lithium methanoate that hcoo is the conjugate base of hcooh lithium methanoate will deliver the conjugate base to this buffer solution know that the conjugate base tends to come in a salt with a cation and you want it to come in a salt with a cation that's highly soluble lithium is a group one metal so that's going to work that one is a great potential buffer hso4 hydrogen sulfate is a weak acid so that would work and so4 is the conjugate base of that weak acid and see it paired once again with sodium a group one metal really soluble so you're going to get the so4 dissociating and you'll get the conjugate base in solution with the weak acid neither of these are acids so this isn't going to work co3 and c204 both work as bases but again there's no acid in there so that's going not going to work hcl is our prototypical strong acid so that's not going to work and then lastly h2po4 is on our weak acid table that's a weak acid and its conjugate base which has one fewer h plus ion and a one lesser charge is hpo4 minus so here's our weak acid and its conjugate base in the same solution notice the weak acid is coming as a salt also it's potassium dihydrogen phosphate but once it dissociates then you've got h2po4 the weak acid in solution along with the k2 hpo4 which dissociates meaning hpo4 the conjugate base in solution also so a little bit of tricky stuff going on there hopefully that worked for you so this next section we'll look at as the title implies how and we'll see the math how it is that buffer systems work why they resist ph changes when a strong acid or strong base is added this first section will be how they work when a strong acid is added to a buffer system i flipped back a page so that you realize that the buffer system we're talking about here once again is acetic acid in solution with its conjugate base acetate and they will be an equal concentration for the next couple of problems here so again we're looking at acetic acid in solution with its conjugate base acetate so if a strong acid is added to this buffer system strong acid completely dissociates generates hydronium ions that's here the hydronium the hydronium from the strong acid will react with the conjugate base hydronium plus base generates that base's conjugate acid a weak acid and water what we're doing is turning a strong acid hydronium into a weak acid that doesn't dissociate much if that hydronium were added directly to water the ph would plummet but with the weak base in there the hydronium and the weak base becomes a weak acid which doesn't dissociate much and doesn't ultimately release anywhere near as much hydronium as was added so again this is how a buffer system will react with an acid the conjugate base in the buffer reacts with the hydronium generating a weak acid and water so we're now contemplating what the mathematical implications are of adding .01 molar strong acid 0.01 molar hydronium to this buffer system that has 0.1 molar acetic acid in solution with 0.1 molar conjugate base acetate well before that strong acid is added the ph of this system equals the pka of acetic acid remember if the weak acid has the same concentration as the base the ratio of acid to base is one the hydronium equals the ka and the ph equals the pka so the beginning ph of this solution is 4.74 and here's the math again explaining what i just mentioned meanwhile if you add 0.01 molar strong acid or hydronium to this buffer system 0.01 molar base will react so 0.1 goes to 0.09 and the amount of base that reacts turns into conjugate acid so the conjugate base goes down by 0.01 the conjugate acid goes up by 0.01 the ratio now of weak acid to conjugate base is no longer 1 to one but rather 0.11 to 0.09 or 1.22 if we plug that in to the expression that we know to exist for hydronium equaling ka over acid excuse me hydronium equaling ka times acid over base hydronium equals the ka of acetic acid 1.8 times 10 to the minus 5 times acid over base 1.22 equaling 2.2 times 10 to minus 5 molar and if we take the negative log of that to get the ph we see that the ph only drops to 4.66 as the ratio of acid to base shifts from 1 to 1.22 the ph only drops by .06 units 0.08 excuse me that's hardly a decrease at all on the other hand and i'm looking down here at this little highlighted yellow in exclamation if you added 0.01 molar strong acid to water the ph would drop from 7 water neutral to 2 because 10 to the minus 2 molar is a ph of 2. that's a drop of 5 units versus 0.08 mathematical proof of how effective buffers are amazing all right i've scrolled down we're still dealing with the same buffer system acetic acid in solution with its conjugate base acetate in equal concentrations of 0.1 well if a strong base is added to this buffer system it will react with the weak acid acetic acid generating that weak acid's conjugate base acetate and water this time we're turning hydroxide a strong base into acetate a weak base when you add a strong base to a buffer system the strong base will react with the weak acid in that buffer generating that weak acids conjugate base and water and because now instead of a strong base it's a weak base the ph will go up but by way less than it would if this were pure water same math applies before the strong base was added the acid and conjugate base where in equal concentration the ratio was one to one the hydronium equals the ka and the ph is what it was before but if the strong base reacts with the weak acid for every 0.01 molar of weak excuse me a strong base that's added the weak acid goes down by that much the conjugate base goes up by that much and the ratio now is 0.09 to 0.11 right the ratio is now less than one it's 0.82 and if we plug that ratio of 0.82 into the equation of hydronium equaling ka times the ratio of acid over conjugate base we get 1.48 times 10 to the minus 5. and if we take the negative log of that the ph of this system that had had a strong base added is 4.83 if you recollect it was 4.74 before the strong base was added now it's only going up by .09 hardly an increase at all figure 621 here and green is a nice summary of what we've just seen when the concentrations of weak acid and conjugate base are equal the hydronium concentration equals the ka of that weak acid or the ph equals the pka if strong base is added that reacts with the strong acid the amount of base that's added is the decrease that the weak acid undergoes and the increase that conjugate base undergoes because weak acid is converted to conjugate base the ratio then of weak acid to conjugate base falls and the hydronium concentration falls ph goes up of course as hydroxide's added meanwhile if strong acid is added that reacts with the conjugate base turning that conjugate base into its conjugate acid acid over base goes up ratio goes up hydronium goes up ph goes down all right so we're looking at the sample problem here on page 352 of our textbook 360 of our ebook and i'm going to walk you through how to how to think about and do this sample problem so we've got a one liter buffer solution comprised of one molar nitrous acid and one molar sodium nitrite that's going to completely dissociate because again sodium is a group one metal so you have nitrite in solution with its conjugate acid here's the acid and its conjugate base they are an equal concentration as well they're both one molar a asks us to write the equation for this weak acid in excuse me for the weak acid equilibrium in this solution and highlight the predominant species in that equilibrium so we have the weak acid plus water on the left in equilibrium with that weak acids conjugate base and hydronium on the right the weak acid as we know donates a proton to water generating the weak base and hydronium now if this were just a weak acid being added to water we know that it doesn't dissociate much and that there's hardly much product there's hardly much conjugate base and hydronium but remember we've added the conjugate base by itself we've delivered the conjugate base from sodium nitrite so the conjugate base isn't coming from the acid dissociating it's coming from another source they're both highlighted because they are both the predominant species they are both in high and in this case equal concentration b asks us to write the net ionic reaction that occurs when hcl a strong acid is added to this solution hcl dissociates into h plus and cl minus h plus reacts with water to generate hydronium chloride spectates so we ignore it so now we see hydronium from hcl reacting with the weak base in this buffer solution which we know is what happens generating the conjugate acid and water so again here's the hydronium from hcl reacting with the base in this buffer system generating that base as conjugate acid and water and why the ph doesn't change much is that hno2 is a weak acid it hardly dissociates much so we're turning h3o plus nothing but hydronium ions into a weak acid that doesn't dissociate much all right c says how will the hno2 to no2 ratio hydronium concentration and ph change following the addition of the acid well if we add acid we know that the concentration of conjugate base goes down and it becomes the conjugate acid so acid concentration goes up base concentration goes down the value of this fraction increases ka is constant therefore hydronium goes up that makes sense you're adding a strong acid the hydronium should go up but only a little bit if hydronium goes up we know ph goes down so again the addition of a strong acid to this buffer system will raise hydronium but just a bit and drop the ph all right flip the page looking at practice problem 621 here and i'd like you to try this problem that's highlighted in red number one so give number one a shot and come on back once you're done and see how your answer compares to mine all right so following the addition about excuse me following the addition of a small amount of strong base to a buffer solution the h a to a minus ratio decreases the hydronium then decreases also and that causes the ph to increase here's the net ionic equation for when sodium hydroxide can sodium spectate so we're just dealing with hydroxide here interacts with this buffer solution the buffer solution is the one from the prior problem with the nitric acid or excuse me nitrous acid and nitrite so the hydroxide will react with the acid generating water and that acid's conjugate base sorry about the particularly poor penmanship i've got a really lousy stylus that is tricky to use meanwhile if that happens the ratio of acid to base will fall because we're turning some acid into base the hydronium will also fall because those are directly proportional causing the ph to go up well done folks all right got one more for you to try same page here practice problem three in green give this one a go on your own write your ideas into your textbook and come on back once again see how your answers compare to mine hopefully your penmanship will be better than mine all right i'm going to do this one verbally the answer to a is no that isn't a good buffer system h i hydroiatic excuse me hydroiodic acid is a strong acid so this is a strong acid and it's conjugate base but it's got to be a weak acid in that weak acids conjugate base so a won't work b also won't work naf will dissociate into sodium ions and f minus ions f minus ions function as a weak base nacn will dissociate same thing cn minus ions also function as a weak base so these are both weak bases won't work as a buffer c though will work k2 c204 will dissociate into potassium ions which spectate and the oxalate anion which is a weak base khc204 will also dissociate into k plus ions and hc2o4 hydrogen oxalate oxalate c2o4 is a weak base hydrogen oxalate is that weak bases conjugate acid and since these are both in high concentration and they're equal or relatively similar this would function well as a buffer system all right so we've turned the page uh we're now looking at this section titled buffering a solution in the basic ph region you're going to hear me say pretty much the same things you've heard me talk about for the last six or seven minutes it's nice to hear things twice but the difference is that this buffer system that i'm referring to will have a ph greater than seven all right so let's take a look we've got ammonia in equal concentration with its conjugate acid ammonium nh3 is the base ammonium nh4 plus is the conjugate acid and their concentrations are equal so this works as a buffer system the reason why this is a basic buffer or its ph is greater than 7 is that nh3 ammonia has a kb that is greater than the k4 ammo ka for ammonium so it's a basic buffer only in that the base is stronger than the weak acid really you don't need to think about whether it's a basic buffer or an acidic buffer you can just let the math work its way out but here's another buffer system with a different base and a different acid again it's a basic buffer because this base is stronger than its conjugate acid well we can do the same thing for this basic buffer that we did for the acidic buffer kb is equal to acid or conjugate acid times hydroxide over base and if we want to isolate hydroxide from this hydroxide equals kb times the base over the acid algebra so we get in general hydroxide equals kb times base over acid but for any buffer whether it's a basic buffer or an acidic buffer you can still go with that old form where hydronium equals ka times base over acid and then if you need hydroxide or poh you can go from hydronium to hydroxide to poh so either of these expressions work because they can be converted between the two like last time the ph of this buffer system or the poh or the hydroxide concentration because they're all interchangeable is based on the kb of the weak base and once that's been decided upon once that's then a constant it's really just the ratio of acid to base once again that will determine the ph or poh if we want to call it that of this basic buffer so you're hearing the same thing similar mathematical proof about how this basic buffer resists ph changes if you have a 0.1 molar buffer system with ammonia and ammonium we find that the ph is 9.25 you can use the o h equation or the h3o equation either works you'll get 9.25 for the ph it's a basic buffer because the buffer is greater than 7. meanwhile if you add some hydronium a strong acid 0.01 molar again it will react with the base turning that base into its conjugate acid base falls by 0.01 acid rises by 0.01 and now the ratio of base to acid is 0.82 if we plug that in to the equation we find that the ph goes down but only a little bit it goes from 9.25 down to 9.17 so proof again that the addition of a strong acid doesn't cause the ph to fall very much this buffer system reaches excuse me resist the ph change same thing except slightly different if now a base is added to this buffer system the strong base hydroxide reacts with the acid in this buffer system ammonium turning that acid into its conjugate base so again before the acid or base was added the ph was 9.25 now that we are adding hydroxide the ratio of base to acid rises from 1 to 1.22 when we plug that into our h3o or oh minus expression we find that the ph shifts to 9.34 so it goes up when hydroxide is added but only a little bit it goes from 9.25 to 9.34 all right we flipped the page looking at figure 622 here it's similar to the figure i explained you prior if you'd like take a peek at this because this summarizes what happens when a strong base or a strong acid is added to this basic buffer system but again it could be thought of as we thought of the prior buffer system we're going to go to sample problem 622 here going to walk you through this one give you a chance to practice a couple practice problems on your own and we're close to the end all right so number one says that hydrazine n2h4 is a weak base with a kb of 1.7 times 10 to the minus 6. again the reason why this buffer would be basic is that this kb is greater than 10 to the minus seven it's greater than the ka of the conjugate acid what can we add to 0.5 molar hydrazine to make a buffer solution well we need to add its conjugate acid in a similar if not an equal concentration well the conjugate acid of hydrazine needs one more hydrogen on it and it would be positively charged so n2h4 plus would be the conjugate acid but you can't deliver a positive ion to solution it has to come in a neutral compound so we say well what negative ion could we deliver this positive ion this n2 h4 plus with well the answer is something that it'll be totally soluble with chloride whenever you're wondering what anion to pair with a cation so it'll dissociate completely pick chloride meanwhile if you're ever wondering what cation to parent anion with so it'll dissociate completely pick sodium a group one metal so nh4 excuse me n2h plus along with chloride or n2 h5cl at .5 molar would make a great buffer system along with this weak base let's extend ourselves so number two says that the concentration of that compound in the final solution were also 0.5 molar which we suggested it should be what would the hydronium concentration be in the buffer solution well if the ratio of acid to base is one to one then the hydroxide concentration equals the kb or the hydronium equals the ka so we got a couple ways we could solve this since we were given the kb we might as well say well that's hydroxide so 1.7 times 10 to the minus 6 being the kb is also hydroxide well we know hydroxide times hydronium is equal to kw so if you divide kw by hydroxide you'll get hydronium the other option was we could have figured out what the ka is of this conjugate acid remember kw divided by kb is ka and then that ka would also equal hydronium it's the same math because again there's multiple routes often to get to the same answer in asset-based problems all right i've flipped my page i'm on 3-5-7 of the textbook or 3-6-5 of the e-book looking at this section the moment you've been waiting for titled the henderson-hasselbach equation and this equation we do end up using quite a bit in in buffer systems in fact most of the math around buffer systems uh uses this equation so we know already that the hydronium concentration in a buffer system equals the ka times the ratio of acid to base hydroxide equals kb over base times acid but we're going with the hydronium version of it meanwhile if you take the negative log of that entire expression and you could look at the derivation it's a beauty we end up with the ph of a buffer system equaling the pka of the weak acid in that buffer system plus the log of the ratio of base over acid it's base over acid no longer acid over base because it's the negative log it becomes the inverse so there it is the henderson hasselbalch and i've just taken you to your equations and constant sheet looking at the section titled equilibrium where the other equations for acids and bases have shown up and here it is second from the bottom there's the henderson hasselbach again the ph of a buffer system equals the pka of the weak acid in that buffer system meaning the negative log of the ka plus the log of the base over the conjugate acid or the conjugate base over the acid either way again there she is henderson hasselbalch this equation works for acidic buffers or basic buffers and again you don't really need to think of it as an acidic buffer or basic buffer let the math work out but if you're given a weak base in a buffer problem and you're given the kb of it turn that kb into a ka turn that ka into a pka and then you can use the henderson-hasselbalch equation and if at the end you need to find the poh of the buffer system same thing use this equation find the ph and 14 minus ph will equal poh so this equation will work for all buffer systems but again no it's the p k a of the weak acid in the buffer system that shows up in this plus the log of concentration of base over acid going back to the definition of a buffer of course it's weak acid and conjugate base in relatively high and relatively equal concentrations well the higher the concentration of weak acid and weak base the greater the buffer capacity often you're asked about which of these buffers has the highest capacity well it'll be the buffer system with the highest concentration of weak acid and base the reason that higher concentration buffers have higher capacity is that when the initial concentration of weak acid or base is greater when a strong acid or base is added that ratio changes less the math would prove it the ratio changes less and if that ratio changes less hydronium concentration changes less and so does ph all right we're now in this last section here titled preparation of a buffer you're doing great hang in there you've heard a couple of times that for a buffer to work has to have weak acid and conjugate base in high concentration and in relatively similar concentration well relatively similar is no more than 10 times more acid than base or no more than 10 times more base than acid the ratio of base to acid can't be less than 0.1 or greater than 10. what that ends up meaning and we'll we'll see the math later is that the pka of the weak acid in the buffer can't be one more than one greater or one lesser than the desired ph of the buffer again we need to pick a weak acid whose pka is within one of the desired ph for that ratio to be no more than ten to one or no less than one to ten and if you think about the math which is tricky it's because the log of 10 is one and then the pka won't be more than one away from the ph and the log of 0.1 is -1 also substantiating that pka can't be more than one less than the ph so again the ratio of base to acid or acid a base can't be more or less than 10 to one and the pka of the weak acid has to be within one integer one of the ph of the buffer system you're trying to design all right so the last thing we're to look at is a sample problem where we figure out what it'll take to prepare a buffer with a desired ph we use the henderson-hasselbalch equation for it the the ph in henderson-hasselbalch gets substituted for desired ph and then i'll tell you what else to do as we look at this sample problem on the page that is just after the one we were just looking at so sample problem six two three here says that an environmental chemist requires a solution buffered to ph five the desired ph is five to study the effects of acid rain on aquatic microorganisms decide on the most appropriate buffer components and suggest their appropriate relative concentrations all right here we go first and foremost we need an acid whose pka is between four and six no more than one lesser or greater than five ideally we want an acid whose pka is as close to five as possible in fact if we can find an acid whose pka is 5 then the ratio of acid to conjugate base can be 1 to 1 and this buffer system can buffer equally well against the addition of a strong acid or strong base so we'd go to our weak acid table and try to find a ka that when we take the negative log of is as close to 5 as possible well it turns out that good old acetic acid fulfills that criteria the ka for acetic acid is 1.8 times 10 to the minus five and if you take the negative log of that or the pka we get 4.74 which is in within one of five 4.74 is pretty close that is the acid whose pka is as close to 5 as you'll find on our table so we've chosen acetic acid we know of course that in this buffer there would need to be the conjugate base acetate acetate's a negative ion it can't be delivered to solution on its own we have to pair it with a positive ion we know to pick sodium so we're going to go with acetic acid and sodium acetate the question is what concentration of each of these do we need to get the ph to be 5 given the pka is 4.74 so we plug in for the desired ph or for the ph in the henderson-hasselbalch equation 5. set that equal to the pka 4.74 plus the log of base over acid so we've got 5 equaling 4.74 plus the log of base over acid subtract 4.74 from 5 and end up with the log of base over acid equaling 0.26 so we've got to do inverse log or raise 10 to the 0.26 which is point excuse me which is 1.8 1.8 must equal the ratio of base over acid that's what the relative concentrations of base to acid must be for the ph to be 5 if the pka is 4.74 so we need 1.8 times more acetate or sodium acetate than acetic acid all right well one option is we have one molar acetic acid and 1.8 molar sodium acetate that's 1.801 maybe you have 0.5 molar acetic acid and 0.9 molar sodium acetate that's also 1.8 to 1. maybe you have 0.1 molar acetic acid and 0.18 sodium acetate that's also 1.8 to 1. all of those work but the higher the concentration of both the greater the buffering capacity so 1.8 to 1 would have the greatest capacity but all three of these options would give you a ph of five if the pka were 4.74 where this lecture not to have gone on as long as it has already i'd probably ask you to do a practice problem but that's going to come your way certainly in some group problems you'll be doing and some review questions so that is it folks congratulations for making it through one of the conceptually more difficult topics hopefully it made some sense um look forward to seeing you soon take care