in this video we're going to talk about buffer solutions a buffer solution is composed of a weak acid and the conjugate weak base so let me give you some examples of that hf hydrofluoric acid that's a weak acid the conjugate base of hf is fluoride all you got to do is remove a hydrogen so we're going to pair up fluoride with sodium so here we have the weak acid and the conjugate weak base together these two components form a buffer solution another example will be acetic acid which is a weak acid the conjugate weak base of acetic acid is acetate c2h3o2 minus and we'll pair it up with the sodium ion so sodium acetate will be the weak base these two combine they create a buffer solution or let's say if we have hydrocyanic acid hcn with its conjugate weak base sodium cyanide this too will create a buffer solution so a buffer solution is composed of a weak acid and the conjugate weak base the purpose of a buffer solution is to maintain a constant ph level throughout the solution it resists changes in its ph now it has the ability to do that because it contains both a weak acid and a weak base so let's say if we have a solution of hf and sodium fluoride so we have a buffer solution if we were to add acid to this solution let's say hydrochloric acid instead of hdl let's use h3o plus so if we were to add acid or anything that would increase the h2o plus concentration the acid is going to react with the weak base component of the buffer solution so it's going to react with fluoride and so we're going to get hf and water if fluoride wasn't there the ph would decrease dramatically when you have a high level of h2o plus or hydronium ions in a solution the ph will be very low and that's what acids do whenever you add an acid to a solution it brings down the ph but because fluoride reacts with the incoming hydronium ions you're not going to have a significant drop in ph because they will quickly consume these ions in the solution create an hf in water now let's look at the other situation let's say if we add a strong base like sodium hydroxide if we immediately increase the hydroxide ion concentration the poh of the solution is going to go up i mean i take that back the poh is going to go down rather the ph is going to go up these two are related to each other remember the ph is 14 minus the poh of the solution so when the poh goes down the ph goes up now once you add hydroxide it's going to react with the weak acid component of the buffer so it's going to react with hf hf is going to quickly consume the hydroxide ions creating fluoride and water and so because hf quickly reacts with hydroxide we're not going to see a sharp decrease in the poh of the solution nor will we see a sharp increase in the ph the ph will increase slightly but not by that much because these are being reacted with hf and so that's how a buffer solution can maintain a relatively constant ph level in the solution it's by having the ability to react with any incoming acid or any incoming base that is added to the solution now let's talk about how we can calculate the ph of a buffer solution so let's say we have a solution with hydrofluoric acid and water and this solution also contain fluoride so in this reaction we have the weak acid and the conjugate weak base so we have a buffer solution everything is in an aqueous phase except water if we were to write the equation that is associated with ka and the acid dissociation constant this would be equal to the concentration of the products divided by the concentration of the reactants so water is not included because it's in the liquid phase now what we're going to do is we're going to take the negative log of both sides of the equation so we're going to have negative log of ka and that's going to be equal to negative log of the right side of the equation so we have f minus times h3o plus divided by hf now a property of logs allows us to split one log into two separate logs so for instance let's say if we have log a b we can write this as log a plus log b so let's say fluoride is a h3o plus is b and h f is c we could also do this if we have log a b over c we can split b from a and c so we can say this is log a over c plus log b we don't have to separate a from c we can keep them together or we can separate them right now i want to take out h2o plus out of this expression so b would correspond to h3o plus i'm going to keep a and c the way it is so i'm going to use this property of the log expression so this is going to be negative log ka and that's going to equal so we need to distribute the negative sign so keep that in mind negative log we're going to have a over c actually let's write log b first so this is going to be negative log h real plus and then we're going to have log a over c but with a negative sign because this has to be distributed and then minus log and then it's going to be the base over the acid so fluoride is the base hf is the acid now let's go ahead and delete some of the stuff that we have here to make some extra space now some formulas that you need to be familiar with are these two you know that the ph is equal to the negative log of h3o plus you've seen that many times before at this point also you know that the pka is negative log ka so we can replace this with the pka we can replace this with ph so we have pka is equal to the ph minus the log of the base in this case the base is f minus divided by the acid hf now you could put this in parentheses to indicate concentration but since you have a ratio it could be in moles as well because moles and molarity they have the same ratio given the same volume of the solution now what we're going to do is we're going to isolate ph so i'm going to take this term and move it to the other side it's negative on the right side but it's going to be positive on the left side so the pka plus log of the base over the acid now i'm going to replace this with a more generic form of the base which is typically a minus and the generic form of an acid would be ha on the right side we still have ph i'm going to reverse these two equations i'm going to put ph on the other side and i'm going to put this on the other side so if we simply reverse these two equations or just reverse the equation rather by switching both sides we're going to get the henderson-hasselbach equation so this equation helps us to calculate the ph of a buffer solution so the ph is equal to the pka of the acid plus log base over acid a minus is the weak base component of the acid h a is the weak acid so that's how you can calculate the ph of a buffer solution it's by using the henderson-hasselbach equation so the ph is dependent on the ratio of the conjugate base to the conjugate acid now there are some things that we can discern from this equation what can we say regarding the ph of a buffer solution when the concentration of the weak acid is equal to the concentration of the conjugate base if h a equals a minus what can we say about the ph if this is 10 and that is 10 10 over 10 is one if this is 0.5 that's 0.5 0.5 over 0.5 is one so whenever these are the same and the one-to-one ratio you get log of one log of one is zero so this disappears thus we have the ph is equal to the pka so whenever you have a buffer with an equal amount of weak acid and weak base the ph will be equal to the pka so let's say if you want to create a buffer solution with or that's centered at a ph of 4.5 if you want to create a buffer solution that's centered at that ph well you want to find an acid with a pka that is very close to 4.5 likewise if you want to create a buffer solution with a ph that's centered around 6.0 you want to look for an acid that has a pka of a value that is close to 6.0 close as possible the closer the better so that's how you can create a buffer solution that's centered at a given ph it's by choosing the appropriate acid with a pka that is very close to the ph now let's talk about the ratio of base to acid and acid base log 10 is 1. let's think about what that means so let's say if a minus is 10 times the value of h a let's say this is one m and h a is 0.1 1 divided by 0.1 is 10. log 10 is 1. and let's say the pka is 4 what's the ph the ph is going to be one unit higher whenever the ratio between acid the base or base to acid when it differs whenever you have a 1 to 10 ratio the ph is going to be one unit away from the pka if you have more of the acid it's going to be one unit lower than the pka if you have more of the base it's one unit higher log of a hundred is two a hundred is ten squared a property of logs allows you to move the exponent to the front so log of ten squared is the same as two log 10 and log 10 is 1 so you get 2 times 1 which is 2. so log of 100 is 2. therefore if the ratio between base to acid or acid the base is 1 to 100 the ph will be two units away from the pka log of a thousand which is log ten to the third that's equal to three so if the ratio of base or base to acid is a thousand the ph will be three units away from the pka so let's say the pka of a certain acid we're gonna say it's 4. actually let's let's make it 6. and let's say the numbers on this line represents the ph of the solution when a ph is six we're going to have equal amounts of base and acid so 50 will be base and 50 will be the weak acid ha under those conditions the ph will equal the pka so we can describe that as a percentage or we could say you know we have one molar of the conjugate base and the one molar solution of the conjugate acid at a ph of seven let's keep the amount of base the same let's say it's 1 molar the amount of acid will be 10 times less it'll be 0.1 m ha so as you can see here we have more base than acid so the ph is going to be higher than the pka what will be the situation at a ph of 8. so if we have the same amount of base amount of acid will be 10 times less is going to be 0.01 m so anytime the ph is well any time you have more base than acid the ph will be greater than the pka so on the right side where a minus that is the weak base component of the buffer if it's greater than h a or will be greater than h a under those conditions the ph will be greater than the pka now what if we had one m h a and 0.1 m a minus what's the ph so here we have more of the acid than the base so the ph is going to be less than the pka and the ratio is 1 to 10 so we're going to be one unit away from the pka therefore the ph here will be it's going to be five likewise let's say if we have point zero one m a minus and one mha the ratio between acid base or base to acid is like one to a hundred so log 100 is two we're going to be two units away from the pka or let's say if we have point zero zero one of the base and the one molar solution of the acid the ratio is one to a thousand one divided by .001 is a thousand so we're going to be three units away from the pka so that's how you can determine the ph of a buffer solution conceptually by looking at the ratio between base acid or acid base and remember if you have more of the acid than the conjugate base the ph is going to be less than the pka and if you have equal amounts of acid and base the ph will be equal to the pka whenever you have a buffer solution now let's go ahead and work on some practice problems number one what is the ph of a solution consistent of 0.75 molar acidic acid and 0.5 molar sodium acetate and we're given the acid dissociation constant of acetic acid is 1.8 times 10 to the minus 5. feel free to pause the video if you want to work on this problem so what we're going to do is we're going to use the henderson-hasselback equation because we have a weak acid and we have the conjugate weak base which means we have a buffer solution the ph is going to equal the pka of the weak acid plus log of the base the base we can write as a minus or you could simply put b for the base and then a for the acid or you can write h a if you want i like to write b over a base over acid now the first thing we need to do is calculate the pka of the weak acid and that's negative log of the ka so we have the ka value which is 1.8 times 10 to the negative 5. so the pka of acetic acid is 4.7447 so now let's plug it into this formula so that we have the pka it's this number and then plus log the base this is the base component of the buffer so the concentration is 0.5 molar and the concentration of the weak acid component is 0.75 molar the unit's molarity will cancel because we're dealing with a ratio here so let's go ahead and plug this in 4.7447 plus log 0.5 divided by 0.75 this is you can round it to 4.57 so that's the answer for this problem so that's how we can calculate the ph of a buffer solution now there's one more thing i want to mention regarding this problem notice that we have more of the acid than the base because we have more of the acid than the base the ph is going to be less than the pka as we discussed earlier in this video here we can see the ph is 4.57 the pka is 4.74 so the ph is less than the pka because we have more of the acid component than the base component of the buffer number two what is the ph of a solution containing .15 moles of ammonium chloride and 1.5 moles of ammonia and we're given the kb the base dissociation constant of nh3 so what should we do here well once again we have a buffer solution we have the weak acid and we have the weak base the conjugate weak base now in this case notice that we have more of the base than the acid we have more of ammonia than ammonium chloride so what can we say about the ph and the pka the ph is going to be greater than the pka because we have more of the base than the acid let's go ahead and calculate the pka we're given the kb so we need to calculate the pkb first pkb is negative log of kb so this is going to be negative log of 1.8 times ten to the minus five and we've seen that value so this is four point seven four four seven three now the pka plus the pkb these two they add up to 14. so to calculate the pka it's going to be 14 minus the pkb so 14 minus 4.74473 that gives us a pka value of 9.25527 so since we have more base than acid we know that the answer the ph of the solution has to be greater than the pka it has to be greater than 9.25527 let's round that to 9.255 so let's go ahead and calculate the ph of the solution but before we do that notice that the ratio of base to acid is ten to one one point five divided by point fifteen is ten whenever the ratio is ten to one or one to ten we know the ph will differ from the pka by one unit and since we have more base and acid one unit higher than this number is going to be 10.255 therefore conceptually we know that the ph of this solution should be this value and we're going to confirm it by calculation so we should get 10.255 so let's write the buffer equation ph is equal to pka plus log base over acid the pka is 9.255 and then we can add the 2 7 as well to get a more accurate result and then plus log now as was mentioned before this could be in units of molarity or it could be in moles both units will cancel so we're going to plug in the moles we have 1.5 moles over point 15 moles so we can cross out these two units now let's plug this into our calculator nine point two five five two seven plus log one point five over point fifteen this will give you ten point two five five two seven so that's the ph of the solution so we could see why it's one unit higher is because the ratio of base to acid is ten to one number three calculate the ph of a solution containing 15 grams of hydrofluoric acid and 21 grams of sodium fluoride and 750 milliliters of solution so once again we have a buffer solution but this time we're given the grams of the acid and the base before we had the molarity and during the second problem we had the moles for this one all we need to do is convert grams to moles and then we could use the buffer equation so let's do that first so we have 15 grams of hf to convert it to moles we need to know the molar mass the molar mass of hydrogen is approximately one for fluorine is 19. so this is approximately 20 grams per mole so one mole of hf has a mass of approximately 20 grams so we can cross out the unit grams of hf it's going to be 15 divided by 20 which is 0.75 so that's how much moles of acid we have in this solution let's start with 21 grams of the base naf and let's convert that to moles sodium has an atomic mass of approximately 23 and for fluorine is 19 23 plus 19 is 42. so one mole of naf has a mass of approximately 42 grams 21 divided by 42 is 0.5 so we get 0.5 moles of sodium fluoride so now that we have the moles of the acid and the base we can now calculate the ph but first let's calculate the pka the pka which is negative log of ka and ka is that number so we have the negative log of seven point two times ten to the negative four and so that's going to be three point 3.1427 so now that we have that let's use the henderson-hasselbach equation to get the ph of the solution so the pka is 3.1427 plus log of the base the base is by the way will the ph be greater than or less than the pka so notice that we have more of the acid than the base so because we have more of the acid than the base the ph is going to be less than the pka so the ph should be less than 3.1427 the solution is acidic so now let's go ahead and continue with this equation so the base is 0.5 moles the acid is 0.75 moles so let's plug this in 3.1427 plus log 0.5 over 0.75 this is going to be approximately 2.97 which is less than 3.14 so that's the ph of the solution so anytime you have more of the acid component than the base component the ph will be less than the pka number four what is the pka of an unknown weak acid if the ph of the solution was measured to be 5.62 when h a is 0.45 and a minus is 0.85 so we want to calculate the pka if we know the ph and the concentration of the weak acid and the conjugate weak base so let's start with the buffer equation ph is equal to pka plus log base over acid now what we need to do is we need to isolate pka so i'm going to take this term move it to the other side so it's positive on the right side it's going to be negative on the left side now i'm going to switch the left side of the equation with the right side of the equation so we could say that the pka of the unknown weak acid is going to be the ph of the solution minus log of the base over the acid so this is how we can calculate the pka of an unknown weak acid all we need to know is the ph of the solution and how much of the acid and base that we've dissolved in the solution so the ph is 5.62 the base is the concentration of a minus so that's 0.85 the acid is the concentration of ha so that's 0.45 so this is going to be 5.34 so this is the pka of the unknown acid now let's see if our answer makes sense so whenever the base is greater in concentration one quantity then the acid we know that the ph is going to be greater than the pka is that the situation here well the ph the ph is 5.62 the pka is 5.34 so we can clearly see that the ph is greater than the pka so this answer makes sense you