okay so welcome to applications of aqueous equilibria part 2 today we're going to be doing calculating the pH at any point on a titration curve so we're going to look at these points here ABCD all the way through H and calculate the pH of each one and we're gonna figure out like a strategy to do for each different point and this is available online so you can print this page out that you're seeing now and have the question and these little templates down here we're gonna look at each what species in the solution at each point and and we keep going on that so first let's look at what species we have in solution at all of these points on the titration curve so we're gonna split this into the species in the solution from the sample also called the analyte and then what species are in solution from the titrant so we know that at the very beginning point a we have mostly the sample like oxalic acid a very very small amount is ionized and that's why the pH is less than 7 but it's essentially all in the form of the oxalic acid or almost all of it the first equivalence point here point D we've deprotonated or neutralized one of the protons one of the two protons in this diprotic acid so at this point all we have is the conjugate base of this which is hydrogen oxalate so at this point we have all this in between these two points we have decreasing amounts of this and increasing amounts of this so I'm gonna leave this as an open circle because at that point we don't have any more rock solid acid left but between those two points we have oxalic acid so I'm gonna connect those with a line right there and then leave that one open this one we're gonna say that we don't really have much at the beginning but as soon as that first drop if titrant goes in I start creating more of the hydrogen oxalate so I have it in this region here and then after I go past the first equivalence point I'm starting to take the the second proton off so the concentration of this is going down but I still have it until I get to the last equivalence point but at the last equivalence point the second equivalence point here we're not going to have any more of this so we have this in this region and I'll leave those two circle open and as we have that's all we have at this point that's all we have at this point here we're increasing the concentration as we go so I'm going to put a blue line here maybe there we go and then at this point that's all we have at Point G the second equivalence point so I'm gonna make that a nice solid dot there we don't have any of this but now we have we only have this here we don't have any of this but we only have this right so that kind of makes sense there and then after this point we actually also have the oxalate all the way to the end because now we have nothing else like to turn into from the titrant before I agony I don't have anything in there so these bots are both going to be open the sodium which is the conjugate acid of the strong base we're using it as a titrant as soon as we put a first drop in there we're gonna have sodium ions so we're gonna have sodium ions here and at this point as well and through here and at this point as well and through here so basically it's after the first drop we're gonna have sodium ions in the solution but they are just spectators and they're neutral in terms of pH right that's why we're using a strong base here because it's conjugate acid is neutral in terms of pH which is why I drew this in green and then finally we have the hydroxide now you would think that looks a lot like the sodium but it doesn't because the hydroxide is neutralizing the protons from the kanji from the sample and so this is being neutralized remembered acid-base neutralization is go to completion so we don't have in here we don't have any here we don't have any here because it is neutralizing the sample even at the last equivalence point we've added a still geometrical mouths we don't have any excess hydroxide but after the last equivalence point there's nothing left and neutralized then we're gonna start accumulating hydroxide so what that's up here where we're now going to the pH of the titrant the sodium hydroxide okay so we have these different regions here and we should always be able to understand what we have in solution at any given point because for example if we know at Point C we have these two species in solution well then I can use the henderson-hasselbalch equation to calculate the pH at Point C but that wouldn't be appropriate at Point E because these are the two that I have in solution from my sample okay so let's do the first point let's figure out the initial pH okay so we're gonna find the initial pH of the solution so we're gonna be the doing point a right there and there's a couple more pieces of information we need to do this we need the KA value something I added those right here to this sheet okay so we're gonna ask ourselves a few questions to figure out what strategy we should use to find the pH at that point so the first question we're going to ask ourselves is is the solution a buffer yes or no because if it's a buffer we can use the henderson-hasselbalch equation if it's not we can't so at this point is it a buffer and the answer is no because I don't have a pre cheb Allah mount of the conjugate base the hydrogen oxalate yet I have almost entirely oxalic acid so it's not a buffer at this point no so that rules out the henderson hasselbalch equation so what's determining the pH at this point well the only thing we have in solution to a large amount is the oxalic acid so it's going to be the ionization of this weak acid of the oxalic acid so ionization of the h2 C 204 but barely got that on there okay so now what is our strategy so if we have a weak acid in solution and we it's not a buffer we can't use the henderson-hasselbalch then we just have to go to an ice table so we're going to do an ice table and use ka1 because it's the first ionization and these two are fairly far apart and our concentrations are relatively high so I probably don't have to consider any h+ from the second ionization and due to ice tables i can probably just do the first one so let's go ahead and set that up so we're going to write the reaction for the ionization of oxalic acid down and do an ice table underneath it okay so I set up that ionization for the weak acid and my ice table initial change equilibrium and my initial row here this is the initial concentration of the oxalic acid which is given to me as one molar and we haven't made any of those yet so those are 0 minus X plus X plus X 1.0 minus X X and X alright the next step is to set up our equilibrium expression using ka 1 so I'm going to say K a 1 equals products over reactants plugging in from the equilibrium line here but my products are H+ and hydrogen oxalate and my reactant is the oxalic acid concentration okay so plugging in mind for my equilibrium line row x x 1.0 minus x now we're going to try to see if this is a if this X is negligible and it turns out it's probably not because this initial pH is 1 molar so 10 to the 0 the KA one is five point nine times ten to the minus two so it's more than one times ten to the minus two so it's less than two order magnitudes smaller than the initial concentration so I'm going to I'm suspicious I don't think it's going to be negligible it turns out that if you did assume it was negligible and then solved for X and then come back and do the check you don't you get more than five percent you get a twenty four percent ionization because this is a fairly large K value so that is not a good assumption so in this case it's not we can't assume that it's negligible so we're gonna have to go on so that equals ka one which is five point nine times ten to the minus two and I will have to bring this up to this side distribute this multiply those and then collect like terms all on one side and when I do that I will get this second-order polynomial okay so that when I bring this over distribute collect like terms this is what I get my coefficient on the x squared is a my coefficient on the X term is B and my coefficient my other number without any X's is C in this case it's negative and that negative part of C I'm going to put that into the quadratic equation and when I get out are these two values from the plus the square root and the minus the square root and this one can't be right because it's negative and we set up the quadratic equation so that our X's would be positive we set up our ice table so the X is we positive so that one can't be right and this one looks like it could be any of those sig figs leftover from the quadratic equation okay so now what we need is we need to know the H+ concentration at equilibrium so this is what we're what we're looking for so we can find the pH and so the h+ concentration at equilibrium is just equal to X which equals this number right here so zero point two one five one eight the sig figs right there molar h+ right now all we have to do is take that and plug it into our ph equation so ph equals minus the log of the H+ concentration so pH equals minus the log of this number right here zero point two one five one eight molar h+ when i put in my calculator i'm just gonna use the whole number we have these two sig figs right here and i get that my ph equals zero point six seven I get more digits in that point six six seven one nine zero zero six forty four but my two sig figs become two decimal places that's gonna be six and seven so that is my initial pH right there okay so a few few hints about doing the second ionization so we said that usually we don't need to consider any contribution to each plus from the second ionization unless there's certain things are true if the KA values are very close together then it might you might need to consider the second ionization in this case there's three orders of magnitude apart which is fairly far apart the other thing is if the concentrations the initial concentrations are very low you might need to consider from the second ionization because it the lower the concentration the higher the percent on ization so normal ionize in a lowers concentrated solution so if I did go ahead and do the second I stable and I put these in as my initial values on the initial line in the second ionization I would have gotten a slightly lower pH you can see this is the pH the unrounded pH from the first I stable and this is the unrounded pH I would have gotten out of the second I stable and you can see it is a little bit lower here but within the sig figs it's not any different see how this rounds up to six seven this also rounds up to six seven so it's not until a couple of digits out that it mattered and so for our for our purposes we didn't need to consider that okay let's go on to point B okay point B so now we are a quarter of the way to the first equivalence point so we're gonna use our same method of fire ground our strategy so is the solution a buffer at this point well I have oxalic acid and I have its conjugate base hydrogen oxalate so yes it is a buffer at this point and yes I do have more oxalic acid than I do have hydrogen oxalate but they're still within an order of magnitude of each other so that's still in the buffering region here so actually the answer is yes that is a buffer so the pH is gonna be determined by the conjugate pair the buffer and the concha pair is the oxalic acid and its conjugate base hydrogen oxalate - again barely really cop that in there alright so if I have a buffer I could do ice tables but a better strategy would be to use the henderson-hasselbalch equation that's a lot easier so we're gonna do that so i'm gonna set that up okay so here I've set up the henderson hasselbalch equation and i'm the specifics so here's the conjugate base here is the conjugate acid on the bottom this ka is ka one because for ionizing the conjugate acid into this conjugate base so those two match so I'm all set up so now I can go ahead and plug in my numbers okay so just leave that as a variable pH that's what we're solving for remember P is just minus the log of so minus the log of the KA one which they gave to us so that number plus the log of this ratio now I could actually calculate those two concentrations if I wanted to and plug those in I would also of course have to account for dilution in those concentrations because I've had a titrant at this point so it doesn't add up to 1 molar anymore that's something less than that but I could also use the same trick we used in the buffer lab i sorry that acid-base equilibria lab so I know that this is a quarter of the way to the first equivalence point here we have all the conjugate acid here we have all this conjugate base right in the middle we have 50:50 equal amounts of the two but here we're not quite there yet we have mostly the conjugate acid remaining and but we have form some of the conjugate base and we're a quarter of the way there so a quarter of the conjugate acid has been neutralized and formed the conjugate base the remaining three quarters are still in the form of the conjugate acid so this ratio was a quarter to 3/4 and so the quarters cancel so this ratio here is just one third so minus the log of the KA one plus the log of one third and that gives me a pH of zero point seven five two zero two six seven three three six from this subtraction here these are this has infinite sig figs because we're saying it's exactly a quarter of the way but there here we have two sig figs so that becomes two decimal places so when I do this addition here it's gonna cut off right there at two decimal places and so I get that my pH equals 0.75 at point B okay Point C now the first half equivalence point using the same logic so is it a buffer yes in fact it's the ideal buffer because it has equal amounts of the conjugate acid and the conjugate base the pH is going to be determined by the buffering pair which is still the oxalic acid and the hydrogen oxalate didn't run out of room that time okay now our strategy is actually gonna be the same but it's gonna be easier than before we're just gonna use the h h equation the henderson hasselbalch equation just like we did before except that whole last term is gonna drop out so let's set that up okay so I said the henderson hasselbalch equation just like before these two have to match that's my conjugate acid my crime to get base so the same as as it was for point B except this time this is a half and a half right I have half of it converted in this form half of it remaining in this form so that's just one another way I could look at that is that the concentration of the oxalic acid is equal to the concentration of the hydrogen oxalate and if those are equal that's one and the log of one is zero so that whole term is going to drop out of there and I'm just gonna have the pH equals the pKa so the pH equals minus the log of ka one so five point nine times ten to the minus two and so the pH equals one point two two nine one and that goes on for seven nine eight eight but we have two sig figs here so that's gonna give us two decimal places here that's not a sig fig and so we get our pH equals one point two that's gonna round up three at the first half equivalence point Point C okay Point D the first equivalence of point so same strategy idea we did before is this a buffer the answer is no we only have this kanji it at this point we don't have this one and we don't have this one so we don't it's not a buffer at this point so the pH is going to be determined by the partial ionization of this weak acid which is H C 204 - so hydrogen oxalate and so our strategy is going to be an ice table like it was at Point a except this time I'm going to be using ka 2 because it's the second proton that's going to be coming off at this point in this ionization and you think it would be just like point a but actually this points a lot more complicated because at Point a we hadn't added any titrant yet but at Point D we have and so we've effectively diluted the sample by adding titrant at these two points we didn't worry about that because we were using the henderson hasselbalch equation the solution was buffered and dilution is not to change the pH of a buffer as long as there's enough of those kanji it's in there but at this point we don't have a buffer anymore so I do need to account for the dilution because the concentration affects the percent ionization and so I'm also going to have to account for and so that's gonna take a little bit of extra work one way I could figure out how much titrant I've added at the equivalence point is by using the neutralization equation another way is by using stoichiometry this was a little easier so I'm gonna go with this in this equation this stands for the number of protons donated the molarity the acid the volume the acid the number of protons accepted by the base and the molarity the base the volume of the base so here even though this is a diprotic acid at this point I've only donated one of the two protons so this is a one the molarity the acid is 1.0 molar the volume of the acid was 10 milliliters B for sodium hydroxide is 1 it can accept one proton with that hydroxide to make a water the molarity of the base is 0.40 molar and the volume of the base is what I'm solving for so I get that the volume of the base equals 25 milliliters so 25 milliliters of sodium hydroxide solution or I could have also done this using stoichiometry so if you prefer that we can run through that as well we have 10 milliliters of this of the acid of the sample and then I can change milliliters to liters so one milliliter is 10 to the minus 3 liters now I have this is of the h2c 204 solution originally that we started with so in now in liters I can use its molarity so 1.0 moles of h2 C 204 for one liter of that solution and I have one mole of h2 C 204 for every one mole of NaOH and I know the molarity of the NaOH so I have 0.40 moles of NaOH for every one litre of that NaOH solution and then I can use milliliters to liters again so 10 to the minus 3 liters on the bottom one milliliter on the top and I get that this equals the same thing we had before 25 milliliters of NaOH I like this way it's a lot easier but this is totally legit as well notice here when I did the molar ratio it wasn't one to two because I haven't pulled off both protons yet I've only pulled off the first one so it's at this point it's just one one stone okay so this part here was to figure out the volume of titrant at the first equivalence point and whichever way we chose to do it we're gonna get the same value 25 milliliters of titrant so now we got to figure out how much the sample has been diluted what's the concentration of the species at this point so that's no longer one molar because we've added a bunch of titrant so the easiest way to do that is using the dilution equation so in the beginning we had 10 milliliters of 1 molar oxalic acid so we can put that in so 1.0 molar and we had 10 milliliters we're solving for M 2 what's the new concentration and V 2 now is a total of 35 milliliters because we have 10 milliliters of sample 2 that we've added 25 milliliters of titrant so the total volume of the reaction mixture at this point isn't 25 mils it's 35 mils the 10 original plus the 25 so this is 35 milliliters and we can go ahead and solve for m2 and m2 equals zero point two eight five seven one four two eight five seven and this is molar H 2 C 2 O 4 right the original species we converted all of that to H C 2 O 4 minus its conjugate base so that's a 1 to 1 molar ratio so we can do that and get the concentration of the h C 2 O 4 minus species that we actually have in the solution at that point then we can do that ok so I just did that I did my conversion factor my old molar ratio 1 to 1 and so it's the same number but now it's the molarity of the conjugate base the hydrogen oxalate ion and we still have these 2 sig figs in here there's another way we could have done that the same part and that's also using stoichiometry and then plugging into the definition equation of molarity I think this way is easier and I think this way is easier so that's how I would go but I'll show you the other way just so that we have all our bases covered so or stoichiometry so again it would be very similar to this where I start with my 10.00 milliliters of my H 2 C 2 O 4 my original sample milliliters to liters so one milliliter is literally 10 to the minus 3 liters so it really stands for and then we have now we're in liters of oxalic acid so we can use its molarity so 1.0 moles of h2 c 204 for every 1 liter of that solution and now I'm gonna use a molar ratio I have one mole just like I did right here H 2 C 2 0 4 is equal to 1 mole of h co 2 o 4 minus and I at this point don't have molarity I have just moles right so these cancel I'm just left in moles so that is zero pulling 0 1 0 moles of H C 2 O 4 minus and to get to molarity I still need to plug into the definition equation for molarity so molarity equals moles of solute in this case the hc2 oh four minus per liter of solution and plugging in this moles of solute and the 35 milliliters but I can't have a milliliters can I so I've got to convert that one milliliter it's 10 to the minus 3 liters I can do that there so point zero three five liters and I get the same value I got up there zero point two eight five seven one four two eight five seven molar hc2 zero four - and again I just have my two sig figs there either way okay so why did we go through all that well we had to figure out the molarity of this Ishii's because it's no longer one molar because we've added so much titrant so now the molarity is down to just a little over a quarter of that and so that's gonna affect the ionization and this is the value we're gonna start with in our ice table so let's go ahead and set that up okay so I've set up my ice table here and I'm using the second ionization right so taking the second hydrogen off of the hydrogen oxalate and going all the way down to oxalate so this value that we spent so much effort getting is our initial value of this species so zero point two eight five seven molar I'll use the I'm around a number of my calculations but it's only two sig figs so if you're gonna round it at least have a couple digits that aren't significant these are zero and zero because I haven't made any of that yet and this would have been neutralized by the hydroxide so we're gonna have minus X plus X plus X here so zero point two eight five seven minus X X and X are our values and like usually we are gonna go on and set up the equilibrium expression this time using ka - okay so I've set up my equilibrium expression with variables so products over reactants just like normal plugging the values from equal line so we have X X and 0.285 7 minus X all right is this X negligible well let's see this is 10 to the minus 1 my ka value is 10 to the minus 5 so this is probably gonna be negligible so I'm gonna go ahead and try that I'm gonna say assume that this is negligible and so this will approximately equal x squared over 0.285 7 those are my sig figs there and that's still equal to ka 2 which is 6 point 4 times 10 to the minus fifth okay so bring this up to this side and then take the square root to solve for X and I get that x equals zero point zero zero four two seven six one seven nine nine you know we had two sig figs in there so let's go ahead and do our check here and I want to see that is this X that I'm neglecting really less than 5% of the number I'm neglecting it from so is x over these this amount here which is the initial concentration of the h c 2o 4 minus x 100% is that really equal to less than 5% let's get this up a little bit so if I do that I get that plug my numbers in here my X and my initial concentration of this times 5% I get one point five percent ionization so that is going to be okay because it is less than five percent so that's still within my threshold for making that assumption so that's okay so now this x-value is gonna go on and be our concentration of h+ at equilibrium that's what I wanted to solve for so I can find my pH so I can do that and I'm just gonna come over here real quick so the concentration of H+ at equilibrium is just equal to X which equals to this number zero point zero zero four two seven six molar H+ now I can plug that into my pH equation pH equals minus the log of the H+ concentration and equilibrium so that value so pH equals minus log at zero point zero zero four two seven six molar h+ two sig figs so my pH equals two point three six eight nine and it goes on for four zero three five but that's gonna be enough because those two figs sig figs are those two decimal places so my pH is gonna equal to point three seven ah the first equivalence point that was by far the hardest one we've done so far but remember it's only about half as much work as I've shown there because we did both ways the neutralization equation and stoichiometry the dilution equation and stoichiometry to find the values we needed to plug in here so doing it it would have been a lot less for you but but you could do it either way all right let's go on to the next point okay I don't know about you but I'm I'm ready for a one that's a lot easier than the last point we just did and so for this one we ran out point e the second equivalence point that's gonna be easy we have a buffered solution we're gonna get to use the henderson-hasselbalch equation when we first threw that on the board you were probably like I don't want to use that equation but now you can't wait to use it right because it's so much easier than the ice tables but yes this is a buffer at this point yay and my conjugate pair in that buffer is the hydrogen oxalate and the oxalate so we're right here and we have equal amounts of the two right so this is an ideal buffer with that kanji pair so the pH is going to be determined by the buffering pair and that is hydrogen oxalate HC - oh for one - and its conjugate base just oxalate C - 0 4 2 - and the strategy is going to be to use the henderson-hasselbalch equation happy to use that so let's set that up okay so I've set that up it's very similar to what we had at Point C except this time this is our conjugate pair so I'm using ka 2 for the second ionization because remember those do you have to match and so I'm gonna plug those in and I get that my pH equals minus the log of ka 2 here which is six point four times ten to the minus five now what about this term right here well those two concentrations are equal right so this whole term is going to drop out because the log of one is equal to zero so we just have the log of this which equals four point one nine three eight and it goes on of course to zero zero to six but we have two sig figs here so we get two sig figs there so my pH at this point is four point one nine so much easier okay so point F getting to the end here so this is still a buffer in here so our answer to our question is the sub uh fur is yes and the pH is then going to be determined by the buffering pair which is now hydrogen oxalate and oxalate - - and so our strategy again is going to be used the henderson hasselbalch equation and so we'll set that up okay so this is just like we did it for Part II except this term isn't gonna drop out but it's not a big deal so pH equals P minus the log of ka - which is six point four times ten to the minus five plus the log and we have this ratio here again we could calculate the concentrations and plug in the actual values or we could use that technique that we used in the acid-base equilibria lab where we know that we're three quarters of the way between the equivalence point we have all this in the equivalence point where we have all this so three quarters has been already neutralized into this form so we have three quarters as the C - oh four two - formed and we have one quarter as the hydrogen oxalate remaining so again the quarters cancel out and this ratio is 3 so minus the log of the KA 2 plus the log of 3 and we get that at 28 our pH equals four point six seven oh nine four one two eight one hour two so this is again infinite sig figs because of the these being defined as exactly if they're quarters of the way we have two sig fig here so in this term we have two decimal places so when we add these terms it's gonna stay at two decimal places so our pH equals four point six seven at Point F right here all right on to Point G okay Point G the second equivalence point now is this a buffer tragically the answer is no because the second equivalence point we only have this kanji and we don't have any of that kanji it anymore and so it's not a buffer and even if we were getting really close to either the equivalence point so we wouldn't want to use a henderson hasselbalch because the ratio would be too far off but in this case there isn't any of that so it's not a buffer pH is gonna be determined to by the species in the solution there which is the oxalate the oxalate sodium ions neutral and so but that doesn't have any more protons to donate so it's not going to act as a bass acid it's gonna act as a weak base so the pH is gonna be determined by the partial ionization of C 204 to minus which is a weak base because it's kanji it was a weak acid so it's a weak base and so our strategy is we're gonna have to use an ice table and we're gonna do the ionization of a weak base as our reaction and so what K value are we gonna use well we're not gonna use ka 1 or ka 2 we're gonna use k d1 k b1 the KB for this going the other way picking up a proton so ka 1 is related to KB 2 and ka 2 is related to KB 1 because this is the same conjugate pair the proton coming off in this case the proton going back on in this case and ka 1 is related to K B to the first front proton coming off or the last proton going back on and so that's the same conjugate pair so when we use that relationship that the ka times the KB equals 10 to the minus 14 we're used we relating ka 2 to find KP 1 so we'll have to do that at some point so before I set up my ice table I need to know what my concentration is the initial concentration is of the oxalate so I'm going to use the die Lucien equation again so this is very similar to over here where we had to figure out that dilution of the sample so I can view that again so here I have M 1 V 1 equals M 2 V 2 in the beginning I know I had 1 molar and I had 10 mils M 2 is what I'm solving for so I'll leave that as a variable and V 2 here at this point we could figure it out again using the neutralization equation we could figure out what the volume is here or we could use stoichiometry but I already figured out that the first equivalence point was at 25 mils so that means that this one has to be at 50 mils so now I'm not gonna plug 50 in here I'm gonna plug in 60 because I have 50 mils of titrant and 10 mils of samples so the total volume v2 is 60 mils so we'll plug that in there and we get that m2 equals point one six six molar and this is repeating molar this was originally H 2 C 2 O 4 but I can just do a molar ratio one mole of h2 C 2 O 4 gave me one mole of oxalic right after both protons were removed so I also have 0.16 six molar C 2 O 4 2 minus there so that's the concentration I'm gonna plug into my ice table and now I can go ahead and set that up okay I'm just doing a note in there for you that that's 60 milliliters where that came from and that's from the dilution sample okay so here is my ionization my partial ionization of this weak base remember for weak bases you have to put water in on the reactant side because you have to pull that proton from somewhere it's gonna be from the water that's surrounding it that's a liquid though so it's not really gonna make any difference in my ice table but it's a part of the equation all right so now this is the value I'm plugging in right here zero point one six six and this is zero and zero so minus X plus X plus X so zero point one six six minus X X and X alright next step set up the equilibrium expression okay so I've set up my equilibrium expression products over reactants and remember the liquids not in there because it's a liquid and I'm gonna pull the values from my equilibrium line here so I have X X and 0.16 - X all right so let's see can we assume that that X is negligible well it's 10 to the minus 1 mmm we got to figure out what KB 1 is so let's go ahead and do that I'm just gonna do that up here so I know that ka times KB equals 10 to the minus 14 for a conjugate pair so I have ka times KB equals 10 to the minus 14 which is kW for water at 25 degrees C but these have to be a conjugate pair so as we mentioned before if I want KB 1 this has to be ka 2 because this is losing the second proton from the acid this is gaining the first proton back from the oxalate right so it's the same kanji pair going this way for the KA or going this way for the KB so those are the two that are related so to get KB 1 that's going to be 10 to the minus 14 over ka 2 which is six point four times ten to the minus five so I get that K be one is equal to one point five six to five times 10 to the minus 10 all right now I can see if this is a good assumption ten to the minus one ten to the minus ten oh that's gonna be a great assumption so we're gonna assume that this X is negligible and I'm going to say that this is approximately equal to x squared over 0.166 and that's equal to KB 1 which is 1.5625 times 10 to the minus 10 and I can solve for I can solve for X there and I have just two sig figs which are these two here and I have two sig figs in this one as well here in here all right so now I'm gonna get my ex and I solve for X and it equals five point one zero three one zero three six three times ten to the minus six that's gonna be a great assumption but we'll go ahead and check it anyway so we're looking for X as a percent of the value I'm neglecting it from which is the initial concentration of the oxalate times 100 percent and so I get my X it's ten to the minus six over this which is 0.16 six times 100 percent and I get this is also my percent ionization is point zero zero three one percent ionization so zero point zero zero three one percent ionization so this is a fantastic assumption it's much less than five percent so we're all good so we can go on now now here's the thing now we don't have h plus in here we have Oh H minus and that's okay we can use that to figure out the pH so this time the hydroxide concentration at equilibrium is equal to X which equals this number here five point one zero three one zero three six three times ten to the minus six and this is now molar hydroxide and so I can now do this in two ways I could do the Poh and then go to the pH or I go from the hydroxide concentration to the H+ concentration and then find the pH so we'll do it both ways just so you can see them both but let's do the way that I think easier first P Oh H equals minus the log of the Oh H concentration in equilibrium so equals minus the log of this number here five point 103 times ten to the minus six molar Oh H minus so the Poh is equal to five point two nine two one six five six one two okay well we don't need all those digits because only two of them are significant so this one in this one right here but that remember that's not the the pH as the Poh but pH plus Poh equals 14 so this is the H+ concentration times the hydroxide concentration equals 10 to the minus 14 but we've taken minus the log of both sides and so I've got to subtract this Poh from 14 and when I do that I get that the pH equals eight point seven one does that make sense yes because we have a weak base in solution so the solution the pH should be higher than seven I would have been a red flag there if I thought that was the pH because that's an acidic pH for a weak base in solution right so that makes more sense the other thing we could have done is we could have done we could have gone from the Oh H concentration to the H+ concentration using this relationship without taking - the log of it so I'll do that too so that we can see the difference between those two so at this point I could have also said that the H+ concentration times the Oh H minus concentration equals kW which is ten to the minus fourteen and so I could plug this in and solve for that so h plus equals 10 to the minus 14 over this number five point 103 103 six three times 10 to the minus six so h+ concentration is one point nine five nine five nine one seven nine four times ten to the minus nine molar h+ and then I can do the pH equation pH equals minus the log of the H+ concentration so pH equals minus the log of this one point nine five nine five nine should be good enough times 10 to the minus 9 molar H+ and I get that the pH equals eight point seven one so either way I hit the same thing I like doing this way better because I think it's easier to subtract from 14 than it is to divide 10 to the minus 14 by something but either way is fine okay so one point left all right I'm so ready for our last point here so now we're gonna be two point eight two milliliters after the second equivalence point so we've added 52 milliliters of the titrant is it a buffer yeah so we only have the the conjugates of the sample the tox light so no conjugate pair there we have the conjugate of the titrant the sodium but that's a neutral and then we are now at this point we actually have excess hydroxide so the pH is going to be determined by the excess hydroxide and we can calculate that we're gonna have to account for a dilution do we need to account for the extra hydroxide we get from the ionization of oxalic acid and the ants are sorry of oxalate the answer to that is no we can assume can assume no ionization from the C 204 to - because of the common ion effect alright in that equation that we had just before with the C 204 - minus plus of water going to h c 204 - plus o h - we have excess hydroxide in there so that's an upward arrow over the hydroxide on the products side so that's going to shift this ionization back towards the reactant side is effectively going to shut down the ionization from the oxalate and even if that was contributing a little bit it's not going to be very much compared to the hydroxide that we're just putting into the solution so our our strategy here is going to be to determine the the pH 1 on the excess hydroxide so pH and it's a diluted strong base so all we could do screw out the hydroxide concentration in there take the Poh and then get the pH from that but we do have to account for the dilution so we'll do that okay we could do this dilution of the excess titrant in various ways we could do the stoichiometry way much like we did earlier but the easiest way is gonna be use m1 v1 equals m2 v2 so m1 here is the molarity of the 2 mils that we're putting in which is 0.40 molar v1 is the excess that we're putting in so just the two milliliters but that's gonna get diluted because I'm not putting it in I'm putting it into 60 mils of solution right the 50 mils of titrant that I've already put in the 10 mils of the original sample plus the 2 mils that I put in so the total that it's going in and it's going to be distributed over is now 62 milliliters so I can figure out that m2 equals zero point zero one two nine zero three two two five eight this is molar naoh but the molar ratio can can do H and O H minus is 1 to 1 so one mole of NaOH is also one mole of O H right just a formula ratio formula molar ratio there and so I get that the concentration of hydroxide at this point is zero point zero one two nine zero just a two sig figs their molar Oh H minus now I can just go on my merry way again I could do this in either the ways I just showed you but I'm just going to do the Poh way cuz that's what I think is easier so P o H equals minus the log the Oh H concentration and this is that so equals minus the log zero point zero one two nine zero three two two five eight molar hydroxide so P o H is equal to one point eight eight nine 301 703 these are my two sig figs right here and I'm gonna say that pH plus Poh equals fourteen subtract this from 14 and I get that the pH equals twelve point one one which makes sense because we're pretty high up there at this point now I'm doing a subtraction we can assume that this is infinite so I'm gonna keep two two columns so I still just have two sig figs in this the one on the 1 the 12 is just the place keeper okay and that's it we've done every point on this titration curve that we had labeled here I'm sorry let's do this again damn it