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# Titration of a weak base with a strong acid (continued)

## Video transcript

in the last video we looked at the titration curve the titration of a weak base with a strong acid and we found the pH at two points on our titration curve we found the pH before we'd added any of our acid so that was right here we did that in Part A of our question and Part B of our question we found the pH after we'd added 20 milliliters of our acid so that point is right here on our titration curve we found the pH in Part B in Part C our goal is to find the pH after we added 40 milliliters of our 0.1 molar solution of HCl so how many moles of acid did we add this time so the concentration of HCl is equal to point one molar all right so that's equal to 0.1 molar molarity is mole's over liters so how many moles of acid did we add we added 40 milliliters which if we move our decimal place one two three that's point zero four liters so we added point zero four liters and we could just solve for moles here so the moles of HCl be equal to 0.1 times point zero four so point one times point zero four is equal to point zero zero four so that's how many moles of HCl we have added so you could think about this as h plus and Cl minus and if you have point zero zero four moles of HCl that's how many moles of h plus you have - or you can think about h plus and h2o giving you h3o plus or hydronium so you could also say point zero zero four moles of h3o plus all right how many how many moles of ammonia did we start with right well our concentration was point one so the concentration of ammonia that we started with was equal to point one molar and once again molarity is mole's over liters we started with 40 milliliters of ammonia which is point zero four liters so we started with we started with 40 milliliters or point zero four liters so to find moles of ammonia it's the same calculation we just did point one I'm 0.04 gives us point zero zero for moles of ammonia moles of nh3 so the acid that we added in our titration is going to react with the base that we started with so let's go ahead and write out the neutralization reaction so one way you could write that would be nh3 plus h3o plus so ammonia nh3 plus hydronium h3o plus so h3o plus is our acid right it donates a proton to ammonia so if nh3 picks up a proton it turns into NH 4 plus and if H 3 o plus donates a proton were left with h2o so that's one way to represent our reaction the other way we could write it would be nh3 plus HCl so let's write it that way too so ammonia plus HCl alright HCl donates a proton to nh3 giving us nh4 plus and once HCl donates a proton you have CL minus right the chloride anion the conjugate base so this would be NH 4 plus CL minus alright let's write down how many how many moles of acid that we have alright so we added point zero zero four moles of our acids let's write down point zero zero four here four moles of acid that we added and four moles of base that we started with that's also equal to point zero zero four let's get some more room down here so we started with point zero zero four moles of our base all right if you look at your mole ratios there right so one to one mole ratio so we have now added enough acid to completely react with the base that we originally had present alright so this represents the equivalence point of our titration right this is the equivalence point for our titration because we've now added enough acid to completely react with our base and so all of this acid is going to react alright so we use up all of that and it uses up all of the base that we had present so all this base is going to react with the asset we're going to lose all point zero zero four moles of nh3 so we're left with zero of that ammonia ammonia turns into ammonium nh4 plus and so if we start with if we start with zero for NH four plus alright whatever we lose for ammonia is what we gain for NH four plus so for losing point zero zero for moles of ammonia that's how many moles of ammonium we are gaining so we gain Oh point zero zero four moles of NH four plus so after we add the acid alright we're left with point zero zero four moles of NH four plus so this is at our equivalence point so we have some ammonium at present what's the concentration of ammonium that we have present well concentration once again is moles over liters all right so we have point zero zero four moles of NH four plus so let's write point zero zero four moles of NH four plus what is the volume right what's the total volume now well let's go back up to the problem right we started with 40 milliliters of our we start with 40 40 milliliters and we added another 40 so 40 plus 40 is 80 milliliters or one to three point zero eight liters so our total our total volume now is point zero eight liters so let's go ahead and write that in here so point zero eight point zero eight liters so point zero zero four divided by point zero eight is just equal to point zero five alright so point zero five molar is the concentration of NH 4 plus that is present all right ammonium ammonium is a weak acid ammonium can react with the water that is present so let's now write out that reaction so nh4 plus is going to react with the water alright and ammonium is going to function as an acid ammonium donates a proton to water so if NH four plus donates a proton to h2o h2o turns into h3o plus or hydronium and if you take away a proton from NH four plus you're left with NH three or ammonia so ammonium reacts with water and that's going to affect your pH right if I go back up here so NH four plus does react with water but the chloride anion does not react appreciably with water so you don't have to worry about this being present but the NH four plus will affect the pH which our goal of course is to figure out the pH all right so let's uh let's get some more room and let's write in our initial concentration of NH four plus which is point zero five so the initial concentration of NH four plus is point zero five and once again for these for these problems these weak acid equilibrium problems we assume that we're starting with zero for the concentration of our products and our change right we're going to lose a certain concentration of ammonium a certain concentration of ammonium is going to react with water so we're going to lose X right and then whatever we lose for ammonium turns into ammonia so if we lose X for the concentration of ammonium that's the same concentration that we gain for nh3 for ammonia right and therefore that's also the same concentration of h3o plus so at equilibrium all right at equilibrium we're going to have point zero five minus X for the concentration of NH four plus for the concentration of h3o plus we have X and for the concentration of NH three we also have X all right next this is functioning as an acid right NH four plus is donating a proton to water so for the equilibrium expression we have to write ka here alright and ka is equal to concentration of products over reactants so we have concentration of h3o plus which is x times concentration of nh3 which is X all over the concentration of NH four plus which is zero five minus X so point zero five minus X all right ka for NH four plus write ka for ammonium we calculated in an earlier video when we found it to be five point six times ten to the negative ten so this is equal to this is equal to x times X is x squared and point zero five minus X if X is a very small concentration point zero five minus X is approximately the same as Oh point zero five so we write point zero five in here to make our lives easier for the calculation and we can get out the calculator and solve for x five point six times ten to the negative ten we're going to multiply that by point zero five and then we're going to take the square root of our answer and solve for X X is equal to five point three times ten to the negative six so let's go and write that down X is equal to five point three times ten to the negative six and we go back up here to see what X represents X represents the concentration of hydronium ions right at equilibrium so this would be this would be the concentration of h3o plus we've done a lot of work and the last thing we have to do is find the pH and now we've found the concentration of hydronium ions so the pH is just the negative log of R what we just found so negative log of five point three times ten to the negative six let's get out the calculator and do it so we have the negative log of five point three times ten to the negative six and we get a pH of five point two eight if we round that all right so our pH is equal to five point two eight so at the equivalence point our solution is acidic right a pH of five point two eight let's find that on our titration curve here so we've we've added 40 milliliters of our acid and our pH is five point two eight let's say that is that is five point two eight here's this is Part C of our question the pH is equal to five point two H at the equivalence point finally in Part D our goal is to find the pH after we've added 60 milliliters of our point one molar solution of HCl so this time how many moles of acid have we added we can find that from the molarity since our concentration is point one molar right of HCl and molarity is mole's over liters so we're adding 60 milliliters which is point zero six liters so that's point zero six liters we solve for moles 0.1 times point zero six is equal to point zero zero six so that's how many moles of HCl we've added now HCl is H+ and Cl minus so you could think about this as being point zero zero six moles of H+ or add your proton onto water to form hydronium h3o plus so point zero zero six moles of acid remember from Part C we needed point zero zero for moles to completely react with the ammonia so at this point we're going to lose point zero zero for moles because that much is necessary to react with our ammonia so how much is left over so after our acid has reacted with the ammonia we're left with Oh point zero zero two moles of acid so that's how much we have present in solution so next since our goal is to find the pH we need to find the concentration of hydronium ions so the concentration of hydronium ions would be equal to the moles which is point zero zero two alright and we need to divide that by our volume well our volume would be 40 milliliters plus 60 milliliters right which would be 100 milliliters so that's our total volume here 100 milliliters would be one point one so we have point one we have point 1 liters here and we can solve for the concentration of hydronium ions so point zero zero two divided by point one gives us a concentration of hydronium equal to this is point zero two zero zero molar so now we're ready to find the pH since the pH is equal to the negative log of the concentration of hydronium so the negative log of point zero two zero zero let's get out the calculator is equal to so the negative log of point two one point six nine nine so our pH here our pH is equal to one point six nine nine or however you want to think about rounding these things you just say one point seven zero just to make things a little bit easier a little bit easier to think about that number so approximately one point seven after we've added 60 milliliters of our acid so on our titration curve alright we've added 60 milliliters of acid so here we are right here alright and our pH is pretty close to one point seven so we're past the equivalent point on our titration curve now