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Current time:0:00Total duration:15:13

we've been looking at the titration curve for the titration of a weak acid acetic acid with a strong base sodium hydroxide and in Part A we found the pH before we'd added any base at all so we found we found this point on our titration curve we also found in Part B the pH f you add 100 milliliters of base so we found the pH at this point in Part C our goal is to find the pH after the addition of 200 MLS of 0.05 molar solution of sodium hydroxide so how many moles of hydroxide ions are we adding here so if the concentration of sodium hydroxide is point zero five molar that's the same concentration for hydroxide so for hydroxide we have a concentration at a point zero five molar and molarity is mole's over liters right so moles over liters we have 200 milliliters if we move our decimal place one two three that's point two liters right so this is point two liters here so we solve for moles so point zero five times point two is going to give us point zero one point zero one moles of hydroxide ions so that's how many moles of hydroxide ions were adding to our original acid solution how many moles of acid did we start with well we had 50 milliliters of a point two molar solution of acetic acid so the concentration of acetic acid concentration of acetic acid was equal to 0.2 molar and that's equal to moles over liters so 50 milliliters would be point zero five liters so this would be point zero five liters once again solve for moles so 0.2 times 0.05 gives us point zero one zero moles of acetic acids so notice we have the same number of moles of as we do of base and the base is going to neutralize the acid let's go ahead and write the reaction let's write the neutralization reaction if we have we start with some C t'k acid alright and two are acidic solution we add our sodium hydroxide so we're adding some sodium hydroxide here the hydroxide ions are going to take the acidic proton right so a hydroxide ion takes this acidic proton right here h+ + o h- give us h2o if you take away the acidic proton from acetic acid you're left with acetate ch3coo - and we're starting with we're starting with O point 0 1 moles of acetic acid right so let's let's color coordinate here so we're starting with O point 0 1 moles over here we'll put moles so point 0 1 moles of acetic acid and that's the same number of moles of base right point 0 1 so we have point 0 1 moles of base notice our mole ratio is 1 to 1 alright so we have 1 - 1 here so all of the base is going to react it's going to completely neutralize the acid that we originally had present right so all of our base reacts and we end up with 0 here and all of our acid has been completely neutralized right we lose all of this so we lose all of that and so we've neutralized all of our acid - so this is the equivalence point for this titration and if we're if we're losing acetic acid right we're converting acetic acid into acetate so if we think about starting with 0 all moles of acetate right and we lose point 0 1 moles of acetic acid that turns into acetate so we have to write plus point 0 1 over here so we're making that so we end up with point 0 1 moles of our acetate anion alright next if we have if we have moles of our acid and we have a concentration all right so if we find the total volume of our solution right we can find the concentration of acetate anions so let's do that next we'll go back up here what is the total volume now we start with 50 milliliters and we added 200 more so 50 and 200 give us 250 milliliters so that's our total volume now and 250 milliliters would be move our decimal place three point two five liters so next we find our concentration of acetate all right so what is our concentration of acetate now it would be moles over liters so point zero one so we have point zero one moles over 0.25 liters over 0.25 liters so you can go ahead and find the concentration right you could do this in your head or I'll just show you on the calculator point zero one divided by 0.25 it's going to give us a concentration of point zero four so the concentration is point zero four molar so we have acetate anions in solution at the equivalence point and acetate reacts with water so let's go ahead and show that reaction so we have acetate which can react with water and this reaction will will eventually come to an equilibrium here so the acetate anion acts as a base and takes a proton from water all right so if acetate picks up a proton it turns into it turns into acetic acid so ch3cooh if we take a proton away from water we're left with hydroxide so Oh H minus so now let's think about our initial concentration of acetate right so our initial concentration of acetate was o point zero four so we write here initial concentration is point zero four molar and we're we're assuming that we don't have any products yet so we write zero here next we think about our change well a certain concentration of acetate is going to react right so going to lose a certain concentration which we call X whatever we lose for acetate we gain for acetic acid so if it's minus X for acetate it must be plus X over here for acetic acid and therefore also plus X for hydroxide alright so at equilibrium alright at equilibrium our concentration would be point zero four minus X for acetate and then we'd have X over here and X over here so if we write an equilibrium expression for this acetate is acting as a base so for our equilibrium expression we would write K be and remember that's concentration of products over reactants so that would be X times X so x times X all over point zero four minus X leaving water out so this is over point zero four minus X and we did all this in weak base equilibrium so make sure that you've watched that video before we watch this one because I have to go a little bit faster here alright so next we need to find the KB value we can get the KB by because we know the KA in the last video the KA for acetic acid was 1.8 times 10 to the negative 5 and we know ka times KB for a conjugate acid-base pair is equal to 1.0 times 10 to the negative 14 again this is from an earlier video so if we plug in ka into here we can solve for Kb and I won't do it to save time on the calculator I'll just give you the answer that KB is equal to five point six times 10 to the negative 10 so we plug this in to our equilibrium expression so we get five point six times 10 to the negative 10 is equal to x squared over here we make the assumption like we did in all the earlier videos that X is a very small number it's a very small concentration so point zero 4 minus X is approximately the same thing as point zero 4 so we write point zero 4 in here and next we need to solve for so let's take out the calculator so we have five point five point six times ten to the negative ten we need to multiply that by point zero four and then we need to take the square root of our answer and so we get X is equal to four point seven times ten to the negative six so X is equal to four point seven times ten to the negative six what does X represent right we go up here and we notice that X represents the concentration of hydroxide ions so this this is equal to this is equal to the concentration of hydroxide ions so four point seven times ten to the negative six molar all right our goal was to find the pH so at this point it makes sense to find the Poh Poh is equal to the negative log of the concentration of hydroxide ions so we plug that into here and we solve for the Poh negative log of four point seven times ten to the negative six gives us a Poh of five point three three so the Poh is equal to five point three three and one more step pH plus Poh is equal to 14 so if we plug in our Poh into here pH is equal to 14 minus five point three three which is eight point six seven so we're at the equivalence point but this is a titration of a weak acid with a strong base and so we have a basic salt solution at the equivalence point right so our pH is in the basic range right it sits above seven so let's find that point on our titration curve so we've added we've added here 200 ml of our base and the pH was eight point six seven so you've added 200 MLS of our base and they equivalence point should be somewhere in there all right so right about there about eight point six seven so that's our equivalence point for a titration of a weak acid with a strong base for this particular example finally we're on to Part D which asks us what is the pH after the addition of 300 milliliters of a point zero five molar solution of sodium hydroxide so once again we need to find the moles of hydroxide ions that we are adding the concentration would be equal to point zero five so point zero five molar is our concentration of hydroxide ions and that's equal to moles over liters so 300 milliliters would be 0.3 liters so we have 0.3 liters here multiply 0.05 by 0.3 and you get moles so point zero five times 0.3 is equal to 0.01 five moles of hydroxide ions in Part C we saw that we needed point zero one moles of hydroxide ions to completely neutralize the acid that we originally had present so we're going to we're going to use up we're going to use a point zero one moles of hydroxide that's how much was necessary to neutralize our acid alright so how many moles of hydroxide are left over after the neutralization well that would just be point zero zero five so point zero zero five moles of hydroxide ions are left over after all the acid has been neutralized so our goal is to find the pH right so we could find we could find the Poh if we found the concentration of hydroxide so what is the concentration of hydroxide ions now after the neutralization has occurred so concentration is moles over liters liters so it's point zero zero five divided by what's our total volume we started with 50 and we have now added 300 milliliters more so three grid plus 50 is 350 milliliters or 0.35 0.35 liters so what is our concentration of hydroxide ions right so this is point zero zero five divided by point three five so our concentration of hydroxide ions is point zero one four all right so let's write let's write point zero one four molar here once we know that we can calculate the Poh so the Poh is the negative log of the concentration of hydroxide ions so it's the negative log of point zero one four so we can do that on our calculator negative log of point zero one four and we get a Poh of one point eight five all right so our Poh let's get a little bit more room here Poh is equal to one point eight five and finally to get the pH we know that pH plus Poh is once again equal to 14 all right so we plug in the Poh into here and the pH the pH would be equal to 14 minus one point eight five so 14 minus one point eight five gives us twelve point one five so we're now past the equivalence point all right so we're past the equivalence point here so our pH is twelve point one five after that at three hundred milliliters of our base so let's find the point on our titration curve we've added three hundred ml's of our base so we're right here all right so we go up we go up to our titration curve and that would be right here so we just found the pH is a little bit over twelve right so approximately we got twelve point one five so that's the pH right here on our titration curve