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Video transcript

here we have a titration curve for the titration of 50 milliliters of 0.2 0 0 molar of acetic acid and 2 are acidic solution we're adding some point zero five zero zero molar sodium hydroxide so once again we're putting pH in the y-axis and down here in the x axis is the milliliters of base that we are adding so in Part A what is the pH before you add before you've added any base right so the addition of zero point zero MLS of sodium hydroxide so we only have to think about a weak acid equilibrium problem here so our weak acid is acetic acid alright and it's in water so acetic acid donates a proton to h2o and so h2o turns into h3o plus and if you take away a proton from acetic acid you're left with ch3coo minus the acetate anion our initial concentration of acetic acid is point two zero zero molar alright so this is point two zero zero molar here and we're pretending like we don't have any of our products yet right so this is what we did in the video on weak acid equilibrium so the change whatever we lose from our concentration of acetic acid since acetic acid turns into acetate right we would gain for the concentration of acetate and we would also therefore gain for the concentration of hydronium h3o plus so at equilibrium right our equilibrium concentration would be point two zero zero minus X and then over here we'd have plus X and over here we would also have plus X so when we write our KA expression or equilibrium expression ka is equal to write concentration of products over reactants and so I'll just go a little bit faster since we've already done a video on all of this and so make sure to watch the video on weak acid equilibrium before you even look at these alright so this would be X right times X so that's these two X's over here so our products over our reactants and that would be 0.2 out of our equilibrium expression so this would be point two zero zero minus X if we assume that X is a really small number right a very small concentration compared to point two zero zero we can approximate it and say that point two zero zero minus X is about the same as point two zero zero so we can rewrite this let's rewrite it over here so we have x squared now over point two zero zero this is equal to ka and ka for acetic acid is 1.8 times 10 to the negative 5 so we can go ahead and solve for X so let's get out the calculator here so we have 1.8 times 10 to the negative 5 and we're going to multiply that by point two zero zero right so we get we get three point six times 10 to the negative 6 and then we need to take the square root of that so the square root of our answer gives us X and we can see X is equal to point zero zero one nine so let's write that down X is equal to point zero zero zero one nine and remember X represents the concentration of hydronium ions right so this X right here right this X represents our concentration of hydronium ions at equilibrium and so if this is the concentration of hydronium ions right so this is equal to the concentration of hydronium ions finding the pH is very straightforward because we just need to plug in the concentration of hydronium into into our pH calculation pH is equal to negative log of the concentration of hydronium so we take this number point zero zero one nine and we plug it in here and we can solve for the pH and so let's go ahead and do that so let's take out the calculator so the negative log of point zero zero one nine gives us the pH and so we get two point seven two so the pH is equal to two point seven - so before we've added any base alright so zero point zero MLS of sodium hydroxide the pH should be two point seven two and so we go over here to our titration curve right and at zero point zero MLS of base we go over to here and so this is our first point right so this is a right here and that should represent a pH of two point seven two alright next let's add some more sodium hydroxide now in Part B our goal is to find the pH athle added 100 point zero MLS of our point zero five zero zero molar solution of sodium hydroxide so let's see how many moles of base we are adding so if the concentration of sodium hydroxide is point zero five zero zero molar that's the same concentration of hydroxide ions in solution so the concentration of hydroxide ions in solution is point zero five zero zero molar and molarity is mole's over liters right molarity is equal to moles over liters so this is equal to moles over how many liters is 100 milliliters we move our decimal place one two three so that's point one so that's zero point one liters alright so to find moles all we do is multiply right so point zero five times point one gives us point zero zero five so we're working with point zero zero five zero zero moles of hydroxide ions alright so that's how many moles of hydroxide ions were adding how many moles of acid did we originally have present let's go back up here to our to our titration curve so we can see those numbers so we were starting with we're starting with 50 milliliters of point two zero zero molar acetic acid alright so let's figure out how many moles of acetic acid we started with here so the concentration was point two zero zero molar so for our concentration of acetic acid our concentration was - 0.200 molar and that's equal to moles over liters we start with 50 milliliters which is point zero five zero zero liters so we multiply point two by point zero five and we get 0.0100 moles of acetic acid so that's what we're starting with here so we have an acid and a base our acid is acetic acid and our base is the hydroxide ions the hydroxide ions are going to neutralize the acid that's present so we get a neutralization reaction let's go ahead and write what happens our acid is acetic acid and we're adding some hydroxide so the hydroxide is going to take the acidic proton from acetic acid the hydroxide is going to take this proton right here and ohy Ness and h plus of course form h2o so once hydroxide takes a proton from acetic acid we're left with the conjugate base for acetic acid which is of course the acetate ion so we also make ch3coo minus here all right what were we starting with for acetic acid right we were starting with point zero one zero zero moles so let's go ahead and write that 0.0100 moles and for hydroxides for hydroxide we were adding point zero zero five all right so we're adding point zero zero five moles of hydroxide so the hydroxide ions are going to neutralize the acetic acid all right so we're going to lose all of our hydroxide so all of it reacts so all of this reacts here and so we're left with nothing we use up all of our hydroxides right and that much that many that much hydroxide is going to neutralize the same amount of acetic acid so we're going to lose that much acetic acid so point zero one minus point zero zero five right this is going to give us point zero zero five zero moles of acid left over so this is how much acid right is how much acid was not neutralized all right so if we lose that much acetic acid we're also going to gain that much acetate anion right so if we start with zero over here we're going to gain the same amount so point zero zero five is how much we gain over here so we're going to finish after neutralization with point zero zero five moles of acetate anion alright next let's figure out the concentration of acetic acid in the acetate anion so we have moles for both alright we have moles for both but we need to find the volume well let's think about that so let's go back up here to the problem and we're adding we're adding a hundred milliliters alright and we started with 50 milliliters of our acid solution alright so 50 plus 100 is 150 milliliters right so now our new volume our new volume is 150 point zero milliliters 50 plus 100 and that's equal to 0.15 liters right we move our decimal place so let's figure out our concentrations we have moles we have volume so let's find the concentration of acetic acid now after neutralization all right so we have point zero zero five zero moles all right so let's go ahead and put that in point zero zero five zero moles and our volume is 0.15 0 right so 0.15 zero is our volume so we can go ahead and do our calculations so I'll just take out the calculator here so we have point zero zero five divided by point one five zero and we get point zero three three so the concentration is point zero right this is point zero three three Moeller alright what about the concentration of acetate alright so the concentration of acetate anions in solution alright would be equal to look at the numbers they're the same we have we have Oh point zero zero five here and once again our total volume is point one five zero so it's the same calculation zero point zero zero five zero over zero point one five and we would therefore also get a concentration of point zero three three molar so we have equal concentrations of a weak acid and its conjugate base and that makes me think about a buffer solution so that's what we form to your as we as we drip base as we drip hydroxide ions into our original acidic solution we're slowly forming a buffer solution and here we have equal concentrations so our goal was to find the pH and since we have a buffer solution now it's easiest to just use the henderson-hasselbalch equation alright so the henderson-hasselbalch equation was ph is equal to the pKa plus the log of the concentration of a minus over the concentration of h a alright and the KA for acetic acid we talked about earlier was 1.8 times 10 to the negative 5 so the pKa would just be the negative log of that so let's take out the calculator let's take the negative log of 1.8 times 10 to the negative 5 and so the pKa is equal to four point seven four so let's write that down here PKA is four point seven 4 so we plug that in to our henderson hasselbalch equation right here so the pH is equal to four point four point seven four point seven four plus log of the concentration of a minus a minus would be would be acetate right which is point zero three three over H a H a would be acetic acid which is point zero three three so with the same concentrations so this is log of point zero three three all point zero three three and that of course is one right so we have we have log of one here and you probably already know what log of one is equal to I'll do it on the calculator so you can see that log of one is equal to zero and so all of this is equal to zero so the pH is equal to four point seven four plus you know this is all be zero over here so the pH is equal to four point seven four and this is the half equivalence point we've neutralized half of the acid right and half of the acid remains and using henderson hasselbalch to approximate the ph we can see that the pH the pH is equal to the pKa at this point so let's go back up here to our titration curve and find that alright so the pH is four point seven four the pH is four point seven four after we've added 100 ml of our base so we go right up here to 100 MLS and so this would be our second point this is what we did in Part B and the pH is approximately four point seven four at this point