If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:31

Video transcript

the party talked about how to write an equilibrium expression so if we have some generic acid H a that donates a proton to h2o h2o becomes h3o plus and H turns into the conjugate base which is a minus and so here's our equilibrium expression and the ionization constant ka for a weak acid we already talked about the fact that it's going to be less than one so here we have three weak acids so hydrofluoric acid acetic acid and methanol and over here are the KA values so you can see that hydrofluoric acid has the largest ka value so even though they're all considered to be weak acids all right 3.5 times 10 to the negative 4 is larger than 1.8 times 10 to the negative 5 so hydrofluoric acid is stronger than acetic acid and acetic acid is stronger than methanol but again they're all considered to be weak acids relative to the stronger ones so let's talk about PKA so the pKa is defined as the negative log of the KA so if we wanted to find the pKa for methanol all we have to do is take the KA and take the negative log of it so the pKa is equal to the negative log negative log of 2.9 times 10 to the negative 16 so let's get out the calculator and let's do that negative log of 2.9 times 10 to the negative 16 and this gives us 15 point 5 4 all right when we round that so the pKa of methanol the PKA methanol is equal to 15 point 5 4 so we could write in a PKA column right here and for methanol it's 15 point 5 4 if you did the same calculation for acetic acid you would get 4 point 7 4 and once again if you did this for hydrofluoric acid you would get 3 point 4 6 so as we go as we go up on our table here right we're increasing in acid strength so out of our three weak acids hydrofluoric acid is the strongest so it has the largest value for ka but notice it has the smallest value for the pKa all right so the the lower the value for PKA the more acidic your acid so three point four six is lower than four point seven four and so hydrofluoric acid is more acidic than acetic acid the problem asks us to calculate the pH of a one molar solution of vinegar which is acetic acid in water so let's start by writing our acid-base reaction so we have acetic acid so ch3cooh plus water and if acetic acid donates a proton to water water would turn into h3o plus the hydronium ion and we would also get the acetate anion ch3coo minus the conjugate base to acetic acid so before we get to this problem let's just pretend that we're starting with a hundred molecules of acetic acid so 100 molecules of acetic acid here and let's say none of them have reacted so this is before any reaction has occurred which means we don't have any of our products right we don't have any hydronium ions we don't have any acetate anions and acetic acid is a weak acid if a strong acid would ionize 100% so all 100 of those molecules pretty much would on eyes but since acetic acid is weak let's pretend like only one of those acetic acid molecules donates a proton to water so we're going to lose we're going to lose one molecule of acetic acid and that's going to turn into the acetate anion so if we lose one molecule of acetic acid we gain all right we gain one acetate anion and therefore we're if we protonate one molecule of water with that proton right we're going to gain one hydronium ion so we would say that we would have ninety-nine molecules of acetic acid and then we have one hydronium Aion and we would have one acetate anion so that's the kind of thinking that we're going to apply to concentration so next we're going to start with an initial concentration here we're starting with a 1 molar solution of acetic acid so I can write here one point zero zero as my initial concentration and just like before we're going to pretend like nothing has happened yet so the concentration of hydronium is zero and the concentration of the acetate anion is also zero so next we're going to write a C here for change alright and let's let's define X as the concentration of acetic acid that reacts so if if we have a concentration of acetic acid as being X let me go ahead and just write X here that means we're going to lose a certain concentration of acetic acid and therefore going to gain the same concentration of the acetate anion and we're going to therefore gain the same concentration of the hydronium ion so when everything comes to equilibrium so our equilibrium concentrations would be 1 minus X for the concentration of acetic acid and for the concentration of hydronium we would have X and for the concentration of acetate anion we would also have X so now let's let's write our equilibrium expressions let's get a little more room down here so our equilibrium expression so ka would be equal to concentration of hydronium ion so the concentration of hydronium ion times the concentration of acetate anion all right and this is all over all over the concentration of acetic acid and we leave water out so all over the concentration of acetic acid here all right let's plug in what we know so the equilibrium concentration of hydronium that's X right so that's X so let's go ahead and put that in here so we can put X in for concentration of hydronium and concentration of the acetate anion that's also X all right so that's the concentration of the acetate anion so I put it in X here too over the concentration of acetic acid at equilibrium that's 1 minus X so I put in here one point zero zero minus X and this is equal to the KA for acetic acid alright so we had that on our table above so let's go back up here to the table that we talked about earlier and here is the KA for acetic acid 1.8 times 10 to the negative 5 so we plug that in we plug that in here so let's plug in 1.8 times 10 to the negative 5 now at this point you could you could solve for X but you need to use the quadratic formula so let's make an assumption at this point and we're going to assume that X our concentration is much much smaller than 1 molar because that makes our life easier for the math because if X is much much smaller than 1 1 minus X is approximately the same thing as 1 right so this is an extremely small number subtracting it from 1 isn't going to do anything is it going to do much of anything so we're going to we're going to make this assumption and say that 1 minus X is equal to 1 and that just makes our lives easier so now we have 1.8 times 10 to the negative 5th is equal to x squared over 1 so all we have to do now is solve for X so let's get out the calculator here turn it on so we just need to take the square root of 1.8 times 10 to the negative 5 and so we get point 0 0 for 2 all right so let's go ahead and write that down here so let's get some more room and X is equal to point zero point zero zero four two and remember what X represents X represents the concentration of hydronium ion right X here represents the concentration of hydronium ion at equilibrium so this is equal to our concentration of hydronium ion and now we can calculate the pH which is the whole point of the problem all right so let's calculate the pH of our solution so the pH should be equal to the negative log of the concentration of hydronium so the negative log of point zero zero four two and we can use our calculator for that so let's go ahead and take the negative log of point zero zero four two and that's going to give us a pH of two point three eight so let me go ahead and write that so our pH is two point three eight so that is the pH of our vinegar solution