 Okay, let's do this particular problem. It's actually a two-part problem It's actually a titration problem and a buffer problem. So the first part says Calculate the pH of a 0.20 molar ammonia slash 0.20 molar ammonium chloride buffer and then the second part says what is the pH of the buffer after the addition of 10 mils of 0.1 molar HCl to 65 mils of the buffer. Okay, so let's do the first part first The cool thing about this is that We have Talked about the Henderson-Hasselbach equation already so we can actually use that. Okay, so what is the Henderson-Hasselbach equation? The pH equals the pKa plus the log of the Concentration of conjugate base over the concentration of the weak acid like that So that's nice because we've got the concentration of both of those things and we've got the kA. So remember the pKa Is going to be the negative log of the kA So that's going to be negative log So when I do that in 0.25 as the pKa Okay, so we have pKa, we have concentration concentration, right? This we can already see is going to be 0.2 over 0.2 Right, so 0.2 divided by 0.2. That's going to be 1. The log of 1 is 0 Okay, so this one's nice because it tells you well At the the concentration are the point when you have the Concentration of the conjugate acid and conjugate base being the same the pH is equivalent to the pKa of the weak acid Okay, so let's just do it together. So the pH is or the pKa is 9.25 plus the log of 0.20 0.20 So 9.25 plus the log of 1 Okay, so the log of 1 is 0 so the pH and the first part is going to equal 9.25 Okay, and you would expect that because The kA is much smaller the kA of ammonium Ion is much smaller than the kB of the of the molecule ammonium So you would expect this to be a basic solution Okay So the second part asks us to add Some HDL to this okay So we're going to make this very basic solution a little bit more acidic. Okay, that makes sense, right off the bat, right? Okay, so what I'm going to do is erase some of this stuff That we figure down Okay, so now it wants to know well, what's the pH after the addition of 0.10 HCl so the concentration of HCl Remember HCl goes all the way to hydrogen ions so Concentration of HCl is going to be the same as the concentration of H plus in order to just think about that, right? strong acid So we get H3O plus Cl minus a quiz, okay So what do we have a one to one ratio of those two things? So why don't we just erase this and put H3O plus like that, is that okay? If we do that, does that make sense? Yeah, okay wonderful. Okay, so I'm going to erase this part here And they gave us the volume of the buffer is 65.0 mils and the volume of the HCl was 10.0 mils that we added to it. Okay, so in order to figure this equation out, we're going to, or this problem out, we're going to have to figure out well How many moles of a H3O plus did we make, okay? So remember the volume of HCl and the volume of H3O plus are going to be the same So What's the number of moles of H3O plus? Now we have well 0.1 Molar, right, that means moles of H3O plus per one liter of solution So we're going to multiply that by well one liter per 1,000 mils times Well, how many mils do we have? 10.0 mils of the H3O plus. Okay, so when we do that, liters cancel with liters, mils cancel with mils Okay, so we should get, if I'm calculating my head, right 1.0, oh, sorry 1.0, that's 10 to the negative 3 moles of H3O plus. So calculate it, make sure I do it right Okay, so we're now also going to have to calculate well What was the number of moles of each of these that we started with? Okay, so Number of moles of ammonia is going to be 0.20 moles of ammonia per one liter of solution We'll do the same thing one liter 1,000 mils 65.0 mils Cancel, cancel, cancel, cancel Well, what do we get? Well, I can't do that one in my head 1.2 I'm 65 1,000, of course 1.3 times 10 to the negative 2 2 moles of ammonia Okay The number of moles of the ammonium ion is going to be the same right because it's 0.2 divided by 1,000 times 35. Okay, so we're just going to forego doing all that and say 1.3 times 10 to the negative 2 moles of the Ammonium ion 2 So in fact what I would do now Since I know this is times 10 to the negative 2 I'm going to change this to 10 to the negative 2 as well because it'll make my ice table a lot easier so If I want to change this to 10 to the negative 2 That's going to be instead of 1 It's going to be 0.10 like that Or 1, whatever you want, 0.5 Okay, so Are you set with what we've done here so far? So now What we're going to do is Think about some ice tables. Okay, so if we have ammonia in solution And we add H3O plus To that solution it's going to react like that And we're going to get some NH4 plus some ammonium ion Plus H2O So let's do a nice table What we're doing So ammonia, what was my initial concentration or my initial number of moles? Sorry 1.3 times 10 to the negative 2 Initial concentration of the ammonium ion the same thing, right? So in the first problem, we didn't have any of this to start out with right but this time we've added So what how much should we add 0.1 times 10 to the negative 2 right 0.1 Whoops, so 0.1 times 10 to the negative 2. So that's going to react completely Right with the ammonia here So it's going to be minus 0.1 times 10 to the negative 2 Why because that number is smaller than that that number so it's all going to react so at the end We're not going to have any okay, so this is going to be minus 0.1 times 10 to the negative 2 as well Okay, does that make sense? So here we're going to get 1.2 times 10 to the negative 2 right And here Well, if we subtract it all here, right, what's going to happen here at it, right? So plus 0.1 times 10 to the negative 2. So what are we going to have down here? 1.4 times 10 to the negative 2. Does that make sense? Okay, wonderful. So those are the new number of moles that we have of these two substances So what I'm going to do is erase all of this stuff and just write down those new number of moles, okay? Is that all right? Can I erase all of this? I don't have any moles of hydronium anymore Only the moles of the other two matter so moles of ammonia We've got 1.2 times 10 to the negative 2 moles And the moles of the ammonium ion 1.4 times 10 to the negative 2 moles Okay, I'll put ammonia there But we if we're trying to use this equation here, we need concentration values, right? Not more or not number of moles. Okay, so how do we get concentration values? Well, we have to know the volume that we're using, right? So the volume is going to be the volume of the acid added plus the volume of the buffer original, right? So the total is going to equal volume of H3O plus or HCl whatever if you still got that written, you know volume of buffer So 10.0 mils plus 65.0 mil Like that So that's going to be 75.0 mils, right? But we want it in liters like that Do you get that? Divide the whole thing by 0.0 750 liters like that Same thing with this one here 0.0750 liters Okay, and that's going to give me my two concentrations Good two So the top one The concentration is 0.16 molar to two sick pigs ammonia Remember that's a a minus in this equation. That's where we're going with this. Okay, so this one here Do the same thing So And this one I get well to two sick pigs, but I'll I'll keep those digits When I do the next calculation 1.9 molar So if you just stick 1.9 in here your number might be a tiny bit Because I'm going to do that 1.8666 Whatever, okay, so we have this this and pk right so we should be able to figure out what our ph is So let's do it. So ph Equals pk which is 9.25 plus the log Of a minus which is 0.16 divided by 0.19 Like that. Okay, so 0.16 divided by That answer we're going to take the log That answer and we're going to add that to 9.25 And like I said, we should expect since we added hcl to this that our ph would go down, right? And in fact it does in my calculation 9.18 is my new ph Makes sense how to do that These are actually just the first two steps of the titration calculation, so question