hi there I'm Jeremy Krug and in this video we're going to take a look at AP Chemistry Unit 8 section 4 which is all about the reaction of acids with bases if you haven't subscribed to my channel yet go ahead and consider clicking that subscribe button that way you'll have instant access to all of my 100 plus AP Chemistry daily videos as well as AP review videos problem walkthroughs all kinds of good stuff let's go ahead and jump into what happens when a strong acid reacts with the strong base now this is a very common uh type of question that's asked on AP Chemistry or in general chemistry courses and this might be a very typical example of that Solutions of hydrochloric acid and potassium hydroxide are mixed together well anytime you have a strong acid reacting with a strong base the net ionic reaction is always the same and this is something that we mentioned way back in unit 4 but this is certainly worth repeating H+ aquous plus hydroxide aquous yields water liquid that's the net ionic reaction for any strong acid and strong base everything else that's there is going to be a spectator ion so in this case that would be your your potassium ions your chloride ions those are going to be Spectators in this process so let's try an example with this a student adds one 50 Mill of 50 mol nitric acid to 300 milliliters of 3000 mol sodium hydroxide it asks us to give the net ionic equation The Spectator ions present after the reaction is over and to calculate the pH of the resulting mixture so for part A the net ionic equation well we just have to recognize that there's a strong acid being mixed with a strong base which which means the net ionic equation has to be this right here H+ plus o Nega equals or yields water and The Spectator ions would be all the other ions that are there that aren't part of that net ionic equation so that would be sodium and the nitrate those are your spectator ions now the harder part of this is part C we're going to calculate the pH of the resulting mixture I would strongly recommend that you calculate the moles of the H+ and O minus that are reacting and result from this reaction using a mole ice box now we've used ice boxes to work equilibrium problems this is not really an equilibrium problem but I think an ice box if we just use moles instead of marity actually uh can save us some time overd doing stochiometry and actually doing all the steps in the stochiometry so let's start with finding out how many moles of H+ we have initially now the H+ of course is coming from the strong acid and it's 150 milliliters of 0.5 molar so to find the moles of the H+ we just have to take the 0.15 L times .5 molar like we see here so that gives us 075 moles of H+ that's going to go in this initial spot right down here next we have moles of hydroxide I see that's 300 Mill of3 molar hydroxide so to find the moles I just multiply those two numbers by each other3 L time3 molar so that gives me 009 moles of hydroxide now there's a lot of water here this is an aquous solution we're not really concerned with how much water is made over the course of this process process we're just going to kind of ignore the water here in this uh in this scenario anyway so what we want to do is notice which one of these two items is going to run out well we have less H+ so it's going to run out first so in our change row this side will go down by 0075 since it's the limiting reactant it runs out first the hydroxide also goes down by 0075 leaving us with zero moles of H+ but when you subtract this we're left with 015 moles of hydroxide and we have to reason that since there's excess hydroxide the resulting solution when you mix these two things together will be basic it's going to have a pH greater than seven let's find out what that pH is so we had .015 moles of hydroxide we have to find out the marity of that so we divide it by liters and the liters will essentially be the sum of the two uh mixtures that were added together so we had over here we had 150 Ms over here there are 300 Ms so that's 450 Ms total so that's why we have to divide this by 450 lers to get the marity so when you divide that hopefully you see where that number came from it's just the sum of the two volumes and when you divide it out it's 00333 moles per liter hydroxide now if we know the concentration of hydroxide then if we take the negative log of that that's going to give us the P so that's 1.48 and if you know the p and the pH is pretty easy isn't it just subtract that from 14 and you'll find that the pH is 12.52 so that's how you solve a pH problem like this let's try another one there's a lot going on here so it's worth doing another problem this one says a student adds 290 milliliters of200 molar hbr hydrobromic acid to 185 Mill of50 molar berium hydroxide boh2 and once again it asks us for the same three things as we had in the last problem so the first part the net ionic equation is pretty straight forward since it's a strong acid being added to a strong base it's just the same as it was before H+ plus o minus yields water now by the way if you write this as h3o+ plus hydroxide yields two water molecules that's fine but this is how this is how most people actually write this uh in in chemistry but if you write it as h3o plus that's okay too now once again The Spectator ions are the ions that aren't doing anything so that would be the barium in the base and the bromide from the acid so those those are the spectators now this this leaves us with part C once again calculating the pH so we're going to use the ice box just like we did before this time it's with m though instead of a marity so the moles of hydrogen ions well we multiply 290 L times the2 molar hbr and when you multiply those out you find that the moles will be 0580 moles of hydrogen ions to find the moles of hydroxide we have to multiply 185 l time50 molar times 2 now you see that there's a time two here do you see why it has to be multiplied by two well this has two hydroxides in it so it's it's essentially a 2:1 ratio we have to account for that so when you multiply that out we get 0.0555 moles of hydroxide on a problem like this that is honestly the problem the most common mistake students forgetting to multiply by two because this is one of those strong bases that has two hydroxides in it like calcium hydroxide or stonum hydroxide it has the two now as we look at these two values we can see that looks like the hydroxide is the smaller value this time so that's the one that's going to run out first hydroxide is our limiting reactant and so that means that we're going to have zero moles of hydroxide but when you you subtract the H+ values we find that we have 0.0025 moles of hydrogen ions and that means that since we have excess hydrogen ions this is going to be not basic like in the last problem this is going to be an acidic resulting solution so we're going to take the .0025 moles of H+ that we just calculated and we're going to divide that by the total volume now do you see where the 475 L come from this time we had 290 milliliters and we added that to 185 milliliters so the total volume when you add these two together gets you 475 milliliters or that's our 475 lers so when you divide that out that's 0.53 molar and if you know the H+ concentration then to find the pH you just take the negative log of that H+ concentration just like we learned earlier on in in section one of this unit and so that's 2.28 and so that's all you have to do for a strong acid strong base problem now that's that's the most common type to be completely honest but you'll also sometimes have cases where you have a weak acid reacting with a strong base this is a fairly common type in fact we'll talk more about this in the next uh section section five about acidbase titrations and if you have a weak acid reacting with a strong base the net ionic equation is going to look something like this where ha is the formula of your weak acid of course hydroxide that's what's reacting out of the strong base the water is a product and this a minus here represents the conjugate base of the weak acid so basically it's just whatever the weak acid was starting with with an H+ taken away from it so that's the that's the template for writing the NIT onic equation for a weak acid strong base problem now usually what's going to happen in these problems is most likely you'll have a certain amount of acid and you'll have fewer moles of hydroxide than you have moles of the weak acid and so that means that all of the hydroxide reacts and what you're left with is a mixture of the weak acid this ha and the conjugate base the a minus so whenever you have a mixture of a weak acid and its conjugate base we call that a buffer and that's an important term to know a a mixture of a weak acid and its conjugate base we'll talk more about buffers toward the end of Unit 8 it's a very important concept something that's that's actually used quite a bit in in uh in chemistry so that's weak acid strong base now you could have another type of problem where a strong acid reacts with a weak base now this this this type is not quite as common in AP Chemistry but every but every now and then it does pop up if that happens this is the basic uh template for that type of equation in this case the strong acid I'm using the h3o plus that the hydronium to act as the strong acid that's aquous of course the weak base is just the B and that's aquous and the products are HB plus now that represents the conjugate acid of the weak base because this is the weak base the conjugate acid is just adding on an H plus to that basee essentially and of course water is uh produced as well so this is the template for writing a strong acid weak base reaction now once again most of the time if you see a problem like this you're going to have acid and of course the weak base and you're probably not going to have enough strong acid to react with all of the weak base and if that happens that means that what you're going to have is all of the hydronium reacting and so it's all essentially gone and you have a mixture of the weak base and its conjugate acid the B and the HB plus well that is also a buffer anytime you have a weak base and its conjugate acid or weak acid and its conjugate base that is a buffer and you probably know a few things about buffers already from perhaps earlier science classes buffers are solutions that resist changes to pH so that means you can add some acid or base to it and its pH is not going to change very much now your blood has a very important buffer in it so your your blood maintains a fairly constant pH we'll talk more about buffers later on in Unit 8 like I said earlier now let's take a look at a couple questions about the strength of acids and bases this one says the dissociation of cyanic acid is given below which of the two bases is stronger the CN negative ion or the water molecule explain your answer now the key part here to recognize is that this Ka is a fairly small number 3.5 * 10-4 that means that this reaction does not proceed forward in any appreciable amount uh since the value for KA is relatively small that means that this equilibrium lies far to the left you're going to have a whole lot more reactants than products so what that means is that this CN negative is very effective at accepting protons and we know that because we have a whole lot of hcn the CN negative has accepted lots and lots of protons evidently whereas water is not accepting a whole lot of protons we know that because you don't have a whole lot of CN negative produced this is a a two-way competing process we have a a forward process that is not proceeding very much but we have a reverse process that is proceeding quite a bit so in a a weak acid situation we'd say that its conjugate base is very good very strong I should say at accepting protons it's a stronger base than the water is let's look at another example which is is a stronger base in this example uh water or the nitrate ion and explain your answer well the giveaway on this one is that we're dealing with a strong acid dissociation on this one there really is no Ka because notice there's a single-headed arrow that means that this reaction essentially proceeds to completion so when that's the case notice that water is is a very good base because it is accepting every single one of the protons that nitric acid can throw at it it is an excellent base in this example whereas nitrate is a pathetic base it's not really acting as a very good base at all and that's because it it can't accept any of these protons that hydronium might be throwing at it and you see that that's essentially what the explanation is saying here water accepts protons very well uh nitrate is very ineffective so water is the stronger base so the takeaway from this is that when a strong acid is reacting with water well water is an excellent base in this example I hope this video has helped you to learn something about how strong acids and strong bases and weak acids and and and strong bases and and and other and that other type strong acids and weak bases react uh sometimes hard to keep it all straight in my head but hope you uh learned something from this if you did please smash that Thumbs Up Button smash the like button and I hope to see you in the next video where we are going to jump right in to talking about acidbase titrations in Unit 8 section five hope to see you then