section 16.9 is about the acid-base properties of salt solutions and when i say salt solution here i mean a soluble ionic compound this section actually encompasses many of the concepts we've been talking about for weeks so to really understand what's going on here we need to understand about the equilibria of weak acids and weak bases we have to understand about the relationship between ka and kb of conjugate pairs we have to understand about equilibrium you know the bad news is this is a difficult section but the good news is if you can understand what's going on here you must have a good grasp on almost all of the previous material in chapters 15 and 16. let's say that i had three solutions a solution of sodium fluoride one of sodium chloride and one of sodium bromide would those solutions be neutral acidic or basic i'm going to give you a little bit of a hint by reminding you that all sodium salts are highly soluble and so therefore what we're really talking about are sodium ions and halide ions disassociated from each other in solution what do you think well to answer this question what we're really looking at would be the chemical behavior of these ions and the sodium ion is in fact a spectator it's not going to do anything of interest when we put it in water but the halide ions the situation is a little bit more complicated let's start by considering the fluoride ion so i've brought back up the table of different strong and weak acids and their conjugates arranged by strength but let's find the fluoride ion on there okay there we see it is and you can see that the fluoride ion is the conjugate base of hydrofluoric acid and hydrofluoric acid is a weak acid therefore the fluoride ion is a weak base and when you put a weak base in water we expect it to make the solution basic in contrast the chloride ion we find at the very top of this table and the chloride ion is the conjugate base of hydrochloric acid which is a strong acid and that means that the chloride ion is not a strong enough base to undergo any protonation in water the bromide ion is the conjugate base of hydrobromic acid which we also know is a strong acid and so therefore we can now answer the question the sodium chloride and the sodium bromide will make neutral solutions because the chloride ion and the bromide ion are conjugates of strong acids and therefore so weak that they do not act as bases at all in water the sodium fluoride solution is different however because the fluoride ion is the conjugate of a weak acid hydrofluoric acid and it will make the aqueous solution turn basic so let's first talk about the neutral solutions and we're going to find neutral salt solutions whenever the anion is a conjugate base of a strong acid in the past example we had the chloride and the bromide ions but nitrate would be the conjugate acid of nitric acid perchlorate is the conjugate acid of perchloric acid and so on and so forth so whenever we see those anions we know that the anion is not going to affect the ph of the salt solution if we look at the cations and we have cations that are also conjugate acids of strong bases like sodium hydroxide potassium hydroxide strontium hydroxide they're also we don't expect the cation to have any reaction with water so we saw sodium chloride and sodium bromide on the previous example but we could talk about barium chloride that salt would give a neutral salt solution it's a little bit more interesting though to think about what's going on with the sodium fluoride which we just saw gives a basic solution so let's take a microscopic picture of what things might look like in that solution here are three possible scenarios three possible schematic diagrams that show what's going on in a sodium fluoride solution which one of these three do you think is the most accurate okay let's take a look at each of them solution a okay contains only sodium ions and fluoride ions this would be the right picture for sodium chloride because we know that gives a neutral solution and we know that the chloride ions do not react with water in order the sodium ions but it's not the right picture here if we look at solution b now we see a couple of hydroxide ions and so this would be a basic solution which is a step in the right direction but it's not the right answer because the question is how did we get those hydroxide ions where did they come from finally when we look at solution c we see still the two hydroxide ions that make the solution basic but now we can see where they came from those hydroxide ions exist because there were fluoride ions that took a proton away from water to make hf and of course when you pull a proton away from water what's left behind is the hydroxide ion so the correct answer is solution c we could take this a step further by quantitatively working out the ph of a sodium fluoride solution here a 0.15 molar solution of sodium fluoride given the ka value of 6.8 times 10 to the minus 4 for hydrofluoric acid so how do we solve this problem this is going to be a very similar problem to the problems we talked about when we were calculating the ph of a weak base let's start as we always do with the right chemical equation and this might be one of the hardest parts of predicting the ph of a salt solution so we have this equation where the sodium fluoride goes into water and dissolves this does happen this is an important equation but this equation cannot help us calculate the ph what we have to recognize is what is going to be the effect of the fluoride ion reacting with water and that would be this equation fluoride plus water goes to hydrofluoric acid plus hydroxide so this is the key equation we need for calculating the ph now that we have our equation we can set up an ice table and the ice table is pretty simple here the only thing present to begin with is the fluoride ion before equilibrium is established through this reaction and we know that the concentration of fluoride is 0.15 molar so its activity is 0.15 before their equilibrium is established we don't have hf or hydroxide present so their initial concentrations are zero then when we take into account the changes minus x for fluoride because it's a reactant plus x for hydrofluoric acid and hydroxide because they're products and we add them together to get the equilibrium line next we write the equilibrium constant expression and we can get that directly from the balanced equation above so kb is going to be the products hf times hydroxide over the reactants fluoride and water but the activity of water is one because it's a pure liquid so it drops out of the equation okay at this point the next thing we need is we need to know the value of kb we weren't given the value of kb in this problem but we were given the value of ka for hydrofluoric acid hydrofluoric acid is the conjugate acid of the fluoride ion therefore we know kb of fluoride times ka of hf will yield 1 times 10 to the minus 14 k water just rearranging a little bit here we see that the kb of the fluoride ion is 1 times 10 to the minus 14 divided by 6.8 times 10 to the minus 4 which was ka for hydrofluoric acid and so we get a kb of 1.5 times 10 to the minus 11. now that we know kb and we know the equilibrium entries in our ice table we can solve for x so here's our equilibrium expression again i'm going to plug in my value of kb and then my equilibrium concentrations from the ice table and notice that kb here is a pretty small number so we can be pretty confident that x is also going to be a small number much much smaller than 0.15 which allows us to use this simplification and that makes the math quite a bit easier so we come up with x equals 1.5 times 10 to the minus 6. and if we were to go back and check you would see that x is indeed much much smaller than 0.15 now that we have this x we can put it back into the ice table to get the concentration of hydroxide that's really what we're after and we see that that is simply equal to x 1.5 times 10 to the minus 6. and so we can calculate the poh from that right that's minus log of that concentration and we get a poh of 5.83 if we subtract that from 14 we would finally arrive at our ph value 8.17 and so let's stop just for a minute and ask ourselves is this a reasonable answer we had already made the prediction that the solution should be slightly basic and a ph slightly larger than seven is exactly that a weekly basic solution summarizing these kinds of salts where the anion is the conjugate base of a weak acid those are always going to give us basic solutions i've listed here just a few of the very many anions which are conjugates of weak acids and so we see that will come up quite often here is another example sodium hypochlorite this is the salt that's actually used to make bleach and when you put that in water the hypochlorite ion is going to take a proton away from water to form hypochlorous acid and the hydroxide ion and it is the presence of that hydroxide ion that makes the solution basic are there circumstances where the cation might affect the ph of the salt solution and so we're looking here for a cation that might act as a weak acid or a weak base and if we go back to our table of weak bases notice the first few entries in all of those cases the conjugate acid of the weak base is a cation and we could have salts or ionic compounds where one of these species was the cation ammonium is probably the most common case we'll run into but we could have one of the others as well so whenever we encounter one of these cat ions in an ionic compound we're going to expect in the salt solution this is going to act as an acid and make this ph somewhat acidic what you want to look for are cations that are themselves conjugates of weak bases okay the ammonium is probably the most common but it's not the only one that can do that but here we can see that if you put ammonium chloride into water you would get ammonium cations and chloride ions we've already talked about chloride ions they are really spectators in water they don't undergo any reactions and so we don't have to worry about those but the ammonium cation can go on to react with water to form ammonia plus hydronium and it is the presence of that hydronium ion that's going to make the solution acidic let's do an analogous problem to the one we did a few minutes earlier let's look at the ph of a 0.22 molar solution of ammonium chloride now to solve this problem we need to be given the kb of ammonia which is 1.8 times 10 to the minus 5. the first step is going to be write a chemical equation that's always going to be the first step in this chapter so let's do it the first equation just shows the dissolution of the ionic compound into individual cations and anions that's not really the important equation for us but it is necessary the important equation is the one that was written on the last slide the reaction of ammonium with water and that's going to give us the hydronium that make the solution acidic now let's set up an ice table which is simple and we've seen a lot of these and the main thing to note is that we initially have ammonium cations present from the salt and their concentration is 0.22 molar everything else follows the same rules we've been using over and over and over again let's write the equilibrium constant expression which we can do from the balanced equation up here the acid-base equation and you know the products hydronium and ammonia and the reactant is ammonium cation plus water but as i've said water is not going to show up because its activity is one so we need to know now the ka before we can go any further and we can get that from the kb of ammonia which is the conjugate base right and we've already done this calculation in the lecture repeating it here the ka for the ammonium cation is 5.6 times 10 to the minus 10. now we have everything we need to solve for x so we go back to our equilibrium constant expression we put in this k a value actually that should be ka not kb there and put in the concentrations that we have in our ice table and we come up with x squared over .22 equals 5.6 times 10 to the minus 10. multiply both sides by 0.22 take the square root and we get an x of 1.1 times 10 to the minus 5. notice that that is many times smaller than 0.22 and so our simplifying assumption that we made above is valid now that we know x we can we can return to our ice table and we see that x is actually equal to the hydronium concentration so if we take the negative log of that concentration we'll have ph and once we do that we see that the ph is 4.96 we had predicted that the solution would be acidic and in fact this is a slightly acidic solution the ph has come down from seven to about five so that's a reasonable answer now there's another example of salts that give acidic solutions that's probably going to be not so obvious to you and this is going to happen when the cation is a small relatively highly charged cation transition metal and main group elements that get up to the plus three or even the plus four charge are going to actually produce acidic solutions so aluminum three plus iron three plus are a couple of good examples we could see this more quantitatively if we look at this table of the ka values for these kinds of cations now notice the first three in all of those cases we have two plus signs and we see that ka values are pretty small so 10 to the minus 10 10 to the minus 11. these are going to produce only very weakly acidic solutions because that's quite a small ka value but when we up the charge to plus 3 you can see that the ka goes up by somewhere between 1 to 10 million i mean look at iron iron 2 to iron 3 when we make that one change we just take an electron away you can see that the ka increases by 2 times 10 to the 7th the chromium and the aluminum are not quite as acidic as the iron but still quite high in acidity now we can see this in practice if we look at some solutions of these ions with acid-base indicator present so you can see in the case of a cation being sodium or calcium that the ph is very close to seven so these are effectively neutral solutions notice that in all cases the anion is nitrate and the nitrate ion being the conjugate base of nitric acid is not going to do anything to the ph either if we go to zinc which like calcium is 2 plus but is considerably smaller we do see the ph drops to about 5.5 so it has become a little bit more acidic and that's consistent with this small but non-negligible ka value we saw for the zinc two plus ion if we now were to go to aluminum three plus now the ph drops down to something like 3.5 right this is equivalent to the kind of ph you might get in vinegar or an acetic acid solution so now the aluminum three plus ion is a pretty potent weak acid now you might be asking what is going on here why is it that as the cation becomes small and highly charged it somehow acts as an acid how is that happening and we don't have quite all the tools to explain this at this point in the semester we're going to talk about this kind of thing a bit more in chapter 23 but here's the basic explanation when you put a small highly charged ion into water you end up getting very strong iron dipole forces between the cation and the oxygen ion of the water molecule and in fact those ion dipole forces are strong enough that you can really call this something pretty close to a chemical bond and when the water molecule makes this bond with the iron it actually weakens the o-h bonds a little bit it weakens the o-h bonds enough that another water molecule can come by pull off a proton to form h3o plus and in process of doing that it converts this bound water to a bound hydroxide that lowers the charge a little bit from plus three to plus two so it simultaneously stabilizes this polyatomic cation and it also creates this hydronium ion so this is what's going on with these small highly charged cations this is how they make the solutions acidic okay let's kind of put a lot of this together at least in a qualitative sense and i'd like you to look at these five solutions and predict for me if those solutions are going to be acidic or basic or neutral once you pause the video while you make these predictions then we'll come back and go through the answers all right rather than me just telling you the answers we're going to go to a video from the demo lab and we're going to see with our eyes what kind of solutions we get for each of these well not actually not for the aluminum perchlorate but for the other four let's take a look at the acidic and basic properties of salts here i have ammonium nitrate sodium bisulfate sodium chloride sodium carbonate sodium fluoride and sodium nitrite in these beakers is yamada indicator which is a mix of four indicators it gives us a rough idea of ph green is neutral yellow and orange are acidic and blue and purple are basic i've provided reference samples for acidic and basic here let's start with ammonium nitrate do you predict this to be acidic basic or neutral acidic is that what you predicted now let's look at sodium bisulfate [Music] also acidic what about sodium chloride did you predict it would remain neutral now let's look at these ones let's start with sodium carbonate basic [Music] sodium fluoride is this going to be acidic basic or neutral [Music] and lastly sodium nitrite at the end of this video there is a chart you can use to predict whether a salt will be acidic basic or neutral all right just to summarize then we see that the ammonium nitrate is acidic because of the action of the ammonium cation we see that the sodium hydrogen sulfate salt is acidic and this is one that's kind of different from anything that we've talked about so far but remember that the hydrogen sulfate ion hso4 minus is itself an anion but also a rather strong weak acid the aluminum perchlorate is acidic because of the presence of this aluminum three plus ion sodium carbonate is basic because of the carbonate ion and sodium nitrite is basic because of the nitrite ion as if this wasn't all hard enough sometimes we encounter a salt where both the anion and the cation can react with water to change the ph so it's not too hard to imagine that you could imagine a case where your anion was the conjugate base of a weak acid and there are many of these and the cation was the conjugate acid of a weak base so here uh iron iii sulfite or or ammonium hypochlorite what about that latter one ammonium hypochlorite would that salt give a solution that was acidic or basic now without any other information it's actually almost impossible to answer that question but if we know the kb of ammonia and the ka of hypochlorous acid we can't answer the question and the way we would do that is we're going to consider the strength of the acid which is the ammonium cation right and we can calculate here its ka value to be 5.6 times 10 to the minus 10. okay and we get that by dividing k water by the kb of ammonia and then we look at the kb or the basicity of the hypochloride ion and we can see that its kb is 3 times 10 to the minus 7 which we get by dividing k water by the ka of hypochlorous acid and when we compare these two you can see that kb for the clo minus ion is larger than ka for ammonium and so that means because it's a stronger base then ammonium is an acid it will have a bigger effect on the ph and the solution will be basic