welcome to applications of aqueous equilibria this is openstax chapters fourteen point six to fourteen point seven and chapter fifteen so just the buffers at the very end of chapter fourteen because that's an application of aqueous equilibria and chapter fifteen which is also talking about the common ion effect and solubility products so those few things alright first we're gonna start off with the common ion effect so the common ion effect is really just the Chatelet principle in action the definition is that adding another substance that contains an ion in an equilibrium expression will cause a shift according to Lachey's principle so it's not really any different than what we already learned but we'll show how this effects a couple of different systems so first we'll look at the solubility of an ionic compound in an aqueous solution and see how the common ion effect affects that okay let's say we have an ionic solid like sodium chloride that's gonna dissolve and form sodium ions and chloride wine ions as it completely dissociates the part that that does dissolve completely dissociates into ions it's a strong electrolyte but even something very soluble like sodium chloride does have a solubility limit there is a point at which the ions are colliding with the surface of the crystal so fast that they are recrystallizing or precipitating as fast as they are dissolving and then you've reached your solubility limit now what if we have another source of one of these ions in solution what if we have say potassium chloride in the solution as well well that's gonna be another source of chloride ions and so that's gonna cause a shift and we're gonna do a stressed shift equilibrium arrow diagram in order to see what would happen so our stress is that we're gonna put potassium chloride also in the mixture and you could add that before it could already be in there you could be adding it as you go but for some reason there's also potassium chloride in the solution that's gonna be an additional source of chloride ions besides just the dissolving of this solid and so that's stress is an upward arrow above the chloride well just like with shot leah is one that we covered if I increase something on this side it's gonna shift away in order to use that up so we have a shift going this way a down arrow down arrow on the tail side of that and an up arrow over here when we add these lines together to hit the equilibrium line how it will have changed when we establish equilibrium nothing in an up up nothing in a down is it down and then here we have an up and a down remember we can only minimize the stress so it's gonna be a little arrow in the direction of the original stress so the chloride Constitution will be higher than it would otherwise but the sodium ion concentration will be lower and there'll be more solid and less ions from this solid in solution so that would decrease the solubility of this ionic compound in water now that's not always the case what if we had something different what if I had something like aluminum hydroxide solid dissolving into aqueous ions now this would have a much lower solubility aluminium salt hydroxide isn't very soluble but the part that is the part that does go in a solution is still completely dissociated but what if I had HCL in the mixture HCL in the mixture so that's completely ionized in two hydrogen ions and chloride ions much like that the tascam chloride that's dissolved would be ionized into potassium ions and chloride ions this time it's not the chloride because that's not in there but the hydrogen ions in there would have an effect what we see like we've seen before hydrogen ions and hydroxide ions are gonna completely neutralize each other so forming water that would actually decrease the hydroxide concentration so in this case if we're gonna lower something on one side we're gonna shift towards that side to replace it and we're gonna be using up some of this to do that so nothing to downs and down nothing in an up isn't up and then a down and up it's gonna be a little down a little down okay so what would happen is the hydraulic concentration will go down a little bit but it would also cause a shift that would make more aluminum hydroxide dissolved than otherwise would so in this case it would actually increase the solubility of this ionic solid in the solution so that's an interesting and go either way all right next we're gonna look at ionization of weak acids okay so here we have a weak acid hydrofluoric acid ionizing into hydrogen ions and fluoride ions unlike this one above where we had solids going into aqueous ions now everything is aqueous but we have this equilibrium between the molecular form and the ionic form so let's cause put a stress on this let's say that we are gonna do this but instead of just having this happen in pure water with just the hydrofluoric acid we're gonna also put in some sodium fluoride into that water so let's look at that stress so if I have sodium fluoride in the mixture the sodium's of spectators not in this equilibrium expression but the fluoride is so that's an up arrow over the fluoride as the stress it's gonna shift away these go down this goes up and at equilibrium we have an up a down and a little up okay so you can see that what's gonna happen here is that the hydrofluoric acid is gonna I extend more of it will be in the form of the neutral molecule and less of it will be ionized in the form of these two ions there will be a little extra fluoride in there but that's from the sodium fluoride but not as much as there is sodium in there from the sodium fluoride and so what happens in this case is that the percent ionization would be lower now also notice that if I have a lower concentration of H+ that corresponds to a higher pH closer to seven so the concentration of H+ at equilibrium would be lower therefore the pH would actually be higher or closer to seven not as acidic as it would have been if the sodium fluoride haven't been in there now in terms of pH the sodium's conjugate base sodium hydroxide is a strong base so this is this is neutral in terms of pH the fluorides conjugate acid is actually hydrofluoric acid which is weak so the fluoride itself in solution is a little basic as well but just looking at the hydrogen ion concentration we can also see that the pH is gonna be higher than it would have otherwise okay so let's do an example we're gonna do that ionization we saw right above with the HF and sodium fluoride and we're gonna calculate what the pH is and percent ionization is of one and a half molar HF in two molar sodium chloride and this is a weak acid so I need to give you a KA value and that's seven point two times ten to the minus four so it's a weak acid we're gonna approach this using an ice table at this point and to do that we need to set up the reaction and this is the ionization of an acid ka that we have the K value for so that equation above the ice table is gonna be the ionization of the weak acid okay set up the ice table with my ionization for that weak acid up there and the initial concentration of HF if I didn't have this I just put in one point five I'm still gonna do the same thing here that's gonna be one point five molar HF normally I put a zero in a zero here but in this case this is zero but this is not because I have two molar sodium fluoride I can do the stoichiometry the molar ratio between sodium fluoride and fluoride if it's in Equus solution as the two molar suggests is one-to-one and so that is going to be also two molar sodium ions which are spectators in this case and two molar fluoride ions which are not spectators they appear right here so this is starting with two molar solution of fluoride now at this point I don't have to compare K and Q to see if there's a little forward or backwards because this is zero I know that it has to go forwards because there hasn't been any H+ made yet from the hydrofluoric acid so if it's going forward my change here is going to be minus X and this is gonna be plus X and plus X adding those two lines together I get one point five minus X X and two point O plus X so those are my equilibrium outs my equilibrium line the next step is to set up the equilibrium expression so as always K equals products over reactants and here they're all equally astir they're all gonna be in there so my ka equals products concentration of h+ at equilibrium concentration of fluoride ions at equilibrium both to the first power over the concentration of H F at equilibrium also to the first power and I'm going to set up my equilibrium expression here with variables because I want that point on the quiz and then I'm going to put in the lines for these values from the equilibrium line so the hydrogen is X this is two point zero plus X and on the bottom I have one point five minus X okay now we have to decide is X negligible can I neglect X and not do the quadratic equation or do I have to do it so if I look my initial concentrations are fairly high right they are 10 to the 0 power so here and my ka is 10 to the minus 4 so it's more than two words of magnitude smaller than my concentrations that I'd be neglecting it from the amount that does ionize and so that's probably a good assumption so I'm gonna assume that this X is negligible like we have before I'm also gonna assume if that's a good assumption then this is even a better assumption I'm gonna assume that this X is also negligible so assume negligible all right so then this is approximately equal to two point O X this two times this x over 1.5 and that's still all equal to the KA which was given to us as 7.2 times 10 to the minus 4 so I'm gonna bring this up to this side divided by that to get X by itself and what I get for X is five point four times ten to the minus four interestingly enough if I did not assume those are negligible and I did go through the quadratic equation I would get the exact same value within the sig figs so but we still have to check to make sure that that's a good assumption we don't want to assume and then not check so and slide this up a bit and let's do that check so our check which is also going to be our percent ionization it's gonna be X the amount were neglected over the initial amount when I'm doing the check I wanna make sure I'm checking it against the smaller of the initial situations which is the 1.5 rikes I wanted to be less than 5% of the smallest thing I'm neglecting it from and luckily for us that is the HF concentration so the check will be the exact same thing as the percent ionization and then the concentration of HF initial times 100% right so that's going to be my check and I get 5 point 4 times 10 to the minus 4 molar over 1.5 molar times 100 percent and I get that my percent ionization is 0.03 6% so 0.03 six percent ionization that's one of the answers they asked for so I'm gonna go ahead and box that and I'm also gonna say that my check is valid because this is less than 5% so I'm good to go that's a lot less than 5% we'll come back and look at that in a minute once we finish the problem but let's go on and do the rest of this and find the pH so to find the e the pH we need to figure out what the hydrogen ion concentration is at equilibrium and that is X so that's kind of a nice H plus at equilibrium equals x which equals this five point four times 10 to the minus 4 molar and this time specifically we're talking about h plus so I'm just gonna throw in my pH equation pH equals minus the log of the H+ concentration at equilibrium which is this so equals minus the log of five point four times 10 to the minus 4 molar H+ and then I can get my PA and that is equal to 3.27 does that make sense well this is more than 10 to the minus 4 less than 10 to the minus 3 so she'd be between 3 & 4 so yeah that makes sense and notice my two sig figs here became my two decimal places there and the 3 is not significant it is the power of 10 okay so that's my other answer ph equals three point two seven all right now just for some comparison because we said that the common ion effect would lower the percent ionization and increase the ph closer to ph seven than it would have been without the common ion effect so just for comparison if I look at the the percent ionization and the pH of one-and-a-half molar hydrochloric acid in pure water so one point five molar HF in pure water the percent ionization is going to be is gonna be higher it would actually be two point two percent ionization which is much more than look how much that was reduced by two orders of magnitude and the pH would also be higher it would be pH equals one point four eight sorry the pH is higher with the common ion effect without it it would have been lower so the pH is one point four five and this is just for comparison but if we solved it putting a zero in here this is what we would get right so it really lowered the percent ionization and it increased the pH to have that common ion in there okay buffered solutions we covered these in the acid-base equilibria lab so we're not going to cover them entirely again but just as a refresher remember a buffered solution is one that contains a weak acid and its weak conjugate base they both have to be weak and you need both of them I've added a little something here it contains those that conjugate pair in sufficient quantities if there's not enough of them it can act as a buffer and in close to equal concentrations so that means ideally you'd have equal amounts of the two but you can be off a little bit one way or the other as long as it's not more than an order of magnitude or so in concentrations because then it won't act as a buffer and you will be very unprotected in one direction versus the other and even the direction that you're more protected in won't be as good as if they were equal so and we found that out in the buffers lab that we just did so and this conjugate pair is gonna shift back and forth according to the châtelet principle so this minimizes the change in pH of the solution according to the châtelet principle because if it's shifting to minimize any stress and H+ is in the equilibrium expression then the H+ concentration isn't gonna change as much as it would have without the buffering pair and so the pH is not gonna change as much as it would have without the buffering pair okay so here we have a couple of stress shift equilibrium arrow diagram set up and this is just generic acid AJ and its conjugate base a minus right you shouldn't lose one H+ to get back and so we have this happy-go-lucky system in equilibrium and then suddenly we knock it out of equilibrium by say putting another acid in there so this buffer verses acid our stress is going to be that we're adding another source of H+ so that would be an upward arrow over the H+ concentration it's gonna shift away to the other side to minimize that this will go down this will go down this will go up and what we'll notice is that there's a net consumption of the conjugate base of the pair of the buffering pair and making more of the kanji acid of the buffering pair and what happens is the hydrogen ion concentration still goes up a little bit but not as much as it would of if this wasn't there to shift back and forth and that was the total increase in my H+ concentration okay now this is shifting back and forth but I want us to realize is that what's happening as we add the H+ is actually an acid-base neutralization so this is the equilibria that's shifting but the chemical reaction that's happening as we add acid to this buffer is that the conjugate base form of the buffering system the two conjugates make up the buffering system is reacting with the H+ that we're adding and acid-base neutralization is go to completion so we're ending up with a CH a more H a so we're consuming the conjugate base we're making more of the conjugate acid okay in this case the same happy-go-lucky system in equilibrium but this time we're gonna set this versus a base so we're just some source of hydroxide so we're adding Oh H - that is our stress now hydroxide isn't in this equilibrium expression but I know that hydroxide and hydrogen ion react to completely neutralize each other so and make water so that's going to be a downward arrow over the H+ concentration as it gets neutralized and that would cause a shift in this direction consuming the conjugate acid of the buffering pair making more of both of these some of it's gonna replace that and more of that conjugate base down down so a little down so the H+ concentration is gonna go down but just a little bit not as much as it would have otherwise and this is gonna go up okay so again just like we had over here but now in the opposite direction the h8 concentration so the conjugate acid the pair's concentration is going down and I'm making more of the conjugate base so I'm shifting the other way from conjugate acid to conjugate base if I was going to show that as a neutralization reaction I could say that the conjugate acid form is reacting with hydroxide in one way ticket to the conjugate base and then also when I neutralize that H+ I'm also getting into water in that case so that would be the neutralization equation associated with this shift in equilibria up here we can see that as he add the acid we neutralize the conjugate base and make more in the conjugate acid so if we started with a perfectly balanced buffer where we had equal amounts of both conjugates I would be consuming this making more of this in one to one ratio and so I would end up shifting the balance in this direction where I end up with more conjugate acid in let's conjugate base adding a base I'm gonna be deprotonating the acid making more the conjugate base right here so I'm gonna be shifting the ratio the other way so you can kind of see why we would want large amounts of the both kanji in pairs in our buffer our ideal situation we have large amounts and equal amounts of the two because the more I have the more I can shift before this ratio gets too far off once this ratio gets to be about more than an order of magnitude difference its ceases to act as a buffer we've exceeded the buffering capacity and then the pH will swing wildly so we want a lot of it in there now why the 50/50 is good and how we're gonna choose a buffer we're gonna come back to so I'm gonna leave the bottom of this page blank because we're gonna come back to it and we're gonna go on to the henderson hasselbalch equation and then we'll filled some more in on this page anderson hasselbalch equation so this equation that we've already come familiar with in lab can only be used for solutions that have reasonable amounts of a weak acid weak base conjugate pair if they don't this this equation falls apart in other words this can only be used for buffered solutions and the reason for that is that it makes a few assumptions and if we don't have reasonable amounts of both kanji it's in the pair it falls apart so but first before we get into why that is let's look at where this equation comes from and what it is and so like many equations such as dalton's law of partial pressures it's a very impressive looking equation that doesn't really bring up something very new so dalton's law of partial pressures was it the total pressure is the sum of the partial pressures and you guys name on an equation for that very similar Henderson and Haase Bach both got their names on equation that comes from something we already know so I'm gonna write down a weak acid weak base equilibrium so like we did before for the buffers and then the liberating expression for that okay so we have h a going to h plus and a minus and if I take the KA the equilibrium expression for this reaction products of reactants I have h plus a minus over the concentration of H a so what I'm gonna do is I'm just gonna take the log of both sides alright so log of ka and I can break this into separate logs right and keep them together an all in one log what I've chosen to do is break out the H+ is multiplied in here and so plus what's left is this ratio here I just left left those together you can kind of see where I'm going with that now I'm just gonna rearrange I can bring this to this side and this to this side and we'll see what we get so I've just brought them to the other sides and now I have minus a lot of the H+ concentration minus the log of the KA the equilibrium constant for the acids ionization and still this term in the end here what do we have here minus the log of something minus the log of something Oh that's just the same as the pH and this is the same as the pKa and then you still have this last log term a minus concentration over H a concentration look at what we have here the henderson hasselbalch equation they just took the log of the equilibrium expression for the ionization of a weak acid they didn't even bother to take minus the log they just took the log and they got their names on equation both of them did but anyway if you ever forget the henderson hasselbalch equation you can just take the log of the equilibrium expression and here it is now something important is that this was the KA for this weak acid so this K a and this conjugate acid have to match this has to be the KA for the conjugate acid in the pair and a conjugate acid in the pair is always in the denominator and the conjugate base concentration is in the numerator and when you're plugging into this and when you're writing this for a particular this is a generic acid base but if you're running for a specific conjugate pair you should always write the conjugate pair in here and these two species should only differ by one h+ you should actually put the conjugate pair in here and not any other salt or any other thing that you may be getting that ion from but the ion itself if it's a anion or that the species itself the conjugate pair okay now you'll notice that in this we are making an assumption we are assuming that there is no ionization involved that the weak acid isn't ionizing to form more of the weak kanji base we're also assuming that the weak base isn't reacting with water and ionizing our forming more of the conjugate sit and that's an assumption that we're making that's only true if we have reasonable amounts of the two and is it a good assumption it is if we consider the common ion effect in our first example when we looked at the hydrofluoric acid in sodium fluoride we found that the percent ionization was Oh point zero three six percent so three one hundredths of one percent was how much denying eyes whereas if that other kanji it hadn't been there it had been working at two point two percent ionization so that common ion effect really does shut down ionization in both directions so it is a good assumption as long as we have reasonable amounts of both kanji it's if not the whole thing falls apart and what we get what we predict for the pH using the henderson hasselbalch equation will be very far off of what we actually see in a real solution so the henderson hasselbalch equation assumes no ionization that means that it's assuming that none of the conjugate acid is gonna form any of the conjugate base and leave us with protons it also seems that none of the kanji at base is gonna react with water and form more of the conjugate acid and we both with hydroxides and again that's only a reasonable assumption if we have reasonable amounts of both species so that the common ion effect is what's keeping this from happening all right say one more time this is only a good assumption if you have reasonable amounts of both conjugates and then the common ion effect is shutting down ionization in both directions like we saw in the previous example okay let's go back and do that previous example one more time except this time instead of using an ice table and equilibrium expression all we're not we'll just skip straight to the henderson hasselbalch equation so let's do that so we're gonna do the same example what if the pH of one-and-a-half molar hydrofluoric acid two molar sodium fluoride and again the KA values given to us I'm going to do that this time so the ice tables using the henderson-hasselbalch equation notice when I've set this up I'm putting in the specifics so up here I just put in generic conjugates here I'm putting the actual conjugates in there and then this has to be the KA for the conjugate acid of the pair so those two have to match okay so now I can say the pH equals PKA so P remembers just minus log of minus log of the KA which they gave me seven point two times ten to the minus four and plus the log of this ratio here okay well what is the concentration of h f minus of fluoride well we have two molar sodium fluoride if it's in solution is completely dissociated so that means we have two molar sodium ions and they have two molar fluoride ions so I'm gonna put in here 2.0 molar fluoride and we're assuming remember no ionization from the HF so all of the fluoride is coming from the salt and then we're also assuming that none of this has I an eyes so we still have one and a half molar HF so one point five molar H half on the bottom and then I can solve this in my calculator and I get loan to hold the pH equals three point two seven just like I did with the ice table except I think you'll agree much much less work but again these two concentrations were reasonably close to each other within an order of magnitude certainly so this was an appropriate use of the henderson hasselbalch equation now the other question asks could we figure out the percent ionization could we from this well actually we could because remember the percent ionization was H+ concentration over the concentration initial concentration of the HF could I get the h+ concentration from the ph sure 10 to the minus ph is the h plus concentration so I could get that concentration in molarity of h+ divide that by the 1.5 molar hf the initial its initial concentration and i would get x i understand i get the percent ionization so i could still get the percent ionization using the henderson hasselbalch equation 2 I would just go backwards from the pH to find what I solve for as X in the I stable okay so remember this slide will bring this one back so if we have the ideal buffer equal amounts of the two let's look what happens now that we know the henderson hasselbalch equation so if I have equal amounts of those kanji pairs then this ratio the ratio of a minus concentration to H a concentration in the henderson hasselbalch equation equals 1 right well the log of 1 is equal to 0 so this whole term is going to drop out and I have that the pH equals the pKa plus 0 the log of 1 is zero and so what I end up with is that the pH of the solution is just equal to the pKa of the conjugate acid of the pair so minus the log of the KA for the conjugate acid that's true in both directions if these are equal the pH equals a PKA if the pKa equals the pH then those two must be equal it works in both directions all right now if I take the same thing and I look at these other cases where I've shifted the balance one way or the other in this case the concentration of a minus is much greater than the concentration of H a and so this ratio is greater than 1 right we have more in the numerator less in the denominators so the fraction is greater than 1 if I take the log of something greater than one it's a positive value so whatever the pH is it's gonna be the pKa equals the pKa plus some positive term so that means the pH is gonna be greater than the pKa because I'm taking a PKA and adding something positive to it so the pH ends up being greater than the pKa which makes sense if these are equal but now I have more base and less acid than the solutions pH is going to be higher or more basic than it was before all right same thing over here the a- concentration ratio to the h a concentration is now less than one so when I'm plugging in here is less than one the log of something between zero and one is going to be negative so here I have the pH equals the pKa plus a negative term so I'm subtracting from the pKa and that's what the pH is gonna be so now the pH is gonna be less than whatever the pKa is right which again makes more sense if they're equal when these are equal here I have more acid less base it's gonna be more acidic so the pH is going to be lower than the pKa was when they were equal right so we can kind of see how we can play with these ratios a little bit the ideal buffer has a PK at the ideal buffering amounts are when these two are equal right so ideally a pH should be of the buffer that you want to maintain should be as close to the pKa of the conjugate acid of the pair as you can as you can possibly find if that's not exactly what you want you want a slightly different pH that's okay you can give it more conjugate base and less acid and that'll raise the pH a little bit you can give it a little bit more conjugate acid a little less base and that'll lower the pH a little bit but you don't want to go crazy you don't want go too far in one way or the other or it will cease to function as a buffer and so a conjugate pair is gonna be chosen based on the pH you want to maintain for your buffer so when you're deciding what conjugate pair to use to make a buffer look at the pH and look at the pka's of the conjugate acids of the buffering pairs whichever one has a PKA closest to the pH you want to maintain that is the one you want to pick because that will ensure that your ratio is as close to 50/50 as possible so write that up when choosing a conjugate pair to make a buffer choose a conjugate acid whose PKA is as close as possible to the pH you want to maintain remember this is the pKa and the conjugate acid of the buffering pair as close as possible to the pH you want to maintain ideally within plus or minus one pH unit if you're the reason I say plus or minus one pH unit is because if this ratio gets off by more than an order of magnitude you don't act as an efficient buffer and an order of magnitude and concentration a power of 10 in concentration is 1 pH here because it's log base 10 and so that's a good rule of thumb if you have to be off by more than a pH unit from the pKa of the conjugate acid you should probably choose a different conjugate pair and once you know the conjugate acid of the pair how do you figure out the conjugate base well just take away an H+ and that's the conjugate base right so if my conjugate acid is acetic acid then my conjugate base is gonna be acetate if my conjugate acid is formic acid then my conjugate base is gonna be formate what if my conjugate acid is ammonium ions mmm well then my conjugate base is gonna be ammonia nh3 right so just take away one H+ once we know the conjugate acid we can just take away H+ to get the conjugate base and then also related to this a buffer has the highest capacity buffering capacity in both directions when the pH equals a PKA and the reason for that is that these two are exactly equal so I'm equally protected against a base as I am against an acid in this case I am more protected against a base because I've already converted some in this case I'm more protected against an acid because I've already converted some right when they're equal I am equally protected in both directions all right next topic solubility product and solubility so let's first define these solubility is the maximum amount that can dissolve under the given conditions now this could be for any species it could be an ionic compound dissolving in water it can be for a gas dissolving in water it could be for anything but the maximum amount that can dissolve under the given conditions the molar solubility is specifically because solubility can have lots of different units the molar solubility is the solubility given in units of moles per liter or molarity so that's the one we're gonna be using there are the units like you know grams per hundred grams of solvent and this and that but we're gonna use molar solubility so we're talking about solubilities so if I say that the solubility is 0.1 molar that means that that's the maximum hilarity I can have at that solution under those conditions that temperature the solubility product constant or what's most oftenly referred to as just a solubility product is KSP and it's an equilibrium constant just like any other K and the SP tells you the reaction for which that is the constant for so for example ka is the ionization of a weak acid KSP is for the dissolution of a solid ionic compound so when you see SP think dissolution naught is the reaction we're gonna be writing if we're doing an ice table that is the reaction that we're gonna be writing so forgive me we're giving the KSP of CA calcium phosphate what would the reaction that you write down be well first of all you have to remember what Cal's phosphate is right so you have to look at see the calcium's in group two and it's got a two plus charge you have to remember that phosphate so polyatomic ion with a three- charge and then balance the charges to make the ionic formula unit so ca3 po4 in parenthesis with a two and that's a solid so that's the solid formula unit and then this is the dissolution reaction so this is the solid going to aqueous ions and so we're gonna put a reversible arrow because it can dissolve and then it can read crystallize and we're gonna have three calcium ions there's gonna be a twist now and we're gonna have two phosphates that into a to do four three - also aqueous notice how these subscripts become coefficients except in this case because this is a polyatomic ion so i stay together as a unit and we have two of those units alright so they can remember a little bit about about ionic compounds okay well i could put this into an ice table ice and does it matter how much calcium phosphate i have how much solid i have well as long as I have enough where I'm getting to the solubility limit no it doesn't and so because the solid isn't gonna appear in the equilibrium expression so what I can do is just ignore those I'm gonna actually leave this line here though so if they in the initial conditions none of this is dissolved yet so we don't have any of this you don't have any of this and it's gonna go forward and some of it's gonna dissolve so what I'm actually gonna put in here even though it's not going to be in the equilibrium expression is I'm just gonna put in minus X and what I'm doing that essentially what I'm saying is let X equal the amount that dissolves okay and so in other words X if these are all going to be in molarity X is the molar solubility all right now for every one formula unit that goes into solution I'm gonna get three of these ions and two of those ions so this is not good this is not gonna be +3 X this is gonna be +2 X and then I don't care about that that column other than to define X and then so this is gonna be 3 X + 2 X okay now I'm going to set up the equilibrium expression just like we do for all of our ice tables except this time it's a KSP and that KSP again equals products over reactants so I have my concentration of calcium ions at this time that's cubed because I have a coefficient of 3 times my concentration of phosphate ions squared so products over reactants but my reactant is a solid so it doesn't appear in the equilibrium expression so this is a helpful way to remember what solubility product is because it's the product the multiplication of the concentrations of the ions to whatever coefficients I have right and the because the in the dissolution reaction the reactant is always a solid it's never gonna be in there okay so now I'm going to plug in the equilibrium line concentrations for these two species and so I have that equals 3x because that's the concentration of calcium ions the square bracket part but it's still cubed and 2x squared okay so now I have 27 X cubed times four x squared right cubing and squaring remember we can't just square the ax we also get a square the coefficient that's also inside the parenthesis right so 27 + 4 and so I get that KSP equals in this case 108 X to the fifth this is the relationship between the equilibrium constant KSP and the molar solubility X now now that I know this relationship if I'm given one I can solve for the other and if I'm giving the other one I can solve the other one so I have that relationship so for all these problems this parts pretty much gonna be the same I'm gonna write the dissolution reaction right which is the solid a the solace ionic solid iron compound going to aqueous ions and then instead of my ice table setup my equilibrium expression and get to this point where I know the relationship between KSP and the molar solubility then again whatever which one they give me I plug it in I solve for the other so in this case if they give me the solubility limit then I can find the KSP so I'm just taking that relationship right there KSP equals 108 times X which is the molar solubility 2.0 for nine times 10 to the minus 7 molar and that's whole thing is gonna be to the fifth power so this KSP is gonna be a really small number and it ends up being three point nine zero one times ten to the negative thirty-two power and this has become unitless even though this had units remember that K values are unitless I'm gonna include that in my answer to give the number of reference point I also notice that here are my four sig figs from the value they gave me this 108 here is actually has infinite sig figs because it came from whole numbers right these were whole numbers from the coefficients so there was no uncertainty in those all right so now that KSP value is very very small so that means that in this equilibrium I have very very few products and by the vast majority reactants right because products will reactant so I get very very few products so this is not a very soluble ionic compound and you can see that already looking by the side the solubility right here right ten to the minus seventh molar is the most concentrated solution error in the half of this ionic compound at this temperature but that also makes sense because look at these high charges two pluses three minuses right you have a very high lattice energy within this ionic solid so it's gonna be very hard to pull these ions apart and get them to go into solution now we have to be careful at just looking at ksps in this case though because remember the solid isn't in the equilibrium expression so this is just the product of the concentrations of the products to some power but this relationship will be different depending on the coefficients in the balanced equation so sometimes we can look at the KSP and say no that's not very soluble but when I'm comparing two I have to be careful so what we're gonna have to do is realize that we can always compare molar solubility x' i can compare the molar solubility of calcium phosphate to say calcium sulfate and whichever one has a higher number well that one's more soluble I can't necessarily compare ksps however because the relationships between the KSP and the molar solubility can be different which means I'm not comparing apples to apples and so that's where we're going next all right so the solubilities of two compounds can always be compared directly like we just mentioned the KSP values however only be compared if the compounds dissociated to the same numbers of ions in other words these coefficients on the product side are the same the order doesn't matter if it's 3 & 2 or 2 & 3 that's fine but as long as it's the same numbers you can compare them because what that means is if those numbers are the same then you're gonna end up with the same relationship between KSP and the molar solubility but if those numbers are different you're comparing apples and oranges so if that's the case if they're not the same then you have to go back in this kind of process and in the end you're gonna be plugging in ksps and solving for molar solubility x' you gotta calculate 2 sub poly the solubilities and then compare those so we'll do a few examples of that example 1 which is more soluble silver sulfate with a KSP of 1.2 times 10 minus v or barium fluoride with a KSP of 1.8 times 10 to the minus 7 well the question we need to ask first is can we compare the KSP values directly so they gave us those if we can that saves a lot of work if we can't well then we're gonna have to do two ice tables but let's see let's look at silver sulfate first AG 2 so4 again this would be the KSP so it's the dissolution reaction the solid going to aqueous ions so I can set up that equation and I get that this is gonna split up into two silver cations and one sulfate anion these are both aqueous so they are in the equilibrium expression and my coefficients are two and one if I look at the barium fluoride solid that's gonna split up into a barium ion so barium two-plus aqueous and two fluorides nucleus again my coefficients are a 1 and a 2 a 2 and 1 again the order doesn't matter just will they split up into the same numbers of ions two of the to one and one of the other two of them on one of the other and so the answer to my question is yes I can compare them directly can compare KSP values directly I can yay so that's easy here I have a larger KSP which means I'm gonna get more products these are the products here I have a smaller KSP I'm gonna get less products these are the products so silver sulfate is more soluble if I had gone through the whole ice tables I would have realized that the relationship that would have gotten would have been the same and it would have been that KSP equals 4x cubed for both of them you can go through us a little bit of practice and see if you get back setting up either one of those with an ice table but again because the relationships the same if this value is bigger this is gonna be bigger this value smaller this is gonna be smaller and I can compare them ksps all right now I'm gonna do another example where unfortunately that is not the case in this one we have the same question but two different ionic compounds so calcium fluoride and it gives you the KSP and lead to sulfate and it gives you the KSP it's gonna be tempting to compare these but I have to check to see if they split into the same number of ions and they're going to so let's check this out let's do the calcium fluoride first how many I'm side by side here so the dissolution reaction for solid calcium fluoride would be to get a calcium ion aqueous and two fluoride ions aqueous so what I'm splitting into is one and two if I do the lead to sulfate solid I get a led to ion and a sulfate ion both aqueous but it's one and one where it's here I have one in two so they don't split into the same numbers of ions I cannot compare cannot compare ksps made that one word okay so now I'm gonna have to go through the whole operation of setting up my ice tables at this point were pros at that though so it's not too bad alright again solids I don't care about except to define X so this is going to be a minus X and we could do this as a Y so we're not comparing the same thing so - why it's total second ax that isn't it alright so here 0 + 0 0 + 0 + 1 X so plus X + 2 X so I have X + 2 X here I'm gonna have + y + y so y + y that's why not an ax okay so when I set up my KSP and this is for calcium fluoride my products over reactants but I left to solid so I have the concentration of calcium ions times the concentration of fluoride ions squared so I have X and I have 2x squared and I get my relationship is that KSP equals to 4x cubed on that one on this one I have KSP equals the end products of reactants so the concentration have led to ions concentration is Sol ion's both to the first power that is why and why so I have KSP equals y squared so here I have an X to the Q X cubed and you have a y squared so not the same relationship now if I plug in the KSP values they gave me here and solve for X I'm gonna get something different so this equals because it's the KSP 3.5 times 10 to the minus 11 and this equals 1.8 times 10 to the minus 8 here I get that x equals two point one times 10 to the minus 4 molar and that's the molarity not of either one of these ions that because well actually is the molarity of this one but not this one that's the molarity of the dissolved aqueous calcium fluoride supposed to be calcium fluoride here now when I solve for this one I get that y equals one point three times 10 to the minus 4 molar lead to sulfate so which one is more soluble look at that it's not too different but this one's a little bit bigger number so it turns out that it is the calcium fluoride that is actually more soluble if I had compared the the KSP values I would have never thought that because look that's a smaller KSP value than that one is but because I wasn't comparing apples to apples doing that I ended up leading being led astray by the KSP values it's actually this one looked a smaller KSP value that is more soluble and I can again see that by actually going back and calculating the molar abilities alright we have one last topic in this chapter and that's being able to calculate the pH at any point on a titration curve remember to go back to your notes that you took an experiment five acid-base equilibria on days 1 and 2 where we covered titration curves we covered indicators polyprotic acids all of those are part of your lecture notes it's also part of these chapter notes but we covered it in lab so go back and look at those and then we're gonna do one more section here on calculating the pH at any point on a titration curve but that's a whole lecture in itself so we'll save that for another day