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first let's calculate the molar solubility of silver chloride in pure water and then we're going to compare the solubility that we get in Part A to the solubility that we're going to get in Part B and in Part B it's no longer pure water we're going to have some ammonia present so we take some solid silver chloride so that's this over here and we put it in some pure water and some of the silver chloride is going to dissolve but most of it's going to stay in the solid form so silver chloride is only a slightly soluble compound so there's my little pile of undissolved silver chloride but some of it does dissolve and so we get a saturated solution of silver chloride so there are silver plus 1 cations in solution and there are also chloride anions in solution and we have our solubility equilibrium to find the molar solubility of silver chloride let's go ahead and write down our ice table so our initial concentrations are change and finally our equilibrium concentrations before we put the silver chloride in we didn't have any ions in solution so the initial concentrations of our product are zero right we have zero for the concentration of our ions some of the silver chloride that we put in dissolved so we lost a certain concentration of our silver chloride and we're going to call that X so we're losing a certain concentration of our silver chloride if you look at our mole ratios for every one mole of silver chloride that dissolves we get one mole of silver plus one cations in solution so for losing x over here we're gaining X we're gaining X for the concentration of silver cations in solution and since our mole ratios are still one-to-one if we dissolve one mole of silver chloride we gain one mole of chloride anions over here we get a plus X so at equilibrium our concentration of silver cations is X and our concentration of chloride anions is X so we write our equilibrium expression and since this is a solubility equilibrium we're going to write KSP our solubility product constant is equal to concentration of our products so that's concentration of AG plus raised to the power of the coefficient so raised to the first power times the concentration of chloride anions CL - raised to the power of the coefficient to the first power so we can plug in our solubility product constant at 25 degrees its 1.8 times 10 to the negative 10 so we put in here 1.8 times 10 to the negative 10 and this is equal to this would be X times X so this is equal to x squared so we can get out the calculator and solve for X so we have the square root of 1.8 times 10 to the negative 10 so we take the square root of that and X is equal to 1.3 times 10 to the negative 5 if we round that so X X is equal to 1.3 times 10 to the negative 5 molar so that's the concentration of let's say our silver plus 1 cations in solution that's also the molar solubility of silver chloride right that's how much we can dissolve so so not very it's a very small molar solubility so silver chloride is only a slightly soluble compound let's see what happens when we have ammonia present instead of pure water so we have a 3 molar all right we have some we have some ammonia present now what happens when ammonia is around silver cations all right we get this reaction so silver cations plus 2 molecules of ammonia give us what's called a complex ion over here and let's uh let's show what's happening so if we have a silver plus 1 cation all right we have 2 molecules of ammonia so let me go ahead and draw in my 2 molecules of ammonia and this is certainly not drawn to scale so I'll draw one in on that side I'll draw another one in on this side remember your definitions for Lewis acid Lewis base right Lewis base is an electron pair donor and that's what's going to happen here the nitrogen has a lone pair of electrons and each ammonia molecule is going to donate a pair of electrons to our silver plus 1 cation right so ammonia is donating as an electron pair donor so this must be our lewis base and the silver plus 1 cation is accepting a pair of electrons so this must be our lewis acid so the formation of a complex ion is a lewis acid-base reaction and notice notice the equilibrium constant for this one point six times ten to the seventh KF is called the formation constant this is a very high value for the equilibrium constant right so the equilibrium lies to the right and this is a stable complex ion so any any silver ions that are in solution if you have ammonia present ammonia is going to pick up that silver ion and essentially remove the silver ions from solution so if we go back up here and think about our equilibrium right if we have ammonia present we're going to be removing silver cations from solution all right so we're decreasing the concentration of silver cations we're decreasing the concentration of one of our products and the shot liaison suppose says if we're decreasing the concentration of one of our products our equilibrium is going to shift to the right to make more therefore increasing the solubility of silver chloride more is going to dissolve because the ammonia is present because of the formation of our complex ion so if we added some ammonia alright if we if we added a solution of ammonia here we could get we could get our silver chloride to dissolve right because it's more soluble because of the formation of a complex ion so let's say we add enough ammonia solution where we get our silver chloride to dissolve here so that's the idea adding adding our ammonia right cause formation of a complex ion which increased the solubility of our silver chloride so we would expect an increased solubility here of our slightly soluble ionic compound silver chloride and next we're going to we're going to do the calculation we're going to I'm the solubility of silver chloride in in r3 molar ammonia solution now we've looked at two reactions we've looked at the solubility equilibrium for our slightly soluble compound and we've also looked at the formation of our complex ion now let's add those two reactions together to get our net reaction for what's happening right so we're going to add our two reactions together to get a net reaction for what happens when you put silver chloride in ammonia all right we have silver cations on our reactant side and on our products side so we can take those out and so for our reactants for our net reaction we have solid silver chloride plus ammonia and for our product side we would have our complex ion and our chloride anion so we have our complex ion and we also have chloride anions so CL minus so that's the net reaction and if we added those two reactions together to get our net reaction we have a new equilibrium constant which I will call K and if you add two reactions together to get your new equilibrium constant you multiply the equilibrium constants of those two reactions so k is equal to k SP x KF so if you did that calculation 1.8 times 10 to the negative 10 times one point six times 10 to the seventh you'll get your new equilibrium constant and I won't do it here just to save some time but if you do that you'll get 2.9 times 10 to the negative third so that's a 92.9 all right our goal is to find the molar solubility of silver chloride in our ammonia so we set up our ice table so our initial concentrations are change and then our equilibrium concentrations and let's pretend like we're starting out with our solution of ammonia and we haven't added any silver chloride yet so let's go back up here and let's see what our concentration was well we started with a 3 molar solution of a mo so let me put that down here so our initial concentration for ammonia is three molar and if we haven't added any of our silver chloride then we don't have any of our complex ion to the concentration the initial concentration of our complex ion is zero and we also wouldn't have any of our chloride anion either so that concentration would be zero too so now we add our silver chloride to our solution of ammonia and some of our silver chloride is going to dissolve and so we make that concentration X so we're losing a certain concentration of silver chloride so we write minus X next we look at our mole ratios so the mole ratio of solid silver chloride to ammonia is one to two so for every one mole of silver chloride that dissolves right we're also going to lose two moles of ammonia so we write minus two X here for our products if one mole of silver chloride dissolves we make one mole of our complex ion all right so if we're losing x over here we'd be gaining X over here for the concentration of our complex ion and same thing for the chloride anion because our coefficient is a one so we're going to gain X for the concentration of chloride anion so at equilibrium our concentration of ammonia is three minus two X our concentration of complex ion is X and our concentration of chloride anion is also X next we write our equilibrium expression so k is equal to concentration of products over reactants so I look at my complex ion over here so concentration of our complex ion and we raise the concentration to the power of the coefficient so to the first power this is times the concentration of our chloride anion so times the concentration of chloride anion raised to the power of the coefficient so raised to the first power we also have some reactants here so we leave we leave silver chloride out because that's a solid right but we have ammonia so this is all over the concentration of our reactants so this is over the concentration of ammonia and notice our coefficient is a so we need to have a to hear next we plug in our equilibrium concentrations so our equilibrium concentration of ammonia is 3 minus 2x our equilibrium concentration of complex ion is X and our equilibrium concentration of chloride anion is X so on the right side we would have X to the first power times X to the let me take that out times X to the first power all right all over 3 minus 2 X to the second power squared and then we can plug in we can plug in our equilibrium constant K which is 2.9 times 10 to the negative third so 2.9 times 10 to the negative 3 is equal to on the right side we would have x squared so let me write this out we would have x squared over 3 minus 2x squared is equal to 2.9 times 10 to the negative third and so what we could do here these are both x squared all right these are both x squared so we could just say that this is all squared on the right sides and then we can take the square root of both sides right so take the square root of 2.9 times 10 to the negative third and take the square root of what's on the right so the square root of what's on the right would be x over 3 minus 2x and let's take out the calculator to do the other one so square root of 2.9 times 10 to the negative third is equal to let's round that to point zero 5 4 so it just makes it easier here to round that so point zero five for our goal is to solve for X our goal is to solve for the molar solubility so we can multiply both sides by 3 minus 2x so when we do that all right we would have this so we need to multiply point zero five four by three and if you round that you're going to get about point one six and point zero five four times two is approximately point 1 1 so we have 0.16 minus 0.1 One X is equal to X so we have point one six is equal to we put both X's on one side this is one X and this is point 1 1 X so 1 X plus point 1 1 X is 1 point 1 1 X and so now we need to solve for X all right so we've done our math here so we have point 1 6 when you divide that by one point 1 1 and that is equal to point 1 4 so X is equal to point 1 4 which is only approximate because of all the rounding that we've done so this this is concentration right so this is our molar solubility if we go back up here X is equal to 0.14 molar all right so that's our concentration of complex ion in solution at equilibrium that's our molar solubility of silver chloride all right so it's 0.14 molar let's go way back up to the to the beginning and compare the molar solubility x' alright so in in pure water the molar solubility of silver chloride is 1.3 times 10 to the negative 5 molar which isn't very soluble at all but now we have ammonia present we've increased the solubility and we just calculated the molar solubility is 0.14 molar and obviously 0.14 molar is a lot larger than 1.3 times 10 to the negative 5 so the formation of a complex ion increase the solubility of our slightly soluble compound silver chloride