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Current time:0:00Total duration:13:19

let's say we have a solution of acetic acid and we know what's going to happen in solution acetic acid is going to donate a proton to h2o to form h3o plus or hydronium and the conjugate base to acetic acid which is acetate ch3coo minus and so there's a concentration of acetate anions in solution what would happen now if you added some sodium acetate right so if you added some sodium acetate let me go ahead and write this out here so ch3 Co minus and na plus so if you add some sodium acetate to your solution right you now have some more acetate anions so you're increasing the concentration of one of your products right you're increasing the concentration of your acetate anion and according to the shot leaves principle if you increase the concentration of one of the products the equilibrium shifts to the left so the equilibrium is going to shift to the left and that means that some of this acetate anion is going to react with some of your hydronium ion right when your equilibrium shifts to the left this decreases the concentration of hydronium ion and if you decrease the concentration of hydronium ion you're going to increase the pH of your resulting solution so the acetate anion is the common ion and this is the common ion effect all right so there are two sources for your acetate anion so one one is the ionization of acetic acid that's one source for your acetate anion and the other source is the sodium acetate that you added in so we have two sources and so we expect a pH that's higher than just than just a solution of acetic acid alone let's go ahead and do the calculation and see that that is true so calculate the pH of a solution as one molar and acetic acid and there's the KA for acetic acid and one molar for sodium acetate and so we're going to our just going to start by rewriting our acid-base reactions so we have acetic acid plus water and so we're gonna have everything at equilibrium with our products h3o plus and acetate ch3coo - and right now let's just for a second for 10 like we have only acetic acid so these are the kinds of problems that we've been doing we started with our initial concentration and we put a 1 point here we have a one point zero zero molar concentration of acetic acid and then for our change we said whatever we whatever concentration we lose for acetic acid we would gain for the acetate anion and therefore we would gain for the hydronium ion so we're going to gain a concentration of the acetate anion here but we have to add in something right because we're also dealing with this one more one molar concentration of sodium acetate so we have we have a source another source for acetate anions right so really we have 1 molar let's go ahead and put in that concentration so the acetate the acetate ions come from two sources once again one is the ionization of acetic acid and one is the sodium acetate that you are that you add in to make your solution that's how I like to think about it so your initial concentrations would be 1 molar for acetic acid 1 molar for the acetate anion and then pretty close to 0 here for hydronium so at equilibrium and equilibrium let's go ahead and write what we would have so the equilibrium would have 1 minus X for our concentration of acetic acid X for our concentration of hydronium and 1 plus X for the concentration of acetate once again 2 sources 2 sources for our concentration of acetate we can write a KA expression so an equilibrium expression using ka because we're dealing with acetic acid donating a proton to water here so K a is equal to concentration of products over reactants so h3o plus times concentration of acetate so times concentration of acetate all right all over the concentration of our reactants leaving water out so the concentration of ch3 ch3 C oh oh H alright so for hydronium at equilibrium for hydronium equilibrium our concentration is X so we put an X into here for acetate for acetate our concentration of acetate at equilibrium would be 1 plus X so let's put that in so we have one point zero zero plus X and then this would all be over the concentration of acetic acid which would be 1 minus X so over here we put 1 minus X alright let's get some more room down here and so we can talk about this we're going to make that same assumption that we've done before that this concentration X is much much smaller than 1 and if that's a really small number 1 plus X is pretty much the same as 1 and 1 minus X is also pretty much the same as 1 so if this is a very small concentration we don't have to worry about these we could we can approximate them and say 1 plus X is equal to 1 let me go and rewrite this here so this would be x times over here 1 plus X is equal to approximately equal to 1 so that's our approximation over here 1 minus X if X is very small is also approximately equal to 1 this is all equal to the KA value which for acetic acid is 1.8 times 10 to the negative 5 and so the the ones obviously would cancel each other out here and so the concentration of hydronium which is X right remember X represents the concentration of hydronium ions is 1.8 times 10 to the negative 5 so to find the pH all we have to do is take the negative log of that so negative log of 1.8 times 10 to the negative 5 and so we can go ahead and get the calculator and do that negative log of 1.8 times 10 to the negative 5 is equal to four point seven four so our PA H is equal to four point seven four now if you go back to the video on weak acid equilibrium we calculated the pH for a one point zero zero molar solution of acetic acid only and the pH for that came out to be two point three eight so because this is because we didn't have we didn't have any any other acetate anions present here so we have different PHS right the pH is higher the pH is higher for for this situation where we have a seat ik acid where we have two sources again for the acetate ion the ionization of acetic acid and also the addition of sodium acetate let's do another common ion problem here so this time we need to calculate the pH of a solution that is 0.15 molar for ammonia and 0.35 molar for ammonium nitrate so ammonium nitrate is NH 4 plus and no.3 minus in solution and so we start with ammonia so let's go ahead and write what's going to happen when ammonia reacts with water so ammonia is a weak base and it's going to take a proton from water so if nh3 picks up an H+ we form NH 4 plus or ammonium and if we take a proton away from h2o take an H+ away from h2o we form Oh H minus or the hydroxide ion so once again we start with our initial concentration and we're going to pretend like it's one of the problems that we've been doing in earlier videos so if we have 0.15 molar concentration of ammonia we go ahead and point put 0.15 here we think about the change whatever concentration we lose for ammonia is the same concentration that we gained for ammonium since ammonia turns into ammonium and then therefore that's also the same concentration we gain for hydroxide so one source for the ammonium ion it would be the protonation of ammonia right so that's one source but we have an additional source because we also have 0.35 molar ammonium nitrate right there's another source for ammonium ions and so we have 0.35 molar so we go ahead and put point three five molar in here for the initial concentration of ammonium and we're saying that we have zero for our initial concentration of hydroxide so when we come when the reaction comes to equilibrium here for ammonia we would have 0.15 minus X for ammonium we have two sources right so this is a common ion here so we have 0.35 plus X and then for hydroxide we would have just X so since ammonia is acting as a weak base here let's go ahead and write our equilibrium expression and we will write K be alright and KB for ammonia is 1.8 times 10 to the negative 5 so this is equal to concentration of our products over reactants so we have concentration of NH 4 plus times the concentration of O h- all over the concentration of our reactants leaving out water so we have just ammonia here so concentration of nh3 so let's go ahead and plug in what we have for the concentration of ammonium we have 0.35 plus X so we put point three five plus X for the concentration of hydroxide we have X so we go ahead and put an X in here and then that's all over the concentration of ammonia at equilibrium and we go over here and for for ammonia equilibrium it's 0.15 minus X so we write over here 0.15 minus X and we can plug in the KB value 1.8 times 10 to the negative 5 so let's go ahead and plug in the KB so we have 1.8 times 10 to the negative 5 is equal to we have okay once again we're going to make the assumption so if we say that X is extremely small number right then we don't have to worry about it when we're adding it to 0.35 and so we just say this is equal to 0.35 so 0.35 plus X is pretty close to 0.35 so this is times X and let me just make sure you understand this X is this X and then once again 0.15 minus X if X is a very small number that's approximately equal to 0.15 and so now we would have this and we need to solve for X so let's go ahead and do that so let's get out the calculator here we need to solve for x so 1.8 times 10 to the negative 5 times 0.15 and then we need to divide that by point 3 5 and that gives us what X is equal to and so X is equal to seven point seven times 10 to the negative six so X is equal to seven point seven times ten to the negative six x represents X represents the concentration of hydroxide ions in solution so this would be the concentration of hydroxide so seven point seven times ten to the negative six molar is our concentration of hydroxide our problem wanted us to calculate the pH so if we know the concentration of hydroxide ions we can find the Poh right by taking the negative log of the concentration of hydroxide so negative log of the concentration of hydroxide ions will give us the Poh so let's do that so we have the negative log of seven point seven times ten to the negative six and this gives us the Poh where we round to five point one one so the Poh is five point one one let's get a little bit more room here so Poh is equal to five point one one and then we're home free because pH plus Poh is equal to 14 so we just plug in our Poh solve for the pH pH is equal to 14 minus five point one 1 which of course is equal to eight point eight nine and so we've calculated the final pH of our solution