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our goal is to find the ph of different salt solutions and we'll start with a solution of sodium acetate so in solution we're going to have sodium ions na+ and acetate anions ch3coo - and the sodium cations aren't going to react with water but the acetate anions will so the acetate anion is the conjugate base to acetic acid so the acetate anion is going to react with water and it's going to function as a base it's going to take a proton from water so if you add an H plus 2 ch3coo - you get ch3cooh and if you take a proton away from water if you take an h+ away from h2o you get Oh H - with the hydroxide ion alright so let's go ahead and write our initial concentrations here so our goal is to calculate the pH of our solution we're starting with 0.25 molar concentration of sodium acetate and so that's the same concentration of our acetate anion here so we're going to write 0.25 molar for the initial concentration of the acetate anion if we pretend like nothing has reacted we should have a zero concentration for both of our products right so a zero concentration for our two products next we think about the change so CH 3 Co - the acetate anion when it reacts it's going to turn into ch3cooh or acetic acid so whatever concentration we lose for the acetate anion we gain for acetic acid so if we make the concentration of the acetate anion X that reacts alright so if X reacts if X concentration reacts we're going to lose X and we're going to gain X over here alright it's the same thing for hydroxide we would be gaining X a concentration for the hydroxide so at equilibrium the concentration of acetate would be 0.25 minus X so we're assuming everything comes to equilibrium here the concentration of acetic acid would be X the concentration of hydroxide would also X alright next we write our equilibrium expression and since this is acetate functioning as a base right we would write KB here so we write K B is equal to concentration of our products over concentration of our reactants so we have the concentration of ch3cooh times the concentration of hydroxide Zoar times the concentration of LH - this is all over the concentration of our reactants and once again we ignore water so we have only the concentration of acetate to worry about here so we put in the concentration of acetate all right so let's think about the concentration of acetic acid at equilibrium alright so at equilibrium the concentration is X so I go over here and put X and then for hydroxide it's the same thing right the concentration of hydroxide equilibrium is also X so I put X in over here and then for the concentration of acetate right an equilibrium concentration of acetate is 0.25 minus X so 0.25 minus X next when you think about the KB value for this reaction and you will probably not be able to find this in any table but you can find the KA for acetic acid so acetic acid right acetic acid and acetate this is a conjugate acid-base pair and the KA value for acetic acid is easily found in most textbooks and the KA value is equal to 1.8 times 10 to the negative 5 and our goal is to find the KB right what is the KB for for the conjugate base and we know that for a conjugate acid-base pair ka times KB is equal to kW right the ionization constant for water so we can go ahead and plug in 1.8 times 10 to the negative 5 times KB is equal to we know this value is 1.0 times 10 to the negative 14 so we just need to solve for KB so we can get out the calculator here and take one point zero times 10 to the negative 14 and then we divide by 10.8 times 10 to the negative five so we get five point six times 10 to the negative 10 so let's get some more space down here and let's write that so KB is equal to five point six times 10 to the negative 10 and this is equal to x squared right is equal to x squared over 0.25 minus X next to make the math easier we're going to assume that the concentration X is much much smaller than 0.25 and if that's the case right this is an extremely small number we can just pretend like it's it's pretty close to zero and so 0.25 minus X is pretty much the same thing as 0.25 so let's make that assumption once again to make our life easier so this is five point six times 10 to the negative 10 is equal to x squared over 0.25 so now we need to solve for X all right so we have we have five point six times 10 to the negative 10 and we're going to multiply that we're going to multiply that by 0.25 and so we get one point four times 10 is any negative 10 so we now need to take the square root of that number right and we get X is equal to this gives us X is equal to 1.2 times 10 to the negative 5 so let's go ahead and write that here so X X is equal to 1.2 times 10 to the negative 5 all right what did X represent we have all these all these calculations written here we might have forgotten what X represents X represents the concentration of hydroxide ions right so X is equal to the constant of hydroxide ions so let's go ahead and write that down X is equal to these so this is molarity this is the concentration of hydroxide ions and if we know that we can eventually get to the pH right that was our original question calculate the pH of our solution so we could find the Poh from here I know the Poh is equal to the negative log of the hydroxide ion concentration so I can take the negative log of what we just got the negative log of 1.2 times 10 to the negative 5 and that will give me the Poh all right so let's go ahead and do that so the negative log of 1.2 times 10 to the negative 5 is going to give me a Poh of four point nine two all right so I'll go ahead and write Poh is equal to four point nine two and finally to find the pH I need to use one more thing because the pH plus the Poh is equal to 14 all right so I can plug in the Poh into here and then subtract that from 14 so the pH is equal to 14 minus 4.9 2 and that comes out to nine point zero eight so the pH is equal to nine point zero eight so we're dealing with a basic solution for our salt let's do another one so our goal is to calculate the pH of a point zero five zero molar solution of ammonium chloride so for ammonium chloride we have NH four plus and Cl minus the chloride anions aren't going to react appreciably with water but the ammonium ions will so let's write our our reaction here so NH four plus is going to function as an acid it's going to donate a proton to h2o so if h2o accepts a proton that turns into hydronium ion so h3o plus and if NH 4 plus loses a proton we're left with nh3 all right so let's start with our initial concentrations well we're trying to find the pH of our solution and we're starting with point zero five zero molar solution of ammonium fluoride so that's the same concentration of ammonium ions right so this is point zero five zero molar and if we pretend like this reaction hasn't happened yet our concentration of our products is zero next we think about the change and since NH 4 plus turns into NH 3 whatever we lose for for NH 4 plus is what we gain for NH 3 so if we lose a certain concentration of X for ammonium right if we lose a certain concentration of X for NH 4 plus we gain the same concentration X for NH 3 and therefore we'd also gain the same concentration for hydronium as well so at equilibrium our concentration of ammonium would be point zero five zero minus X for the hydronium ion it would be X and for ammonia for NH 3 it would be X as well so we're talking about ammonium acting as an acid here and so we're going to write an equilibrium expression we're going to write ka so ka is equal to concentration of products over reactants so this would be the concentration of h3o plus times the concentration of NH 3 alright all over the concentration of NH 4 plus because we're leaving water out so all over the concentration of NH four plus alright the concentration of hydronium ions at equilibrium is X so we put an X in here same thing for the concentration of NH 3 and right that would be X so we put an X into here this is all over the concentration of ammonium which is point zero five zero minus X so over here we put zero point zero five zero minus X next we need to think about the KA value right so finding the KA for this for this reaction is usually not something you would find in a table in a textbook but we know that we're talking about an acid base a conjugate acid-base pair here so NH four plus and NH three our conjugate acid-base pair we're trying to find the KA for NH 4 plus and again that's not usually found in most textbooks but the KB value for NH 3 is it's 1.8 times 10 to the negative 5 so for a conjugate acid-base pair ka times KB is equal to kW so we're trying to find ka we know KB is 1.8 times 10 to the negative 5 this is equal to 1.0 times 10 to the negative 14 so we can once again find find ka on our calculator so 1.0 times 10 to the negative 14 we divide that by 1.8 times 10 to the negative 5 and so the KA value is 5 point 6 times 10 to the negative 10 so if we get some room down here we say ka is 5 point 6 times 10 to the negative 10 this is equal to so be x squared over here and once again we're going to assume we're going to assume that X is much much smaller than point 0 5 and 0 so we don't have to worry about X right here if it's an extremely small number point 0 5 0 minus X is pretty much the same as point 0 5 0 so we plug this in and we have point 0 5 0 here so we need to solve for X so we get out the calculator and we're going to take five point six times 10 to the negative 10 and we're going to multiply by point zero five and then we're going to take the square root of that to get us what X is so X is equal to 5 point 3 times 10 to the negative 6 so X is equal to five point three times 10 to the negative 6 X represents the concentration of hydronium ions so this is a centration right this is the concentration of hydronium ions so to find the pH all we have to do is take the negative log of that alright so the pH is equal to the negative log of the concentration of hydronium ions so we can just plug that into here five point three times ten to the negative six and we can solve and let's take the negative log of five point three times ten to the negative six and so we get we get five point two eight if we round up here so let me get a little bit more room so we're rounding up to five point two eight for our final pH the pH is equal to five point two eight so we got an acidic solution which is what we would expect if we if we think about the salt that we were originally given for this problem