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Buffer solution pH calculations

Video transcript

let's do some buffer solution calculations using the henderson-hasselbalch equation so the last video I showed you how to derive the henderson hasselbalch equation and it is pH is equal to the pka plus the log of the concentration of a minus over the concentration of H a so we're talking about a conjugate acid-base pair here H a and a minus and for our problem H a the acid would be NH 4 plus and the base a minus would be nh3 or ammonia so the first thing we need to do if we're going to calculate the pH of our buffer solution is to find the pKa all right and our acid is NH 4 plus so let's say we already know the KA value for NH 4 plus and that's five point six times 10 to the negative 10 to find the pKa all we have to do is take the negative log of that so the pKa is the negative log of five point six times 10 to the negative 10 so let's get out the calculator and let's let's do that math so the negative log of five point six times 10 to the negative 10 it's going to give us a PKA value of nine point two five when we round so PKA is equal to nine point two five so we're going to plug that in to our henderson hasselbalch equation right here so the pH of our buffer solution is equal to nine point two five plus the log of the concentration of a minus our base our base is ammonia nh3 and our concentration in our buffer solution is 0.24 smaller so we're going to write point two four here and that's over the concentration of our acid that's NH four plus and our concentration is point two zero so this is over point two zero here and we can do the math right so let's find the log the log of point two four divided by point two zero so that is point zero eight zero so nine point two five plus point zero H is nine point three three so the final pH or the pH of our buffer solution I should say is equal to nine point three three so remember this number for the pH because we're going to compare what happens to the pH when you add some acid and when you add some base and so our next problem is adding base to our buffer solution and we're going to see what that does to the pH so now we've added point zero zero five moles of a strong base to our buffer solution let's say the total volume is 0.5 zero liters so what is the resulting pH so we're adding we're adding point zero zero five moles of sodium hydroxide and our total volume is point five zero so if we divide moles by liters that will give us the concentration of sodium hydroxide point zero zero five divided by point five zero is 0.01 molar so that's our concentration of sodium hydroxide and since sodium hydroxide is a strong base right that's also our concentration of hydroxide ions in solution so this is our concentration of hydroxide ions point zero one molar so we're adding a base and think about what that's going to react with in our buffer solution so our buffer solution has nh3 and NH 4 plus the base is going to react with the acid so hydroxide is going to react with NH four plus let's go ahead and write out the buffer reaction here so NH four plus ammonium is going to react with hydroxide and this is going to go to completion here so if NH four plus donates a proton to O H - o h - turns into h2o right so we're going to make water here and if NH four plus donates a proton we're left with NH three so ammonia alright let's think about our concentrations so we just calculated we have now point zero one molar concentration of sodium hydroxide for ammonium right that would be point two zero molar so zero point two zero molar for our concentration and for ammonia it was point 2 4 so it's going to write zero point two four over here so if point zero one if we have a concentration of hydroxide ions of point zero 1 molar all of that is going to react with the ammonium alright so we're going to lose we're going to lose all of this concentration here for hydroxide and and that's going to neutralize the same amount of ammonium over here so we're going to be left with all right this would give us zero point 1 9 molar for our final concentration of ammonium right hydroxide we would have zero after it all reacts and then the ammonium since the ammonium turns into the ammonia right if we lose this much we're going to gain the same concentration of ammonia so over here we put plus zero point zero one so the final concentration of ammonia would be 0.25 molar and now we can use our henderson hasselbalch equation right so let's go ahead and plug everything in so pH is equal to the pKa we already calculated the pKa to be nine point two five and then plus plus the log of the concentration of base right that does that be nh3 so the concentration of 0.25 so this is 0.25 molar for our concentration over the concentration of our acid right that's ammonium so that's over point one nine so this is all over point one nine here so if we do that math let's go ahead and get out the calculator here let's do this calculation log of 0.25 divided by point one nine and we get we get point one two so nine point two five plus 0.12 is equal to nine point three seven all right so let's get a little bit more down here and we're done the pH is equal to nine point two five plus point one two which is equal to nine point three seven so let's compare that to the pH we got in the previous problem right for the buffer solution just starting out it was nine point three three so we added a base and the pH went up a little bit but a very very small amount so this shows you mathematically how a buffer solution resists drastic changes in the pH next we're going to look at what happens when you add some acid so we're still dealing with our same buffer solution with ammonia and ammonium nh4 plus but this time instead of adding base we're going to add acid so we add point zero three moles of HCL and let's just pretend like the total volume is point five zero liters and our goal is to calculate the pH of the final solution here so the first thing we could do is calculate the concentration of HCl right so that would be moles over liters that's zero point zero three moles divided by our total volume of point five zero liters and point zero three divided by 0.5 gives us zero point zero six molar right that's our concentration of HCl and HCl is a strong acid so you could think about it as being H+ and Cl minus and since this is all in water H+ and h2o would give you h3o plus or hydronium so point zero six molar is really the concentration of hydronium ions in solution and so the acid that we add is going to react with the base that's present in our buffer solution so this time our base is going to react and our base is of course ammonia so let's write out the reaction between ammonia nh3 and then we have hydronium ions in solution h3o plus all right so this reaction goes to completion and if ammonia picks up a proton it turns into ammonium nh4 plus and if h3o plus donates a proton we're left with h2o as we write h2o over here alright for our concentrations right we're going to have point zero six molar for our concentration of hydronium ion so zero point zero six molar and the concentration of ammonia is 0.24 to start out with so we have point two four and for ammonium it's point two zero so we write zero point two zero here so all of the hydronium ion is going to react so we're going to lose all of it all right so we're left with nothing after it all reacts so it's the same thing for ammonia right so that we're going to lose the exact same concentration of ammonia here so we're going to lose zero point zero six molar of ammonia because this is reacting with h3o plus and so after neutralization right we're left with we're left with 0.18 molar for the concentration of ammonia and whenever we lose for ammonia we gain for ammonium since ammonia turns into ammonium so we're going to gain zero point zero six molar for our concentration of ammonium after neutralization so we get zero point two six for our concentration and now we're ready to use the henderson-hasselbalch equation to calculate the final pH so let's do that alright get some more space down here so the pH is equal to the pKa which again we've already calculated the first problem is nine point two five plus the log of the concentration of the base all right that's point one eight so we put zero point one eight here divided by the concentration of the acid which is NH four plus so that's zero point two six so zero point two six and we go ahead and take out the calculator and we plug that in so log of 0.1 eight divided by point two six is equal to is equal to negative point one six right so let's go ahead and write that out here so we have our pH is equal to 9.25 minus 0.16 and so that that comes out to nine point zero nine so the pH is equal to nine point zero nine so remember for our original buffer solution we had a pH of nine point three three so we added a lot of acid the pH went down a little bit but not a not an extremely large amount so once again our buffer solution is able to resist drastic changes in pH