If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Titration of a weak base with a strong acid

## Video transcript

this time let's look at the titration curve for the titration of a weak base with a strong acid so let's say we're starting with 40 milliliters of a point 1 molar solution of ammonia and to our ammonia we're going to add a strong acid we're going to add some HCL in Part A they want us to find the pH before we've added any acid so we have ammonia present so let's go ahead and write nh3 right here we have a solution of ammonia so we write water here since ammonia is a base ammonia is going to take a proton from h2o and if you add an H plus 2 NH 3 you get NH 4 plus so we're going to make some ammonium here and if you take away an h+ from h2o you get a chai drock's I'd our initial concentration of ammonia is 0.1 molar so right here we write initial concentration of ammonia and that's point 1 molar and this is a weak base equilibrium problem right so we're going to assume an initial concentration of 0 for our products here next we think about the change so we're going to lose a concentration of ammonia so a certain concentration of ammonia is going to react and we call that concentration X and since that concentration of ammonia nh3 turns into NH 4 plus whatever we lose for ammonia we gain for ammonium all right so we write a plus X over here so we're going to gain a concentration of ammonium and we would gain the same concentration of hydroxide so at equilibrium right our concentration of a of ammonia is going to be 0.1 minus X for ammonium it would be X and for hydroxide it would also be X next let's write our equilibrium expression since this is a base we would write Kb and we have the concentration of products over reactants right so for our products we would have the concentration of ammonium so that's x times the concentration of hydroxide which is also x over the concentration of ammonia which is 0.1 minus X so make sure you've seen the video on weak base equilibria em before watching this one here alright next we know that the KB for ammonia let's say they gave it to us and it's 1.8 times 10 to the negative 5 so that's equal to x times X which is x squared right over point 1 minus X but here's where we assume that this concentration is really small compared to point 1 and therefore point 0.1 minus X is approximately the same as point 1 and it just makes the math a lot easier so now we have to solve for X so let's get a little bit more room here and take out the calculator we have 1.8 times 10 to the negative 5 we need to multiply that by 0.1 and then take the square root of that so we take the square root of our answer and we get X is equal to point zero zero one three so X is equal to point zero zero 1 3 now if we go back up here we can see that X X represents the concentration of hydroxide ions at equilibrium so let me go ahead and write that here this is equal to the concentration of hydroxide ions that's point zero zero one three molar if we have the concentration of hydroxide ions we can find the Poh because the Poh is equal to the negative log of the concentration of hydroxide so the Poh is equal to negative log of point zero zero one three and let's get out the calculator and do that negative log of point zero zero one three is equal to two point eight nine so the Poh the Poh is equal to two point eight nine and our question wanted us to find the pH so remember pH plus the Poh is equal to 14 so I can take the Poh and plug it into here and solve for the pH the pH be equal to 14 minus two point eight nine which is equal to eleven point one one so we found our pH before we add any of our acid it's 11.11 so on the titration curve right I can see down here just be zero point zero milliliters of our acid added and that's where we are and the pH should be 11.11 so somewhere right about here and so I'm saying that's a pH at the pH of 11.11 for Part B of our question what is the pH after the addition of 20 milliliters of our point one molar solution of HCl so the concentration of HCl is equal to point one molar and we know molarity is mole's over liters so that's moles over liters we're adding 20 milliliters so if you take 20 milliliters and move your decimal place one two three that's point zero two liters so we're adding point zero two liters here and so we solve for moles point one times point zero two is equal to zero point zero zero two so that's how many moles and so many moles of HCl we are adding so HCl you can think about this as H+ and Cl minus or if you think about adding your proton to water h+ and h2o give you h3o plus so you could consider this to be the number of moles of h3o plus that we're adding next what's the concentration of ammonia that we started with so the concentration is 0.1 molar so the concentration of ammonia is equal to 0.1 molar all right and we want to find moles of ammonia what's the volume of ammonia well 40 milliliters would be one to three point zero four liters so we have point zero four liters here and we solve for moles so that's point one times point zero four and point one times point zero four is equal to point zero zero four so that's how many moles of ammonia that we have alright next let's think about what happens to the ammonia when we add the HCL alright so the acid that's present is going to react with the base that's present let's go ahead and write what would happen the ammonia nh3 is going to react with the h3o plus that is present so this reaction goes to completion here and let's think about what would happen this is a base and this is an acid so we're going to protonate nh3 to form NH 4 plus so we add a proton on to nh3 to form NH 4 plus and we take a proton away from h3o plus so we get h2o so we have h2o over here all right next let's look at what we're starting with so this is our neutralization reaction so let me go and write that this is the neutralization neutralization reaction we're adding point zero zero two moles of our acid here point zero zero two so let's put that right here so we're adding point zero zero two moles of acid and we're starting with point zero zero for moles of our base so point zero zero for moles of our base so all of our acid is going to react alright we're going to lose we're going to lose all of the acid that we added it all reacts so we're left with nothing here we're going to lose the same concentration of our base all right so the acid is going to react with that much of our base so we're going to lose point zero zero two moles of base so we're left with point zero zero four minus point zero zero two which is equal to point zero zero two so half of the base has reacted with the acid and half of the base is left over so this is how many moles of ammonia we have left over after the acid has reacted alright if we're losing if we're losing this many moles of ammonia all right since our ammonia turns into NH 4 plus if we started with zero for NH 4 plus that's how much of ammonium we are gaining here so we're gaining point zero zero two moles of NH four plus and so we're going to end up with point zero zero two moles moles of ammonium nh4 plus all right so we have moles of ammonia we have moles of ammonium and next when you think about the concentration of ammonia and ammonium that's present all right so if we're thinking about concentration we need to know volume so let's go back up to here and let's see what the total volume is now we start off with 40 milliliters of our ammonia and we've added 20 milliliters of our acid so 40 milliliters what we started with we added 20 milliliters so our total volume is now 60 milliliters or point zero six liters all right so this is 60 milliliters we move our decimal place one two three and that's point zero six liters so now we have we have liters and we have moles so let's find the concentration all right so let's start with the concentration of ammonia so the concentration of ammonia is equal to moles over liters we have point zero zero two moles of ammonia so point zero zero two moles of ammonia we divide that by our our volume right our total volume is now 60 milliliters or point zero six liters and so we can find our concentration all right so we could go ahead and do that here point zero zero two divided by point zero six is equal to point zero three three three all right so we get this is equal to point zero three three three this is a concentration so this is molar next we do the same thing for ammonium right so what's the concentration of ammonium here well we had the same number of moles right that's point zero zero two moles of ammonium and the volume is the same point zero six liters and so therefore it's the same calculation we get the same concentration of ammo ions in solution point zero three three three molar so now we have we have equal concentrations of a conjugate acid-base pair right NH four plus and NH three is a conjugate acid-base pair so we have a buffer solution all right so now now we have a buffer solution so our goal is to find the pH and since we have a buffer solution the easiest way to do that is to use the henderson-hasselbalch equation alright so we go ahead and write out the henderson hasselbalch equation which is the pH is equal to the pKa plus the log of the concentration of a minus the conjugate base over H a alright so let's let's think about what the pKa is how can we find the pKa well we know the KB for ammonia right we know the KB for ammonia and we know that ka times KB is equal to 1.0 times 10 to the negative 14 all right so if we plug in the KB 1.8 times 10 to the negative 5 all right this is equal to 1.0 times 10 to the negative 14 we can solve for the KA and to save time I won't show this on the calculator I'll just go ahead and write the KA is equal to five point six times 10 to the negative 10 so remember this from an earlier video if you multiply the for a conjugate acid-base pairs multiply a times KB you get this and so now we've just solved for the KA is five point six times 10 to the negative 10 this is for NH 4 plus so next we can find the pKa because the pKa is the negative log of the KA so the negative log of five point six times 10 to the negative 10 so let's get out the calculator and do that negative log of five point six times 10 to the negative 10 gives us a pKa of nine point two five so the pKa is equal to PKA is equal to nine point two five so we plug that in to our henderson hasselbalch equation so now we have the pH is equal to nine point two five plus the log of the concentration of a minus a minus is your conjugate base all right so that's what we're talking about up here for ammonia so that's point zero three three three so this is equal to point zero three three three and that's over the concentration of your acid which is NH four plus which is also equal to point zero three three three so that's equal to point zero three three three all right so this is on the right all of this this is equal to the log of one and the log of one is equal to zero right log of one is equal to zero so therefore your pH is equal to your PKA which is equal to nine point two five so the pH is equal to nine point two five since all of this is equal to zero so now we can find our point on our titration curve alright we've added twenty milliliters of base let's go up acid I should say let's go back up here and look at our titration curve alright we've added we've added twenty milliliters of our acid so right down here 20 milliliters of our acid so let's find this point on our titration curve so we're right there and we calculate the pH should be nine point two five so if I go over here the pH should be equal to nine point two five so we're still before the equivalence point