Chapter 16: Titration of strong acid with strong base

Added: 2026-05-23

[00:00:03]hello okay I am on chapter 18 slide number six we are going to start learning about acid-base titrations a titration is a way that we can accurately determine the amount of an acid or base so what would you do let's say we're going to titrate an acid so I'm going to take some fluid it's a water based solution it's got H+ in it I'm gonna add probably some indicator to this so I'll add like a phenolphthalein indicator which is colorless an acid however phenolphthalein would be pink in base okay so I want to know exactly how much H+ is in the solution so I'm going to do what's called a titration and what I'm gonna do is I'm going to add hydroxide to this solution probably sodium hydroxide so I had drops of a solution of a known concentration of aqueous sodium hydroxide so when this drop hits here it's going to add hydroxide to the solution well that's going to react with any H+ and make h2o I'm gonna continue to add drops of sodium hydroxide known volumes of this we'll just say every drop has a hydroxide in it so every drop is going to convert each plus to water and I'll add one more drop and that'll convert this h plus 2 h2o and at that point the solution has become neutral at that point where we've added precisely the amount of hydroxide that exactly equals the amount of hydronium or protons present we call this the equivalence point and notice this definition is listed in chapter 18 slide six the equivalence point is when we have added precisely the amount of hydroxide to exactly equal the amount of protons or hydronium that's present in the solution this is what we want to know we want to know where the equivalence point is because if I know the concentration of hydroxide that I've added and I know its concentration then I know how much H+ is present through just reaction stoichiometry well that's not what we get to see what we basically see is the end point the end point of a titration is a signal which we try to make it some kind of signal that you have reached the equivalence point so you want these two things to be equal you want the endpoint of a titration to be the same as the equivalence point but they rarely are and in this really simplified example for example I would have to add another drop of sodium hydroxide to make this solution basic and turn it pink so the endpoint would be when you view this solution turning pink whereas the equivalence point that's exactly when you've exactly added just the right amount of hydroxide to react with the H+ that's present so this is this is generally a theoretical thing that you can't really you can't really see but you can see the endpoint all right let's move on we're going to do the first part of chapter 18 slide 7 I'm going to show you how to answer sort of these questions and the first part of chapter 18 slide 7 so what we're going to do is we're going to titrate 50 mils of 0.1 molar strong acid will call it hydrochloric acid with 0.1 molar strong base so we'll just call it sodium hydroxide we want to know the starting pH we want to know the pH after 45 mils of base have been added we want to know the pH at the equivalence point and we want to know the pH at 55 mils of the sodium hydroxide solution added okay first thing we're going to do is we're gonna calculate the starting pH well that's easy I've got 50 mils of a solution that 0.1 molar in HCl well we've got HCL at 0.1 molar that means all of this has dissociated 2 H+ ions and chloride ions which means regardless of the constant or the volume of this I have 0.1 molar h+ than 0.1 molar chloride because all of this dissociates well that's easy I can find a pH that's going to be the negative log of the H+ concentration raishin which is going to be the negative log of 0.1 if you do the math you're gonna find out that that's going to be equal to one so the pH the starting pH is gonna be one point zero zero you can do that in your calculator and show it to be the case okay the second thing we want to check is the pH after 45 milliliters of base have been added now this is gonna get a little bit more complicated alright so I have 50 mils of 0.1 molar HCl and to that I'm gonna add 45 milliliters of my point 1 molar sodium hydroxide this is Part B plus 45 milliliters of 0.1 molar sodium hydroxide so this is going to be the reaction between a strong acid and strong base let's figure out how many moles of HCl well let's write the equation first HCl plus sodium hydroxide is gonna react with a one-to-one stoichiometry we're gonna get water and we're gonna get sodium chloride this is all dissolved okay so I've got 0.1 moles of HCL for every liter and I had point O five liters 50 mils that's gonna be point zero zero five moles of H plus what I write h+ cuz that's the part that's going to react with the hydroxide I'll have to worry about chloride it's just a spectator ion how much hydroxide would have been added at this point well I've got point one moles of my sodium hydroxide solution for every liter and in this case I have point O four or five liters if I do the math that comes out two point zero zero four five moles of hydroxide when these two things react what's going to happen while I'm going to have point zero zero five moles of HCl and point zero zero four or five moles of sodium hydroxide this is an excess it's going to react with all of that converting it to point zero zero four or five moles of water at point zero zero four or five moles of dissolved sodium chloride how much of this is left oh four five sorry that's messy point zero zero five minus point zero zero four five right this much H+ react to that much hydroxide that's gonna be 5 times 10 to the negative 4 moles of h plus left well what's the concentration hmm well for that I need moles and liters right so let's look at this I have 5 times 10 to the negative 4 moles of h plus left in 50 plus forty five liters so that's gonna be 0.09 five liters because I started with 50 mils but I added 45 well let's see how much is that gonna be 0.005 one two three four yeah divided by 0.095 that's gonna be point zero zero five two six molar h+ now i take the negative log of that one two three it's gonna be about a three let's do the actual thing let's just take the log of this let's see hit the log well negative log make it positive there you go negative log of the H+ will be the negative log of this which is going to be 2.28 and we're done with Part B that's my pH all right where are we supposed to calculate and Part C that's going to be easy what's the pH at the equivalence point that's going to be seven all right that one's a piece of cake cuz here that's the equivalence point you'll have added just enough HC excuse me just enough hydroxide to react with the HCL that's present so you know as long as you got the carbon dioxide out of the water it's going to be neutral you've exactly balanced them now we're going to do Part D what is the pH when we have an excess of sodium hydroxide at it check this out let's do the calculation again let's see I got 0.1 moles of HCL and point O five liters to start with that's per liter times 0.05 liters now to give me point zero zero five moles of H Plus that I start with now I'm going to have added fifty five mils of the sodium hydroxide and the sodium hydroxide is point one molar so I got point one mol of the sodium hydroxide in every liter times point zero five five liter that gives me point zero zero five five moles of hydroxide well once again we've got this reaction basically h plus plus hydroxide goes to water that's what happens when these two are added so the hydroxide gets direct into the 50 mils of HCL and it consumes it and it's going to consume all of it right there naught point zero zero five five moles of hydroxide that'll get consumed by point zero zero five zero moles of h plus this is an excess so all of the h+ gets consumed but notice I'll be left with one zero zero zero five moles of hydroxide in a total of a hundred and five milliliters so I'm going to divide this by 0.1 oh one liter because I need the concentration of hydroxide so let me figure out what that is point zero zero zero five divided by point one zero one I do that right did I do that right no that's not right let me just check this here real quick 105 yep 50 plus 55 that's 105 I hope you caught that all right let's get this again point zero zero five curses point zero zero zero five divided by point one zero five there we go that's my hydroxide concentration so my concentration of hydroxide is point zero zero four seven six molar what's the question asks for it asks for the pH well shoot I got the hydroxide concentration well let me just find the Poh that's gonna be the negative log of the hydroxide concentration negative log of this come on phone there we go let's see point zero zero four seven six take the log base ten of that and make it positive so that's going to be a Poh be careful now a two point three - that's my Poh remember pH plus Poh is 14 so 14 minus this will give you my pH just make it negative and out of the 14 that'll work look at my pH shot all the way up to eleven point six eight that is the answer to Part D and that's the first half of slide seven