- Titration questions
- Acid–base titrations
- Worked example: Determining solute concentration by acid–base titration
- Titration of a strong acid with a strong base
- Titration of a strong acid with a strong base (continued)
- Titration of a weak acid with a strong base
- Titration of a weak acid with a strong base (continued)
- Titration of a weak base with a strong acid
- Titration of a weak base with a strong acid (continued)
- Acid-base titration curves
- Titration curves and acid-base indicators
- Redox titrations
Calculating the pH for titration of acetic acid with strong base NaOH before adding any base and at half-equivalence point. Created by Jay.
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At7:25where he states that we started with 0.01 moles of CH3COOH, didnt we actually start with 0.01 - x moles of CH3COOH, where x is the number of moles that dissociated into H+ and CH3COO- ions?(25 votes)
- Yes, technically, but at about2:40he explains that x is so small that in comparison with the initial concentration/amount that it hardly makes a difference. In other words, 0.01-x is almost equal to 0.01, to the point where subtracting x unnecessary, especially given the low precision of the data. The % ionization of acetic acid, by the calculations in this problem, is around 1%, so only a tiny amount of the acetic acid is dissociated at any given time. Thus, it makes more sense to write the amount of acetic acid in solution as 0.01 moles– it is 99% accurate and it simplifies the problem immensely. However, if you are dealing with the equivalence point, where all of the available acetic acid has been neutralized, the amount of conjugate base (CH3COO-) becomes significant, since it is the only species left in solution other than water, and you will want to calculate that x to find out the pH of the solution at the equivalence point, or what have you. I hope this helps!(19 votes)
- how do you know this a buffer solution?(8 votes)
- Why did you not account for the reaction of acetate and water to calculate the pH in part b like you did in part c (next video)?(10 votes)
- for the 1st section on initial pH of the solution before adding any base, why is it incorrect to estimate the pH by using the concentration of acetic acid and plugging it into the pH formula?(5 votes)
- Because pH = -log [H3O+]. The initial concentration of CH3COOH is not equal to [H3O+]. You need to do ICE in order to figure out how much H3O+ has been formed by the reaction of CH3COOH with water.(7 votes)
- For problem b, could we just have looked at solely the H3O+ and OH- and figured out how much H30+ was left after adding NAOH, then using that to find the pH of the solution? We can find the amount of moles of H30+ from the calculations we made in a).(7 votes)
- Can the moles of h30+ be used to determine the ph instead of moles of weak acid. It's seems to be more direct.(2 votes)
- You need to know the moles of the weak acid and the pKa to determine how much of the acid dissociates and forms H3O+, which is ultimately used to determine pH. You need to go through this route because acid will (rarely) dissociate completely, and therefore 1M of [acid] will not always give you 1M of H3O+.(8 votes)
- why did we not account for the dissociation of acetic acid in water to form some deprotonated acetic acid. With initial concentration of 0.019 similar to the concentration of the hydronium ion(5 votes)
- At12:10it talks about the Henderson-Hasselbach equation. What other equation could be used to find the pH of a buffer solution?(5 votes)
- There are alternates, but this is the most useful. It won't be any easier than with this form of the HH equation.(1 vote)
- neutralisation of ch3cooh is reversible reaction ,isn't?(1 vote)
- In question a) you calculate the concentration for H+. Why don´t you use the 0.0019M concentration of H+ in question b) instead of 0.2M concentration?(2 votes)
- Question a) asked for the concentration of H⁺ in 0.200 mol/L acetic acid.
Question b) asked for the concentration of H⁺ in 0.200 mol/L acetic acid after the addition of NaOH.
Some of the acetic acid was neutralized by the NaOH.
We have to calculate the concentration of the remaining acetic acid before we can calculate the H⁺ that it produces.(3 votes)
- Here we have a titration curve for the titration of 50 milliliters of 0.200 molar of acetic acid, and to our acetic solution we're adding some 0.0500 molar sodium hydroxide. So once again we're putting pH in the Y axis, and down here in the X axis is the milliliters of base that we are adding. So in Part A, what is the pH before you've added any base? Right, so the addition of 0.0 mLs of sodium hydroxide. So we only have to think about a weak acid equilibrium problem here. So our weak acid is acetic acid, right? And it's in water, so acetic acid donates a proton to H2O, and so H2O turns into H3O plus. And if you take away a proton from acetic acid, you're left with CH3COO minus, the acetate anion. Our initial concentration of acetic acid is 0.0200 molar. Alright, so this is 0.200 molar here, and we're pretending like we don't have any of our products yet, right? So this is what we did in the video on weak acid equilibrium. So the change, alright, whatever we lose from our concentration of acetic acid, since acetic acid turns into acetate, right, we would gain for the concentration of acetate. And we would also therefore gain for the concentration of hydronium, H3O plus. So at equilibrium, right, our equilibrium concentration would be 0.200 minus X, and then over here we'd have plus X, and over here we would also have plus X. So when we write our Ka expression, our equilibrium expression, Ka is equal to concentration of products over reactants. And so I'll just go a little bit faster since we've already done a video on all of this, and so make sure to watch the video on weak acid equilibrium before you even look at these. Alright, so this would be X, right, times X. So that's these two X's over here. So our products over our reactants. And that would be 0.200 minus X, leaving water out of our equilibrium expression. So this would be 0.200 minus X. If we assume that X is a really small number, right, a very small concentration compared to 0.200, we can approximate it and say that 0.200 minus X is about the same as 0.200. So we could rewrite this. Let's rewrite it over here. So we have X squared now, over 0.200 is equal to Ka. And Ka for acetic acid is 1.8 times 10 to the negative 5. So we can go ahead and solve for X. So let's get out the calculator here. So we have 1.8 times 10 to the negative 5, and we're gonna multiply that by 0.200, right? So we get 3.6 times 10 to the negative 6. And then we need to take the square root of that. So the square root of our answer gives us X, and we can see X is equal to .0019. So let's write that down. X is equal to 0.0019. And remember X represents the concentration of hydronium ions, so this X right here, right, this X represents our concentration of hydronium ions at equilibrium. And so if this is the concentration of hydronium ions, right, so this is equal to the concentration of hydronium ions, find the pH is very straightforward because we just need to plug in the concentration of hydronium into our pH calculation. PH is equal to negative log of the concentration of hydronium. So we take this number, 0.0019, and we plug it into here, and we can solve for the pH. And so let's go ahead and do that. So let's take out the calculator. So the negative log of .0019 gives us the pH. And so we get 2.72. So the pH is equal to 2.72. So before we've added any base, alright, so 0.0 mLs of sodium hydroxide, the pH should be 2.72. And so we go over here to our titration curve, right, and that 0.0 mLs of base, we go over to here, and so this is our first point, right? So this is A right here, and that should represent a pH of 2.72. Alright next, let's add some more sodium hydroxide. Now in Part B, our goal is to find the pH after we've added 100.0 mLs of our 0.0500 molar solution of sodium hydroxide. So let's see how many moles of base we are adding. So if the concentration of sodium hydroxide is 0.500 molar, that's the same concentration of hydroxide ions in solution. So the concentration of hydroxide ions in solution is 0.0500 molar, and molarity is moles over liters, right? Molarity is equal to moles over liters. So this is equal moles over, how many liters is 100 milliliters? We move our decimal place one, two, three, so that's point 1. So that's 0.1000 liters. Alright, so to find moles, all we do is multiply. Alright, so 0.0500 times 0.1000 give us 0.005, so we're working with 0.00500 moles of hydroxide ions. Alright, so that's how many moles of hydroxide ions we're adding. How many moles of acid did we originally have present? Let's go back up here to our titration curve so we can see those numbers. So we were starting with 50.0 millileters of 0.200 molar acetic acid. Alright, so let's figure out how many moles of acetic acid we started with here. So the concentration was 0.200 molar, so for our concentration of acetic acid, our concentration was equal to 0.200 molar, and that's equal to moles over liters. We started with 50 milliliters, which is 0.500 liters. So we multiply 0.200 by 0.0500, and we get 0.0100 moles of acetic acid. So that's what we're starting with here. So we have an acid and a base. Our acid is acetic acid, and our base is the hydroxide ions. The hydroxide ions are going to neutralize the acid that's present. So we get a neutralization reaction. Let's go ahead and write what happens. Our acid is acetic acid, and we're adding some hydroxides. So the hydroxide is going to take the acetic proton from acetic acid, the hydroxide is gonna take this proton right here. And OH minus an H plus, of course, form H2O. So once hydroxide takes a proton from acetic acid, we're left with the conjugate base for acetic acid, which is, of course, the acetate ion. So we also make CH3COO minus here. Alright, what were we starting with? For acetic acid, right, we were starting with 0.0100 moles. So let's go ahead and write that, 0.0100 moles. And for hydroxide we were adding 0.00500. Alright, so we're adding 0.00500 moles of hydroxide. So the hydroxide ions are going to neutralize the acetic acid, alright, so we're going to lose all of our hydroxide. So all of it reacts. So all of this reacts here, and so we're left with nothing. We use up all of our hydroxides, alright, and that much hydroxide is going to neutralize the same amount of acetic acid. So we're gonna lose that much acetic acid. So 0.0100 minus 0.00500, right, this is gonna give us 0.0050 moles of acid left over. So this is how much acid was not neutralized. Alright, so if we lose that much acetic acid, we're also going to gain that much acetate anion, right? So if we start with zero over here, we're going to gain the same amount. So 0.00500 is how much we gain over here. So we're gonna finish, after neutralization, with 0.00500 moles of acetate anion. Alright next, let's figure out the concentration of acetic acid in the acetate anion. So we have moles for both, alright, we have moles for both, but we need to find the volume. Well, let's think about that. So let's go back up here to the problem, and we're adding 100 milliliters, alright? And we started with 50 milliliters of our acid solution. Alright, so 50 plus 100 is 150 milliliters. Alright so now our new volume is 150.0 milliliters, 50 plus 100, and that's equal to 0.15 liters, right? We move our decimal place. So let's figure out our concentrations. We have moles, we have volume, so let's find the concentration of acetic acid now, after neutralization. Alright, so we have 0.00500 moles, alright, so let's go ahead and put that in, 0.0050 moles. And our volume is 0.1500, right, so 0.1500 is our volume. So we can go ahead and do our calculations. I'll just take out the calculator here. So we have 0.005 divided by .150, and we get .033. So the concentration is 0.033 molar. Alright, what about the concentration of acetate? Alright, so the concentration of acetate anions in solution, alright, would be equal to- Look at the numbers, they're the same. We have 0.00500 here, and once again, our total volume is .1500, so it's the same calculation. 0.0050 over 0.1500, and we would therefore also get a concentration of 0.033 molar. So we have equal concentrations of a weak acid and its conjugate base. And that makes me think about a buffer solution. So that's what we've formed here. As we drip base, as we drip hydroxide ions into our original acidic solution, we're slowly a buffer solution. And here we have equal concentrations. So our goal was to find the pH, and since we have a buffer solution now, it's easiest to just use the Henderson Hasselbalch equation. Alright, so the Henderson Hasselbalch equation was pH is equal to the pKa, plus the log of the concentration of A minus over the concentration of HA. Alright, and the Ka for acetic acid, we talked about earlier, was 1.8 times 10 to the negative 5. So the pKa would just be the negative log of that. So let's take out the calculator. Let's take the negative log of 1.8 times 10 to the negative 5. And so the pKa is equal to 4.74. So let's write that down here. PKa is 4.74. So we plug that in to our Henderson Hasselbalch equation right here. So the pH is equal to 4.74, plus log of the concentration of A minus. A minus would be acetate, right, which is 0.033, over HA. HA would be acetic acid, which is 0.033. So we have the same concentrations. So this is log of 0.033 over 0.033. And that of course is one. Alright, so we have log of one here, and you probably already know what log of one is equal to. I'll do it on the calculator so you can see that log of one is equal to zero. And so all of this is equal to zero. So the pH is equal to 4.74 plus, you know this would all be zero over here. So the pH is equal to 4.74. And this is the half equivalence point. We've neutralized half of the acids, right, and half of the acid remains. And using Henderson Hasselbalch to approximate the pH, we can see that the pH is equal to the pKa at this point. So let's go back up here to our titration curve and find that. Alright, so the pH is 4.74. The pH is 4.74 after we've added 100 mLs of our base. So we go right up here to 100 mLs. And so this would be our second point. This is what we did in Part B. And the pH is approximately 4.74 at this point.