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A 25.0 -mL solution of $0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}$ is titrated with a $0.200 \mathrm{M}$ KOH solution. Calculate the pH after the following additions of the KOH solution:(a) $0.0 \mathrm{mL},$ (b) $5.0 \mathrm{mL},$ (c) $10.0 \mathrm{mL},$ (d) $12.5 \mathrm{mL}$(e) $15.0 \mathrm{mL}$

a) $\mathrm{pH}=2.87$b) $\mathrm{pH}=4.57$c) $\mathrm{pH}=5.35$d) $\mathrm{pH}=8.79$e) $p H=12.10$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

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Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

24:14

In chemistry, a buffer is a solution that resists changes in pH. Buffers are used to maintain a stable pH in a solution. Buffers are solutions of a weak acid and its conjugate base or a weak base and its conjugate acid, usually in the form of a salt of the conjugate base or acid. Buffers have the property that a small change in the amount of strong acid or strong base added to them results in a much larger change in pH. The resistance of a buffer solution to pH change is due to the fact that the process of adding acid or base to the solution is slow compared to the rate at which the pH changes. In addition to this buffering action, the inclusion of the conjugate base or acid also slows the process of pH change by the mechanism of the Henderson–Hasselbalch equation. Buffers are most commonly found in aqueous solutions.

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A $25.0$ -$\mathrm{mL}$ so…

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A 10.0-mL sample of $0.250…

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Consider the titration of …

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Hello. So today we'll be looking at a tie tray shin when you have 25 milliliters off 250.1 mil. Araji acetic acid, which is a weak acid. And what tight trading it with 0.2 polarity. Good potassium hydroxide, which is a strong base. So first, let's look when man of the acidic acid has been Thai traded, so no potassium hydroxide has been added yet. Well, first porn piece of information is that the K F acetic acid is 1.8 times 10 to the negative fists. Now let's draw our reaction. So we're going to have acetic acid, and some of it is going to react with water, and it's going to form. It's conjugated base and the Heidrun you my own. Conversely, you could just say that it's ah, dissociating into its ah conjure git base and just the proton. But in reality it would ah, the proton. It would attach itself to the unto the water molecule two point form the hydro knee mild. So let's take a look at our initial state. So initially we have, ah, 0.100 Moeller and we don't have any of this, but of the conjugate base because it hasn't reacted yet. And the height? Joni, um, ion or the H plus the protons. It's AC in water. It's a 10 to the negative seventh, but this is very small, so we're gonna approximate it as zero. Now we're gonna look at how much it changes, so we'll have. Ah, the acetic acid reacting in roll form one molecule off its conjugated base and one molecule of the height. Geranium, My on. And what? It doesn't matter. So now, at the equilibrium, we'll have 0.100 Mueller, we're going to subtract acts and access, probably going to be very small, because if you look at the k A, the K is very small, so it's going to stay mostly on the acetic acid side so we can actually approximate this as just 0.1 moller and then these will both be axe. So now you see, So the K A would be conjugated base times that had drawn Yamile over acetic acid. So in this case, it's 1.8 times 10 to the negative. Fifth is equal Teoh X squared or zero point ones. And if you just solved this so X squared equal to 1.8 times 10 to the negative six and then you square root it. We will see that X will be 1.34 times tend to the negative three, and this is equal to the concentration of the hydro knee Mayan. So to find the pH. That would be the negative long of the hydro in your mind, or the negative log of 1.34 times 10 to the negative three, which in this case is to 0.87 So that is when we have added zero mill leaders of our base. So what happens if we do add milk some over base? So let's say we've got five milliliters of her base. Well, first, let's find out how many moles of our acetic acid that we have. So 25 milliliters, we're going to convert that to leaders and then we can use our malaria T to see that we have 0.100 moles of acetic acid in one leader and we will see that we have zero point literal 0 to 5 malls of acid. Now let's see how many moles of our base. We added, We've got five milliliters going to convert that toe. Leaders gonna use thumb polarity to find out how many moles we have. And since potassium hydroxide associates completely, really, we're looking at the high drawn you my own, the concentration of the hydro knee wild. And so we see that we have a zero point zero zero one mold of hydroxide. And so this Molesworth hydroxide will react with the acid. And so we just subtract and we will see we have 0.15 moles of acetic acid left since 0.1 moles has reacted. And so now let's find out the concentration. So if we've got 0.0 015 moles of acetic acid and then we see that we had 25 milliliters, we added five. So have a total 30 milliliters. Converting milliliters is 0.3 leaders. We will see if we do some math, we will see that that's a total of zero went 05 polarity. So that's our new concentration of the acetic acid. Now let's take a look at the concentration of the country base because so 0.1 Mom has it of the acetic acid has reacted to form the conjugal base. And we're gonna divide it by the new volume and we will see that we have 0.0 three repeating polarity of the country. Get base. So now let's draw a new ice table. So we still have our Sita casted reacting with water. It's going to form the conjugated it pais and the hydro new mild. So initially, we now have 0.5 hilarity of acetic acid. And now we have 0.3 polarity of its conjugal base. And we have about zero more polarity of hydro knee, um, ions. So the acetic acid is going to react and reacting. One mole will create one mole of its country get base and one mole. If I draw on your minds and at equilibrium when have 0.5 minus X and since 0.5 is so high in the peaks and then que is going to be very small, well, we can approximate. This is just Europe 105 and you'll notice that this concentration will also be a lot larger than X so we can also approximate this as 0.3 and then the Hydra millennium ion concentration will just be acts. So let's take a look. So we're going to have 0.3 repeating times x over 0.5 equal to the K A, which is 1.5 times 10 to the negative five. We will find that X is 2.7 times 10 to the fifth negative five and this is the concentration of hydro knee. Um so to find the pH is just a negative log in the P H will be 4.57 So that's only added five milliliters. Now say we added 10 milliliters of base. Well, let's do something similar. We already found out how many moles of acetic acid we have. 0.25 So now let's find out how many moles of base we added. So 10 milliliters of hydroxide going to convert that to leaders and then use the polarity and we will see that we have zero 0.2 bones Oh hydroxide. So we will have 0.0 to most of the contract your base and we're gonna divided by the new mall volume. So 25 plus 10 is now 35 mil leaders, which is 0.0 35 leaders. And the new concentration that will get is zero Clint zero 57 polarity. No, let's take a look. That's the concentration of the acetic acid. So originally are moles. Waas 0.25 But we serve Chung's Europe 100 to sell means we have 0.5 moles of acetic acid. And now we divide by the new volume and we will see that's our new concentration. 0.143 polarity. So now how about Lee racing this? And now let's draw new ice table. So Sita Kassid reacting with water to form it's conjugated base and Heidrun you miles. So initially, our concentration was zero point 0143 and the concentration of vast 80.57 and so from massive Tate is going to form that it's con trick it base and I draw on your minds. So we already addressed that. It's going to be very little is going to it associate, and this is a fairly high polarity compared to the other. The concentration that's being reacted so we can just approximate it by its initial concentration. And then And so now we're just going to do some math. So we've got 0.57 times acts over 0.143 and that's going to equal to look a, which in this case is 1.8 times 10 to the negative five in. So if we do some ass, we will see that ax is equal to 4.5 times 10 to the negative six. And if we take the negative log of that, we'll see that the P. H is five point 35 So now we've done 10 milliliters and five milliliters as well. A zero no leaders. So how about what if we reach the equivalents point? So we have 25 milliliters of zero 0.1 polarity. Acetic acid were tight, trading it with 0.2 similarity hydroxide. So in order to reach the equivalents point, we would need half of 25 which in this case is 12.5 milliliters of potassium hydroxide. So this would be the equivalents point. So all of that I So all the civic asset has been turned into its conjugal base. So that means that instead of having acetic acid associate, we are going to have the Contra give base reacting with water, and it's going to form Placido kassid and hydroxide. So now how about we try to figure out the concentration of the conjugal base of acetic acid as it hate? So let's take a look. So we've got 0.0 25 moles. So conjugal base and we had 25 milliliters. We added 12.5. So that's 37.5 milliliters, which is zero point zero 375 leaders. And so we will get that our concentration initial concentration of acetate is 0.6 repeating. And then we have no ascetic acid right now and the hydroxide is around zero because 10 to the negative fifth tend to the negative. Seventh is very small, so now we're going to react acetate to give acetic acid and hydroxide and the K B for the Kontic It at the base of acid Tate is five 0.6 times 10 to the tent to the negative 10th. So K B is very small. So we're going Teoh. So very little will be of the conjugate base will be converted toe acid Tate so we can approximate this a 0.6 repeating. And then we have these two so we just need to do a little bit of math. So x squared over 0.6 well, equal, but KB, which in this case is five 0.6 times 10th and negative. And after doing some as we will see, that X is he quoted 6.1 times 10 to the negative six. And if we take the log of that, we'll get the P O. H. To take the negative log, and the peel H in this case is five point to one. Now, if you remember, the pH is some plea 14 minus the p o. H. So if we have appeal wage who confined the PH very easily, we will see that it is eight 0.79 So we have reached the equivalents point. But what happens if we go beyond the equivalence point? So we had had added we have added 12.5 meal milliliters of potassium hydroxide. What if we add 15? So if we 15 well, we subtract 12.5 that we used to reach the equivalents point. That will be 2.5 milliliters off of hydroxide I owns. And those hydroxide ions aren't being used to convert the acetic acid to its contra get base anymore. So those hydroxide ions just to go basically to making the solution more basic. So basically, let's find out how many moles of hydroxide we added. So you've got 25 extreme a leaders just of hydroxide that's being added. Now let's convert two moles. I'm leaders, I mean and then we can find the moles if we use the polarity and we will see that we have added 0.0 five moles of hydroxide and this is a pretty big amount. We're gonna we have, like, 6.1 times 10 to the negative six. So that's quite small. So we're gonna assume that it's negligible. So now let's find the Malek of Similarity Concentrations. We've got 0.5 moles. We're going to divide that 25 plus 15 is 40 criminal leaders, so we're gonna divide that 0.4 leaders, we're going to see that we have zero 0.0 1 to 5 polarity of hydroxide. We will see that this is indeed much larger than the concentration we had before. So it in fact, is negligible and weaken just to use this. And if we plug this in, will get that R P O. H is 1.90 and we confined our Ph from that and we will see that are pH is 12 0.10

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