Part B first video AP chem revision qs1-10

[00:00:01]Hello guys. How are you today? I hope that everyone's doing well. So today I'll be explaining the uh part B for the AP chemistry. Okay, we have um chapter one. We have only CQDs 1 to 20.

[00:00:14]So these are the only 20 CQDs I will be doing. So without further ado, let's get started. Chapter one, it is about chemical equilibrium. Okay, 1.1 is about calculating and measuring the pH of a solution.

[00:00:27]Now, question one, what does it say?

[00:00:28]They said calculate the concentration of H+ and the concentration of OH- of water at a temperature where KW is um equal to 4.0 * 10 to the power of minus 14. Now, let's write the general equation. The general equation is what? It's H2O liquid, right? It will dissociate into what? Into H+. Let me not write concentration. It will dissociate into H+ and OH-, right?

[00:00:55]Now, they are both what? They are both aqueous because they are ions. Now, they do thus KW, which is the equilibrium constant of water, it's equal to 4.0 * 10 to the power of minus 14. But we know it's equal to the concentration of products to the power of their coefficient times concentration of products to the power of coefficient divided by one. Why? Because here I have a liquid.

[00:01:17]So what would this mean? This would mean that KW it's equal to um 4.0 * 10 to the minus 14 and it's equal to what? It's equal to the concentration of H+ times the concentration of OH-.

[00:01:32]Now, something very important you need to know. Every time water dissociates, it produces equal amounts of H+ and OH- ions since pure water is neutral. So because pure water is neutral, what do we mean by neutral?

[00:01:50]We mean it's not acidic, it's not basic.

[00:01:52]The pH is equal to a seven in most cases, but here in this case it's not seven.

[00:01:57]Maybe it is seven, maybe it's not seven, but just in general, you need to think because pure water is neutral, the concentration of H+ it's going to be equal to what? It's going to be equal to the concentration of OH- So, in order to find the concentration concentration of H+, which is the concentration of OH-, you should do what? You should do a radical KW. So, to find my answer over here, I need to do a radical KW, which is the radical 4.0 * 10 ^ -14. Put it on your calculator, I got an answer of 2.0 * 10 ^ -7 molarity.

[00:02:30]So, this is what? This is the concentration of H+ and the concentration of OH- of water at this specific temperature. So, when this question comes in the exam, to find them, you just do a radical what? You do a radical KW, okay? We'll move on now to question two. What does it say in question number two?

[00:02:47]It says, "Calculate the concentration of H+ and OH- of a 0.20 molarity HCl aqueous solution."

[00:02:55]Now, let's take the HCl aqueous solution. Now, we know it will dissociate it into what? It will dissociate into H+ and Cl- and these are both what? They are both aqueous, right?

[00:03:07]If you balance them out, always remember to balance your equation, you will get one HCl, one H+ and one Cl-. That means the HCl aqueous and the H+ aqueous are in a one-to-one ratio. What does this mean? This means that the number of moles of HCl it's equal to the number of moles of H+, okay? And since we have the same volume, volume HCl equals volume H+, right?

[00:03:31]Then what can we conclude? If we divide number of moles by volume, number of moles by volume, we can say that the concentration of HCl, so I'll just write it like this, concentration of HCl it's equal to the concentration of H+, right?

[00:03:44]So, it's going to be what? So, it's going to be 0.20 molarity. Now, what about the OH-?

[00:03:51]What is the concentration of OH-? We go back to the previous equation, which was what? Which was H2O liquid dissociates into H plus, right?

[00:04:01]And OH minus, right?

[00:04:05]Now, if I'm dealing with H2O and H plus and OH minus, then I need to look for KW. If I look at my given and I don't see that KW is given, so KW is not given, what do I do? I'm going to assume that KW it's what? It's 1.0 * 10 to the power of minus um 14, okay? So, if KW is not given, then what do I do? I assume that KW is 1.0 * 10 to the power of minus 14.

[00:04:30]Okay, so now what do I do? I know that KW it's equal to the concentration of product. This aqueous, this is aqueous.

[00:04:36]So, the concentration of product to the power of coefficient times concentration of product divided by one, why? Because this is liquid. So, I have the concentration of H plus, which is 0.20.

[00:04:47]So, I got KW, which is what? Which is 1.0 * 10 to the minus 14. It's equal to 0.20 * the concentration of OH minus.

[00:04:55]So, the concentration of OH minus it's equal to 1.0 * 10 to the minus 14 divided by 0.20. Put it on your calculator, I got an answer 5.0 * 10 to the power of minus 14 molarity. So, this is the concentration of H plus and the concentration of OH minus, okay? We move on to question three, what does it say?

[00:05:14]It says the following table lists several solutions at 25 degrees. They're all at the same temperature. When they tell you that the temperature is 25 degrees, then you're going to assume that this is room temperature that they're talking about in most scenarios.

[00:05:26]So, you'll assume that KW is 1.0 * 10 to the power of minus 14, okay? Just to keep in mind. Now, look what they gave us. They gave us the concentration, right? What does this mean? This means that all of my substances have the same concentration, which is what? Which is 1.50 molarity. They said choose the solution that is expected to have the lowest pH. What does it mean if it has the lowest pH? That means it is the strongest acid. That means it releases H+ the most. So, uh greatest concentration of H+ released, right?

[00:06:00]So, our job now is to cancel out everything until we find the strongest acid. First of all, let's remove all bases. This is clearly a base because it has OH. NH3, you should know it is a base. Here, because this has NH2, it does not have the the ability to release a hydrogen. So, this is going to be what? It's going to be a base. So, now we're left with two acids, HNO3 and H2SO4.

[00:06:24]But, how do I compare between HNO3 and H2SO4? Remember, we said the one with the lowest pH would be the strongest acid, and it has the greatest uh So, now here, this is very important.

[00:06:36]HNO3, it's what? It's a monoprotic. That means it has one hydrogen. That means it's going to release one hydrogen, for example. But, then H2SO4, it has two hydrogens. So, while each HNO3 releases one hydrogen, H2SO4 releases two hydrogens. So, it releases a greater concentration of hydrogens. That means which one is the stronger acid? H2SO4 is the stronger acid. So, H2SO4 is expected to have the lowest pH. So, so this one has the lowest pH, okay?

[00:07:07]That is it for question number three.

[00:07:08]We'll move on to question four. What does it say? It says, "Calculate the concentration of H+ concentration of OH- the pH and the pOH of a 0.20 molarity NaOH aqueous solution." So, we have a list of things to find. Let's get started. First of all, we have NaOH aqueous, right? NaOH aqueous. Now, by the way, I think I did a mistake before. I wrote HCl partially dissociates. Please excuse me. HCl is a strong acid, so it fully dissociates.

[00:07:36]NaOH is a strong base, so it fully dissociates into Na+ and OH-. And these are both what? These are both aqueous.

[00:07:43]Excuse me for the mistakes. I'm very tired. I'm so sorry.

[00:07:47]So, what did they give us? They give us the concentration of NaOH, which is 0.20 molarity. Now, what is the ratio of NaOH to OH-? It's a 1:1 ratio. What does this mean? This means, first of all, that the number of moles of NaOH it's equal to the number of moles of OH-. Okay? The volume of NaOH, because they're in the same container, is equal to the volume of OH-. Okay? So, if I divide number of moles by volume, number of moles by volume, I get that the concentration of NaOH is equal to the concentration of OH-, which is what? Which is going to be 0.20 molarity. Okay? So, that's it for the concentration of OH-. How do we find the pOH? The pOH, very simply, what is the difference between pH and pOH? Do you remember how to calculate the pH? pH it's the minus log the concentration of H+ ions, right? Now, pOH it's the same exact thing, but this H stands for H+. So, instead of H, what do I put? I put OH-. So, it's going to be what? It's going to be minus log concentration of OH- ion, right?

[00:08:52]So, let's find the pOH. The pOH it's going to be what? It's going to be minus log concentration of OH-, which is 0.20.

[00:08:59]So, I got my pOH to be I think we're doing it to how many significant figures? To two significant figures. So, to two significant figures I got 0.70.

[00:09:09]Is there a unit for it? No, there's no unit for pOH and pH. So, I got my pOH to be 0.7.

[00:09:16]Now, they we still have to find the concentration of H+ and the concentration of OH-. There are many ways to do this.

[00:09:23]But, I prefer a single method. Okay? I'm going to try to explain as much as possible the many different methods, but I prefer this method because it's like very easy and very simple.

[00:09:33]They said calculate the concentration of H+, OH-, pH, and pOH. Let's find the pH.

[00:09:39]If you read, they did not specify anywhere for you KW. Okay, if they did not specify for you KW, so you will assume it to be 1.0 * 10 to the power of minus 14. Now, remember, we have H plus gives me pH, OH minus gives me pOH, KW, what will it give me? It will give me pKW.

[00:09:58]pKW is equal to minus log the KW. Okay, so you can put it on your calculator and you will get an answer of what? You will get an answer of 14. So, now, I'm looking for what?

[00:10:09]I'm looking for the pH. You should know that the pH plus the pOH it's equal to what? It's equal to the pKW.

[00:10:16]So, I'm looking for a pH. pH is equal to the pKW minus the pOH. So, it's equal to 14 minus 0.70.

[00:10:24]So, I got an answer of 13.30. This is what? This is my pH. Now, how do I find the concentration of my H plus ions if I have my pH? I know pH is what? pH is minus log concentration of H plus. But, concentration of H plus it's equal to what? It's equal to 10 to the power of minus pH. So, 10 to the power of minus my pH, which is 13.30. I put it on my calculator. 10 to the power of minus 13.30. I got an answer of what? Of the two decimal places, 5.0 * 10 to the power of minus 14 molarity. Um there you go. This way I found the concentration H plus, I found pH, I have uh pOH, and I found the concentration of OH minus.

[00:11:06]Now, another way you could do it is that you could use KW and you can say that uh KW is equal to the concentration of OH minus times the concentration of H plus.

[00:11:15]Then you find the concentration of H plus, which you will get to be 5.0 * 10 to the minus 14. And then using this and you do minus log H plus, uh you will get what? You will get the pH. There's so many ways to do it, but as long as you understand the main concept, you're going to be fine. Okay? That's it for question four. We'll now move on to question five. What does it It tells us a solution R contains nitric acid HNO3 and a pH of 1.800.

[00:11:38]A 60 ml, this is the volume, so sample of solution R is diluted with distilled water to a final volume of 180 ml. They told us to calculate the concentration of hydrogen ions in the solution R. So, the solution R is the original solution.

[00:11:52]It has a pH of what? It has a pH of 1.800, right? They want what? They want the concentration of H+. How do I get the concentration of H+ if they gave me the pH? Very simply, the equation is 10 to the power of minus pH. So, the concentration of H+ is going to be 10 to the power of minus 1.8, right? Put it on your calculator and I got an answer of 1. Here they're doing three decimal places because of here. So, 1.

[00:12:20]58 * 10 to the power of minus two molarity. This is the concentration of H+. Okay?

[00:12:26]This is part A. Part B, they said determine the new concentration of H+ in the solution. Look what I have. I have This is concentration of H+ initially.

[00:12:34]Okay?

[00:12:35]Um then, I have volume initially and volume finally. And I'm dealing with what? I'm dealing with the same exact acid. I did not change it.

[00:12:44]So, uh what does this mean? They want me to find the new concentration. So, I have concentration initially times volume initially. This is my initial number of moles. It must be equal to because I'm using the same acid and I did not change the number of moles. It must be equal to concentration finally times volume finally. Okay? This is after distillation. So, if I'm looking for concentration finally, concentration finally is equal to concentration initially times volume initially divided by volume finally. So, it's going to be concentration initially, which is 1.58 * 10 to the power of minus two times volume initially, which is 60 * 10 to the power of minus three because it's in milliliters, you need to change it to liters. Then, you're going to divide it by what? You're going to divide it by the uh final volume, which if you divide by 1,000 0.180 L 0.1800 L.

[00:13:34]Uh, So you put it on your calculator and I got an answer of 5.27 * 10 to the power of minus three molarity. That's it for part B. Now on part C they said calculate the pH of solution S. Remember pH is equal to minus log concentration of H plus. So it's going to be minus log the concentration of solution S which I found three significant figures 5.27 * 10 to the minus three molarity. Uh, now here I forgot to mention something important. You need to count the number of significant figures 1 2 3. Because I have three significant figures my pH should be to three decimal places. DP means decimal places. So you put it on your calculator minus log 5.27 * 10 to the power of minus three. I got uh, So I got 2.27.

[00:14:19]Oh, excuse me I didn't see this. 2.278 as my pH to the correct number of decimal places. Now part D it says explain how dilution affects the pH of a strong acid solution. Let's look.

[00:14:31]Initially I had 1.800. This was my pH.

[00:14:35]Finally after dilution I got 2.27. So what happens? Um, the dilution of an acidic solution after we added the water after we diluted it what happened to my pH? It increased. So the dilution of an acidic solution increases the pH of the solution keeping the solution acidic.

[00:14:51]Okay? The concentration of the proton ions will decrease upon adding water which will result in an increase in the pH of the solution.

[00:14:58]That's it for question number five.

[00:15:00]We'll move on to question six. What does it say? A mass of 0.600 g of potassium hydroxide KOH is dissolved in 250 ml of solution.

[00:15:11]Okay? What do they want? They said that given that the molar mass of KOH. So let's write the given. Okay, I can do it like this that's fine. We have space we can start.

[00:15:20]They said calculate the concentration of OH- in the solutions. They said KOH is dissolved. So, let's write the because KOH is a strong base that will completely dissolve. So, KOH completely dissolves into OH- aqueous plus K+ aqueous, okay? So, again, 1 KOH 1 OH- they are in a 1:1 ratio. So, the number of moles of KOH is equal to the number of moles of OH- right?

[00:15:54]And because the volume of KOH is equal to the volume of OH- and you do number of moles divided by volume, you get that the concentration of KOH is equal to the concentration of OH-. So, let's find the concentration of KOH. First of all, we need to find the number of moles of KOH.

[00:16:10]We have the volume, right? This is our volume which is 0.250 L, okay? Excuse me. I need to move this 200 and 0.250 L.

[00:16:20]Now, my number of moles, we know that it's mass over molar mass. If it's not concentration times volume, okay? So, I have my mass which is 0.600 divided by my given molar mass which is 56.1. It's going to be what? It's going to be 0.0107 mol to three significant figures, okay?

[00:16:39]So, now that I have the number of moles and I have my volume, I'll be able to find the concentration of KOH which is the concentration of OH- which is what?

[00:16:46]Which is 0.0107 divided divided by 0.250.

[00:16:51]So, you will get an answer of 0.0428 molarity.

[00:16:55]Okay, that's it for part A. In part B, what does it say? It says calculate the concentration of H+ ions using the relationship between OH- and H+. Since they did not specify KW for us, we will assume that it's 1.0 * 10 to the power of -14. We calculated the concentration of OH- to be what? To be 0.0428 molarity, right?

[00:17:20]And we know that Kw is equal to the concentration of H+ times the concentration of OH-. So, if I want to find the concentration of H+ using the relationship between OH- and H+, it would be Kw divided by the concentration of OH-. So, it's 1.0 * 10 to the minus 14 divided by 0.0428.

[00:17:42]And it's going to be what? It's going to be I got an answer of 2.34 * 10 to the power of minus 13 molarity, okay? That's it for the question 6b.

[00:17:52]Now, see, they said they want the pH of the solution. You should know that pH is equal to minus log concentration of H+.

[00:18:00]So, it's going to be minus log. What's the concentration of H+? 2.34 * 10 to the minus 3. Because this is to three significant figures, my answer will be to three decimal places. So, I will get 12.631 as my pH. That's it for the question 7.

[00:18:16]We'll now move on to question Sorry, that was for question 6. We'll now move on to question 7.

[00:18:21]What does it say? The table below lists acids with their Ka value, okay? Write the expression for the acid dissociation Ka for the HCOOH and give its unit. So, HCOOH. First of all, because they gave us a Ka, that means we're dealing with what?

[00:18:36]We're dealing with a weak acid. So, it will partially dissociate into H+ and COO- H or No, excuse me, into H+ and HCOO-.

[00:18:48]Yeah.

[00:18:49]Excuse me, I was thinking of the compound. So, HCOOH partially dissociates into H+ and HCOO-.

[00:18:55]So, Ka, because this is aqueous and this is also aqueous and this is also aqueous, it's going to be the concentration of products. So, concentration of H+ times concentration of HCOO- to the power of their coefficients, of course, divided by the concentration of the reactants to the power of its coefficient. So, HCOOH, okay? They said, "Give its unit." So, this is basically m * m over m. m and m cancel, I get m.

[00:19:18]What's the unit of m? It's mol per decimeter cubed, okay? So, that's it for part A. What do they want in part B?

[00:19:25]They said, "Arrange the acids in order of pH from lowest to highest." So, the lowest pH it's going to be the strongest um acid. The highest pH it's going to be the strongest base, or let me write it in terms of acid because we're dealing with um KA, so it's going to be the weakest acid.

[00:19:41]So, here's something very important to know.

[00:19:43]The one with the lowest pH, which is the strongest acid, it should have what? It should have the highest KA value.

[00:19:52]The weakest acid, the one with the highest pH, it should have what? It should have the lowest KA value. So, basically, the greater my KA value, the stronger my acid, the more hydrogen it releases, the lower my pH, right?

[00:20:06]And this is the same for bases. The um stronger my base, the higher the KB value, the higher the pH, okay?

[00:20:16]So, let's look at the KA values and let's rank them from the um highest to lowest. So, this is the highest, then this. Please pay attention to the minus two, minus four. These are very important. This is how I noticed immediately. Now, I have to compare between these two, then comes this one, then comes this one. So, that means my strongest acid is H2C2O uh four, okay? It has the lowest pH.

[00:20:41]What comes after it, which is my my second strongest acid?

[00:20:45]It is the HCOOH.

[00:20:48]Then, what comes after it? Here we're comparing in terms of pH. So, this has the lowest pH, so that means it's the strongest acid, okay? Then, after it comes this one, then after it comes C6H5COOH.

[00:21:00]Then, after it comes what? Then, after it comes CH3COOH.

[00:21:06]Okay, I hope you understand that the higher the KA value, the stronger the acid, the more hydrogen it releases, the lower the pH, okay?

[00:21:14]Uh now, what does it say? Give the chemical formula for the conjugate base of the listed acid that's expected to have the highest KB value. Now, here you have to know that KB is directly or sorry, inversely proportional to KA.

[00:21:26]That means the higher my KA, the lower my KB. That means the highest KB value must have what? Must have the lowest KA value. The lowest KA value. So, let's look for the lowest KA value. This is the one that has the lowest KA value.

[00:21:41]So, this would have what? It has the highest uh KB value, right?

[00:21:49]So, let's update them into their appropriate one. First of all, let's draw the dissociation of the acid in general. It's CH3COOH.

[00:22:00]It partially dissociates. Of course, this is aqueous into H+ and CH3COO-.

[00:22:06]If this was my acid, that means this is what? This is my conjugate base. So, now let's write it in terms of my conjugate base. So, my conjugate uh conjugate base is CH3COO-. Of course, it's aqueous.

[00:22:16]Now, remember, whenever I'm starting with bases in my equation in my dissociation equation, I need to include the water. It's impossible for me to not include the water. I need to. It's a must to include the water.

[00:22:27]So, then what happens? The water will act as the acid. It will give a hydrogen to the CH3COOH. So, it partially dissociates and it will give me CH3COOH, right? Which is aqueous. And then, I took away a hydrogen from this. I'll be left with OH-, which is also an aqueous.

[00:22:44]Now, don't forget that this is water, which is liquid. So, KB would be equal to the concentration of products. So, concentration of OH- times the concentration of CH3COOH divided by the concentration of CH3COO-.

[00:23:00]Okay, we don't include the water because it's liquid. So, this is the value of my KB, okay? We move on now to question eight, grade. What does it say? It says, "Use the information below to answer part AI."

[00:23:11]So, look at the our acid. We have HOI, which has a KA of 2.0 * 10 to the minus 11. And then HOBR, which has an a KA of 1.0 * 10 to the minus nine. Just because they gave us a KA, this means that they are weak acids to start with, okay?

[00:23:26]Which of the following two acids is weaker, HOI or HOBR? Now, we know that the stronger the acid, okay?

[00:23:35]Stronger the acid, the higher the KA value, right?

[00:23:42]So, what does this mean? This means the opposite. So, the weaker the acid, the lower the KA value. So, let's look at the two values. You need to know and understand the numbers. So, between these two, this is the smaller number, okay? This is the lower KA value. That means HOI is what?

[00:24:04]HOI is the weaker acid. Why? Because its KA value is smaller than the KA value of HOBR. Now, II, it says, "Draw a complete Lewis dot I Sorry, a complete a complete Lewis electron dot diagram for the acid you identified."

[00:24:17]So, they want us to draw HOI.

[00:24:19]So, my central atom is going to be oxygen. Now, let's start. Hydrogen is which group? Hydrogen is in group one, so that's going to be one.

[00:24:27]Plus oxygen is in which group? Oxygen is in group six, plus iodine, which is in group seven, okay? 1 + 6 + 7 So, I should have a total number of 14 electrons, but pay attention. I used two electrons for this bond and two electrons for this bond. So, in total, that's minus four electrons. So, I'm only left with 10 electrons. Now, what do I do? You have to remember, hydrogen, it cannot exceed two electrons. It only goes up to a duplet. So, I have to spread out 10 electrons between what?

[00:24:56]Between iodine and oxygen. Now, I can them into their lone pairs. So, 10 divided by two I get five lone pairs.

[00:25:02]So, let's distribute. You always start with the surrounding atoms. So, this gets 1 2 3 4 5 6. That's already six gone. So, three lone pairs. I still have two, so I'm going to add them over here.

[00:25:13]So, this is what this is the Lewis electron dot diagram for the acid that you identified in part A I.

[00:25:18]Okay, we'll move on to triple I. What does it say? It says hypochlorous acid has the formula HOCl. Predict whether HOCl is a stronger acid or weaker acid than HOBr. Justify your prediction in terms of the chemical bond.

[00:25:30]Now, remember what we said before, right?

[00:25:33]Um that HCl is a weaker acid than HBr is a weaker acid than HOI. Because as we go down the group, the bond length increases, right?

[00:25:44]Because the bond length increases, why?

[00:25:45]Because the the iodine has a great number of electrons, so its electron cloud is really big, so it has a great uh atomic radius. Because of this it has a really long bond length, which makes it easier to break. If it's easier to break, then it's easier to release the um H+ ions, right?

[00:26:00]But then, we said when we enter introduce oxygen, it's flipped. It's the other way around. Why? Because oxygen, when we have oxygen in whatever we're dealing with, we're only going to be talking about the electronegativity. So, whenever we have oxygen, the only thing we talk about is the electronegativity.

[00:26:17]Okay?

[00:26:18]So, that means HOCl it's going to be what? It's going to be a stronger acid than HOBr. Why? First of all, let's read what they said. They said HOCl is a stronger acid than HOBr because the OH bond in HOBr is stronger than the OH bond in HOCl.

[00:26:32]Why?

[00:26:34]This is due to the lower electronegativity of Br compared to chlorine. So, with oxygen we're looking at electronegativity. Chlorine is more electronegative than Br. So, what's it going to do? It's going to attract the electrons to itself, okay? More strongly than uh the hydrogen. So, that what does does make? It makes the bond the OH bond very weak. Thus, I can release hydrogen very easy, which makes it a strong acid, okay?

[00:26:56]Now, Br is less electronegative, so it's harder. So, it's like less electron-withdrawing, okay? As a result, the electron density between H and O atoms in HOBR is higher, right? Meaning I have more electrons in between H and O, which makes it um a stronger and less likely to dissociate. However, with the chlorine, because the chlorine pulls the electron density away from the OH bond, it will weaken the OH bond and make it easier to release the proton, thus increasing its acidity. So, it's very logical. You just need to pay attention.

[00:27:24]When I have oxygen, I directly switch my topic to electronegativity, okay?

[00:27:28]Uh now, we'll move on to part B. What does it say?

[00:27:31]They said, "Write the equation uh that occurs between hypoiodous acid and water." So, they want the equation that occurs between HOI and H2O.

[00:27:40]Now, because it's a weak acid, it will undergo um partial dissociation to give us what? First of all, this is an acid.

[00:27:47]If that's an acid, then water acts as a base. So, the HOI will donate its proton, so it will turn into OI minus, okay? And then the H+ because it's a base, it will gain the proton, so it will turn into H3O plus. That's it for B.

[00:28:00]Uh C, it says, "A one molarity NaOI solution is prepared by dissolving NaOI in distilled water up to 198 K.

[00:28:09]The hydrolysis reaction, they gave us the hydrolysis reaction, OI minus aqueous plus H2O liquid.

[00:28:15]I believe this is equilibrium. And equilibrium will give us HOI aqueous plus OH minus aqueous occurs, okay?

[00:28:21]Write the equilibrium constant expression for the hydrolysis reaction that occurs between OI minus and H2O liquid.

[00:28:28]Very simply, they want us to write the um equilibrium constant. So, first of all, equilibrium constant is K. I need to know am I dealing with an acid or a base? This over here is the con conjugate base. How do I know? This was my equation. Acid, then this is the result of my acid. The result of my acid is the conjugate base, and the result of my base is the conjugate acid. So, OI minus is the conjugate base, that means I'm dealing with KB. It's equal to concentration of product, so HOI times concentration of OH- divided by what?

[00:28:55]Divided by concentration of reactants that are not liquid or solid, only aqueous or gaseous. So, OI- aqueous.

[00:29:02]So, that's it for part C. Okay? We'll move on now to question number nine.

[00:29:06]What does it say question nine?

[00:29:09]It says very simply, uh calculate the pH of a 0.30 HNO2 solution and they gave us the value of KA.

[00:29:16]>> [clears throat] >> So, you should know that HNO2 is what?

[00:29:18]It's a weak acid, so it will dissociate.

[00:29:21]HNO2, when it dissociates, it dissociates partially of course, it will give us what? It will give us H+ and NO2-, right? H+ + NO2 -.

[00:29:32]Okay?

[00:29:33]Uh now, look what they gave us. They gave us the initial concentration of HNO2, right?

[00:29:39]By the way, how did I know it was a weak acid? Because they gave us KA. Whenever they give you KA, then you know immediately it's a weak acid.

[00:29:46]So, let's do the ICE table. This is something very important topic. I C E, right? What's the initial concentration of HNO2? They gave it to me 0.30 molarity. For H+ it was zero, for NO2- it was zero. Okay?

[00:30:00]KA, which is my equilibrium constant, okay?

[00:30:04]Uh we need to compare it to Q. What is Q? Q is the uh initial constant, okay? So, it's going to be 0 * 0 over 0.30, which is what? Which is zero. And here you can see that Q it's what? It's smaller than KA. Because Q is smaller than KA or KA is larger than Q, then we can say what?

[00:30:25]Then we can say that the forward reaction will proceed. So, my reaction will move forward. That means I'm going to lose some of my concentration here and I'm going to gain concentration here. Now, of course, you have to pay attention to the coefficient. This is one, this is one, this is one, that's why I always multiply my X by one.

[00:30:43]So, here I'll end up with what? I'll end up with a 0.30 - X with 0 + X or + X and here I'll end up with + X. So, at equilibrium, K it's going to be equal to what? It's going to be equal to X squared divided by 0.30 - X and it's equal to what? It's equal to 4.0 * 10 to the power of -4. Now, before we do anything, we need to remember something that we can do in the AP, which is that we can assume this X to be negligible. Let's see what happens if we assume to be negligible.

[00:31:13]If we assume this X to be negligible, we would before that we'd need to find the percent dissociation, percent alpha.

[00:31:19]Percent alpha is equal to X, which is X divided by the initial concentration C * 100.

[00:31:25]>> [snorts] >> So, let's find X first of all to start with. KA, okay, sorry, X squared over 0.30 now because we're assuming that this is zero is equal to 4.0 * 10 to the power of -4. So, 0.30 * 4.0 * 10 to the power of -4, I will get that X squared is equal to 1.2 * 10 to the power of -4.

[00:31:43]So, you do radical on both sides and I got X to be to the correct significant figures, which is two, 0.011 molarity. Now, what do I do? I'll find the percent dissociation, okay? So, 0.011 divided by the initial concentration, which is 0.3 * 100, okay?

[00:32:02]So, 0.011 divided by 0.3 * 100, I got my percent dissociation to be 3.67 and because it is less than 5%, so yes, I can say that my assumption is valid. I can assume that it is negligible. Why? Because my percentage is less than 5.

[00:32:18]So, now what do I do? I have the the the value of X. I found X to be 0.0 11 molarity, right?

[00:32:26]They want me to find the pH. I look again at my given. I know that H+ the constant the equilibrium concentration of H+ is equal to what? It's equal to X.

[00:32:34]So, X is equal to 0.011 molarity. Now that I have the concentration of H+, let me find the value of my pH. pH is equal to minus log concentration of H+. So, it's going to be what? It's going to be minus log the concentration of H+, which is 0.011. You put it on your calculator and I got my concentration to be correct number of decimal places because I'm dealing with how many significant figures? Two significant figures, so it should be to two decimal places. I got it to be 1.96, okay? So, that's it for question nine.

[00:33:05]We'll move on to question number 10.

[00:33:06]Okay, so this is question 10. So, after this I'm going to take a break. So, I'm going to post the first half and then I'll come back with the second half because this took a long time. I'm so sorry.

[00:33:15]Anyways, um it's not required, but I will still do it. Find the pH of 0.20 molarity H2SO4 solution given that KA is that HSO4 minus.

[00:33:24]H2SO4 is a strong diprotic acid, so it completely dissociates into what? Into its um ions, which are H+ and HSO4 minus, okay?

[00:33:35]Now here, this is going to have a concentration of 0.20, 0.20, and 0.20.

[00:33:39]Why? Because this is in a 1:1:1 ratio and because it completely dissociates.

[00:33:43]Now, HSO4, they gave us a value uh HSO4 minus. They gave you a They gave us a value of the KA, that means it's a weak acid. So, it will dissociate into H+ and how much? And uh the SO4 2 minus. Now here, we need to apply the ICE table, okay? So, ICE. Initially, I have 0.20 H2SO4 and pay attention, initially I have 0.20 H+, but I have initially 0 SO4 2 minus. So, what happened?

[00:34:13]I have KA, it's 1.2 * 10 ^ -2. Let me find Q. Q, it's going to be the initial equilibrium or sorry, the initial uh constant. So, the 0 * 0.20, the concentration of products divided by the concentration of reactants, which is 0. So, I got Q is less than KA, that means my forward reaction will take place.

[00:34:31]That means here this loses. Okay? Pay attention to the coefficient. This loses value, it loses concentration, this gains concentration, and this gains concentration. So, I'll end up with 0.20 - X, 0.20 + X, and then + X.

[00:34:46]So, KA it's equal to what? It's equal to concentration of SO4 2- which is X * 0.20 + X which is the H+ divided by 0.20 - X. Okay? And it's going to be equal to what?

[00:35:01]It's going to be equal to the KA which is 1.2 * 10 ^ of -2. Now, what we could do is we could assume that this X is negligible. So, let's assume this X is negligible. So, then it's like I have X * 0.20 + X / 0.20.

[00:35:16]So, how would I solve it? I'd bring this to the other side, and I would be left with X * 0.20 X. So, 0.20 X + X ^ it's going to be equal to 0.2 * 1.2 * 10 ^ of -2. So, I got 2.4 * 10 ^ of -3. So, I'll move this to the other side, and then I do multiply 3 1 0.20 and then -2.4 * 10 ^ of -3. So, I got my first value of X to be to two decimal places 0.011.

[00:35:47]And then my second value it's what? It's negative, so I reject it. I don't take the negative value. So, this is a possible option. So, 0.011 / 0.2 * 100. If it's greater than 5, then my hypothesis is or like my my assumption is rejected. Okay?

[00:36:03]And I got it to be 5.5. It is greater than 5, so I will reject my assumption.

[00:36:08]I'm going to say bye-bye to this. I cannot assume that X is 0 anymore. So, I need to include X. Okay? So, all this work I did here, gone. Okay? Imagine as if I did not um uh remove X. Okay? So, now I'd have to move this here. I'd be left with 0.20 X + X ^ it's equal to 0.2 * 1.2 * 10 ^ -2 that's 2.4 * 10 ^ -3 -1.2 * 10 ^ -2, right?

[00:36:39]Oh, please don't forget the x. So, now you can move them to the other side, okay? So, then you'll have x ^ 2 + 0.20 + 1.2 * 10 ^ -2x, okay? Uh you bring this to the other side, this will turn into so, plus uh all of it x 2.4 * 10 ^ -3. So, on my calculator I do mode 5 3. I have 1 + open bracket 0.20 + 1.2 * 10 ^ -2.

[00:37:11]And then for c I have what? For c I have 2.4 * 10 ^ -3. So, I get I get two values, one which is negative, so I immediately reject it. And I get another value of x. I got my value of x to be 0.0107.

[00:37:26]Now, take it exactly as it is, 0.0107.

[00:37:30]So, actually 0.11 again. So, all of this process is the same as if you neglected the zero, but it's fine. You still need to do this whole process. Imagine that came in the written part or in the actual AP, you need to do this whole entire process. You need to show your work. You need to show that you understand, okay?

[00:37:46]Now, actually why is it important I did this? Because when it when I come back here and they ask me to find the concentration of HSO4 minus, I cannot say that I found it and then I considered this to be negligible. No, that would be very wrong. Uh here I'd need to consider the x when I am finding the concentration of HSO4 minus, okay? Anyways, this is not what they're asking. They're asking for the pH. So, to find the pH I need the concentration of H plus. The concentration of H plus is 0.20 + 0.011.

[00:38:15]So, 0.22 significant figures 21 molarity. So, the pH it's what? It's it's a minus log concentration of H plus, which is 0.21.

[00:38:24]So, put it on your calculator and I got 0.68 as my pH. That's it for question 10. I'll see you in the next video. I'm going to cut it