AP chem Ams part 2 qs (20-38)

Added: 2026-06-04

[00:00:01]Hello guys, how are you? So, this is the second part of the um quiz. Hopefully, I'll be able to finish in uh less time than the first part. So, without further ado, let's get started. So, first thing we notice that they told us that the nitrate ion, NO2-, has a KB of 1.4 * 10 to the -11. And they're asking for the pH. Okay, when they give us the KB, first thing we should know that they're telling us that the NO2 will dissociate but as a base. And when it dissociates, then we need to add water to it. So, NO2, which is aqueous, we add to it water, which is H2O. Sorry, let me fix it. H2O, which is liquid, it will dissociate partially because there is a KB value, and it will give us what?

[00:00:43]First of all, with the base, the water will turn into OH-, and it will give its H+ to what?

[00:00:50]My writing, this is NO2-. Yeah.

[00:00:52]And it will give its H+ to the NO2-, so it will turn into NO2H.

[00:00:59]Okay? Uh so, now, what are you going to do? You're going to do the ICE table.

[00:01:03]So, this is ICE.

[00:01:06]Initially, this was 0.10. This is negligible because um it's water. Okay, it's liquid. Okay? Here, this is zero and zero. Here, this - X, this + X + X.

[00:01:18]Here, this turns out to be 0.10 - X.

[00:01:21]This turns out to be + X. This turns out to be + X.

[00:01:24]Now, remember, because they did not give us pH, we can assume this to be zero.

[00:01:29]But, of course, you need to check the percentage dissociation.

[00:01:33]So, now, anyways, what would KB be? KB would be equal to the concentration of NO2H * concentration of OH- / concentration of NO2-, which is X * X / 0.10, which is equal to 1.4 * 10 to the power of -11. So, it would be what? It would be um 0.10 * 1.4 * 10 to the power of -11. This is equal to what? This is equal to x squared. So, x squared is going to be equal to 1.4 times 10 to the power of minus 12. So, then x is equal to what?

[00:02:06]x is equal to a radical answer. So, that's going to be 1.18 times 10 to the power of minus 6. Now, how do you find the percentage alpha? Percentage alpha is equal to x over c, where c is the initial concentration. Okay, so it's going to be 1.18 times 10 to the minus 6 divided by 0.10. They gave it to us over here.

[00:02:27]10 to the minus 6 divided by 0.10. And this would give me And don't forget to multiply it by 100. So, this would give me what? This would give me 1.18 times 10 to the power of minus 3%. So, this is for sure less than 5%. That means yes, my assumption is what? My assumption is valid. I can resume.

[00:02:47]I can move on from this step now.

[00:02:50]So, now that I have the value of x, this x because I am dealing with KB, will it give me the concentration of H+ or OH-?

[00:02:58]No. This x would give me the concentration of OH-, but they want the pH. So, what I could do is I could find the value of pOH. pOH is equal to minus log concentration of OH-. So, that would be minus log 1.18 times 10 to the power of minus 6. So, that would be 5.

[00:03:20]You needed to two significant Sorry, to two decimal places. So, 5.92.

[00:03:25]Or let's say 5.93 if you round it properly. Now, this is not what they want. They want pOH. Since they did not tell us the value of KW, they did not say that it's a different temperature, then we can say that the the p of KW, okay?

[00:03:44]It's going to be what? It's going to be 14, okay? So, pKW is 14. What is pKW? We know first of all that pOH it's minus log concentration of OH minus. pK pKw is minus log concentration of Kw, which is what 1 * 10 ^ -14. So, then it would give you 14, right?

[00:04:03]And anyways, what's the relation? The relation in order to find pH, pH would be pKw minus pOH.

[00:04:11]So, that would be pH is equal to 14 minus 5.93.

[00:04:16]Put on your calculator 14 minus 5.93 and I got this as an answer 8.07 and yes, this is my answer. That's it for question number 20. We'll move on to question 21. What does it say?

[00:04:29]So, what does it say in question number 21?

[00:04:32]Hypoiodite ion, IO minus. Oh, sorry.

[00:04:35]Excuse me. Let me go back. Has a Kb. So, hypoiodite ion Yeah. Sorry.

[00:04:42]Has a Kb of 4.3 * 10 ^ -4. They want the pH of what? 0.80 molarity of IO minus.

[00:04:50]Now, IO minus, we know it's a base because they gave us the Kb. And when they give us Kb, then they want us to write the dissociation reaction. So, IO minus, you add to it water. And remember, this is aqueous, this is liquid. It will partially dissociate and the water will give you OH minus and it will give the H plus to what? To the IO minus to a plus um HOI or HIO, it does not matter, okay?

[00:05:12]Uh and this is also aqueous, this is also aqueous. So, yeah. Now, you need to do the ICE table. Okay, so 1 2 3.

[00:05:20]Uh ICE table.

[00:05:22]Okay?

[00:05:23]My initial concentration of IO minus was given to us as 0.80.

[00:05:28]Here, this is negligible because it's water. Here, we don't need to do anything cuz it's the sign.

[00:05:33]Okay?

[00:05:34]Um here, OH minus, initially it's zero.

[00:05:37]HIO, initially it's zero. Here then, it's plus X plus X. Here, it's minus X to plus X plus X. Okay? plus X plus X.

[00:05:46]Okay?

[00:05:48]because they did not give us the OH, we could assume this X to be negligible, this X to be zero. Okay? So, now KB it would be what? It would be concentration of HIO times concentration of OH minus times Sorry, divided by concentration of IO minus. So, it would be X times X over 0.80 equal to 4.3 times 10 to the power of minus four. So, X squared is equal to 4.3 times 10 to the power of minus four times 0.80.

[00:06:15]0.80.

[00:06:17]I got 3.44 times 10 to the power of minus four. Then you do a radical on both sides. So, I got X to be what? X is equal to 0.018.

[00:06:30]185. Okay?

[00:06:32]Now, first of all, before I proceed, I need to find my percentage um wait.

[00:06:41]My percentage Oh my goodness, I forgot the word.

[00:06:45]Oh yeah, my percentage dissociation.

[00:06:47]It's going to be what? X over C. X in this case is 0.0185 over 0.80 times 100. Don't forget to multiply by 100. So, 0.0185 divided by 0.80 times 100. It gave me 2.3%, which is less than five. That means yes, this is a valid assumption. I can proceed.

[00:07:09]So, now they want to find the pH. I have X and X is the concentration of OH minus. So, the concentration of OH minus it's equal to X, which is 0.0185 molarity.

[00:07:22]You could do it like the way I did in the previous one, which is that you find pOH. Okay? And because they did not give you pKw, you will assume that it's 14. Okay?

[00:07:32]You will find pOH and then pH is equal to pKw minus pOH. Or you could do it in this way. You know that Kw, if they don't mention it to you, it's 1 times 10 to the power of minus 14. You have concentration of OH- you can find concentration of H+ because you know concentration of H+ * concentration of OH- is equal to KW. So, concentration of H+ is KW. What's KW? It's 1 * 10 ^ - 14 divided by concentration of OH- which is what?

[00:07:59]Which is um 0.0185.

[00:08:03]So, 1 * 10 ^ - 14 / 0.0185.

[00:08:08]This gave me an answer of 5.4 21 * 10 ^ - 13. Now, you're going to do minus log answer and on my calculator I will get what? I will get 12.27 very simply and this is my answer, okay?

[00:08:28]That's it for question 21. Let's move on to question 22.

[00:08:32]Okay, now we can done with these 10 trillion questions that took us like 5 hours each.

[00:08:38]Now, the following table lists several solutions at 25°. First of all, we know they're at the same temperature all of them and they all are at 25°. They said which of the following 1 mol per decimeter cubed? So, they all have the same concentration is expected to have the lowest pH. The lowest pH means that it is the strongest acid. So, we basically need to look for the strongest acid. First of all, anything with OH we cancel it out uh except for the COOH, except for the organic ones because these are acid. So, this is an acid and um this is what? This is for example a base a base um a base because they are with OH. Now, this is an organic acid.

[00:09:19]Um it's a carboxylic acid CH3COOH compared with HCl and you should know that HCl for sure is a stronger acid than CH3COOH. You should know that all all organic acids are weak. They are not strong at all. So, which one would have the lowest pH? It would be the strongest acid which is HCN. Okay, we'll move on now to question 23. What does it say in question 23?

[00:09:43]Okay, X is a solution of HNO3 aqueous that has a pH of 1. A sample of 50 cm cubed, so this over here is what this over here is our volume, is diluted with water, so now this is volume initial and this is now volume final to form solution Y. What is the pH of solution Y? So very simply, look what we can do.

[00:10:04]We can find the concentration of H+.

[00:10:07]What's the concentration of H+ using pH?

[00:10:09]It would be 10 to the power of minus pH.

[00:10:12]So it would be 10 to the power of minus 1, which is what? 0.1. So the concentration of H+ is 0.1 molarity. I have volume initial, I have concentration initial, I can find concentration final using the equation C1V1 = C2V2. Why? Because we diluted it with water.

[00:10:28]So it will be 0.1 * 50 over 100 or no, 50 over 1,000.

[00:10:34]So it's going to be C2. What is C2? C2 is what we need to find times 200 over 1,000 because you need to put them in the proper units. So then it will turn 0.1 times 50 over 1,000 divided by 200 divided by 1,000. So then I will get an answer of 0.025.

[00:10:53]So this is concentration final of H+.

[00:10:56]Okay, so concentration H+ final.

[00:10:58]So now I can find the pH again by doing minus log concentration of H+. So minus log 0.025 on your calculator, I got an answer of 1.602.

[00:11:11]Okay, why did I put it to uh three decimal places? Because um if you look at the smallest significant figure, it's three significant figures. So I should put it to three decimal places, not three significant figures, but three decimal places, okay? We'll move on now to question 24. What does it say in question 24?

[00:11:30]Question 24, what does it say? An aqueous solution of nitric acid HNO3 aqueous has a pH of 3.5. So now we can find the concentration of H+. This would be minus 10 to the power or sorry, 10 to the power of minus pH. So 10 to the power of minus 3.5. This gave me an answer that the concentration of H+ is 3.16 * 10 to the power of minus four, right? Now, they told us that the solution is diluted 10 times.

[00:12:00]So what does this mean? This means that the volume Okay, hold on. That means concentration one volume one equals concentration two volume two.

[00:12:10]So concentration one over concentration two is equal to volume two over volume one. So concentration one is equal to 10 times concentration two.

[00:12:20]So we need to find concentration two.

[00:12:22]Concentration two is equal to the initial concentration which is 3.16 * 10 to the power of minus four divided by 10. So it's going to be I think it's 3.16 * 10 Yeah. So it's going to be 3.16 * 10 to the power of minus five. And then what can I do to find the pH? To find the pH, I'm going to do minus log concentration of H+. So minus log 3.16 * 10 to the power of minus five and then I got 4.5. Now, another way you could do this is because it's diluted 10 times.

[00:12:57]Whenever the solution is diluted by 10, then the pH increases by one. So every time the solution is diluted by 10, then my pH will increase by one. So I just add one to my pH to get the answer.

[00:13:09]Okay, we'll move on now to question 25.

[00:13:11]What does it say in question 25? So we'll move on now to question 25. It has a mass of 0.600 g of KOH is dissolved in 150 cm cubed. This is the volume. And the molar mass of KOH is given to us.

[00:13:25]They want the concentration of H+. Very simply, using this I will be able to find the concentration of OH minus because KOH is a base, right? Now, how do I find the concentration of OH minus?

[00:13:35]Very simply, I need to find the number of moles. I know number of moles is mass over molar mass, but I also know number of moles is concentration times volume.

[00:13:41]So, I could get rid of this and I could say concentration times volume is equal to mass over molar mass. So, concentration is equal to mass over molar mass times volume. Now, I substitute my number, so 0.6 over my molar mass, which is 56 times my volume.

[00:13:55]Of course, you need to put it in the proper units. So, this is 150 cm cubed.

[00:13:59]It turns in It turns into 150 divided by 1,000. So, I get um the concentration of OH minus to be 0.0714.

[00:14:11]Okay? They want the concentration of H plus though. So, you should know that KW is equal to concentration of OH minus times concentration of H plus. So, concentration of H plus is equal to KW, which is 1 * 10 to the minus 14 divided by the concentration of OH minus, which is 0.0714.

[00:14:30]So, then 1 * 10 to the power of minus 14 divided by 0.0714.

[00:14:35]I got an answer, which is 1.4 * 10 to the power of minus 13 mol per decimeter cubed. And this is my answer over here. We're going to move on now to question 26. What does it say in question 26? A mass of 0.350 g. This is the mass KOH. This is the volume, which is 200 * 10 to the power of minus 3 dm cubed and this is the molar mass.

[00:14:59]Again, they want Oh, you didn't want the concentration of H plus, they want the pH. So, let us find the concentration of H plus in order to find the pH. First of all, I can find the concentration of OH minus. How do I do that? Very simply, um you know number of moles is mass over molar mass and it's also concentration times volume. So, you get rid of this.

[00:15:17]So, concentration times volume is equal to mass over molar mass. So, you move this down. So, now concentration is equal to mass over molar mass times volume. You substitute the proper units 0.350 / 56 * 200 / 1,000, and I got the concentration of OH- to be 0.03125.

[00:15:36]Okay? But, as you can see, the smallest is to three significant figures, so you should change it to three significant figures, so 0.0313.

[00:15:43]Now, how do you find the concentration of H+? KW is equal to concentration of OH- * concentration of H+. So, concentration of H+ it's equal to 1 * 10 to the power of -14 / 0.0313.

[00:15:57]So, I got concentration of H+ to be 3.1 uh 9 * 10 to the power of -13.

[00:16:06]Okay?

[00:16:07]So, now what am I going to do? Because it's to three significant figures, I need to find the pH. pH is -log concentration of H+.

[00:16:15]So, -log 3.19 * 10 to the power of -13, and this would be uh 12.495.

[00:16:28]I got 12 I got 12.496.

[00:16:31]It doesn't matter as long as you get the appropriate answer, and you put it to three decimal places cuz you have three significant figures. So, the correct answer is 12.495 very simply, okay? We'll move on now to question 27. I'll move on to question 27. What does it say? A solution is prepared by dissolving This is the number of moles. This is our volume.

[00:16:50]But, pay attention. They gave us KA, right? But, they gave us KA of HSO4-.

[00:16:56]So, so, please pay attention. H2SO4 it starts as a strong acid that should not have a KA. It starts and it fully dissociates into uh H+ and uh HSO4-.

[00:17:09]What did I do? Very simply, I took the H away because it's an acid that's an electron donor, okay?

[00:17:16]Now, because it fully dissociates, the uh concentration here should be the same here and should be the same here. So, here it would be the number of moles, which is 5 * 10 ^ -2 / the volume.

[00:17:26]Okay? So, it's going to be 0.333 molarity. So, this is also 0.333 molarity and this is 0.333 molarity.

[00:17:34]But, after being consumed, there will be no more HSO4 left, so the concentration here is going to be what? It's going to be zero. Now, the second dissociation, which is the weak acid dissociation, that's why it has a KA, is the HSO4 minus, which would give me what? Which would give me the H+ because it's a proton donor and the SO4 2- Now, we can apply the ICE table.

[00:17:58]So, ICE, right?

[00:18:00]Initially, H2SO4, we knew it was how much? We knew it was 0.333, right?

[00:18:06]And we know initially as well, the concentration of H+ is 0.333, whereas SO4 2- it is zero. So, here we subtract X here, we add X. Uh this, by the way, it's not an auto, excuse me, it's partial dissociation. So, -X +X +X.

[00:18:21]So, this turns into 0.333 -X and 0.333 +X and then here we have +X. Now, KA, you should know that it's concentration of SO4 2-, which is X, times concentration of H+, which is 0.333 +X over concentration of HSO4 -, which is 0.333 -X.

[00:18:43]Um which would give me what? Which would give me 1.2 * 10 ^ -2.

[00:18:50]Now, remember something really important over here. Because we need to find the pH, we can assume that this over here is what? This over here is negligible.

[00:18:58]Okay? So, then, look what we could do.

[00:18:59]We could bring this upwards. Okay?

[00:19:03]So, now I will have 0.333x + x^2 is equal to 1.2 * 10 ^ -2 * 0.333.

[00:19:12]That would be 3.996 times 10 to the power of minus three. So, then I move it to the other side and I solve quadratic. So, mode five three. Okay, this would turn into minus over here equals zero.

[00:19:25]Uh mode five three, my first value is one, my second value is 0.333, my third value is minus 3.996 times 10 to the power of minus three.

[00:19:36]So, then I will get uh 0.0115 Actually, no. Let me round it to 0.0 116. Okay, 0.0116.

[00:19:49]So, now I have the value of X. Let me check the percent dissociation of alpha.

[00:19:53]So, 0.0116 divided by 0.333 times 100% to make sure that my uh assumption is valid. Oh my goodness, it's okay.

[00:20:03]0.116. No, 0.0116 divided by 0.333 times 100. I got 3.48%.

[00:20:11]It is less than 5%. So, yes, I can assume I can say that my assumption is valid. So, now I can say that my X is 0.0116.

[00:20:22]So, this is the concentration Is this Sorry, is this the concentration of H+?

[00:20:26]No. The concentration of H+ is 0.333 plus X. And what's X in this case? It's 0.0116. So, 0.333 plus 0.0116.

[00:20:38]That would be uh 0.03 or 0. Sorry, 0.3446.

[00:20:45]Now, this over here is the concentration of H+. And how do I find the pH? pH is equal to minus log concentration of H+.

[00:20:54]So, that would be minus log 0.3446.

[00:20:58]Put it on your calculator. Minus log 0.3446.

[00:21:02]I got 0.463.

[00:21:06]And the most closest value to this is 0.400. I don't know why they got 0.400, but what I got and what AI got was 0.463, okay? Now, we'll move on to question 28.

[00:21:20]What does it say in 28?

[00:21:23]A student is asked to calculate the concentration of H+ and the concentration of C3H5O3- in a 0.200 molar concentration solution of this, and they gave us the value of KA, right?

[00:21:38]So, here we need to we need to write the dissociation. So, HC3H5O3 partially dissociates into H+ and C3H5O3- okay?

[00:21:50]Uh we partially dissociate it like this.

[00:21:53]So, we have ICE, right?

[00:21:56]Um here the initial concentration of this was 0.200. Here, this was initially zero, initially zero. So, plus X plus X minus X. So, 0.200 minus X plus X plus X, right? Now, we can assume that this is negligible because they did not give us the pH, okay? So, we are going to assume that it is negligible, okay?

[00:22:18]Uh so, now what can we do? We can find the value of X. How do we find it? It's concentration of C3H5O3- um which is X times concentration of H+, which is X over concentration of this, which is 0.200, is equal to KA, which is 1.38 * 10 to the power of minus four.

[00:22:37]You move it up, and you will get that X squared is equal to uh 1.38 * 10 to the power of minus four times 0.2. So, it's equal to 2.76 * 10 to the power of minus five. So, X is equal to a radical of this value, which is what? Which is 5.25 * 10 to the power of minus three. This is the value of X. Now, is this value of X the concentration of H+? Yes, it is. There you go. And is this value of X the concentration of C3H5O3-?

[00:23:08]Yes, it is. There you go. Okay, now we'll move on to question 29. What does it say?

[00:23:15]What is the percentage dissociation of solution of 1.50 mol dm to the power of -3 HCOOH?

[00:23:23]Okay, now I'm going to stop doing the full explanation. I explained enough.

[00:23:27]It's been like 28 29 questions. I've explained the full process on how to do it. Now, I'm going to give you the very quick brief explanation. When they ask you for X, which is the concentration of H+, okay? It would be radical KA times the concentration the initial concentration that's given to you. If they ask you for X, which is the concentration of OH-, it's going to be radical KB times the concentration that they gave you initially, okay?

[00:23:56]Please, of course, pay attention that you're dealing with the proper terms.

[00:24:00]So, you're dealing with an acid, you need to find H+, you use this. You're dealing with a base, you need to find OH-, you use this. Now, to find the percentage dissociation, okay? It's going to be radical KA times the concentration over the concentration times 100 or radical KB times the concentration over the concentration times 100. It depends on what you are dealing with.

[00:24:22]And then, of course, in order to find the pH, you have to do minus log concentration of H+. And in order to find the pOH, you have to do minus log concentration of OH-. This is all you need to know. These few equations, they will help you. For example, look here.

[00:24:36]They want the percentage dissociation of a solution 1.50 mol per dm cubed HCOO2H, right?

[00:24:43]What is the equation I will use? I have KA, I have concentration. I will use this equation for the percentage dissociation. It is equal to a radical KA, which is 1.8 * 10 to the power of -4 * 1.5 / 1.5 * 100. Let's put it on our calculator. So, radical 1.8 * 10 ^ -4 * 1.5 / 1.5 * 100. And I got 1.10 So, 1.10%.

[00:25:10]So, this is it for question 29. We'll move on now to question 30.

[00:25:14]What did they ask from us? Look here, very simply.

[00:25:18]What did they say? They said consider 0.40 molarity NaOH aqueous solution. So, we have NaOH aqueous solution, right?

[00:25:27]NaOH fully dissociates, right? Whatever happens to NaOH, it will fully dissociate into OH- and Na+. That means the concentration here is the same as the concentration here is the same as the concentration here. So, 0.40. Here now this is 0.40. This is here also 0.40. But since we consumed that, this is now what? This is now zero. So, they want the concentration of OH- at 0.40.

[00:25:49]Now, they want p OH.

[00:25:53]How do you find pOH? Very simply, it's minus log of OH+. So, minus log 0.40.

[00:25:59]Put on your calculator minus log 0.40.

[00:26:02]You will get as well again 0.397, which is approximately 0.40 to two decimal places.

[00:26:09]Now, they want the concentration of H+.

[00:26:10]How do you find the concentration of H+?

[00:26:12]They did not specify for us Kw. So, we'll assume that Kw is 1 * 10 ^ -14.

[00:26:19]So, concentration of H+ * concentration of OH- is equal to Kw.

[00:26:25]So, Kw over concentration of OH- is equal to concentration of H+.

[00:26:31]So, uh the concentration of H+ it's 1 * 10 ^ -14 / 0.40, which is the concentration of OH-. I got 2. uh 50. Where'd they get the one from? Wait.

[00:26:47]Actually, you know what? Since they wanted to be very accurate, let Let find the pH first. How do I find the pH?

[00:26:53]Okay, I know that pKw - pOH is equal to what is equal to pH. pKw is always 14 if they don't mention it to you. - pOH we found it to be 0.40. It will give us 13.6.

[00:27:05]Now I have my pH. How do I find the concentration of H+?

[00:27:09]Uh it will be 10 to the power of - pH, which is 10 to the power of - 13.6, which is approximately 2.51 to three significant figures, 2.51 * 10 to the power of - 14. So that's it for a question 30. We'll now move on to question 31. What does it say?

[00:27:27]Calculate the percentage dissociation.

[00:27:29]What did we say the equation was?

[00:27:30]Percentage dissociation in one step.

[00:27:34]It's uh radical. They gave us Ka. So Ka * initial concentration over initial concentration * 100. So it's going to be radical 3.5 * 10 to the minus eight times four times 10 to the power of minus six over four times 10 to the power of minus six times 100. Don't forget that. So radical 3.5 * 10 to the minus eight times four times 10 to the minus six divided by four times 10 to the minus six times 100. This gave me 9.35%.

[00:28:02]There you go, all in one step. Now question 32.

[00:28:06]What do they want? They said a sample of 0.0720 mol of HOCl is dissolved to make 100 ml solution. Okay, calculate the concentration of all species present in the solution. Uh because they gave us a Ka, we know that this is not a It's not a fully dissociation. So we know that HOCl will partially >> [clears throat] >> Excuse me. Dissociate into H+ and OCl- So we need to apply the ICE table.

[00:28:37]So ICE The initial concentration of HOCl, we can find it because they gave us the number of moles and the volume. So the concentration is equal to number of moles over volume. So, 0.0720 / 100 * 10 ^ -3 cuz you need it in liters. So, 0.0720 / 100 * 10 ^ -3. This is 0.72 molarity.

[00:29:02]This is zero. This is zero. Then here, -x +x +x, right?

[00:29:08]So, here you get 0.72 -x. Here you get +x +x. Here you take this to be negligible. Now, instead of continuing and doing the whole whole process, you should know one very simple step, okay?

[00:29:20]That first of all, you found the concentration of HOCl. It's whatever you found over here, 0.72. But, they wanted to three significant figures, so it would be 7.20 * 10 ^ -1.

[00:29:34]Now, we need to find x. x is the concentration of OCl- and the concentration of H+. You should know that x is equal to radical KA * C directly. So, radical 3.5 * 10 ^ -8 * C.

[00:29:50]What is C? C we found it to be 0.72, right? So, radical 3.5 * 10 ^ -8 * 0.72.

[00:29:58]It gave me an answer of 1.59 * 10 ^ -4 to three significant figures for both of them, so that's the answer. That's the value of x, okay? We'll move on now to question 33. What does it say? They want us to calculate the value of KB.

[00:30:14]Okay, let's see.

[00:30:16]The pH of 0.50, this is the concentration, is found to be this much, okay?

[00:30:21]Uh so, we can now find the concentration of H+.

[00:30:25]Okay, actually here they helped us. They gave us the dissociation equation.

[00:30:29]So, they want us to find KB. Let's find it. We need to do the ICE table.

[00:30:32]I'm so tired of doing ICE table.

[00:30:35]If I do the ICE table one more time, I'm actually so exhausted, okay? So, uh let me just make it like this. All right, let me separate them now. Water is negligible as always because it's a liquid, okay?

[00:30:48]Now, the initial concentration of this was 0.50. This was zero, this was zero.

[00:30:53]This is minus X, this is plus X, this is plus X, right?

[00:30:57]So, now this is 0.50 minus X. The X is negligible because they did not give us Oh, no, never mind. They gave us the pH, so the X is not negligible. There we go.

[00:31:07]Look, that's a very good example. The X is not negligible because they gave us the pH. And this is plus X, this is plus X. Now, look what I'm going to do. I'm going to play it smart. I can find the pOH and then I can find the concentration of OH minus, right? They didn't mention They mentioned to me that it's at 25°. Perfect. So, that means pKw is equal to how much? It's equal to 14.

[00:31:27]Now, pOH is equal to pKw minus pH. What is pKw? 14 minus 12.22.

[00:31:36]14 minus 12.22, that's going to be 1.78.

[00:31:40]So, pOH is 1.78.

[00:31:44]Now, how do you get the concentration of OH minus? It's very simply 10 to the power of minus pOH. So, 10 to the power of minus 1.78. 10 to the power of minus 1.78. So, it's going to be concentration of OH minus is equal to 0.166, okay?

[00:32:02]Molarity. So, that means X is equal to 0.166 molarity, okay? So, then let me put it over here. Let me substitute 0.166.

[00:32:11]0.166.

[00:32:13]0.5 minus 0.166, that's going to be 0.

[00:32:17]334.

[00:32:19]So, now how do you find Kb? Kb it's going to be concentration of OH minus, which is 0.166 * 0.166 / 0.334.

[00:32:29]So, I got Guys, I just noticed I did a mistake.

[00:32:33]That's why my answer was wrong. So, look what I'm going to do. 0.15 - X here, this is + X here, this is + X. Excuse me, X is not this. X is 0.0166.

[00:32:45]So, basically 0.01 Yeah, 0.0166.

[00:32:49]So, 0.0166.

[00:32:51]Excuse me, that's a mistake because I copied it wrong. So, substitute it here.

[00:32:55]0.0 166. 0.0166.

[00:33:00]Now, 0.15 - 0.0166, I got 0.4834.

[00:33:07]So, now what's KB? KB is going to be concentration of OH- which is 0.0166 times concentration of whatever this is.

[00:33:14]So, 0.01 66 divided by 0.4834 which is the concentration of this. And I got Now, to two significant figures, I got it to be 5.7 * 10 ^ -4.

[00:33:31]I think because they want it to two significant figures, I should change this to two significant figures, right?

[00:33:38]But no, my pH was to two decimal places.

[00:33:40]That means yes, to two significant figures.

[00:33:43]Uh that means that 0.017, 0.017.

[00:33:47]Then 0.50 - 0.17, that's 0. Sorry, 0.017.

[00:33:55]That is 0.400 for 0.48.

[00:33:59]And this is 0.017, 0.017. I'm trying to get the exact answer. I'm trying to figure out how they got the exact answer. So, 0.17 * 0.17 / 0.48.

[00:34:11]Um Sorry, I keep on repeating the same mistake. 0.017.

[00:34:17]Okay?

[00:34:19]No, it still did not work. I think they used the exact value. So, 0.01654, 1657 for example.

[00:34:29]Um and then 0.01657.

[00:34:32]And then here 0.3 0.483.

[00:34:36]Yeah, I'm getting close to the approximate answer. So, look, in the exam, use the exact values. Please do not do not What's it called?

[00:34:48]I forgot what's Do not round them up.

[00:34:50]Use the exact values, okay? Otherwise, you will get it wrong. So, here it's 5.5 * 10 ^ -4. We'll move on to question 34.

[00:34:58]Okay, an aqueous solution of potassium hydroxide. So, here look what they gave us. They gave us C, they gave us KW.

[00:35:08]Oh, no, look, this is KOH. Okay, KOH will fully dissociate into K+ and OH-.

[00:35:15]So, this concentration is the same for all of them, except here because it's zero, that means it's fully dissociated.

[00:35:20]So, here it's 0.010 0.0010, okay? So, what would be the concentration of H+? If they gave me KW to be this, okay? Very simply, concentration of H+ is equal to KW over concentration of OH-. So, here it would be what? It would be 1 * 10 ^ - 14 / 0.0010.

[00:35:40]So, I got 1 * 10 ^ -11. Now, how do I find the pH? Very simply, the pH is minus log concentration of H+. So, minus log uh whatever this is, and you will get what? You will get um 11 as an answer, and of course, put it to the correct decimal places.

[00:35:57]Question 35. The table lists several solutions at 25°, all the same temperature, and all the same concentration. They want the lowest pH.

[00:36:04]That means they want the strongest uh acids, right? Now, let's see. This is not an acid, this is not an acid, this is not an acid. We're left with this and this. Now, you should know you should if you want look up the list of strong bases Sorry, not strong bases, strong acids, okay? You will see this is not an option. The only diprotic strong acid is H2SO4. Remember that the only diprotic strong acid is H2SO4 and that's what we are left with HNO3. Now, why is it our choice in part A? First of all, because it's a strong acid. Second of all, it donates one proton per formula unit. This is just us explaining our choice. Why is it First of all, why did we choose it? Because it's a strong acid and our choice donates one proton per formula unit, okay?

[00:36:48]We'll move on now to 36.

[00:36:50]What does it say in question 36? At 50° C, the dissociation constant of water is the So, now they changed Kw, okay?

[00:36:58]They want us to calculate the concentration of H+ and OH-.

[00:37:01]Okay, look. When you are dealing with water, you should know that concentration of H+ of water, this is only only only in water. It's equal to the concentration of OH-. Why? Because the pH of water is 7, right? So, concentration of H+ equals concentration of OH- equals radical Kw. What is Kw?

[00:37:20]It's 5.5 * 10 ^ -14. So, then two two significant figures, concentration of H+ and concentration of OH- are 2.3 * 10 ^ of -7. That is 36. We'll move on to 37.

[00:37:34]Again, same thing. They gave us Kw for water. They changed it, though.

[00:37:38]Concentration H+ in water equals concentration OH- equals radical Kw. So, radical 4.0 * 10 ^ of -14, it will give you 2.0 * 10 ^ of -7. So, um basically, this is it for question 37.

[00:37:58]And last but not least, the last and final question. Thank God, we finished. Okay, consider 0.0300 molarity. This is the concentration of HNO3. We can get the concentration of H+. Now, we know that HNO3 from the previous question is a strong acid. That means it fully dissociates into H+ and NO3-. Um so, here this has a concentration >> [clears throat] >> which is the same as this of uh as the concentration of HNO3 because it's fully dissociated which is 0.0300 right? And this is also 0.0300. So the concentration of H+ is 3 * 10 to the power of minus 2.

[00:38:36]Now how do you get the concentration of OH- since they did not mention the temperature or anything we'll assume KW is 1 * 10 to the power of minus 14. So concentration OH- is equal to KW over concentration of H+. So it's going to be 1 * 10 to the power of minus 14 divided by 0.0300 and it gave me 3.33 * 10 to the power of minus 13. Perfect. That's it for this video. Thank you guys for watching. I hope you enjoyed and have a wonderful day.