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A 10.0 -mL solution of $0.300 \mathrm{M} \mathrm{NH}_{3}$ is titrated with a $0.100 \mathrm{M} \mathrm{HCl}$ solution. Calculate the $\mathrm{pH}$ after the following additions of the HCl solution: (a) $0.0 \mathrm{mL}$ (b) $10.0 \mathrm{mL},$ (c) $20.0 \mathrm{mL},$ (d) $30.0 \mathrm{mL},$ (e) $40.0 \mathrm{mL}$

(a) 11.36 (b) $9.55 .(\mathrm{c}) 8.95 .(\mathrm{d}) 5.19 .(\mathrm{e}) 1.70$

Chemistry 102

Chapter 16

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria

Aqueous Equilibria

University of Central Florida

Rice University

University of Kentucky

University of Toronto

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.

29:13

Consider the titration of …

11:59

18:40

in this problem, we are asked to consider the titillation of 30 ml of a 0.030 moller ammonia. And that's the ammonia With 0.25 molar hydrochloric acid. The amount of hydrochloric acid varies. For each of the six trials for which we have to do a calculation. Our job is to calculate the ph For each of six situations and I'm just going to start with these and we'll do them one at a time letter A. We have what we were told here are 30 0.30 moller ammonia And we have 0 mm of hcl. Before we get going too far I would like to get the concentration the number of moles for each trial of harmonia. We're going to take our leaders of ammonia times are concentration to get nine times 10 to the -4. Should have written this somewhere else. I'm gonna write this right here. 9.0 times 10 To the -4 malls. And that concentration will be the same for each and every trial that we do. Okay, I wrote with a really light pen here. So I hope this shows up for trial number one, we are going to write down our chemical equation. There we go. And since we have no I hcl added, we're gonna go ahead and do a quick ice and we're just going to go with more clarity on this one, 030 and zero and 0 oops. So our equilibrium concentrations will be 0.030 -1 X. And X. Going to make a comment for this and all of our equations. We're gonna ignore this because there's going to be relatively small. Which you can prove if you'd like. Okay, so we're good there. Now I'm going to write our KB expression are whoops. Our KB expression for this reaction will be NH4 Ammonium concentration times the hydroxide concentration divided by the ammonia concentration. We can look up the KB for this And it is 1.8 times 10 to the -5. From our equilibrium concentrations we get X squared over 0.030. I might have one more zero there solving for X. I got 7.348 Times 10 to the -4 Mueller. And that is my ohh concentration. To find my ph That will equal 14 minus the negative log of 7.348 Times 10 to the -4. And I got 10.87. And that's our first of six long calculations, let's do let her be for let her be we are adding 10 mL of our 0.025 Mueller hcl Let's figure out our concentration. We're gonna take our 0.0100 leaders times or more clarity and that will give you my concentration of my acid Which was 2.5 times 10 to the -4. And that's for my hcl So next I'm gonna write my acid base reaction. I'll put my our states on this one. But I'm not gonna put my states on all of them. And let's go ahead. Our concentration for this, Every trial is 9.0 Times 10 to the -4. We just calculated that our initial concentration, It's 2.5 Times 10 to the -4 and we have zero here. My change will be the smaller of these two numbers. So it'll be business 2.5 Times 10 to the- for us -2.5. That was 10 to the -4 and plus 2.5 Times 10 to the -4. So my equilibrium concentrations will be 6.5 times 10 to the -4, Zero and 2.5 times 10 to the- for us. From here. My next step will be to determine the polarity. So determine the modularity of the NH three the Ammonia. I will take 6.5 Times 10 to the -4 balls. I'm going to divide that by my total volume, which will be my initial volume, Which was 30 ml and my volume of asset. So this one will equal where is it? 1.6, 2, 5 times 10 to the -2. And then I'm gonna do the same calculation 2.5 times 10 to the -4, divided by this time, I'm just gonna write 0400 liters. And this week. Well 6-5 times 10 to the 3/3 moles or polarity. Okay now I'm going to write down the What we're going to use for many of these calculations here. We're gonna use this equation and h. four our K. Expression for this will be rearranging this to solve for H. Plus. So we can find ph we will have our K. A. Times are and I should have a plus right there. Okay and then last but not least we have to figure out what our K. A. Will be. Arcadia is going to equal R. K. W. Divided by R. K. B. Which we know. So our K. A. Will equal one times 10 to the minus 14 divided by R. K. B. Which we already looked up 1.8 times 10 to the -5 That will equal 5,56 Times 10 to the -10. That's our K. A. Now I could do a nice table for these but I wanted to mention that um this value here is so low that we're not going to be considering our changes that are defined as plus X and minus X. So for the first one I'm going to write in what we would get for an ice for the initial change equilibrium and then I'm not going to do that again. Okay, so for this problem let's go ahead and substitute whatever he used. So I've got my In age four and the numerator is 6.5 Times 10 to the -4. I'm not going to close that parentheses yet And I've got by N. H. For plus And when we're looking at our initial and um Change items here. My NH three Which is my 6.5 will be plus X. And this will be a minus X. But we're going to ignore that the plus X. And the minus X. In this and every other calculation. So from now on I'm going to ignore these two. And solving for my ex. And this problem let her be was at 6.25 or 6.5 I think it's 6.25 here I think I copied that down. Run 6.25 times 10 to the -4. We verify I made a couple little airs here. I copied down the wrong numbers. This one should be 1.625 Times 10 to the -2 And this one should be 6.25 Times 10 to the -3. There we go. Now I got the right numbers down here and when we're ready to solve this I got 2.1 385 Times 10 to the -10. It's too many sick things but that's okay and then I'm gonna take the negative log of that number To find my ph so my ph is 9.67. That's two out of six. Okay lets do see And let her see in this one we have 20 millilitres of our hydrochloric acid added. So our concentration Or are M.olds will be zero 200 leaders Time 0.025 balls per liter. And that will give me 5.0 times 10 to the -4 malls. We're gonna set this problem up pretty much the exact same as your last one. I'm gonna leave my units off this time are my states off. And In every trial this is 9.0 Times 10 to the -4. We just calculated that this is 5.0 times 10 to the -4, zero and not applicable. Our change will be the lower of these two numbers. So to be negative By zero times 10 to the -4. And our equilibrium concentrations will be as I'm writing right here. Now for each of these I'm going to switch colors into a purple. I'm going to divide each of these numbers by the total volume And that will be 20 plus 30. So 0.0500 leaders, 0.0500 leaders. And this will give me my molar itty. This is 8.0 Times 10 to the -3. And here we have 1.0 Times 10 to the -2. And those are polarities for these two substances. Now we're going to use our same calculation that we used on the previous page. Let's go to something different um about a blue. So we're going to calculate Rh plus calculus concentration. And first we'll take the K. A. Which was 556 Times 10 to the -10. Then we're going to take our ammonium concentration. We're gonna leave the units off. And then in the denominator I'm going to have my moughniyah concentration and my H plus concentration when I saw for this is 6.95 Times 10 to the -10 Mueller. Take the negative log of this number And I got 9.16 as my ph three out of six. Done. Okay let's switch here for deed. We have 35 mm of hcl. So our concentration for this one will be 0.0350. I hope so. I should probably put my leaders on there Time 0.025 moles per liter. And my concentration for letter C. Was this letter dino. This is letter deep For letter d. 875 Times 10 to the -4 moles. And again this time I'm going to abbreviate even more. I'm gonna have my n. 0. 3 concentration. My Hcl. Gonna leave my water off and just put my NH four plus. That's all I care about As every trial. This is 9.0 times 10 to the -9 or tomatoes. 4th This one is 8.75 Times 10 to the -4. And this is zero. So my change will be the smaller of these two numbers. Just eight point seven -8.75 Times 10 to the minus forests -8.75. It was meant in the mines 4th and one more time with the plus Which gives me an equilibrium concentration of 0.25 times 10 to the minus forest, zero and 8.75 Times 10 to the -4. These are balls. So to get my polarity, I'm going to divide each of these by my total volume. Which will be 35 plus 30 Should be 0.065 and 0.065, 0.065. And that one equals three 846 Times 10 to the -4. And this one equals 1.346 Times 10 to the -2. And that's more clarity. Now we're gonna do the same calculation we've been doing for h plus concentration that will equal are okay, multiplied by our and each four it was too and divided by our and each street 3.846 Times 10 to the -4. And this time I got 1.946 Times 10 to the -8. Taking the negative log of this number. I got 7.71 as my ph two More. We're gonna make it for let her e We have 36 mm And let's go ahead and do our 0.0360 leaders Time 0.025 moles per liter And I got 9.0 Times 10 to the -4. But you can see where this one's going. These are the only species I care about right now. And we have the same concentration for each of these. which means that I'm gonna have zero and 9.0 times 10 to the -4. Okay then we're going to divide this quantity by my total volume to get our concentration. And I got about one 364 times 10 to the -2. And we can put that into our calculator. This will be quantity X. And my O. H. Will be X. So let's do this for this one. I'm just going to use my K. A expression and this will be equal to X squared And we'll put that over our one 364 Times 10 to the -2. For this one I saw for X. I got 2734 Times 10 to the -2. And I did the negative log of this number And I got a ph of 5.53. We have one more to do and I think I can do this one on this page. So I don't have to do a whole another page. So for app. For this one we had 37 mm And that will be 0.0370 Leaders Times 0.025. And that gave me 9.25 times 10 to the -4. Okay, so NH three Hcl and at age four Here as every trial I have 9.0 times 10 to the -4 Here I have 9.25 Times 10 to the -4 and zero. So we're going to decrease by our lower of the two numbers and I'm going to have and excess of 0.25 Times 10 to the -4 Hcl. If I divide that by my volume of zero point 067 leaders, I will get a concentration at 3.731 3.7 31 Times 10 to the -4. More clarity. And that's in terms of HcL So they've got an excess of a strong base. This time I'll just take a negative log of 3.73. 1 Times 10 to the minus 4th did I say, Yep. And that equals 3.43 for my ph congratulations, you made it to the end

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