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UrUpSasked to prepare a pH = 3.00 buffer solution start- 17.25 You are from 1.25 L ofa 1.00 M solution of nitrous acid (HNO,) ing amount you need of sodium nitrite ...

Question

UrUpSasked to prepare a pH = 3.00 buffer solution start- 17.25 You are from 1.25 L ofa 1.00 M solution of nitrous acid (HNO,) ing amount you need of sodium nitrite and any (NaNO)): (a) What is the pH of the nitrous acid solution prior to add: ing sodium nitrite? (b) How many grams of sodium nitrite should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium nitrite is added.

urUpS asked to prepare a pH = 3.00 buffer solution start- 17.25 You are from 1.25 L ofa 1.00 M solution of nitrous acid (HNO,) ing amount you need of sodium nitrite and any (NaNO)): (a) What is the pH of the nitrous acid solution prior to add: ing sodium nitrite? (b) How many grams of sodium nitrite should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium nitrite is added.



Answers

You are asked to prepare a $\mathrm{pH}=3.00$ buffer solution starting from $1.25 \mathrm{~L}$ of a $1.00 \mathrm{M}$ solution of hydrofluoric acid (HF) and an excess of sodium fluoride (NaF). (a) What is the $\mathrm{pH}$ of the hydrofluoric acid solution prior to adding sodium fluoride? (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

We are asked to prepare a buffer solution with a ph of 2.5. We are starting with 1.50 leaders of a 0.75 Mueller solution of Hydrofluoric acid. We can use any amount of sodium fluoride. Our first question. I should probably move this down a little bit too right. There is what is the ph of the initial solution? So that is the 0.75 moller H. F. And that one's pretty easy to figure out. I actually figured out using quadratic and not using quadratic and got the same number. So I think I will do the non quadratic. So the K. Let me go ahead and write the equation. We look at the K. A. For this. The K A. Is 6.8 times 10 to the -4. Okay. The initial concentration as we were given is Where is it? 0.75 archangel B minus x plus x and plus X. So you get 075 minus x. X. And X as my equilibrium concentrations. So our K A will be equal to these two concentrations divided by this concentration. So we will get 6.8 Times 10 to the -4 equals X squared over 0.75 minus x. I'm going to highlight the portion that we are going to basically take off and I can do this. I did this as a quadratic. I got the same answer both ways. So for this equation solving for X, I get X equals Going to find out where I wrote. I've got so much writing I can't read my own writing. 2.26 times 10 to the -2. And that is the Mueller more clarity of my hydrogen I on. So if I take the negative log of 2.26 Times 10 to the -2 more clarity, I get 1.65 p And that's my answer. For part A part B. Let me get over here. What is, how many grams of sodium fluoride Are required to get the ph that we were given of 2.5. I also did this 1-2 ways. I did it with Henderson, Hasselbach and I did it like the old fashioned way. I don't know what to call it, the old fashioned way. I got the same answer within four hundreds of a gram. This was answered in your text and they only reported to one Sigfig. And I'm looking at my answers. I should have had this reporter to to sig figs And the two sig figs. It was the same answer. So I'm just gonna go ahead and do this. I'm trying to figure out which way I'll do it with Henderson. And for that I'm going to rearrange the equation for a moment and I'm going to end up with the log of my sodium fluoride concentration over my HF concentration. Yeah will equal my ph this is a bit rearranged plus the log of my k. So I've got the log of this again equals 250. Was a given ph Plus the log of 6.8 times 10 to the -4. And this site equals zero point negative negative zero point 6675. This is the same right there. So now if I um get rid of that log and go like this I get 0.215 equals And I know my HF was 0.75. So my concentration of my sodium fluoride Well 0.215 time zero point 75. And that equals 0.1613 Mueller and A. F. So now we're very close. Um 0.1613. So 0.1613 moles. Any F. Times of the Molar mass was 41.99 grams per mole. And we had I believe won 50 bulls per liter. Let me verify. It was 1.5. Looking for all my work here, 1.5. And here I got 10.15. And to to sig figs. That's one 0.0 Times 10 to the first g of any F. And often you'll see that written like that we're done

In this question, We are hops to prepare. Um, A P h, um, three points total buffer. So ph off offer. He's 3.0 Starting from walk 0.2 liters. Online to fight liter off one by 00 Moeller I don't flooring Carson on in here months. You need all of us. City Council. Right. Okay, so x sodium. All right, So what is the ph off the Hydrofluoric acid solution trial? Too hot in the sodium fluoride. So the ph of the 1.25 leaders off one point Darrell Mohler each have so so 1.0 more. That means we have ah. Won't find zero mole in ah, one child's on meal or one leader. Then put it that way. One liter. So therefore we have ex mole in ah, won't want you five liters show Daddy sell. We're going to survive more, right. So Texas won't want to five more. So the more clarity all the solution is, uh, of the solution is of course, one finds a role. Um, Lulu and ah, this solution is, uh, the forecast. A solution. So when it's higher Nice. You have Ah h boss uh, also half minus. So if you have ah, one finds you no more off. Uh, I don't Florrick Acid. Let's do it like this. So we have the concentration we are in this year. Oh, wait. We have, Ah, a studio highs. And this year, concentration on duh. They find our concentration. So initially we have Ah, we have ah, won't find you Don't know. And then we have zero off this. And now we have zero days. So then the concentration of the minor sex breathe for sex and this river plus X then the total will be Won't find you told minus X And this is X and this is fixed. Silly K movie. Each loss concentration F minus concentration divided by concentration of Hydrofluoric acid and the cable he each have 613 times turn restore power. Miners fall, and I'll be concentration of ex off h boss's ex. Much of Iran concentration of have minors, which is, uh, ex. If I didn't buy war, find zero minus x. So if we saw for hex. So the hex. Well, give us, uh, it's going to be square. Okay. So has to give us, uh, miners 010 to 54 and, uh, 0.0 248 So we're gonna go with days because concentration cannot be negative. Therefore, we can calculate speech. Since we noted concentration of H boss pH. It's going to be negative. Longer name off h plus. And from here we said that ah x is age villas and every stage Balsillie, whichever tea uh so excessive wasn't hexes have minus or the concentration of beach blows on F Miners are the same. So we find extra visit opens their old 2 to 4 hate and my nose there above the ridge UF off 254 and would be cause I don't 0.24 heads because concentration cannot be negative to that music concentration off H policies 0.248 So dumb it's ah Ph. Remember, negative longer them off 0.0 to 4 age and I will give horse what's want 1606 So this is the pH before hardy in sodium fluoride. Be said are many grams absolute floor. I should be handed to her fear to prepare the buffer solution off PS 3.0. So we know the pH is, uh, peak it here, plus logon name off the, um, fluoride. I don't Florida concert for the study on full. Right again. But, Allie All right. So the peach will want his 3.0. It's inquest in Topeka. He off sodium off the buffer. He saw three point year plus longer team off. We're looking for the concentration of sodium fluoride we don't know. And, uh, concentration of Hydrofluoric acid. He said I want my 25 more because, uh, explaining here that, uh, concentration is, well, 0.0 and, uh, 1.0 moller. And that's what by more than one liters. And we have I won't 10.25 liters. So that 1.25 later, we contain 1.25 more. So since we're trying to determine the amounts of the studio, all right, and needs to be hearted. We need to know we need you used a number of more off the Hydrofluoric acid eyes present. So if we solve this, go ahead. And, uh, this will be my nose at a 0.2 is because to log off X will 1.25 and, uh, this to be, UH, 0.631 He's in quest of ex who won't want to. Fine the f o X will be equals to 0.7 89 Oh, and that is the number of more off sodium fluoride. It's just a number of more off the city in full right. Therefore, we can calculate a master sodium fluoride by multiplying this number off more biting Mama's oxygen flow. Right? So well, I must study on Fluoride is 42 grounds for more careful must off sodium fluoride. Maybe you close to 42 grounds for more sign 0.789 more and that will be 33 points. Want three heads? Graham, also the uncle right?

We are asked to consider the preparation of a buffer solution with benzoate acid and sodium benzoate. Yeah. Our goal is to prepare a ph ph 4.0 solution. We have 1.5 zero leaders of A 0.0200 Mueller's solution of Ben's OIC acid. Which I'm going to abbreviate B A. For Ben's OIC acid because I was out a room there. Step # one. The first thing we're asked to do is find the ph of the solution of the acid before the salt is added. And to do that we're going to do excuse me. A rice table. In my first reaction, I'm just going to write Ben's awake acid cheat a little bit here and we're going to produce our H plus and benzoate fuck ion. My concentrations are my initial concentrations 0.020000. Our change will be plus X and minus X. Giving me these values the K. A. In your text for Ben's OIC acid I believe was Let me see if I can find this here. 6.2 times 10 to the-. No that's wrong. What is it? 63 times 10 to the -5. Okay we'll need that value and recall that the K. A. I'm sorry about that is equal to my H. Plus concentration which was X. In my anti and concentration which is also X. Divided by the concentration. All right. Of the acid. Okay so I'm just gonna plug my values in here and I have 63 times 10 to the -5 equals X squared over 0.0200 -1. And then I'm going to put this into a nice quadratic format and I will have that's my X. Right there minus 1.2. 6 times 10 to the -6. I started solving these my graphing calculators so I never worry if I have to skip them anymore. I have X. Was equal to 0.001127 Mueller. That's my more clarity. And then even take the negative log of 0.00 1127. And I got a ph of 2.95. 2.95 is the ph so looking for the ph our answer was 2.95. 2.95. Okay. Part two. How many grams of sodium benzoate do we need to add for my bumper in order to do that? We're going to use the whole Henderson Hasselbach thing. Yes. And then write the base over the acid concentrations. So the ph that we were looking for let me see if I can find that again. Here was 40 I think I wrote the wrong ph don't nobody brought the right ph down. So I'm gonna substitute. 400 equals the negative log of. That was 6.3 times 10 to the -50. What over? Plus the log of my base concentration Over my acid concentration. Which we were given add 0.0 Mueller. Excellent. So I'm going to get the log of the base Over 0.0200 Equals 4.00 plus the log of 6.3 Times 10 to the -5. Yeah. Okay. So that got me see if I've got my number here. This part We got .201 Equals 0.201. And that'll be the log of my base concentration Over .0-00. And then to get rid of the log here, I'm gonna have my base concentration Over 0.0200 equals tend to the negative 0.201, solving for the concentration of my base. I got 0.126 mol. Aren? T. Then it was a simple calculation to convert that more clarity into how many grams I needed. And let's switch colors to do this. Let's go with the red. So the mass needed is equal to or more clarity. Forgot to zero there. And I'm gonna write my more clarity as moles per liter. I multiplied that by my 1.50 leaders. Which when we run to the beginning page, we were given right here times the molar mass of my benzoate sodium bit. Ben's weight was 1 44.11 grams per mole. And I ended up with 2.72 g. That was it.

So now we'll work on problem 26 from Chapter 17 in this problem, were asked to prepare a P H equals four buffer starting from 1.5 leaders of 0.0 to Mueller solution of Ben's OIC acid and any announed we need of sodium benzoate. So the first thing they want to know is what is the pH of Ben's OIC acid solution prior to adding sodium benzoate? So first, let's go ahead and draw the K ey for Ben's OIC acid. This is equal to 6.3 times 10 to the minus five and now we know that benzo gas It is a weak acid so we can draw nice table generi guys. Table were the acid reacts with water to form the contra git base and the Hydra Roni um Ion. So we have an initial value of 0.2 00 and no initial acid or germanium base or hydro. So we subtract X ad X and X. Can you get 0.2 minus x x and X? So, we said, are equally real constant equal to, uh, ex squared divided by 0.2 minus x And so if we cancel out this negative X and we solve for X, we get X is equal to 1.12 times 10 to the minus three. And from there we can take the negative long to get a pH is equal to 2.95 So the second part of this problem in part B, they ask us how many grams of sodium benzoate are needed to prepare this butter with pH of 4.0 And we can neglect the volume change from adding so the event. So we can go ahead and set up our hands and hostile buff equation, with the pH being equal to 4.0 which is equal to P K, which is 4.20 plus the log of base, which we don't know over acid, which is 0.2 So we can arrange this prop question here to be equal. So to be negative 0.2 is equal to the log of base over 0.2 And so then we have 0.63 If we continue on here, is equal to the concentration of base over Sarah points there to this is after we took the anti log. And so the concentration of base So Ben's away is equal to 0.13 more. Yeah, So this is the concentration here. So we're going to multiply this concentration more per liter times the volume of solution, which is 1.50 leaders. And then we can move supply by Muller Mass to get Graham. So 1 41.11 grams per mole is the Mueller massive sodium benzoate. So we see that most will cancel, leaders will cancel and we'll get grams. And the value for that mass is 2.73 grands.


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