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5.(12 pts) A buffer was prepared by dissolving 0.40 mol of benzoic acid (Ka 6.3 10-5) and 0.50 mol of sodium benzoatc in sufficicnt purc watcr tO form 1.00 L soluti...

Question

5.(12 pts) A buffer was prepared by dissolving 0.40 mol of benzoic acid (Ka 6.3 10-5) and 0.50 mol of sodium benzoatc in sufficicnt purc watcr tO form 1.00 L solution. To this solution was added 30.00 mL of 2.00 M HCI solution_ What was thc pH of the ncw solution? CsH,COOH(aq) + H_O() < H,0' (aq)+ C,H,COO (ag)

5.(12 pts) A buffer was prepared by dissolving 0.40 mol of benzoic acid (Ka 6.3 10-5) and 0.50 mol of sodium benzoatc in sufficicnt purc watcr tO form 1.00 L solution. To this solution was added 30.00 mL of 2.00 M HCI solution_ What was thc pH of the ncw solution? CsH,COOH(aq) + H_O() < H,0' (aq)+ C,H,COO (ag)



Answers

You prepare a buffer solution by dissolving $2.00 \mathrm{g}$ each of benzoic acid, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH},$ and sodium benzoate, $\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO},$ in $750.0 \mathrm{mL}$ of water. (a) What is the pH of this buffer? Assume that the solution's volume is $750.0 \mathrm{mL}$ (b) Which buffer component, and how much (in grams), would you add to the $750.0 \mathrm{mL}$ of buffer solution to change its $\mathrm{pH}$ to $4.00 ?$

Okay, We've got a buffer solution that contains Benzel like acid and sodium benzoate. Equal masses. We need to convert the masses into moles so we can solve for the pH. So P h is equal to p. K um benzo kassid plus a log of the moles of Ben's away, which would be its mass divided by its moulder mass. Divided by the moles have been to look acid, which is its mass 1.5 grams divided by its molar mass 122.12 grams. Get a pH just above 44.13 So more acid is required more than Selic acid would be required in order to reach a pH of four toe lower just a little bit. So we'll take the moles that this value is right here and moles that have been to look acid that we have and figure out how much more so Cassidy need to add to get a ph of four. They will carry out her math, and X will be, in this case, 0.4 to 4 moles of Benzel Kassid, which we could convert to massive required. But it didn't ask for math. Um, or another option would be to add HCL and convert some of the benzoate into themselves. Kassid. And so we would minus the moles eventuate we have by the moles of HCL we add and on the balls of and Zola Kassid for every mole of HCL that we had And again we need to math We got what X is X will be in moles. This is the ratio of moles right here. Point there. There are 25 moles a CEO or converted to mill leaders 12.5 milliliters.

Okay, This is a really long problem. Um, first thing is, when you solve for the ph of the buffer solution to do this, pH will be equal to the negative log of the base that is President. I'm sorry. The acid that is president, which is Benzo Kassid. We can look up the K value of Bensel kassid 6.3 times 10 to the negative five and then plus the log of the moles of benzoate over the moles of Ben's OIC acid. The moles of benzoate can be calculated by taking the massive of sodium benzoate divided by the molar mass of sodium benzoate. And then divide that by the moles of Ben's like acid, which will be the massive benzo kassid divided by the molar mass of Ben's OIC acid. And we get an initial ph of 4.13 now for the second part, which buffer component must be added in order to and in what quantity to change the pH to four with the pH is 4.13 and we want to lower the pH. We need to add an acid, and the acid component is Ben's OIC acid, so we'll start over with our Henderson hassle bolt equation. Now, the target pH is going to be equal to the negative log of K A for benzo kassid plus the log of the moles of benzoate that we start with divided by the moles of the moles of the, um Benda Look, acid we started with plus the amount that we're adding. So what I did is I recognize that we started with 1.5 grams of benzo kassid. I'm gonna add on an unknown amount of Ben's Ilic acid that we need to add in order to achieve our ph of four and then divide that whole thing by the molar mass of Ben's OIC acid. Now I've set up my equation to solve for X. I'll take the negative log of this value subtracted from both sides and then take the anti log of both and I get 10 to the negative. 100.20 is equal to miss ratio here after taking the anti log of both sides. Then what I'm going to dio is I'm going to combine these, uh denominators into a single value and then multiply that by both sides. Well, it'll divide that by both sides. So 10 to the negative point to is 100.0.63 And then when I take my, uh 1 44 and my 1 22 I get 0.847 and then I still have. After I pulled those out, what would be left? 1.50 divided by 1.50 plus X. Now I'm gonna divide both sides by 0.847 and multiply both sides by this denominator 1.5 plus x, I get this 1.12 plus 0.743 X equals 1.50 and my ex is 0.52 grams. If we need to add 0.52 grams of benzo CASS it in order to achieve a Ph of four instead of a pH of 4.13 So we're still not done as I mentioned, this is a very lengthy problem, Phil. Part C. Part C asks us to calculate the amount of hydrochloric acid or sodium hydroxide we need to add to achieve a ph of four. We just established that we need to add an asset Yes, before we added the Benz OIC acid, but Now we're going to add a different asset HCL in orderto lower the pH from the 4.13 to the fore, so we'll still have our target Ph four, which will be equal to the negative log of Benzo Kassid, plus the log of the moles of benzoate. Now the moles of Ben's await we start with is 1.50 divided by the molar mass of sodium benzoate. The moles of Ben's OIC acid we start with is 1.5 divided by the molar Mass have been so look acid. But if we add HCL to drop the pH, we're going to decrease the moles of benzoate when we add the acid by the amount of HCL we add, which will be an unknown volume multiplied by its concentration. Then we will add to Ben's OIC acid the amount of HCL that we add because every mole of HCL we add will convert a mole of benzoate into a mole of Enzo Kassid. So that's why we're subtracting the moles of HCL. We adhere and adding on the moles of HCL we had here. So then what I'll do is I'll take the negative log of this value subtracted from both sides and then take the anti log. So when I take the negative log of this and I subtracted from both sides, I get negative 0.20 I then take the anti log and it's 10 to the negative 100.20 And then the anti log of what's left over here is just going to be what's in the parentheses. So I'll rewrite again. What's in the parentheses? Okay, so then if I go 10 to the negative 100.20 I get 0.63 and then this ratio right here is 1.4 times tend to negative two minus two X. This ratio right here is 1.23 10 to the negative two and then plus two X. Now I'm gonna multiply both sides by this denominator down here and I'll get this. Then I'll collect my like terms and divide both sides by negative 3.26 and I get 8.2 times 10 the negative four leaders or 40.82 milliliters.

We are asked to describe how we would prepare a 3.00 buffer solution that consists of Ben's like acid and sodium benzoate. We will start with 2.00 leaders. Uh 0.025 moller Ben's OIC acid. This split the page so I got to go up to the top and any amount of sodium benzoate that we need for part A. We are asked to find the ph of the initial solution. Okay so for the initial solution, let's first go ahead and write down our chemical equation. There's my initial equation. And we had zero 0- five Mueller and zero and 0. Then my change will be minus X. Plus X. Plus X. So 0.025 minus X. And X. My K. A. Which I looked up at 6.3 times 10 to the -5 we're equal my H plus concentration time is my benzoate concentration divided by my Benz OIC acid concentration. Which will equal X squared on the top. Because both those are X. My denominator will be 0.025 minus X. And we are going to ignore the part I'm underlining. I quite solved this. Both quadratic with and without and it wasn't significantly different. So X equals where did they get my ex 5.02 I Believe, Times 10 to the -2. Mhm. I could read my own writing. And if I take the negative blog of that number I got my ph Equal to 1.30. That's the initial ph question B says how many grams of sodium benzoate should be added to prepare the buffer. And we're told to neglect the small volume changes change that will occur when we add the sodium benzoate. And if you will recall we were told that we wanted a ph of three. Now 10 to the negative three equals one times 10 to the minus third as our age plus concentration. I don't think I use that because I decided to use Henderson also by checking, checking. Yes I used Henderson Hasselbach which I will rearrange as follows. Well actually no I'll just do this. Uh The log of my concentration of the sodium benzoate. The benz OIC acid is equal to my ph plus the log of my K. A. And we were given 3.00 Plus the log of 6.3 times 10 to the -5. And that equals negative 1.2 zero. So now if I take 10 to the negative 1.20 That will equal the concentration of my c. six H. Five C. O. N. A. Over the concentration of my c. six h. 5 C. 00. H. So that will equal I'm sure I didn't write this down. Oh yes I did 6.3096 times 10 To the -2. And then I'm going to multiply that by this concentration which was 0.0 25. And I get 1.575 Times 10 to the -3 Mueller. And that will be my that concentration. Finally 1.575 Times 10 to the -3 more, which is bowls per liter. And we had to leaders of solution. And we know that the molar masses 1 44.117 grams per mole To my math. And I got 0.36 g uh my sodium benzoate. And we're done.

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.


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