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Question 67 (1 point) What is the range of = buffer made from NH3 and NHACI? '(pKb of NHz = 4.75).6.00 - 8.008.25 - 10.257.95 - 9.953.75 - 5.75...

Question

Question 67 (1 point) What is the range of = buffer made from NH3 and NHACI? '(pKb of NHz = 4.75).6.00 - 8.008.25 - 10.257.95 - 9.953.75 - 5.75

Question 67 (1 point) What is the range of = buffer made from NH3 and NHACI? '(pKb of NHz = 4.75). 6.00 - 8.00 8.25 - 10.25 7.95 - 9.95 3.75 - 5.75



Answers

A buffer is created by combining 3.55 g of NH3 with 4.78 g of HCl
and diluting to a total volume of 750.0 mL. Determine the pH of
the buffer.

So we are given 550 g of ammonium chloride. And we also know that we have .0188 moles of ammonia. So it looks like we're gonna have a buffer. So let's go ahead and change grams to moles here. Mhm. So the more mass is 53.5. So we'll get .103 moles of ammonium chloride. And that's the same as the moles of Ammonia. Right? Because there's one NH 4 plus in there. So I have a buffer. So, a convenient equation to use when looking for concentrations. And phs of buffers is the concentration of H plus is ca times the concentration of the acid divided by the concentration of the base. You can also use CA is moles of acid over moles of base. So let's look at why that's true. First of all, so concentrations are moles per liter. So if we're gonna divide moles per liter by moles per liter, well, the volumes are the same, right? Because it's the same solution. So those will cross out and you can just get the same equation and you can do it in terms of moles or more clarity, as long as you're consistent. Uh and you just choose based on the information you're given. So I'm going to need the K. A. For ammonium ion here. So That's going to be 5, 6 Times 10 to the -10. And then The moles of my acid, that's .13. We'll divide that by the malls of the base 0188. So that will give us 3.1 Times 10. The -9 More H Plus. So if we just go ahead and take minus the log of that, we get our ph So our ph is 8.51. Mhm. So then they're asking us what happens when we double the volume. So all we're doing is adding more water and making this volume that we were referring to up here large, larger. Right, So this volume here is going to change. But remember they cancel out? So it doesn't matter if you double the volume simply by diluting it with water. Your ph doesn't change. Okay, The Ph still remains 8.51.

So we are given 550 g of ammonium chloride. And we also know that we have .0188 moles of ammonia. So it looks like we're gonna have a buffer. So let's go ahead and change grams to moles here. Mhm. So the more mass is 53.5. So we'll get .103 moles of ammonium chloride. And that's the same as the moles of Ammonia. Right? Because there's one NH 4 plus in there. So I have a buffer. So, a convenient equation to use when looking for concentrations. And phs of buffers is the concentration of H plus is ca times the concentration of the acid divided by the concentration of the base. You can also use CA is moles of acid over moles of base. So let's look at why that's true. First of all, so concentrations are moles per liter. So if we're gonna divide moles per liter by moles per liter, well, the volumes are the same, right? Because it's the same solution. So those will cross out and you can just get the same equation and you can do it in terms of moles or more clarity, as long as you're consistent. Uh and you just choose based on the information you're given. So I'm going to need the K. A. For ammonium ion here. So That's going to be 5, 6 Times 10 to the -10. And then The moles of my acid, that's .13. We'll divide that by the malls of the base 0188. So that will give us 3.1 Times 10. The -9 More H Plus. So if we just go ahead and take minus the log of that, we get our ph So our ph is 8.51. Mhm. So then they're asking us what happens when we double the volume. So all we're doing is adding more water and making this volume that we were referring to up here large, larger. Right, So this volume here is going to change. But remember they cancel out? So it doesn't matter if you double the volume simply by diluting it with water. Your ph doesn't change. Okay, The Ph still remains 8.51.

Solution for the above question you by using Henderson hassle balls equation we have. BKB is equal to minus log in two 1.8. Multiply 10 rest to dip. Our minus five is equal to four point 74 in two PK. Is it well too 14 minus 4.74 is equal to 9.26 and P. H is equal to 9.12 is equal to 9.26 plus log concentration off in his three divided by concentration all and it's for positive. So here, concentration off in his three is equal to 0.53 more clarity.

So now we're going to work on Problem 56 from chapter 17. This question asks us Mhm Which buffer system is the best choice to create a buffer with a pH equal to nine, and then it gives us three systems. So since we need a ph of nine, we need something, uh, an acid base player which has a base and the conjugate acid. We do not want an acid and a consequent base. So if we're looking at, are possible answers. We have three acids, so H f h n 02 mhm and also h c l o all of the all of these are acids and then the other pair and for each of them is a conjugate base. But the acid will dominate in this situation. The basic one that we do have is NH three. Yeah, Now this can give us a buffer with its contract acid ammonium chloride as being a as being a, uh, a basic pH. So this year is the one that we want to look at. So the second part of the problem is to calculate the mass ratio that we need to create a buffer of Ph nine. So we're going to use the Henderson household back equation. Here we go. Nine is equal to. And then if we find the K A for ammonia, we can get the K for ammonium chloride by finding the KB for ammonia and dividing, UH, one time sent to the minus 14, divided by the KB. And when we do that, we get the K and take the negative log, which is equal to 9.25 for the P K A of sodium chloride, plus log of, of course, based over acid. So from this, we get that the log of the base over the acid is equal to negative 0.25 And then if we take the anti log, all right in a different color on the second line, we take the anti log of this. We get 0.56 now. This year is a mole ratio. The question asks us for a mass ratio. So to get the mass ratio, all we have to do is multiplied by the Mueller masses. So we have base over acid so we can multiply by the molar mass of the base, which is ammonia, which is 17.3 divided by 53.49 which is the molar mass of ammonium chloride. And when we do that, we get 0.18 as being the mass ratio of ammonia, two ammonium chloride necessary to produce the buffer of Ph nine.


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