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Which ane of the following aqueous solutions , when mixed with an equal volume of 0.10 mol L aqueous NHz, will produce buffer solution? Oa 0.10 mol L-I NaCl0b. 0.20...

Question

Which ane of the following aqueous solutions , when mixed with an equal volume of 0.10 mol L aqueous NHz, will produce buffer solution? Oa 0.10 mol L-I NaCl0b. 0.20 mol L-I KBrc.0.15 mol L- NH3d. 0.040 mol L - HCI0.050 mol L - RbOH

Which ane of the following aqueous solutions , when mixed with an equal volume of 0.10 mol L aqueous NHz, will produce buffer solution? Oa 0.10 mol L-I NaCl 0b. 0.20 mol L-I KBr c.0.15 mol L- NH3 d. 0.040 mol L - HCI 0.050 mol L - RbOH



Answers

Which of the following gives a buffer solution when equal volumes of the two solutions are mixed? (a) $0.10 \mathrm{M}$ HF and $0.10 \mathrm{M} \mathrm{NaF}$ (b) $0.10 \mathrm{M} \mathrm{HF}$ and $0.10 \mathrm{M} \mathrm{NaOH}$ (c) $0.20 \mathrm{M}$ HF and $0.10 \mathrm{M} \mathrm{NaOH}$ (d) $0.10 \mathrm{M} \mathrm{HCl}$ and $0.20 \mathrm{M} \mathrm{NaF}$

So we're going to try to figure out which of these things are gonna be buffers. So let's start by finding out what we know about our strontium hydroxide. So we take the polarity times of leaders and see that I have .26 moles of strontium hydroxide. We're gonna multiply that by two and see that we have 52 moles of O. H minus. Okay so these things are going to be added to that much O H minus. So in our first example we're adding some HF that's gonna react with R. O. H minus. It's going to make some H20 and some f minus. We'll do a little table initial change in final. I start with one mall here And .52. I don't care about the water. Obviously the O. H minuses are limiting reactant So -15 to of all these things plus .52 here. So I end up with .48 moles of my weak acid and .52 moles of my week base. So since I have a weak acid and its conjugate weak base, this is a buffer. Our second example, we're simply going to do the same thing. We're just gonna use less O H minus and see what happens. So the same. I see final. So we're adding 75. This time five to zero oh H minus is still my limiting reactant. So I'm just keeping track of what's happening here. So I end up with 23 moles of my weak acid and 5 2 moles of my conjugate weak base. So again this is a buffer. This next example we're going to use even less HF and see if that matters. So point oops, So .3 moles Still .52. So this time my H. F. Is my limiting reactant. So I end up with .22 moles of a strong base and .3 moles of a weak base. Okay, so this is not a buffer. Mhm. So this one was a buffer and this one is not a buffer. Okay, then we're gonna take .3 moles of f minus To add to our .52 moles of o h minus. So we have a weak base and a strong base. So this is not a buffer. Okay? And then finally we're adding a strong acid to our strong base. So this is not a buffer. We need a weak acid and its conjugate weak base, so only A and B. Right, A and B were buffers.

So the basic facts that you need to know in order to answer these questions or that a buffer solution will have a weak acid and it's conjugal base and or it could be a weak base with its cons. Get acid. Things that won't become a buffer solution would be just some strong acid, some strong base, a weak acid and a base that is not its conjugal base. And so these will not work. Let's work through, and I'll explain why you explain how you going to use and it's the 1st 1 we have. N a, C L and A. C. L is non Nason on the base. It's a neutral solution in water. There's just no way that this could buffer pH. We need asked basis going on, not a buffer solution. Next one, we have an A. C. L. Again and manage for C. L, which remember image for a C L breaks into NH four plus and C o minus. NH four plus is weak acid, But again, N a C l is still just a ah week not weak. It's a neutral solution. See, all minuses Neutral ion. So really, we just have an acid, as I said before, and acid is not gonna be a buffer solution. This is not a buffer solution. We need both acid and contrary base. Next, we have, uh, c h three and h two methyl amine. And then we have C H three and H three c l. Which really we can say that this is a C l minus. And then CHT initially plus so C l minus three nose in neutral I on It's the contrary base of HCL. So that's why now? No, this is a base hartman that has acted as a base and gone extra protons. So that 2 to 3 here. And so this is it's kontic it acid, So we're in good hands. Now let's take a look at the constellations. We're told that we have 0.1 and 0.15 and so this is also already looking good. Moller, let me be clear It We're gonna have a certain amount of weak base and its contacted acid. This is gonna be a buffer solution. Great. Okay, next one, we have, uh, 0.1 Moeller hcl and 0.5 Moeller. It's sodium. Uh, and I tried, but really. It's two miles bruising of sodium spectator. Ion mutual doesn't play a role here, so n o to minus is the conjugal base of H and 02 which is a weak acid. And so, if we had both No. Two minus here and h n 02 we will be in luck would have offices. So let's see what happens. The other thing that's given to us is that we have HTM Remember, H sales are really strong acid in old react with any base that it's in contact with. For instance, here we have this, you know to minus. It's a weak base. And so what happens is all this 0.1 molar HCL is going to react with this 0.5 Moeller and two miles, and it's going to form H N 02 and water. In fact, I'm gonna write out this express exploded, see equation explicitly. It's going to HCL, and I know two minus becoming. It's really non equilibrium and pushes really hard. That right becomes a geno to and see all minus now. This is when the concentrations become important. We have 0.1 more HDL and 0.5 Miller and two months just to say that we have mawr hcl than we do n ot minus. So the limiting re agent here is n o to minus it's so whenever we do end up with a Chenault to, we're going to be out of n ot minus and so we're not. We're gonna have acid right here. We're still gonna have the HCL. We're gonna have asked right here, but we're not gonna have any of that contra bass And so this is going to be not a buffer solution. We need weak acid and it's constant base next one similar situation, but the result is different. We have 0.1 Moeller hcl and 0.2 Moeller Ah ch three c o minus. Now the same thing is gonna happen. This is the Kandahar base of ch three Seo Age City kassid. And if when it reacts with HCL in a similar way as we did above here, we're going to get like HCL plus ch three c o minus gives this acid which the weak acid. And so now, for the conscientious we have we appoint one Wolesi help and point to more ch three c Oh, my So it's gonna happen. Is that this one more 10.1 Mueller is going to react and still leave around some ch three c O minus as it goes toe become stationary CEO H and so at equilibrium. We're going to have this and this because all of the HCL will have reacted with this to become that. And so this is exactly what we're looking for. We're looking for a weak base and its conjugate acid or all the way around a weak acid, and it's constant base. Either way, this is a buffer solution. Finally, last one. This is an easy one to get tripped up on ch three c o age and ch three ch two c o minus. Well, we have a weak acid and we have a base that, you know, it seems like it's pretty similar. It's a weak base. It could be it's confident base, but it's not, and that's a very important do not mess this up. We need a weak acid, and it's conjugated base to be here. If you were placed. If you got rid of the CH two right here and we had ch three c e o minus then this would be a buffer solution. But we don't. We just have a weak acid and some other weak base, and that's not going to be a buffer solution, because what needs to happen is we need to have an equilibrium where one of these changes into the other, that's not going to happen here, and so this is not.

So now we'll do problem 52 from chapter 17. So here we are asked to determine if the mixing of pairs of solution will result in a buffer. So for each problem, we can look at it step by step. So for a we can see easily right away that will form a buffer because we have a conjugate acid base pair. So if we have a conjugate acid based pair, we will be able to form a buffer. So for B, we don't have this, but we need to look closer. So when we add HCL two HF um, the contract it uh, acid country base of H F F minus will form. So we will form the country based by adding a strong acid. But we need to make sure that we don't add too much acid because if we had too much strong acid, we're going to overwhelm the buffer we're trying to create. So we just need to quickly look at the number of moles of acid and of base, so the moles of H F is equal to the concentration times the volume. So if we multiply 0.1 by 0.150 we get 0.0 15 moles. Yeah. Now, if we look at the moles of HCL, we want to multiply the concentration times the volume. And if we do so we get 0.23 uh, 236 So since we have more molds of HCL, it will completely consume the strong acid, the weak acid. So no, we do not have a buffer form. So for part, excuse me. So for part C, we have, uh, h f and sodium hydroxide so we can see here. Um, just by comparing the concentrations and the volumes, we have the same concentration of both HF and sodium hydroxide, and we have a larger volume of HF, so we're going to have a larger volume of of weak base than we do strong acid. So we will form a buffer and this is because nuh forms contra bass without consuming all the acid. So now if we look at D, we have metal mean and method co ammonium chloride. So since we have an acid base in a conjugate acid, we say that we have the acid base pair, So yes, and then finally we look at e mhm, which is methyl amine with HCL. Now, once again, if we're just looking at concentrations and volumes, we can see that we have a larger volume of a higher concentrated base in methyl amine. So we're gonna have more moles of base because the volume and concentration is higher. And we have, uh, more, um, less moles of acid. So we're gonna have a situation similar to part C where we say yes, because HCL forms the contra get acid without consuming all the week base, so a buffer is formed. If in any of these cases, we can't obviously tell from the concentration. For instance, if we have a lower volume but a higher concentration, we can always just multiply the two values to get the number of moles, as we did in part B here. To see directly compare the number of

So we have a number of solutions here and we're gonna try to see if we have a buffer. So I'm gonna start by looking at how many moles of H plus we have. So I've got its polarity and its volume and leaders so I can see that I'm working with 0.112 moles of age plus. And this is first example, we're going to add some weak acid to it. All right. So we're going to add some ammonium chloride. So what we're adding is ammonium ion, which is a weak acid two H plus, which is a strong acid. In order to have a buffer, we need to have a weak acid in its conjugate base. So this is not a buffer. In our second example, we're going to add some F minus some chaos so that each plus is going to react with the f minus and the calf K. Is just a spectator. I am and make some age. F keep a little table here. Just keep track of what was happening. So 0.112 moles of this and I'm adding 0.33 moles of this. So all of the f minus will react at the same time. The same amount of each plus will react and we'll make that same amount early Jeff. So when this reaction is done, all of the f minus is gone. I still have some H plus and I've made some H. F. So in this solution we have a strong acid H plus and we have a weak acid HF. So this is not a buffer. In the third example we've got our strong acid and to it we're going to add some strong base. Okay, we take a strong acid and a strong base. There's no way to get a buffer so no buffer, not a buffer. And then Indy again we've got our H plus and to that we're gonna add some night tripe. We added some K. N. 02 K. Plus as a spectator around. So we'll just look at what the nitrate iron is doing. Okay? So we appoint 112 moles of this and we're adding 0.279 moles of this. So all of the H plus is going to react this time and we'll make that same amount over here. Mm So when we're done we won't have any age plus. You'll have 167 moles, no two minus and you'll have 0.112 moles of H. N. 02 So we have a weak base nitrate and it's congregant, weak acid H. N. 02 So this is a buffer in this final example we've got our H plus onto that, we're going to react some K. C. L. O. So the K plus as a spectator I am and the c l O minus will react to each plus make Hochul and again, let's keep track of what's happening here. I start with 0.112 moles of this. I'm adding 0.112 moles of this. So all of this is going to react and all of this is going to react and we'll make some weeks acid. So when we're done I have none of this, None of this. Then all I have is a weak acid so there's no conjugate base present. So this is not a buffer.


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