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A buffer solution contains 0.260 M ammonium bromide and 0.474 M ammonia_If 0.0279 moles of hydroiodic acid are added to 150 mL of this buffer; what is the pH of the...

Question

A buffer solution contains 0.260 M ammonium bromide and 0.474 M ammonia_If 0.0279 moles of hydroiodic acid are added to 150 mL of this buffer; what is the pH of the resulting solution ? (Assume that the volume change does not change upon adding hydroiodic acid)pH =

A buffer solution contains 0.260 M ammonium bromide and 0.474 M ammonia_ If 0.0279 moles of hydroiodic acid are added to 150 mL of this buffer; what is the pH of the resulting solution ? (Assume that the volume change does not change upon adding hydroiodic acid) pH =



Answers

A buffer solution is prepared by adding $0.125 \mathrm{~mol}$ ammonium chloride to 500. mL of 0.500 -M aqueous ammonia. Calculate the pH of the buffer. If 0.0100 mol HCl gas is bubbled into 500. mL buffer and all of the gas dissolves, calculate the new $\mathrm{pH}$ of the solution.

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.

So now we'LL work on Problem forty four from Chapter seventeen. So this question asks us, what mass of ammonium chloride would we need to add to a solution for the two point five five litres of zero point one five five Mueller ammonia solution to obtain a buffer with a pH of nine point five five if we assume the volume doesn't change, so to work on this problem? First of all, we need to find the K just tow, make the problem easy. So we're looking for the K of ammonium chloride. So to find that we divide by, we divide Kate W W divided by K B, which is one times ten to the minus fourteen, divided by one point seven six times ten to the minus five, which is the KP for ammonia. And we get five point six eight times ten to the minus ten for as our K for ammonium chloride. So we can go ahead and use hundreds and hustle back equation now. So we go and put nine point five five because that's our desired pH. And then we put the K ey the P K, which is just the negative log of the K A value. So we'LL go ahead and write it out his nine point two five and then plus the log of base overpasses so we can go ahead and subtract nine point two five from both sides and we get zero point three zero is equal to this log of this ratio. Sit so we can go ahead and take the anti log of both sides. And what we get is, uh, two point zero zero is equal to, uh, base, which we know to be zero point three nine five moles. And we know that because we can calculate it from the amount that they give us a beginning, which is two point five leaders of a zero point one five five mole ammonia solution. And if we multiply those two together we get zero point three nine five miles base and then so on the bottom, we have moles of acid so we can switch malls of acid with two point zero zero and we get the moles of acid, which in this case is ammonium chloride is equal to zero point one nine nine mils. So the question asks us about the mass so to get the mass. All we need to do is multiplied by the Mueller mass of the entire compound ammonium chloride. So the more mass for ammonium chlorate is fifty three point four nine grams. Ramon. And when we do this calculation, we get ten point six to grams of an age three seen her?

To calculate this. PH. We simply need to take the moles of base divided by the moles of acid and plug it into the enders and acid belt equation. PH will be equal to P. K A. We know that k b of a mony of ammonia, so the K be divided into K W 1.0 tend to negative. 14 will give us K. A negative log of that will give us P k A plus the log of the moles of the base. The base, in this case is ammonia. We have 500 milliliters or 5000.5 leaders at 0.5 Moeller. Multiply those together that gives us moles of ammonia within divide by the moles of ammonium, which they gave us as 0.1 to 5, and we get a pH of 9.56 If we then add 0.1 moles of HCL to this same solution, it's going to react with the base, so it'll decrease molds. The base will decrease by 0.1 Moles and the moles of acid ammonium will increase by 0.1 moles, and we get a pH of 9.51


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