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CaAnnUse UUt Raltitdte deceli >ordt Tllntaperded [orItudont the following Iexction tha Isbontory 669 K:ZNH;() FM(d) JH,()Whan the tntroduced 6.4JxIQ" tolet ...

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CaAnnUse UUt Raltitdte deceli >ordt Tllntaperded [orItudont the following Iexction tha Isbontory 669 K:ZNH;() FM(d) JH,()Whan the tntroduced 6.4JxIQ" tolet of NHy(e) mto 100 Iiter containe the found the equilibriutWeteeakon OENHj(e) 62J,7J10 'MLCalculste tke cquilibrium €oastnt Kc she obtarned fot thut raxction

CaAnn Use UUt Raltitdte deceli > ordt Tllnta perded [or Itudont the following Iexction tha Isbontory 669 K: ZNH;() FM(d) JH,() Whan the tntroduced 6.4JxIQ" tolet of NHy(e) mto 100 Iiter containe the found the equilibriut Weteeakon OENHj(e) 62J,7J10 'ML Calculste tke cquilibrium €oastnt Kc she obtarned fot thut raxction



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Contents of Soft Drinks. Refer to Exercise 2.68 a. Round each observation to the nearest $10 \mathrm{ml}$, drop the terminal $0 \mathrm{s},$ and then obtain a stem-and-leaf diagram of the resulting data. b. Truncate each observation by dropping the units digit, and then construct a stem-and-leaf diagram of the truncated data. c. Compare the stem-and-leaf diagrams that you obtained in parts (a) and (b) with each other and with the one obtained in Exercise 2.68

Chinese solution when directed with stool and avoided give some right sector own additional access this compound. So could you inform general which hard place here's the correct terms.

So the solutions mutually dilute one another. And because the volumes are equal, the concentration are half in the final solution. So we have seen a two plus that is equal to not point not not 6 to 5 mola s 04 to minus people do not point. Not not 760 mola So we cannot assume that either concentration remained constant during precipitation reactions. And instead we need to assume that the precipitation proceeds until all of one re agent is used up and this point would be equilibrium. So the reaction is C. A s 04 at C. A two plus is an equilibrium with C A two plus at S 04 to minus. So we construct an ice table when we saw for acts. So we know that X is equal to 2.4 times 10 to the minus three. So percent on precipitated is as follows. See a two plus that's equal to 2.4 times 10 to the minus three, founded by nine point not not 6 to 5 mola initial. This was the Mola Valley final see a two plus multiplied by 100 that gives us 38%. That is, um precipitated

Considering this mixture, let's determine the volume of 0.1 Mueller nitric acid that must be added to achieve a final pH of 7.21 We have to consider all the neutralization is that take place? So let's start with our strong base. Strong base is K O H. So from here, which we can calculate the moles of K o. H. It's a strong base which is equal to the moles of H minus. So we get zero 500 Mueller times 100 mL 0.1000 Leaders will find that this is equal to a 0.500 moles of oh H minus and will then look at our strong acid from HCL. The malls of HCL Zico, TV moles of H Plus and Armel Arat E Here is 0.750 Mueller volume is 200 mL or 2000.2000 leaders and we'll find that the moles are equal 2.150 moles of H plus. Now we'll have some neutralization here We see that the H plus is greater than the O. H minus. Therefore the we can calculate the malls of H plus in excess should be equal 2.150 moles of H plus minus 3.500 moles of the H minus the minus would be completely neutralized, and we'll be left with 0100 moles of each plus left over in excess. Now let's look at our next two bases. They're both weak bases on DWI air told that we have p 043 minus, uh, ignoring the sodium spectator from any three p 04 and we have C n minus. Ignoring these again the sodium spectator so r k A of the contra get acid here, H p 04 to minus is equal to 4.8 times 10 to the negative 13 and R K for the conjugal H C N would be equal to 6.2 times 10 to the minus 10. Comparing these values here, we see that the K A for H P 04 to minus is, uh, much less so. This would be the weaker acid or the stronger base so the acid would first react. Eso the H plus well, first react the excess H plus well. First react with the P 043 minus. So the malls of N A three Peel four is equal to the moles of Peel 43 minus. And we have 0.100 Mueller and 50 mL Their 500.500 leaders in this workout. 2.500 moles p 043 minus. And so now we compare this to our H plus and uh H plus is still greater. So moles of H plus in excess is equal. 2.0 100 moles of the H plus minus 3.500 moles of the peel for three minus, and we're left with 30.500 moles of the H plus in excess. And now second, the H plus in excess will react with the C N minus. And so the moles of the in a CNN sodium spectator so moles CN minus and this is equal to 0.150 Moeller times 0.500 leaders, and this is equal to a 0.750 moles of CN minus and now this time we can see that the sea end minus will not be the same here. So h plus plus c n minus will produce H c N Andi. We'll set up a nice table and moles here. The H Plus is 0.500 CN minus is 0.750 This is zero minus 0.500 minus 0.500 plus 0.500 This would work out to zero 0.250 point 00500 Using the results of this, let's go ahead and way have excess CN minus so the C and minus will now react with be added nitric acid. And we use our Henderson Hasselbach equation. We have all of our values here on D P H desire to 7 to 1 p. K A for H CNS negative log 6.2 times 10 to the minus 10. We don't know how much of the nitric acid so plus log 0.0 to 50 Now I'm gonna add some nitric acid that would shift this to the right so minus X plus x so this will be minus X and 00500 plus x by adding the nitric acid solving this for X, you can solve this and plugging this into my equation solver. I find that X is equal 2.243 So that tells me that the moles of H N 03 added Z equals 2.243 Moles of a Channel three. And so the volume of a Channel three added a Z we told were given the majority 0.243 moles. And we know that as one moles per leaders that similarity and we consol this and it'll yield 0.243 leaders, or 24.3 mil leaders of H and 03 that needs to be added to achieve a Ph of 7.21 to this mixture.

Before we perform many calculations associated with a weak base being titrate ID with a strong acid. Let's review all of the equations that are used to determine the pH in this type of titrate Asian. Initially, we just have some weak base in solution. We have not yet added any strong acid, so to calculate the pH of a solution that could just just contains a week base, we can determine the hydroxide concentration first, I said in it, equal to the square root of K B multiplied by the initial concentration of the week base. If, however, for some reason this initial concentration ends up being greater than 5% I'm sorry the hydroxide concentration ends up being greater than 5% of the initial concentration of the week base. Then we need to use a different equation. Many would say we now need to use the quadratic where K B is equal to the hydroxide concentration squared, divided by the initial concentration of the week base, which is typically given to U minus hydroxide. Then we have one equation with one unknown hydroxide, and we solve for the hydroxide concentration. Because this is a quadratic, there will be two answers. One answer will be plausible, typically positive, while the other is negative. If for some reason you have a question as to which answer is correct, it will be the one that is closest to the hydroxide concentration you determined up here Our pre Oh, and then we take the negative log. The hydroxide concentration divided indicate W In order to get our pH now pre equivalent pre equivalents, we have added some acid, but not enough to completely neutralize all of the week base. So we have converted some of the week base into a weak acid. Thus, we have a buffer solution for a buffer solution. We determine the pH by using the Henderson Hassle Baulch equation. P H is equal to p k, plus the log of the moles of base in the solution divided by the moles of acid. Another form would be molar ity of base over molar ity of acid. But for these hydration, zit ends up being easier to focus on just the moles of the base and the moles of acid. So we don't have to concern ourselves with a changing volume. The molds of base will simply be the initial volume of base before we have added any of the Thai trahant before, we've added any of the strong acid multiplied by the initial concentration of the base. This will be the initial moles of base that we start with, but we need to subtract off the moles a base that was converted into moles of weak acid. This will be equal to the moles of strong acid. Added to calculate the moles of strong acid will take the volume of strong acid added multiplied by the concentration of the strong acid. This will be the molds of base that's left in solution. After some strong acid has been added, the moles of weak acid formed will actually be equal to the moles of strong acid added which is simply the volume of the strong acid multiplied by its concentration. Peek A can be determined by taking the K B value, dividing it into kw to get K and then taking the negative log of K or show you ph So p okay will be equal to the negative log. Okay, w divided by Okay. Oops. The be so k be divided into K. W. Is K a negative log of K is PK. This is how you'll get PK from K B. Now, at the equivalence point all of the week base has been converted into a weak acid. So to calculate the pH at the equivalence point, all we have in solution is a weak acid. So we perform a weak acid pH calculation where the hydro knee, um concentration is equal to the square root of K, which will be KB divided indicate w multiplied by the concentration of the weak acid. Now in solution, the concentration of the weak acid in solution will be equal to the moles of weak base. We started with because the moles of week based we started with is now the moles of weak acid. We have to calculate the moles of weak base we started with will take the initial volume of weak base multiplied by its initial concentration. But we need so this now gives us the moles of weak acid we have, but we need a concentration, so we divide it by the new volume, which will be the initial volume of weak base we started with, plus the volume of strong acid that we have added. This will then give us our hydro knee, Um, concentration. Once we have the hydro knee, um, concentration, we can determine pH by taking its negative log now. Post equivalents. We have an excess amount of strong acid that we have added, and the pH will be determined solely based upon the excess strong acid that is present to calculate the hydro knee. Um, concentration. We'll take the volume of acid that we have added post equivalents. So, for example, if the equivalence point volume is 25 mL and we've added 26 mL, then we've added one Miller Leader post equivalents. So we'll take this volume in units of leaders. So when I'm taking volume times concentration, this volume is always in leaders so that when we multiply it by the concentration and moles per leader, we get moles. So we'll take the volume post equivalents in leaders. Multiply that by the concentration of the strong acid that we're adding. This will give us moles of strong acid post equivalents in excess, and we'll divide that now by the total volume, which will be the initial volume of the week based. We started with, plus the total volume of strong acid that we have added. Once we have the hydro knee, um, concentration will determine pH by taking its negative log appear if for some reason that the equivalence point Ah, hydro knee, um, concentration ends up being greater than 5% of concentration of the weak acid at the equivalence point. Then we need to use the quadratic formula and do something similar to what I discussed in a previous problem or what was mentioned over here. Little B k is equal to the hydro knee, um, concentration squared, divided by the concentration of the acid, which is essentially all of this right here, minus the hydro knee. Um, concentration, because this is a quadratic will get two answers for the hydro knee. Um, concentration. The correct answer will be the one that is closest to the hydrogen concentration we calculate with this simpler equation. Now that we know the equations, I took the data that was provided the initial volume of the ammonia 25 mL, or 250.25 liters, its initial concentration. I took the KB value and determined the PK of ammonium when this has accepted a hydrogen and the initial concentration of the HCL. The equivalence point volume is also going to be 25 mL, or 250.25 leaders, because thes two concentrations air the same and the stoy geometry is 1 to 1. So initially, all we have in solution is ammonia. Using the equation that I described earlier, we can calculate the hydroxide concentration. Once we have the hydroxide concentration, we can divide it into K W in order to get the hydro knee, um, concentration, and then take the negative log of that in order to get pH. This value slightly different than what's found in the back of the book. Probably because they used the quadratic equation, which I described earlier, depending on your instructor, whether or not the use of the Adriatic quadratic equation is necessary, you may get this answer or 11.11 if you use the quadratic. Then I took all of the, uh, pre equivalents volumes expressed in middle leaders and converted them to leaders and did the same with the remaining volumes. Using these volumes and the Henderson hassle bolt equation, I then calculated the pH pre equivalent at the equivalence point. All of the weak base has now been converted into a weak acid. So I used the weak acid equation, the square root of K A multiplied by the acid concentration to get my hydro knee, um, concentration. And then I took the negative log of that in order to get the Ph. Post equivalent. I have excess strong acid, so I determined the moles of excess strong acid post equivalents divided that by the new total volume, which will be the volume I started with. Plus, the volume might have added. Divide that into the moles of excess strong acid to get the concentration of hydro knee, um, and then take the negative log of that in order to get my pH. Once I've done that, I can then plot the pH as a function of the volume of HDL added, and this is the curve I get. If it's helpful, I can show you the actual equations that I used for each of these particular calculations. This is the little equation square root of K B, multiplied by the week based concentration. This is the Henderson Hassle Bolt equation. PH equals P K plus the log of moles base, which will be the moles of base. I start with minus the moles of base that reacted, which is the moles of asset added, divided by the moles of acid added, which is now the moles of weak acid formed all the way through. Then at the equivalence point, all I have is the weak acid. So I'll take the square root of K A multiplied by the concentration of the weak acid that is present. This raise to the 0.5 is the square root part. And then take the negative log of the hydrogen concentration to get my Ph. And then this is how I carried out the calculations for the post equivalents. I drove him concentration, which I then took the negative log of in order to get Ph.


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