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Conskler Ihe tollowng btrabon curve of whch showe the litration ol Fut Qck roprosent Iha ucl tonm Which Iottor bast Incicatos tho oquivalonca polnt? 6) Which lotlor...

Question

Conskler Ihe tollowng btrabon curve of whch showe the litration ol Fut Qck roprosent Iha ucl tonm Which Iottor bast Incicatos tho oquivalonca polnt? 6) Which lotlor bosl indicalos Knenu (na pH pKa? Which lottor indcados wnora molos 0f NaOH addod Oroaten Unn molus 0l wunk achd praseni? Which lotter indicated whoro bufforing #ould be 0est? Which lattor Indicales where (A ] is prosent; IHAJ = Which lotter indicates where [HA] > [A]?Noh: Lal HA ropresant the acd Iomm andConsider tha cunva again;

Conskler Ihe tollowng btrabon curve of whch showe the litration ol Fut Qck roprosent Iha ucl tonm Which Iottor bast Incicatos tho oquivalonca polnt? 6) Which lotlor bosl indicalos Knenu (na pH pKa? Which lottor indcados wnora molos 0f NaOH addod Oroaten Unn molus 0l wunk achd praseni? Which lotter indicated whoro bufforing #ould be 0est? Which lattor Indicales where (A ] is prosent; IHAJ = Which lotter indicates where [HA] > [A]? Noh: Lal HA ropresant the acd Iomm and Consider tha cunva again; Whal reaction or solulion composition would b8 employed l0 determine the pH at the following points? Choices: mL of NaOH weak acid Ionization buffor solution excess slrong acid eak base ionization excess strong base



Answers

A $0.200-\mathrm{g}$ sample of a triprotic acid (molar mass $=165.0$ g/mol) is dissolved in a 50.00 -mL aqueous solution and titrated with $0.0500 \mathrm{M} \mathrm{NaOH}$. After $10.50 \mathrm{mL}$ of the base was added, the $\mathrm{pH}$ was observed to be $3.73 .$ The $\mathrm{pH}$ at the first stoichiometric point was 5.19 and at the second stoichiometric point was 8.00 a. Calculate the three $K_{\text {a }}$ values for the acid. b. Make a reasonable estimate of the pH after $59.0 \mathrm{mL}$ of $0.0500 M$ NaOH has been added. Explain your answer. c. Calculate the pH after 59.0 mL of $0.0500 M$ NaOH has been added.

Yeah. Okay. Let us start with the first part. The in Asian P O H. Concentration for 0.100 Moeller and a watch. Yeah, is given. Then the P H will be 13.0 zero after edition off 24 mil off any which we will be after addition off 24 ml ending which the concentration will be. That's right multiplied 25 divided by 49 it will be 0.0 51 seyval. Next the concentration off HCL will be equal toe 0.100 multiplied with 24 million leaders divided by 49 million liters equals zero point for my zero. This gives us the concentration off which negative which is 0.0 20 Moeller. From this concentration, we can determine the p o. H value just by multiplying it with negative law off 0.0 the any which is in excess quantity, which gives us the P which value equals two points 70 That's the answer

So in this problem, we're given a block like thiss, which is a tight gration, a bladder carb which shows the patrician ofthe week mon a pro take acid with a strong base and here we have to answer some questions. So our first question asked, What is the pH and what is the volume of a red base that the equivalence point. So as we know that the equivalence point of arbitration car is the steeper portion off the car, that means at this point, and also the coolants point is, um, just a half way between the steeper portion. So we can consider, um, this point in the middle of the steeper portion as they given spined knife, he draw a straight line or apartment will align to the volume access we'LL see that it's our own turkey mill. So the volume and day equivalence point is thirty ml. And if we draw another straight line, tow the beach access. Who will say that it's around nine. So are you B h. And they go on this point is nine now in question be You're just Ah, what volume ofthe edit bays At what volume? Off that base is the pH calculated by working and equilibrium problem based on the initial concentration and k of the weakest. So the answer is zero ml. So when we are so when we have not added any of the basis or when the morning of the basic zero. That's the point r R There's a time when, um, the pH is calculated by working in equilibrium. Problem based on the initial concentration and k all the way. Gets it now in question. See your ass that what volume off that base does the Ph physical peak. So the page is equal to peek, eh? Ah, there's happens on we are halfway through. They given spine. That means since our equivalence born volumes thirty ml, the pH will be called Toe Piquet. And halfway off this volume, which is fifteen ml sarah in roughly fifteen ml, we will have our beaches ableto piquet and and in the question D, we were asked at what volume off that base is the peach calculated by focusing on the concentration and K b of the contract at base. So that question is so The answer to this question is tardy ml So this is the equivalence point at which we calculate the pH best on the concentration and Gaby off the consequent base and finally and question E. We have to go say what volume off that base is the peach calculated, um, by focusing on the amount of excess strong base at it. So this is beyond the equivalence point. So there's beyond. They give us point or target mill so beyond this volume on the pH is calculated by focusing on the amount off excess strong base at it.

So our thai tradition is between a strong acid and a strong base. So our net ionic equation it's just gonna be H plus plus O. H minus gives us water. And at the equivalent point other than water we'll just have our two spectator ions also present. Okay and then we're gonna see if we can figure out how much. Anyway H. We need to get to our equivalence point. So let's start with our information about our strong acid. 5.1988 Moller multiply that by its leaders And we'll see that we had 7.36 Times 7 -3 Malls of our strong acid. Therefore we must have also reacted that same amount of our strong base to R. O. H minus. And then if we take that and divide it by the polarity will get our volume. Okay because polarities moles per liter. So that will give us point 03457 leaders or 34 .57 ml. Okay so before the titillation occurs, all we have is our strong acid R. H. Plus concentration Was given as .1988 Moller. So if we take minus the log of that, we'll get our ph Which is 7016. Yeah, mm. Yes, halfway to the equivalence point one, half of the moles of each plus have reacted. So we're left with half of our moles left and we've added half the volume of R. O. H minus. Okay, so the moles of Rh plus that are going to be present Is going to be 73, 6 Times 10: -3 divided by two. So we'll have 3.68 Times 10 to the -3 moles of O of H. Plus that remain. They haven't reacted yet. And we used 34 .57 ml. Well, at this point we have only added half of that. So 17.28 ml. So if we add that to the 37 ml that we had of Rh I who will get our total volume at that point? Which is 54.28 ml or .054- eight leaders. Okay, so our concentration of H plus halfway to the equivalence point Is our moles 368. I'm Santa -3. And we'll divide that by our total volume and we'll get our more clarity as 0678 Moller. That's our age plus. So we take minus the log of that. We'll get our ph as 1.17. Okay. And then finally they ask us, what's the ph at the equivalence point, while the ph is seven, Okay, anytime you have a strong acid and a strong base, the ph of the equivalence point is going to be seven

So I would just sketch, um, A through C A. On the same graph values different lines. So to for six a turn 12 and this is volume 0.1 and a ohh added, That's ph. This line will start here, and I about 10 0 shoot up. You going to get more rounded? But you get the idea and this one will star a little bit above that each year and then shoot up this well on the last one starts at four and really curves up. There's yourself like that. So then for a and Decatur would be from denial time all, uh, blue for be to be time will blue and first see we would use I was there in


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