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IULTIPLE CHOICE Choosc the one altemative that bestcompletes lhe stalement OF answers Ihe question: The titration Nutve" for I0.0 mL of 0.100 HylOalaq) with 0....

Question

IULTIPLE CHOICE Choosc the one altemative that bestcompletes lhe stalement OF answers Ihe question: The titration Nutve" for I0.0 mL of 0.100 HylOalaq) with 0.100 MNaOHaq) ucn belot2 410121416 18202224Eslimate lhe pKaz of H3FO4B122 C148 D) 72Titratiun ofa 250mL aliquot of the acid solution Tequinu cnlulior unknown acid had PHof 3.70, Assuming that the acid monoprotlc; what i hvdroxide for completee teaction; 217mLOfO,OAMeodium its ionization constant? 4127 * 10-Il B) 3.6 * 10-9 C14: 110-7 D

IULTIPLE CHOICE Choosc the one altemative that bestcompletes lhe stalement OF answers Ihe question: The titration Nutve" for I0.0 mL of 0.100 HylOalaq) with 0.100 MNaOHaq) ucn belot 2 4 10121416 18202224 Eslimate lhe pKaz of H3FO4 B122 C148 D) 72 Titratiun ofa 250mL aliquot of the acid solution Tequinu cnlulior unknown acid had PHof 3.70, Assuming that the acid monoprotlc; what i hvdroxide for completee teaction; 217mLOfO,OAMeodium its ionization constant? 4127 * 10-Il B) 3.6 * 10-9 C14: 110-7 DJ 9.0 *10-2 E1 20 _ :10-$ aftet th" addition 0.Q10 mol HCllg) For NHz L.00 Lsolution 0.100 M NHzlaq) 3 Calculale FHof 4 Pkb 47 11.46 BJ 9.26 C) 1021 DJ 8.31



Answers

Consider the following four titrations (i-iv): i. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)$ by $0.2 \mathrm{M} \mathrm{HCl}$ ii. $150 \mathrm{mL}$ of $0.2 \mathrm{M}$ HCl by $0.2 \mathrm{M} \mathrm{NaOH}$ iii. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HOCl}\left(K_{\mathrm{a}}=3.5 \times 10^{-8}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$ iv. $150 \mathrm{mL}$ of $0.2 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=7.2 \times 10^{-4}\right)$ by $0.2 \mathrm{M} \mathrm{NaOH}$ a. Rank the four titrations in order of increasing $\mathrm{pH}$ at the halfway point to equivalence (lowest to highest $\mathrm{pH}$ ). b. Rank the four titrations in order of increasing $\mathrm{pH}$ at the equivalence point. c. Which titration requires the largest volume of titrant (HCl or $\mathrm{NaOH}$ ) to reach the equivalence point?

Okay. This problem is exceptionally long, with six parts and one part having four. So we're almost a 10 parts on this problem, but we'll go through it slowly, take our time and make sure you understand. First thing is, we want to determine the pH before any hydrochloric acid is added. Because ethanol amine is a weak base and we have a cave be value for it. Then we can solve for the hydroxide concentration by taking the square root of the concentration of the week base multiplied by its kay bee value concentration is given to us a point on one, and the K B value is given to us at 3.2 times 10 the negative five. So this thing gives us a hydroxide concentration of 5.66 to the negative. For Moeller, we can calculate pH by taking the negative log of the hydro knee um, concentration, which we can get from the hydroxide concentration by taking the hydroxide concentration on dividing it indicate W This right here is the hydro knee, um concentration, which will take the negative log to get Ph. 10.75 and were asked to determine the pH at the equivalence point. To do this, we first need to figure out what the equivalent point volume is. So we know what delusion is occurring s so that we can correctly calculate the concentration of the weak acid that has now formed. So we'll start with the volume of weak base 25 milliliters or 250.25 Leaders both apply it by its concentration to get the moles of the week bays recognize it's a oneto one stroke you metric reaction with HCL right here we now have moles. Hcl well then use similarity of hcl 0.595 Moeller in order to get the leaders of HCL required to reach the equivalent point 0.2632 leaders are 26.32 mil leaders but we casted concentration then is going to be equal to the molds of weak base. We started with which will be the volume of the week based multiplied by its concentration because all of the week base has become weak acid. Once we reach the equivalent point, well then divide that by the new volume he started with 25 millimeters which will convert to leaders 250.0 to 5. And then to reach the equivalent point, it was 26.32 Mill leaders of 0.0 2632 leaders some of those together to get the new volume Divide that into the moles and we get a concentration of our weak acid in the equipment point the 0.487 So to get the hydro knee, um, concentration. What? I need to take the square root of the concentration of the weak acid, which we just determined here, multiplied by the K A value que value was not given to us. However, we can calculate the came value by taking make a value 3.2 time send the negative five and dividing it into K W. And then again, multiply that by the concentration of the weak acid that we just determined right here. Take the square root of that. We get a hydro knee, um, concentration of 1.23 times 10 to the negative six pH then is going to be the negative log of this value, which is 5.91 Then for part C, we're asked to figure out the pH at the halfway point and the halfway point P H is equal to P. K, so pH will be equal to the negative log of K. A value in the K A value is going to be DKB value. Divided indicate W This is our K value will take the negative log of that. That will be PK and again at half equivalents. We have an equal amount of weak acid in a weak base. So when using the Henderson hassle Bolt equation, the log term is cancelled and pH is just equal to P K 9.51 The indicator that would work best for this particular Titra Asian is one that's going to have the pH at the equivalents point, which was back in Part B, a 5.91 within its transition range. There's multiple ones that you could choose, but the one that looks like 5.91 is about right in the middle of its transition range is broke, wrestle purple. Now we'll move on to Part E, which has four parts to it. 1st 1 is calculating the pH after the addition of five milliliters. We know the equivalents. Point volume is 26. All right here, 26.32 mil leaders. So five milliliters were pre equivalent and pre equivalents we used Henderson Hassle Baulch equation pH is going to be equal to p. K PK again is the negative log of K, which is K w divided by Katie plus the log of the moles of the week basically started with which will be the volume time similarity minus the moles of weak base That reacted which will correspond to the moles of HCL added which will be its fallen you by Bill leaders is what we have added so far multiplied by the concentration of 0.95 were then divide that by the total volume which is uh, no. We then divide that by the moles of weak acid that we formed and the moles of weak acid that reformed is going to be equal to the molds a strong acid. We added five milliliters multiplied by the concentration of H CEO. Essentially every mole of strong acid. We add every mol of H sale we had We made a mole of a weak acid. I remember with the Henderson hassle Baulch equation. We don't need concentrations. We could use concentrations, but really, we just need moles of base over Mel's of a weak acid. Get a pH of 10.13 Now for the next part of party. Let's go to 10 milliliters. It's going to be the exact same set up, except we're now going to have 10 million leaders here instead of five milliliters and then 10 milliliters here instead of five milliliters, that would get a pH of 9.72 We should expect the pH to continue to drop as we add more and more strong acid. Now we go to 20. Milliliters were still pre equivalents, so we're still in the buffer region. So we used Henderson passable equation again. Same as before. Except now we're going to include our 20 milliliters right here and our 20 milliliters right here. We'll get a pH of 9.22 now at 30 milliliters. Need to be astute and recognize that we're now post equivalence equivalents. Point was 26.32 so 30 milliliters were 3.68 milliliters. Post equivalents, that's all we need to do is figure out the excess hydro knee. Um, so it'll be the mill leaders of HCL that I added post equivalents multiplied by the concentration of HCL. This will be the moles of HCL and moles. I hydro knee, um, that are gonna react with anything that just gonna go straight into the solution. And this is going to dictate the hydrogen concentration. And then we'll divide by the new volume, which is going to be our 25 milliliters, plus the 30 milliliters, which we just added. I would get a hydro name concentration of 6.36 10 the negative five. We'll take the negative log of that in order to get our ph 3.20 So if we take all of these Ph. Is that we just determined at all of the particular volumes, we should be able to develop a tie tray shin curve that is very similar to this one right here.


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