5

Consider the titration of 50.00 mL of 0.025M U*+ with 0.1000 M Cet+ In 1.0 M HzSO, (assume that [H + ] is also about 1.OM): a) Write a balanced litralion reactlon_ ...

Question

Consider the titration of 50.00 mL of 0.025M U*+ with 0.1000 M Cet+ In 1.0 M HzSO, (assume that [H + ] is also about 1.OM): a) Write a balanced litralion reactlon_ b) Calculato E at (he following volumes of Cett 5 mL, 25 mL, 25. ML Note: UOZt 4H+Ze H U"' 2Hz0 0.273 V Ce"t + #Ce E? =14v

Consider the titration of 50.00 mL of 0.025M U*+ with 0.1000 M Cet+ In 1.0 M HzSO, (assume that [H + ] is also about 1.OM): a) Write a balanced litralion reactlon_ b) Calculato E at (he following volumes of Cett 5 mL, 25 mL, 25. ML Note: UOZt 4H+Ze H U"' 2Hz0 0.273 V Ce"t + #Ce E? =14v



Answers

Consider the titration of $100.0 \mathrm{mL}$ of $0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}$ $\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)$ with $0.200 \mathrm{M} \mathrm{HNO}_{3} .$ Calculate the $\mathrm{pH}$ of the resulting solution after each of the following volumes of HNO $_{3}$ has been added. a. $0.0 \mathrm{mL}$ b. $20.0 \mathrm{mL}$ c. $25.0 \mathrm{mL}$ d. $150.0 \mathrm{mL}$ e. $200.0 \mathrm{mL}$ f. $250.0 \mathrm{mL}$

Again. These tight rations problems could be quite lengthy because there's multiple parts. Essentially, you're performing for this problem. Seven calculations. You've got the equivalence point volume and then six calculations of pH. So the first thing we need to do then is, as mentioned is calculate the equivalence point volume. If we have 100 mL of the week base, then we have 1000.1 leaders multiplied by the concentration at 0.1. Moeller gives us the moles of weak base. We know that the reaction is one toe one, so the moles of strong acid nitric acid that we need to add is equal to the molds of weak base we start with. Then we convert the moles of nitric acid into leaders of nitric acid by dividing by its similarity of 0.2. And then we go from leaders to Miller leaders and the equivalents. Point volume is 50 mil leaders of nitric acid at zero mill leaders, we haven't added any nitric acid, so all we have is the Week Basin solution to calculate the pH of a solution that just contains the week base. We calculate the hydroxide concentration first by rearranging the KB expression so that it is hydroxide is equal to the square root of Cabe, which was given to us at 3.0 times 10 to the negative six multiplied by the concentration at 60.1. Moeller also given to us. We then get a hydroxide concentration after square rooting of 5.477 times 10. The negative four to get the pH will take the negative log of the hydro knee um, concentration, which will be this hydroxide concentration divided in decay W at 1.0 times 10 the negative 14 and we get a pH of 10.74 at 20 million. Leaders were pre equivalents in the buffer region, so we can use the Henderson Hasselbach equation. PH will be equal to P k A, which will be the negative log of the K, a value which we don't have. But we can calculate by taking the K B value and dividing it into K. W. And then plus the log of the moles of base over the moles of acid. The moles of base still in solution will be equal to the moles of base We start with at 0.1. Leaders multiplied by a 0.1 concentration minus the moles of strong acid added, which will be the 20 mil leaders. Two leaders, multiplied by its concentration of point to for every mole of strong acid, we add, will decrease the moles of weak base by the same amount and increase the moles of weak acid by the same amount. So the moles of weak acid formed will be equal to the 0.2 leaders multiplied by the point to Moller night nitric acid. This then gives us a pH of 8.65 at 25 mL. We recognize that we are at half equivalent, so pH equals P k A. And we have a pH of 8.488 at 40 million liters. We are still pre equivalents because equivalence is 50 mL, so we can use the Henderson hassle. Baulch equation will set it up similar to what we did appear by taking the negative log of the K A value which we get by dividing KB indicate w. And then it will be the log of the ratio of the moles of base over the moles of acid, the moles of base. We start with is still the 0.1 times the 0.1, but now we've added 40 mL, so the moles that have been consumed will be the 400.4 leaders multiplied by the point to Mueller concentration of nitric acid. And then the moles of weak acid formed will be the moles of strong acid added, which is its volume 40 mL, or 400.4. Leaders multiplied by the concentration point to, and we get a pH of 7.88 at 50 mL were a T equivalents point. So all of the week base has been turned into a weak acid we can solve for the pH of a solution containing just a weak acid By rearranging the K expression for hydro knee, um, and has setting hydro knee um, equal to the square root of the K, a value multiplied by the concentration of the weak acid. The K A value has mentioned is K W or K B, which was given to us divided into K W. This is K A. Then we multiply it by the concentration of the weak acid that we have. The concentration of the weak acid is going to be equal to the moles of weak acid that we have, which corresponds to the moles of weak base we started with. So if we start with 0.1 leaders of 0.1 Moeller than we multiply those together to figure out the moles of weak base we start with which is now the moles of weak acid we have. We divide that by the new volume, 100 mL plus 50 mL is the 500.15 leaders. After square rooting, we get a value of 1.49 times 10 the negative five for the hydro knee. Um, concentration. We take the negative log of that in order to get our Ph of 4.83 now at 100 million. Leaders were 50 mL post equivalent, so we have 50 million leaders excess nitric acid. So we'll take our 50 mL excess multiplied by the concentration of nitric acid to give us the moles. Nitric acid in excess. Divide by the volume now 200 mL or point to leaders to give us the concentration of nitric acid in excess, which corresponds to the concentration of hydro knee um, in excess. We take the negative log, then of the concentration of hydro knee, um, in excess, and we get a pH of 1.3

Okay, so this will be the windiest caution in the whole chapter. You have to do one thing over and over again. Seem exactly so. We have few details in the given caution. Those are represented by this using the ice stable. We need to get a mind the change in initial and final equilibrium. Over here, the equilibrium constant gave me for the reaction has been given. We can calculate the equilibrium composition as follows, calculating the equilibrium concentration we have be Ohh value for the solution, which comes out to be three point 262 Using that, we can calculate the pH and the formula for Jacqueline Ph is 14.0, minus the value off P O. H, which turned out to be 10.74 in but be off the ocean. We have few details given over here. We can calculate the number off moles using the formula McGarrett in tow, the volume in Ledo's calculating the concentration, using the formula off hilarity, which turned out to be 0.5 mm uh, using the ice table again as the part we need to calculate the initial and finally change in the equilibrium off the solution. Getting the value off you. It's negative. We can find the value off pH using the same formula as earlier. Same is with the bod. See, over here. We need to construct a nice table. Sorry. The image was not provided. We can calculate the Gaby value to determine the concentration. Equally broom concentration for the solution. Using the same formula as earlier. We need to find the value off p, which and further down we can calculate the value off pH. The same is with process. The same is with the option D. The same is with the option e exact same steps you have to follow. Still auction if

You're part of the problem. The very first thing that we will need to do is you will have to find the initial moles of any Ohh. And how we do that is simple dimensional analysis going to take the value of any wage that we're adding and we're gonna multiply it by one leader is every 1000 milliliters multiplied by the moles of H. C O per every one liter of solution. And when we do this, we will get 0.58 moles. Now we need the volume of eight. CEO need it in our inner solution. So we're gonna take the moles we found before and want to play it by our 1 to 1 molar ratio of our reaction and multiply it for everyone. Leader of a solution. We have 0.75 moles of HCL and we will get 70 77.3 middle leaders And now we have to find the moles of HCL that we have consumed. So we're going to take five milliliters of H. C. O multiplied by everyone. Leader is 1000 with the leaders Times 0.7 rules for every one liter of solution and then again multiplied by our 1 to 1 ratio of our reaction. And now we need to find the sodium hydroxide that is remaining. So we take our initial molds that we found subtract the molds that we just found that is consumed in our reaction. And we will get our remaining bulls 0.542 molds. And now we have to find the polarity of sodium hydroxide. In this first part, she will take the molds that we just calculated, divided by our total volume 55 milliliters. Multiply it by 1000 for everyone. Leader to get this into leaders, and we will get our concentration or polarity 0.985 Moeller And now we're one step closer to finding our pH. Sir pH in our first part is 14 plus the log of our hydroxide concentration. And that's simple. Now we have all the information we need will take 14 plus the log of our calculation from before, and you will be able to get her pH, which is 12.993 now on department. You're gonna follow some of the same steps so we will find the moles of sodium hydroxide that's consumed. Who will take our 50 milliliters of HCL solution. Multiply it by everyone. Leader is 1000 milliliters times 0.75 Moles of H C E o bided for everyone leader and then again multiplied by our 1 to 1 ratio in our reaction and we will get 0.38 moles of an a O. H that is consumed. And now again, we have to find our house or sodium hydroxide that is remaining so you would take our initial moles. 0.580 Divide it. I'm sorry. Subtract it by the most we just found that were consumed and we will get 0.2 moles that is remaining in this part b of the reaction. But then again, we'll find the mole Larry of sodium hydroxide. We will take the moles we just previously found 0.2 Divide that for divide that for every 100 milliliters of our solution. And we have to multiply that by 1000 with the leaders for everyone leader, because remember Mole Aridjis and leaders and then we will get our mole Arat AEA's 0.2 to find the pH. We will take 14 plus the log of 0.2 and we will find that are ph for part B is 2.3 now on to part C. Again, we're gonna follow very, very similar procedure. We first need the moles of each seal that we add. So we will take 0.1 leader of HCL and multiply it by 0.1 mole of eight seal for everyone. Leader of solution we have. So you will get the molds that we add or 0.1 Then we find the moles of HCL that are consumed. And again we've already calculated this. This is 0.58 Now again, same step is the other parts we take the moles of HCL that is remaining. And how do we find that we find the initial 0.1 Subtract the most consumed and we will get the most remaining 0.4 And then we can calculate the molar ity of H CEO With this specific edition a volume we will take the mold. So we just found 0.4 Divide that by the volume of our solution you will get the polarity as 0.4 Now we can find the pH. All we have to do now is take the negative lock of our hard Droney. Um, concentrations will take the negative log of 0.4 And we seem to get 1.4. And this is how we find ph in three different scenarios volume.

So now we'll work on Problem 46 from Chapter 17. In this problem, we're asked to consider the Thai tradition of a's 30.0 millilitres of sample of 0.50 Mueller Moughniyah with 0.25 Mueller Hcea. So let's calculate the pH after the volumes of Titan has been added. So first, let's calculate the amount of ammonia we have at the beginning, and so we multiply the concentration times the volume to get a value of 0.0 15 moles of ammonia. So in part, they were supposed to calculate the pH with zero millilitres added. So we only have ammonia, so we make a nice table in H three plus H 20 is an equilibrium with N h for plus plus O. H minus. So our initial concentration of ammonia 0.0 five and we subtract X and addicts to both sides. So we get an equilibrium of 0.5 minus x x, so we can set this equal to the K B, which is equal to 1.8 times 10 to the minus five, which is equal to X squared over 0.5 minus X which we disregard and we get X is equal to 9.49 times 10 to the minus four. And if we take the negative log of this, we get P. O. H is equal to 3.2 and pH is equal to 10.98 which is our value of the pH for part A in part B. Ah, we've added 20,000,000 milliliters of acid. So we calculate the moles of acid that we've added by multiplying the concentration times the volume that gives us 0.5 moles of HCL which we can subtract from the moles of N H three that we initially had to find the most of an issue that we have left. So this gives us a value of 0.1 Moles, uh, HC n h three remaining. So we can go ahead and calculate a concentration from this by dividing by our new volume, which is 50 milliliters, and we get 0.0 to Mueller and then we look at the we change our expression for ke me to be cool to one point intense 10 to the minus five. And then we modify our initial concentration of ammonia to be 0.2 a minus. X and X is now equal to six times 10 to the minus four, which gives a P O. H of 3.22 and a pH of 10.78 Yeah, for part C, we do the same process. This time we've added 59 Miller leaders, so we calculate the added H C E O to be equal to 0.0 one for eight moles. So now we calculate the moles of any age three left to be equal to 2.5 times 10 to the minus five, uh, models of an age three. So then our K B is unchanged 1.8 times 10 to the minus five. Our initial concentration of N H three has changed to be 2.5 times 10 to the minus five oh minus X, and that gives us a value of X is equal to 2.12 comes 10 to the minus five, which gives a P. O. H equal to 4.67 and they pee age equal to 9.33 In party, we calculate the moles of HCL added as being equal to 0.15 mall's. If we look back at the beginning here in this first calculation, which we did, we see that this is the equivalence point. So now all of our ammonia has been converted to ammonium chlor ammonium. So we calculate our concentration of ammonium by dividing the moles by volume and we get a value of 0.167 moles so we can calculate a K A as being equal to 5.56 times 10 to the minus 10 and we can set this okay equal to the equilibrium expression. We had X squared over 0.167 minus x We disregard this minus X and we get X is equal to 3.5 times 10 to the minus six. And since we're dealing with an acid now, we can directly calculate pH as being equal to 5.52 In party, we now have excess acid. So the malls of HCL, which are added in this case when we have 61 millimeters, is equal to 0.15 three. And so we calculate the remaining after equivalence to be zero point 00003 malls which we can put into, Um, we can divide by the volume, which is your 0.91 leaders, and we get 2.75 times, 10 to the minus for Mueller. And if we take the negative log of that value will get a pH is equal to 3.56 in the last part of this problem here asks us about 65 Miller leaders. So this time it's possible for us to directly calculate the excess. Since we know that the equivalents point occurs at 60 milliliters, we can calculate the amount of moles that's in the excess five milliliters directly, and we get a value of 0.0 125 Moles of excess HCL. We could do that if we know the equivalents point for any volume above the equivalence point, and so we can't tell you the concentration by dividing these moles by the volume, which is 0.95 leaders, that gives us a value 0.132 holer, and this gives us a pH value when we take a negative log of 2.88


Similar Solved Questions

5 answers
Mapolnt TanApCalcBr1O 6.4.028. Evaluate the definite integral: J" es t :3 + 1) dtSubmit AnswerSave Progress
Mapolnt TanApCalcBr1O 6.4.028. Evaluate the definite integral: J" es t :3 + 1) dt Submit Answer Save Progress...
5 answers
In the sodium chloride unit cell, the chloride ions torm= cubu , which each side is arranged like the following figure The circles represent the positions of the chloride ions on one square face of tne cube All the other faces are Ihe same. What is the name of this unit cel?body-centered cubic x-lace cuDic face-centered cubic simple cubic chloride-centered cubic
In the sodium chloride unit cell, the chloride ions torm= cubu , which each side is arranged like the following figure The circles represent the positions of the chloride ions on one square face of tne cube All the other faces are Ihe same. What is the name of this unit cel? body-centered cubic x-l...
5 answers
Given the vectors, & = 4i - j,and V = -3i+2j , find 2u_ 4v(-20-10)(-20,10}(20,-10}(20,10}
Given the vectors, & = 4i - j,and V = -3i+2j , find 2u_ 4v (-20-10) (-20,10} (20,-10} (20,10}...
5 answers
CknsIFi244JcOn the graph are points that vou can raise and lower to draw sketch the histogram; histogram bars, Move these polnts accordingly in order to
Ckns IFi244Jc On the graph are points that vou can raise and lower to draw sketch the histogram; histogram bars, Move these polnts accordingly in order to...
5 answers
Use the proper Test; to show that the following series are convergent o divergent_ () Eaeo%" (6) >A03+7"
Use the proper Test; to show that the following series are convergent o divergent_ () Eaeo%" (6) >A03+7"...
5 answers
Pts ) Explain what meant by the Doppler effect? Why does the Doppler elect create more band broadening flame than in hollow cathode lamp?5. (2 pts ) Explain how thc dcutcrium lamp background correction technique works.(2 pls ) Explain #hat Jucnant spectral chemical . ad ionization interferenee in atomic spectroscopy
pts ) Explain what meant by the Doppler effect? Why does the Doppler elect create more band broadening flame than in hollow cathode lamp? 5. (2 pts ) Explain how thc dcutcrium lamp background correction technique works. (2 pls ) Explain #hat Jucnant spectral chemical . ad ionization interferenee i...
5 answers
Evaluate the integral: 3 5 arctanidx
Evaluate the integral: 3 5 arctani dx...
1 answers
Find the inverse of each $2 \times 2$ matrix using matrix multiplication, equality of matrices, and a system of equations. $$\left[\begin{array}{cc} 5 & -4 \\ 2 & 2 \end{array}\right]$$
Find the inverse of each $2 \times 2$ matrix using matrix multiplication, equality of matrices, and a system of equations. $$\left[\begin{array}{cc} 5 & -4 \\ 2 & 2 \end{array}\right]$$...
5 answers
Consider a biochemical reaction A --> B which is catalyzed bya oxidoreductase. Which of the following statements is true?The reaction will be most favorable at 0°C.The free energy change (ΔG) of the catalyzed reaction is thesame as for the uncatalyzed reaction.A component of the enzyme is transferred from A to B.The reaction will proceed until the enzyme concentrationdecreases.
Consider a biochemical reaction A --> B which is catalyzed by a oxidoreductase. Which of the following statements is true? The reaction will be most favorable at 0°C. The free energy change (ΔG) of the catalyzed reaction is the same as for the uncatalyzed reaction. A component of the enzym...
5 answers
I need all codes and answers for this problem using RStudioyields:43,52.5,46.5,52,47.2,44.1,47.8,49,52.6,52.4,57.3,53,48.7,59.4,48.1,59.6,52.2,50.5,58.5,55.1write all codes for finding 25th and 57th percentile.
i need all codes and answers for this problem using RStudioyields:43,52.5,46.5,52,47.2,44.1,47.8,49,52.6,52.4,57.3,53,48.7,59.4,48.1,59.6,52.2,50.5,58.5,55.1write all codes for finding 25th and 57th percentile....
5 answers
In the best Lewis structure, The central atom in BrFS has6 sigma bonds with no lone pair3 sigma bonds 2 lone pairs4 sigma bonds, 1 double bond; with 2 lone pairs4 sigma bonds 1 double bond, with no lone pair5 sigma bonds and 1 lone pair
In the best Lewis structure, The central atom in BrFS has 6 sigma bonds with no lone pair 3 sigma bonds 2 lone pairs 4 sigma bonds, 1 double bond; with 2 lone pairs 4 sigma bonds 1 double bond, with no lone pair 5 sigma bonds and 1 lone pair...
5 answers
8 Shone Let: Ifx Assume: Let: 27.Ix Statement: V Show ich Statement: 9+ Proof technique: and allou) the irrational. Con and y Ktx 8 1 Contradiction 1 T0 irrational 8 { 1 irrational or y then xy rallonal their exist irrational. ] Icr afionlx 7 or
8 Shone Let: Ifx Assume: Let: 27.Ix Statement: V Show ich Statement: 9+ Proof technique: and allou) the irrational. Con and y Ktx 8 1 Contradiction 1 T0 irrational 8 { 1 irrational or y then xy rallonal their exist irrational. ] Icr afionlx 7 or...
5 answers
#-Test; Two-sampwinCo Walmart 0,42 0,561 0,.58 0.98 0.98 148 2.67 1.18 1.78 3.83 4.118 3,48 4.12 0.88 0,88 0.,88 1,06 12.68 12.84 498 2.741 5.98 7.48 1,58 1.98 2.28 2.78 0.78 0.78 2.18 2.58 227 227 5.98 7.56 2.38 2.42 0.68 0.73 0.78 1.23 1,18 136 254 2.72 1.68 1.,98 1,28 1,38 3.28 4.78 0.88 112 198 1.18 | 126
#-Test; Two-samp winCo Walmart 0,42 0,561 0,.58 0.98 0.98 148 2.67 1.18 1.78 3.83 4.118 3,48 4.12 0.88 0,88 0.,88 1,06 12.68 12.84 498 2.741 5.98 7.48 1,58 1.98 2.28 2.78 0.78 0.78 2.18 2.58 227 227 5.98 7.56 2.38 2.42 0.68 0.73 0.78 1.23 1,18 136 254 2.72 1.68 1.,98 1,28 1,38 3.28 4.78 0.88 112 198...
5 answers
A ten coulomb charge is located at the origin. Find the electric potential at the point (2,0) (two meters from the origin on the positive X-axis) due to this charge:IOC7,,)What is the potential energy of the system if a - 5 coulomb charge is located at the point (2, 0)2
A ten coulomb charge is located at the origin. Find the electric potential at the point (2,0) (two meters from the origin on the positive X-axis) due to this charge: IOC 7,,) What is the potential energy of the system if a - 5 coulomb charge is located at the point (2, 0)2...
5 answers
H9.3) 30 pts) Consider the parametric curve defined byI(t) 3t2 y(t) = + 3tOn the interval _2 < t < 2Sketch the general shape of the curve and find its total length: Find the area of the surface generated by revolving the asteroid around the T-axis.
H9.3) 30 pts) Consider the parametric curve defined by I(t) 3t2 y(t) = + 3t On the interval _2 < t < 2 Sketch the general shape of the curve and find its total length: Find the area of the surface generated by revolving the asteroid around the T-axis....
5 answers
1.3. Sets 102: The Idea of a FunctionShow that if both f and g are one-(O-one, then $o is g 0 f. (This is (i) of Theorem [.1O , page 25.) (b) Show that if g 0 f is onto, then so is g. (This is (V) of Theorem L.lO page 25.) (c) Consider the functions f () et + and g(x) T2 . both with domain R and codomain R Show that g 0 f is one-to-one, but g is not_
1.3. Sets 102: The Idea of a Function Show that if both f and g are one-(O-one, then $o is g 0 f. (This is (i) of Theorem [.1O , page 25.) (b) Show that if g 0 f is onto, then so is g. (This is (V) of Theorem L.lO page 25.) (c) Consider the functions f () et + and g(x) T2 . both with domain R and co...
5 answers
12. Suppose the salaries of people that work at wideo gamecompany are normally distributed. If n 64 people are chosen at random and the average of their salaries is > S6U,(00 with & standard deviation 86.O00 , what is & 95% confidence interval for the true average salary of people that work for this COIQAnv?
12. Suppose the salaries of people that work at wideo gamecompany are normally distributed. If n 64 people are chosen at random and the average of their salaries is > S6U,(00 with & standard deviation 86.O00 , what is & 95% confidence interval for the true average salary of people that wo...
4 answers
Find (a) the slope of the curve at the given point P; and (b) an equation of the tangent Iine at Py = 3x2 +3; P( - 2,15)(a) The slope of the curve at P( - 2,15) is (Type an integer or a decimal:)(b) The equation of the line tangent to y = 3x2 +3 at P( - 2,15) is y = (Simplify your answer: Do not factor:)
Find (a) the slope of the curve at the given point P; and (b) an equation of the tangent Iine at P y = 3x2 +3; P( - 2,15) (a) The slope of the curve at P( - 2,15) is (Type an integer or a decimal:) (b) The equation of the line tangent to y = 3x2 +3 at P( - 2,15) is y = (Simplify your answer: Do not ...

-- 0.024435--