5

An acid base buffer is able to resist changes in pH due to the addition of small amounts of strong acidor base to the system An acid-base buffer typically contains ...

Question

An acid base buffer is able to resist changes in pH due to the addition of small amounts of strong acidor base to the system An acid-base buffer typically contains weak acid and its conjugate base_Ist attemptPart 1 (1 point)J See Periodic TableSee HintAbuffer is prepared by mixing 26.2 mLof 0.209 MNaOHwith 105.3 mLof0.172 Macetic acid. What is the pH ofthis buffer? (The pKa for acetic acid is 4.75.)Part 2 (1 point)See HintCalculate the pH ofthis buffer after the addition of 0.179 g of solid NaOH

An acid base buffer is able to resist changes in pH due to the addition of small amounts of strong acidor base to the system An acid-base buffer typically contains weak acid and its conjugate base_ Ist attempt Part 1 (1 point) J See Periodic Table See Hint Abuffer is prepared by mixing 26.2 mLof 0.209 MNaOHwith 105.3 mLof0.172 Macetic acid. What is the pH ofthis buffer? (The pKa for acetic acid is 4.75.) Part 2 (1 point) See Hint Calculate the pH ofthis buffer after the addition of 0.179 g of solid NaOH:



Answers

A buffer is prepared by mixing $50.0 \mathrm{mL}$ of $0.200 M \mathrm{NaOH}$ with $100.0 \mathrm{mL}$ of $0.175 M$ acetic acid. a. What is the pH of the buffer? b. What is the pH of the buffer after $1.00 \mathrm{g}$ of $\mathrm{NaOH}$ is dissolved in it?

So we're being giving a buffer solution here and we're going to try to find its ph and then find the ph after adding some strong acid or some strong base. So first let's go ahead and look at the two components and find the moles of each. Don't make it easier for us to work with them. So I've got the polarity and the leaders. If I multiply those I'll get my moles 447 moles. Mhm. This is the H. C. Four age four oh six which is our weak acid. Mhm. And then our weak bases, its conjugate weak base. So we'll take its polarity times its volume and make sure you're in leaders. And we'll get the malls 0395 malls. Well this conjugate face. Mhm. Mhm. So we're gonna we definitely have a buffer. Right? We have a weak ass and its conjugate base. So we're going to try to find the ph of that. So I'm going to use H plus is K. A. Times the concentration of the acid over the base. But since I have moles, I'm gonna go ahead and rewrite that as ca the moles of my acid over the malls of my base. And we can do that because concentration is moles per liter, so moles per liter over moles per liter. You can see that the volumes are going to cross out and we're just left with moles. So we'll go ahead and plug in here. We'll use the K. A. For our weak acid right here, which is 455 Who was given to us on the problem, Time to send the negative five and then we'll multiply this by moles of acid, which is point 0447 Over base was 0395 So this will give us 5.15 times 10. The negative five. And this is the molar itty of R. H. Plus. So if we just do minus the log of that number will get our ph which is 4.288 So that's the ph of our buffer before we've added anything to it. Mhm. In part B though we're going to add some strong acid. Well the strong acid is going to react with the weak base so the H plus will react with our weak base. See for H 406 to minus. And it will make some weak acid. Okay, it's congregant weak acid here. So we'll use up some of the week base and will produce some more weak acid. So I'm gonna just write a table here where I do the initial change and final. So I'm adding 0.25 moles of H plus to these malls that we found above. Okay, so we have both of these to start. These are all in malls here. I hope you can see that the H. Pluses are limiting reactant. So we're gonna use up all of this. So we must use the same amount of our weak base and produce the same amount of our weak acid because it's all in a 1 to 1 to one ratio. So I'll end up with none of this. And I went 1.145 walls here. Mm 0.697 moles here. Mm Well we have a weak base and its conjugate weak acid. So we have a buffer. So again we'll go ahead and plug into our buffer equation. Or H plus is R K A 455 I'm still negative five. And then we're gonna do moles of acid over moles of base. So point 0697 moles over 6970.145 moles. You multiply that out. We've got 2.19 I'm telling negative four. And this is the molar itty of R. H. Plus. So if we take minus the log of that, we'll get our ph so minus the log of that number. Gives us 3.660 in the final step here, we're adding hydroxide. So a strong base. So that's going to react with our weak acid. So let's write that reaction. We've got the hydroxide reacting with our weak acid. Mhm. So that's going to make water and its conjugate base. So we're gonna use up some of our acid and produce some more base. So let's keep track of our initial change and final conditions here, I'm adding 0.25 moles of O. H minus to our original moles of our two substances here in our buffer. Okay 0395 We can see that the O. H. Minuses are limiting reactant. So this is all going to react. So I'll react the same amount of my weak acid and produce that same amount of my week base since the ratios are all 1 to 1 to one. So I'll end up with none of this and 0.197 malls of my weak acid 1.645 moles of my week base. So this is a weak acid and this is its conjugate weak base. So we have a buffer. So let's go ahead and plug in. Our equation H plus is R. K A. Which is 455 I'm sending negative five and then we're going to do the moles of acid over base with 0197 moles divided by 0.645 moles. So this will give us 1.39 I'm sending negative five. This is the molar itty of r h plus. So if we just take minus the log of that number, we'll get the ph The ph is 4.8 five seven.

Take a look at a buffer solution and calculating the ph based on some given information. It's a propane OIC acid and propaganda. Wait um, of our solution. We're going to Henderson Hasselbach equation basically looks like this says we can calculate the ph of a buffer solution. If we know the Peca of the acid and we know the ratio of the conjugate base to the acid that's used um, in creating the buffer solution. Will they have all of that information here? The only problem is they provided for us the K A value rather than the peak a value for the assets. But we know this that the peak A is minus the log of the K value. So if we utilize our calculator here and take The log of 1.4 times 10 to the -5. We had negative 4.85 but we need to take the negative of that value. And the PKK that is 4.85 here for that propane OIC acid. So let's plug in the information that we have now to calculate the ph of this buffer solution 4.85 for the PKA. And the concentration of the base they told us is .30 Mueller. The Asset .20 Mueller. All right. So this is basically a 3-2 racial. So if you take the log of 3/2, that should help us find the adjustment factor. We need to determine the ph of this buffer. So when we take that log we get a value of .18 we add that to the Peka. We end up getting a ph of our buffer solution of 5.03. So that's part a of the question to determine the ph of the buffer solution. Makes sense that a little greater than the PK value because we have a concentration in the buffer solution of a little more bass than we have acid. So that's going to make for a little bit higher ph buffer solution than the peak A value that we have. We had a very small concentration of our small volume rather of concentrated acid, one millimeter of 10.1 old Mueller hydrochloric acid to 10 million liters of the buffer solution were asked to calculate how the ph changes here. And since we're dealing with very small volumes, it's oftentimes a little bit easier to think about this in terms of instead of moles per liter for more clarity, million miles per mil a leader. And so these two concentrations First the acid at .2 old Mueller can be thought of as .20 million miles per millimeter. Same thing with the base 0.30 millimeters per millimeter. And if we take a 10 mil sample of the buffer solution as we're told that we do in part B, That means that we have two million moles fasted three million moles of base to that. We are going to um add .10 million moles of basically hydrogen ions from the hydrochloric acid. and so that's going to impact both the concentration of the acid and based components and my buffer for the acid component, We're now going to have an additional .10 million moles of hydrogen ion to increase its concentration to 2.1 Moeller or we'll we'll say 2.1 million moles of acid that we don't have available for the base. We're going to react away .1 million moles As it reacts with the hydrogen ion. So it's going to decrease its concentration to 2.9 million moles. And so we're going to compare the ratio now instead of the 3-2, ratio 2.9-2.1 and see how that impacts again the same PK value that can't change but all the log becomes instead of .3, as it originally was. Um we're gonna compare it now as 0.29. It's a point to one and let's go ahead and calculate that and see how that affects the ph the buffer. So when we complete the math, we find that For that little 10 sample of the buffer. The log value here is .14. And so um we add that to the original PK value. The new ph the buffer solution drops from 5.03 to 4.99. So we add that a small amount of concentrated acid to the small alec wat I guess of the buffer sample um we see a bit of a decrease in the ph the buffer. That's to be expected because we've added an asset to the buffer. Well, there's one more part to this question we're gonna take a look at now. In the final part of the question, we're adding a much larger concentration and amount of hydrochloric acid to the same 10 sample of our buffer solution. So we still have two million moles of our propane OIC acid and three million moles of the propane weight base available to react this time. What's a three mil sample of one molar concentration of hydrochloric acid That gives me then three million moles of hydrogen ions that I can react. And look, that's the same amount of base. Um Well, Mosul base that I have available to react. These are going to fully react and I'm going to basically consume all of the base so that I have none of it left. And it's all going to get converted to acid. So I'm going to have a total of 5.0 million moles of the propane OIC acid. And if I want to look at this in terms of the concentration that five million moles of acid is going to be floating around in a 10 million mL sample Plus the additional three ml Vol of the acid that I added. So it's going to be floating around essentially in a 13 miller liter um volume solution. So let's go ahead and calculate the molar itty of the acid. After performing the calculation here we have a .3 molar solution now of propane OIC acid, that's basically what we have not to calculate the ph of this solution. We would set up and I see table and used the K expression Kaye was given to us at one point as 1.4 times 10 on the negative five. We would know that this, it dissociates into propaganda weight and hydrogen ions. So um, X squared term on top and 0.38 minus X on the bottom. This is big enough concentration of smaller enough K value should be able to ignore the X. And when we solve for X, then we end up getting um a concentration then of um of the hydrogen ion .0023 Mueller. So to find the ph based on that hydrogen ion concentration, we're going to take the negative log of that value. And that gives us a ph 2.64 then for the solution. So what we've done in this final step is reacted away, um all of the base and so we've reached the capacity for the conjugate base component to buffer the solution. And so we see that it made a huge difference in ph once this was no longer a buffer solution. Once it no longer had some conjugate base available to buffer the solution. And so we dropped way down from a ph of five to below three, um based on the fact that again, we've reached and reacted away the entire buffer capacity of the conjugate base available here in that final step.

Okay. So in this problem we're going to have a buffer. We're gonna find its ph and then we're going to find its ph after we add either strong acid or strong base to it. So first let's go ahead and find the malls of each of the components. So I have .5 moller and .300 leaders to give me 150 malls Of our acid are weak acid which is H. two p. O. for negative. And then 317 more multiplied by its leaders. And that will give us the moles of our conjugate weak base. Okay. And those are gonna come in handy as we go forward here. So in the first one. Mhm. We're looking for the ph just of the buffer before we've added anything else to it. So that's gonna be each plus A. K. A. Times the concentration of the acid over the concentration of the base. Mhm. So we've got our K. Which is going to be 6.2 Time soon, the -8. And then we'll just multiply by the concentration of our Acid which was five moller and divided by the concentration of our base. And then we'll get our concentration of hydrogen ion which is 9.8, um 78 minus eight moller H plus. And then if we go ahead and take minus the log of that, we'll get the ph So for this ph is 7.01 for the buffer before we do anything to it. Okay. In the next part though, we're going to go ahead and add some strong acid while the strong acid H plus Is going to react with the component. That's the weak base. So it's going to react with the h. p. 0. 4 two minus to make some more weak acid To. p.-. So some of the weak bases used up and some of the weak acid is produced. Going to write a little table okay. Of initial change and final just to keep track of what's going on here. So I added put 05 malls of each plus To the .095 moles of hydrogen phosphate that I have. And I started with .15 moles of the di hydrogen phosphate. So you can see that h pulses are limiting reactant. So this is gonna be -15. So I must use up the same amount of this since it's 1-1 reaction And I'm gonna add .05 again cause it's 1-1 ratio. So I'll have none of this. And At the after the reaction has taken place I'll have .45 moles. And for our weak acid those up and we're going to have .2 00 moles. Okay? So if you look you can see that we have a weak acid, a weak base and its conjugate acid. So we have a buffer. So we'll go ahead and plug into our buffer equation again. All right. So H plus is a. K. A. Which is 62 times. Send the -8. And then we can do moles of acid over base in the same way as we did concentration because concentration is moles per liter over moles per liter. So we'll see that the volumes cancel out. And you can just use moles if you want to. So I'm gonna go with the moles of acid Which is .200 Over .0545. So that will give us our hydrogen ion concentration Which is going to be 28. I'm saying the -7. So that's the polarity of R H plus. So now we'll just go ahead and do minus the log of that And that will give us a ph of 656. So in our final example we're going to add hydroxide to our buffer. So the hydroxide is a strong base. So it's going to react with our weak acid. So the hydroxide will react with the di hydrogen phosphate to make some water and the conjugate base hydrogen phosphate. So again, we're going to use up some of the weak acid and produce some more weak base. We'll keep track of this with initial change and final table. We're adding .05 moles of this to the .15 moles of this. I started with and I had .095 moles of this. So hopefully you can see the hydroxide as are limiting reactant. So go ahead and use it all up. Use the same amounts here and add the same amount because it's all a 1-1-1 ratio. So you'll see we have none of this. Okay. and we end up with .10 moles of at age two P 04-. And we end up with .145 moles of the conjugate base. So again we've got a weak acid and a weak base. So we have a buffer so this is a buffer so we can go ahead and plug into our buffer equation. Arca is still 6.2 Time's Sermon -8. Again, we'll just do moles of acid over moles of base .100 malls Over .145 moles. And that will give us 4-8. Time to turn the -8 and that is the molar itty of r h plus. So if I take minus the log of that as I did earlier, I will get the ph the ph is 7.37

Okay. So in this problem we're going to have a buffer. We're gonna find its ph and then we're going to find its ph after we add either strong acid or strong base to it. So first let's go ahead and find the malls of each of the components. So I have .5 moller and .300 leaders to give me 150 malls Of our acid are weak acid which is H. two p. O. for negative. And then 317 more multiplied by its leaders. And that will give us the moles of our conjugate weak base. Okay. And those are gonna come in handy as we go forward here. So in the first one. Mhm. We're looking for the ph just of the buffer before we've added anything else to it. So that's gonna be each plus A. K. A. Times the concentration of the acid over the concentration of the base. Mhm. So we've got our K. Which is going to be 6.2 Time soon, the -8. And then we'll just multiply by the concentration of our Acid which was five moller and divided by the concentration of our base. And then we'll get our concentration of hydrogen ion which is 9.8, um 78 minus eight moller H plus. And then if we go ahead and take minus the log of that, we'll get the ph So for this ph is 7.01 for the buffer before we do anything to it. Okay. In the next part though, we're going to go ahead and add some strong acid while the strong acid H plus Is going to react with the component. That's the weak base. So it's going to react with the h. p. 0. 4 two minus to make some more weak acid To p. 04-. So some of the weak bases used up and some of the weak acid is produced. Going to write a little table okay. Of initial change and final just to keep track of what's going on here. So I added put 05 malls of each plus To the .095 moles of hydrogen phosphate that I have. And I started with .15 moles of the di hydrogen phosphate. So you can see that h pulses are limiting reactant. So this is gonna be -15. So I must use up the same amount of this since it's 1-1 reaction And I'm gonna add .05 again cause it's 1-1 ratio. So I'll have none of this. And At the after the reaction has taken place I'll have .45 moles. And for our weak acid those up and we're going to have .2 00 moles. Okay? So if you look you can see that we have a weak acid, a weak base and its conjugate acid. So we have a buffer. So we'll go ahead and plug into our buffer equation again. All right. So H plus is a. K. A. Which is 62 times. Send the -8. And then we can do moles of acid over base in the same way as we did concentration because concentration is moles per liter over moles per liter. So we'll see that the volumes cancel out. And you can just use moles if you want to. So I'm gonna go with the moles of acid Which is .200 Over .0545. So that will give us our hydrogen ion concentration Which is going to be 28. I'm saying the -7. So that's the polarity of R H plus. So now we'll just go ahead and do minus the log of that And that will give us a ph of 656. So in our final example we're going to add hydroxide to our buffer. So the hydroxide is a strong base. So it's going to react with our weak acid. So the hydroxide will react with the di hydrogen phosphate to make some water and the conjugate base hydrogen phosphate. So again, we're going to use up some of the weak acid and produce some more weak base. We'll keep track of this with initial change and final table. We're adding .05 moles of this to the .15 moles of this. I started with and I had .095 moles of this. So hopefully you can see the hydroxide as are limiting reactant. So go ahead and use it all up. Use the same amounts here and add the same amount because it's all a 1-1-1 ratio. So you'll see we have none of this. Okay. and we end up with .10 moles of at age two P 04-. And we end up with .145 moles of the conjugate base. So again we've got a weak acid and a weak base. So we have a buffer so this is a buffer so we can go ahead and plug into our buffer equation. Arca is still 6.2 Time's Sermon -8. Again, we'll just do moles of acid over moles of base .100 malls Over .145 moles. And that will give us 4-8. Time to turn the -8 and that is the molar itty of r h plus. So if I take minus the log of that as I did earlier, I will get the ph the ph is 7.37


Similar Solved Questions

5 answers
19 J (x + 1)3x 2) dx
19 J (x + 1)3x 2) dx...
5 answers
Point) Find all points of intersection (T; 0) of the curvescos(0) ; T = 2 sin(0) .Note_ In this problem the curves intersect at the pole and one other point: Only enter the answer for nonzero in the form (>, 0) with measured in radians_Point of intersection856, tan^-1(5/2))Next find the area inclosed in the intersection of the two graphsArea
point) Find all points of intersection (T; 0) of the curves cos(0) ; T = 2 sin(0) . Note_ In this problem the curves intersect at the pole and one other point: Only enter the answer for nonzero in the form (>, 0) with measured in radians_ Point of intersection 856, tan^-1(5/2)) Next find the area...
5 answers
Find the coordinates of the center of mass of the following solid with variable density:R={{xyz): 0 sxs8 0sys1,0szs5}; p(xyz)=3+
Find the coordinates of the center of mass of the following solid with variable density: R={{xyz): 0 sxs8 0sys1,0szs5}; p(xyz)=3+...
5 answers
(10 points) If this equation is correct; what is the base x?246, + 32, = 311,
(10 points) If this equation is correct; what is the base x? 246, + 32, = 311,...
4 answers
Pcove that rowi Opei Util 2) matrit of the form Partitioned (A I1] matrif of the form to the final I|aj] and I is Qn identity an akn matrix ~A here A Ts A is inverti Lle: matrix Assume that
Pcove that rowi Opei Util 2) matrit of the form Partitioned (A I1] matrif of the form to the final I|aj] and I is Qn identity an akn matrix ~A here A Ts A is inverti Lle: matrix Assume that...
5 answers
Question 6Tell whether the infinite series converges or diverges (Justify) If it converges, find its sum2o-2'(b)32*
Question 6 Tell whether the infinite series converges or diverges (Justify) If it converges, find its sum 2o-2' (b) 32*...
5 answers
Find the distance between the two points. Round the result to the nearest hundredth if necessary. $$(-6,-2),(-3,-5)$$
Find the distance between the two points. Round the result to the nearest hundredth if necessary. $$(-6,-2),(-3,-5)$$...
3 answers
Create and solve six (6) non-linear equations using NewtonRaphson Method.
Create and solve six (6) non-linear equations using Newton Raphson Method....
5 answers
5.For the following reactions, complete the reactions by giving what is missing This can be either the starting material, reagent(s) O product: Draw the intermediates for each major step for full credit: (5 pts ea) [NaOMeCicces H;OEICIAclKMnO4 MMl- X(CHAMLAHPCCNaOEt21,0-Brz FeBryHINO;, H-SO= Mg CO: H,o 6SOCI; 7,EtOH Pyridine
5.For the following reactions, complete the reactions by giving what is missing This can be either the starting material, reagent(s) O product: Draw the intermediates for each major step for full credit: (5 pts ea) [NaOMe Cicces H;O EICIAcl KMnO4 MMl- X(CHAM LAH PCC NaOEt 21,0- Brz FeBry HINO;, H-...
5 answers
Which of the following matches the general description andfunction of the mitochondria?Multiple ChoiceSite for synthesis of proteins, lipids andcarbohydrates. Organelle that breaks down nutrients.Organelle that processes, sorts and sends information.Control Center. Contains DNA and is involved in InformationProcessing.Organelle that synthesizes proteins.It is the organelle where ATP (energy) production occurs.
Which of the following matches the general description and function of the mitochondria? Multiple Choice Site for synthesis of proteins, lipids and carbohydrates. Organelle that breaks down nutrients. Organelle that processes, sorts and sends information. Control Center. Contains DNA and is involve...
5 answers
Let S be the solid obtained by rotating theregion shown in the figure about the y-axis.(Assume a = 3 and b = 1.)Sketch a typical approximating shell.What are its circumference c andheight h?c(x) =2Ï€xh(x) = ?Use shells to find thevolume V of S.V = ?
Let S be the solid obtained by rotating the region shown in the figure about the y-axis. (Assume a = 3 and b = 1.) Sketch a typical approximating shell. What are its circumference c and height h? c(x) =2Ï€x h(x) = ? Use shells to find the volume V of S. V = ?...
3 answers
Match thc followingCorynebacterium diphtherizeWunginection prcvalcnt In AIDS putlerits CarrinoYcrsinla pestisMycobac terium_ 4iumtoxin Uln # torIbosoinecatchaenedcontDacNelssoria gonorthoc c
Match thc following Corynebacterium diphtherize Wunginection prcvalcnt In AIDS putlerits Carrino Ycrsinla pestis Mycobac terium_ 4ium toxin Uln # torIbosoine catchaenedcontDac Nelssoria gonorthoc c...
5 answers
QusedcnQ7) a)Simplify (n-2)4 ("+1) 6) In how many ways can six people be arranged in FOW for a group picture? Attach File Browse Local Files Bror se Conteni Collection
Qusedcn Q7) a)Simplify (n-2)4 ("+1) 6) In how many ways can six people be arranged in FOW for a group picture? Attach File Browse Local Files Bror se Conteni Collection...
5 answers
Supsc Ia+ Ak + 0 ~ox K= | n:Use Aasrange maltcplierS to Fd +he maxi um und Minimum o ~re Furtiin Varribls Yuj ~"1 Xn S Cx _,XJ = Ov Xz K=j9 : Y;2 + + Xp2 = ( Observc tc muy #-Qler 4 CAn net be 0
Supsc Ia+ Ak + 0 ~ox K= | n: Use Aasrange maltcplierS to Fd +he maxi um und Minimum o ~re Furtiin Varribls Yuj ~"1 Xn S Cx _,XJ = Ov Xz K=j 9 : Y;2 + + Xp2 = ( Observc tc muy #-Qler 4 CAn net be 0...
5 answers
Point) Find the slope of the tangent line to the curve Jr cos(Sx 4y) Xetr e-7nf2at the point (71/2,6r) -
point) Find the slope of the tangent line to the curve Jr cos(Sx 4y) Xetr e-7nf2 at the point (71/2,6r) -...
5 answers
Ldlceon Wnllmg LAlLhua Mcchuuhn DeculunlLuiklialiFhr tullanIntAEublabu Tu[tu4 InhulcadnenntaWbecmahieHNcHONte dc Dinltion |Whut utbc mcchante used lo ! TKispans ued Iauh parts ? Waang oicich an " Ilow do Ihe Fana JIn 4erult thcu functu "MI SICAL CIHACTEKAS Aahan_IVe cokx #aralIMTLICATotWhv dr tlic Inc Iinisn Luc 04 hanc_Vefur cle ? Mmd 7Wtcn /Wite" Why "
Ldlceon Wnllmg LAl Lhua Mcchuuhn Deculunl Luikliali Fhr tullanInt AEublabu Tu[tu4 Inhulcadnennta Wbecmahie HNcHONte dc Dinltion | Whut utbc mcchante used lo ! TKis pans ued Iauh parts ? Waang oicich an " Ilow do Ihe Fana JIn 4erult thcu functu " MI SICAL CIHACTEKAS Aahan_IVe cokx #aral IMT...

-- 0.019352--