Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H 2 A + : K a1 = 4.6 × 10 –3 , K a2 = 2.0 × 10 –10 ) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (b) 25.0 mL (c) 50.0 mL

Well everyone's in this problem being told that a thiamine is a compound that contains two amino groups and it can accept two protons since each nitrogen in the compound can accept a proton. So properly and I mean which is also known as one to die amino propane with a molecular formula of H. Two, N C three, H six and H two example of a diamond. So here we are given two different equilibrium reactions as well as the respective kB values. So we have kb one KB two. So and the problem is being told to determine the ph of the solution after the addition of the following volumes of 0.300 molar hcl to mL of the 300.200 molars of the properly in diamond. So we are given six different scenarios here. So let's go ahead and start before we get started on each specific scenario we have to go and calculate for the number of moles of H two and C three, H six and H two. So this equals two. First given our concentration of 20.200 moles or moller, we're gonna go ahead and multiply this by the units of volume. So we're given the 30 ml of course we know that the units of concentration is at least for volume as leaders. So we want to convert this ml into leaders. So for every one male leader here we have one times 10 to the negative three leaders potential calculator. We get the numerical value to be 0. units being moles, solving for the volume of the 0.3 molar hcl. The first equivalence point Is equal to the recently calculated moles that 0.006 moles divide by 0.300 moles per liter. We can tell this is units of concentration. We're gonna multiply this and convert our leaders into milliliters. So for every one times 10 to the negative three leaders, you get one male leader. Once you put this into the calculator, we get the new american value to equal to 20 units being milliliters. Alright, now we go ahead and start off with the first volume of this Hcl that's being added, which is just zero. So let's go ahead and create a nice table for this first reaction. So, here doing scenario one with 0 ml. Alright, so the equation here we have is just the one right over here. So just go ahead and rewrite that. So we have the H 2, 3, H six and H two. Which is this rocks with H 20. To give us a church three, N C three, H six and H two plus. This is a caddy on Now as well as one mole of our hydroxide. All right, so we have ice table again, I for initial C for change and a for equilibrium H oh here is a liquid. So we go and ignore everything here. So, from my initial concentration of my starting material then this is 0.200 molars from the two products here, this is both starting at zero for our change this is all unknown. So for my study materials that I will always be negative and my products that will always be positive. And since this unknown and in math we know that unknowns are just denoted as X. So that's what we'll put in our ice table. So for myself materials, this is a negative X. R. Two products here will be a plus X. And plus X equilibrium is just going to be combining values without we have an initial and change. So for Mr materials this is 0.200 molars minus X. And then my two products is just X. And X. School school and screwed over more space here. All right. So for my KB one value, this is equal to the concentration of my products. So we have H. Three, N. C. Three, H six and H two plus multiplied by the concentration of O. H minus divided by the concentration of my starting materials of H two, N. C. Three, H six and H two. All right. So, we'll go ahead and plug in some of the values that we do know. Starting with our KB one value that's given to us is equal to 6.61 times 10 to the negative five. From ice table. We have the concentration of this cast iron here to be X. The concentration of hydroxide to be X. And then for the concentration of master material, this is just 0.200 minus X. So this is known as equation one. So let's determine the value of the concentration of our product divided by RKB- one value. If this big this value is bigger than 500, then there's no need to use the quadratic formula. So let's do this over to the side maybe in green. So we have a concentration of H 2, 3 H six. NH two divided by R K B one value. This is equal to 0.200, divided by 6.61 times 10 to the negative five. Once you put this in your calculator, we see that this value is equal to 3.03 times 10 to the third power. So since This question here is bigger than 500, there is no need to use the quadratic formula. So X. In the denominator of the right side of equation one can be neglected. And we can see here that this is the equation one. We're seeing that this X here can be ignored. We'll continue our math here we have 6. times 10 to the negative five. You're going to X squared over 0.200. Here are basically solving for X. We multiply both sides by .2. This gives us 1.32, 2 times 10 to the negative five equaling two X squared and just square rooting both sides. We know that we have a positive value and a negative value but we can go ahead and disregard the negative value of X since there is no negative concentration. So my X value then will equal to just the positive value which is 3.6359 Times 10 to the -3. So we set the X value is equal to the concentration of R. O. H minus hydroxide. So we can go ahead and use this to calculate for our P O. H. P. O. H is equal to the negative log of the concentration of O. H. So plugging in this numerical value into our formula again, this is 3.6359 times 10 to the -3. Putting this into my calculator, I get the value to be 2.44. Of course this question is asking us for ph so what we can do to get ph is 14 minus R. P. O. H. Value. This value again is 2.44. This gives me a ph value to be 11.56. So my final answer is that my ph is equal to 11.6. So this is my final answer for our first scenario here, scroll and scroll down for the second. I will do a little black line here to kind of separate each scenario. Alright so for scenario two here we have 10 ml being added. The first equivalence point is at 20 ml and therefore 10 ml is going to be halfway to the first equivalence point. So my P. O. H. value is equal to the PK. B one value, which is equal to the negative log. That's right. This better to to the negative log of our KB one value. So this equal to the negative log Of 6.61 times 10 to the -5. Once you put this into a calculator we get that this value is equal to 4.18. Now getting my ph value then is just 14 minus R P O H. Which we calculate for to be 4.18. Once you put that into the calculator, get the ph value here Is equal to 9.82. So this is my final answer for the second scenario that we have. Again, we'll do this black line to separate each scenario here. Alright, so for my third scenario here is when we add 20 mL, the first equivalent point is at 20 mL, therefore the p O. H. Is equal to. Let's actually write this out. So again, we're doing the third scenario Where we have the 20 ml and we said that the p O H. Is equal to the pkB one value plus the p K B two divided by two. What I'm gonna go ahead and do here is expand our formula. So we have the negative log of our KB one value plus the negative log Of our KB two value. Again, this is divided by two, Plugging in these KB one KB 2 values have negative log of 6.61 times 10 to the - plus our negative log 4. times 10 to the -8 five x 2. Once you put that into the calculator, we get that the PO. H is equal to 5.79. Of course this problem is asking us for ph so ph nine is equal to 14 minus a P O. H. That we just solved word to be 5.79. Once you put this in the calculator, we get that the ph value is equal to 8.21. This is my final answer for our third scenario here. Starting off on the fresh page. Now, Our 4th scenario here is going to be our 30 ml. So 30 ml is halfway between the first equivalence point and the second equivalence point. So our P O H then is equal to the P K. B to value. And just expanding how we get this PKB to value. Using the formula of the negative log of this KB two value. So, plugging in this KB two value, we get the negative log Of 4.07 times 10 to the -8. Once you put that into the calculator, we get the value of 7.39 to be the POH. Now to get the ph this is equal to 14 minus R P O H. Which we just solve for to be 7.39. Once you put this in the calculator we get the ph value to equal to 6.61. So this is my final answer for scenario where we add 30 million years. All right now, moving on to the next scenario, we have 40 ml. Alright, so the second equivalence point is at 40 millimeters. So the concentration of our H three, N C three H six and H 32 plus is equal to actually let's write this out. So again we have this H three, N C three H six NH 32 plus this equals two. The moles of our H3 NC three H 6 At age 3 2 plus, divided by the initial volume of bass plus the volume of hcl that's being added. So again the concentration of H three N C three H six and H 32 plus. We're basically playing in these numerical values here. So for the moles, this 0.6 moles, Divide by 30 ml plus the ml. We actually want the smell layer units to be converted into leaders. Since that's the proper units for volume. For our concentration we do a quick dimensional analysis here. So for every one times 10 to the negative three liters, we have one millimeter. So once you put that into a calculator, we get that. The concentration of this cut ion is equal to 8.5714 times 10 to the negative to moller. So at the second equivalence point the solution will be acidic. So let's go ahead and create an ice chart for this second reaction. And of course we can't see this reaction in Morrison's all the way up there. But I will write our second reaction that's given to us in the problem. This location is when we have our H three NC three H 6 and H 3 2 Plus. This is reacting with one mole of H 20. And its liquid state. This will go ahead and yield H3 and C3 H 6, 2 Plus. As well as our hydro ni um H 30 plus. So of course in swimming. I see. Well we have the eye for initial C. For change and E for equilibrium HBO here will be ignored for ice table because it's in liquid. Alright, so for initial concentration here for my study materials, we already calculated this and this value is 8.5714 times 10 to the - molars from the two products here, there will both be zero. Now from a change here. These are all unknown, how would be known and unknown is of course by X. And this is how we do it in math as well. For my start materials this will have a negative sign for my two products will always be our positive. So then for our change here for this ice table, this product is going to be a negative X. Again my products will be positive. This is a plus X and a plus X. For equilibrium, we're just basically combining our values of the initial and our change starting off with this product here we have 8.5714 times 10 to the negative two molars. Sorry, I meant to start off with the certain materials. So we have this value of minus X. For my products. Now they will just be of course are positive X. Alright, let me go ahead and scroll down so far que a value is equal to the concentration of my products. So that's H three N C three H six, N H two plus, multiplied by the concentration of our H 30 plus divided by the concentration of our starting materials. This is H three, N C three H six and H 32 plus. So for this kitty on here, the K for this is equal to the kw over the KB to value. So then this is equal to one times 10 to the -14 divided 54.7 times 10 to the negative eight. This gives me that my K. Value is equal to 2. times 10 to the negative seven. So continue on with this. Right over here we have our K. Value that would just calculate it for it. This is 2.4570 times 10 to the negative seven. And from our rice table we have X. Multiplied by X divided by 8.5714 times 10 to the negative two molars minus X. And this is what we call equation too. We have to first determine the value of the H. Three, N. C. Three H. Six and H. 32 plus divided by the K. Value to see if this is bigger than 500. To see if we need to use the quadratic formula. So let's go ahead and do that. We can go ahead and just and green right over to the right. So again we need the concentration of H. three and C. Three H. Six and H. 32 plus divided by R. K. A value Is equal to 8. times 10 to the negative to moller Divided by 2.4570 times 10 to the -7. This value that we get, once you put this into your calculator is 3.49 times 10 to the fifth. So since this quotient here is a lot bigger than 500 there's no need to use the quadratic formula. So X. In the denominator of the right side of equation two can be neglected. So we're seeing here is that this X can be neglected. Alright, so continuing on with our math here. All right. So then after we consult this X. We have 2.4570 times 10 to the negative seven equaling two X squared divided by 8. times 10 to the negative two. So here is basically simplifying for X. So we multiply both sides by the denominator. So this big value here and once we do so we get the value to be 2.1060 times 10 to the negative eight equal to X squared. We go ahead and square root both sides. We do. So we know that we get a positive value and negative value for X. We can go ahead disregarding negative value of X since there is no negative concentration. So this means that my ex n. Is equal to my positive value which is 1.451 to 10 is 10 to the negative four. X. Is equal to the concentration of H. 30. Plus. So now we can go ahead and directly calculate for a P. H. So P H. Is equal to the negative log of a concentration of H 30. Plus. So this is just negative log on the concentration HBO plus which is X. And that is 1.4512 times 10 to the negative four. Once you put that into calculator received that my ph then is equal to 3.84. So this right here is going to be my final answer for this scenario. Alright, moving on to the next scenario and the last scenario, this scenario six. So this is where we have the 50 ml being added since 15 ml is already past the second equivalents points there will be an excess of our hcl. So We have our 50 ml -D. 40 ml multiplied by one times 10 to the negative three leaders divide by one millimeter. And this basically us taking this deduction here and changing the units of milliliters into leaders. And then we're gonna go ahead and multiply this by the concentration so that 0. moles per liter. So then this gives me a value of 0.3 units being moles of R H. C. L. Where the same thing as the moles of H 30. Plus. All right, So the concentration of H three L plus is equal to 0.003 moles divided by mL plus our 50 mL. Not converting this denominator millionaire unit into liters every one times 10 to the negative three leaders. We have one millimeter. So this gives me that my concentration of H +30 plus is equal to 3.75 times 10 to the negative two. We go ahead and now plug this into our ph equation. So there's that ph is equal to negative log of the concentration of H 30 plus is equal to the negative log, then of 3.75 times 10 to the negative two. Once you put that into the calculator, we get that my front of ph is equal to 1.43. So this right here is going to be my final and last answer for this problem.

Ch.17 - Applications of Aqueous Equilibria

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McMurry 8th Edition

Ch.17 - Applications of Aqueous Equilibria

Problem 95b,c

Chapter 17, Problem 95b,c

Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H₂A⁺: K_a1 = 4.6 × 10^–3, K_a2 = 2.0 × 10^–10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (b) 25.0 mL (c) 50.0 mL

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Determine the initial moles of H2A+ in the solution by multiplying the volume (50.0 mL) by the concentration (0.100 M). Convert the volume to liters before multiplying.

Calculate the moles of NaOH added by multiplying the volume of NaOH added (25.0 mL for part b and 50.0 mL for part c) by its concentration (0.100 M). Convert the volume to liters before multiplying.

For part (b), compare the moles of NaOH added to the initial moles of H2A+. Since 25.0 mL of NaOH is added, it is half the equivalence point, and the solution is a buffer. Use the Henderson-Hasselbalch equation to find the pH, considering the first dissociation (Ka1).

For part (c), since 50.0 mL of NaOH is added, it reaches the first equivalence point where all H2A+ is converted to HA. Calculate the pH by considering the hydrolysis of HA using the second dissociation constant (Ka2) and the concentration of HA at this point.

Use the appropriate equilibrium expressions and assumptions to solve for the pH in each case, ensuring to check if any simplifications can be made based on the relative magnitudes of Ka1 and Ka2.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration

Titration is a quantitative analytical technique used to determine the concentration of a solute in a solution. It involves the gradual addition of a titrant (a solution of known concentration) to a sample until a reaction is complete, indicated by a color change or a pH change. In this case, the titration of alanine with NaOH will help determine the pH at specific volumes of base added.

Acid-Base Equilibria

Acid-base equilibria involve the balance between proton donors (acids) and proton acceptors (bases) in a solution. The dissociation constants (Ka) of the amino acid alanine indicate its ability to donate protons at different pH levels. Understanding these equilibria is crucial for calculating the pH during titration, especially when determining the point at which the amino acid transitions from its protonated to deprotonated forms.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical formula used to calculate the pH of a buffer solution based on the concentration of the acid and its conjugate base. It is expressed as pH = pKa + log([A-]/[HA]). This equation is particularly useful in titration problems involving weak acids and their conjugate bases, such as alanine, as it allows for the determination of pH at various stages of the titration process.