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. (e) 100.0 mL

Hello everyone today. We have the following problem. What is the ph for the tight rations of 55 mL of a $550. solution of the protein in the form of valine. That has the following K values with 110 mL of 1100.15 molar potassium hydroxide. So this symbol is to show that veiling is di protic and this is why it has to K. A. Values. So now that that's out of the way we can go ahead and solve for this problem. What we're gonna do is we're gonna solve for our molds of acid and base. So first we will start with our modes of acid and in this case that's the protein ated form of veiling. So how do we do that? Well we are going to take our polarity of alien that we have which was 0.15 moller however similarities and units of moles per liter. So we're going to say this is 0.15 moles per one liter and then we're going to multiply by the volume that was given to us which was a 55 miLA leaders. However, we cannot keep it as male leaders and we must convert to leaders by using a conversion factor that 1000 mL is equal to one liter. And when our units all cancel out, we're left with 0. moles of our veiling. Next we're gonna solve for our moles of our base and we're gonna do the same procedure there. We're going to use. And we are going to use our polarity Which was the 0.15 moles per liter. Most of potassium hydroxide over one leader And then we're going to take this and multiply by our volume which was ml. Of course we have to convert this and we're gonna do that by using the conversion factor that one leader is equal to 1000 militia leaders. Are you gonna cancel for that? However? We're gonna also need to multiply by our multi mode by using the multiple ratio because we need most of hydroxide not potassium hydroxide. So what we can do is we can put one mole of potassium hydroxide in the denominator here and then one mole Of hydroxide in the numerator that were most potassium hydroxide cancel out and we are left with 0.0165 moles of hydroxide. So we have our moles of asset, we have our moles of base. If we compare the two closely we can see that they're at a ratio of about one mole of our acid per two moles of our base. And this is gonna be the second equivalent point. And so at the equivalent when only the basic salt of the protein in the form of alien is present. So if we let Our a minus or conjugate base be our basic salt. What we can do is we can find the polarity of this or the concentration. And so how can we do that? Well from our veiling, we can take the number of moles that we have which was 0.00825 moles and divide that by the total volume, which was 110 mL plus our 55 mL. However, malaria is in terms of most per liter. So we have to convert this into leaders by using the conversion factor that one leader is equal to 1000 militia leaders. Are units cancel and we're left with 10000.5 Mueller for our concentration. Next, what we're gonna do is we're gonna show this conjugate base hide relies ng were essentially reacting with water and then we're gonna draw an ice table for it. So we're going to have our conjugate base there Reacting with our H 20. Water. And this is going to form our acid and our hydroxide ions. Then we're gonna form our ice chart. The initial concentration as we solved for for our conjugated base was 0.05 moles. It was zero for each of the products and water does not get anything because it is a constant. Since we are tight trading this, we're going to use our for our change, we're going to use X. And we're going to use minus X. For this. So first subtracted from the reactant we have to be adding to our products until the equilibrium is just the I plus the C. That's going to be 0.5 minus X. Then X. And X. For our products. Next we must solve for our KB or our constant for our base R K B. Let me go too soft for it. 20 people to our equilibrium constant over the KA four hour acid. In other words, that is our equilibrium constant divided by the second or the K. At the second equivalence point When we do those values together in a reference text is the equilibrium constant which is one times 10 to the negative 14th and then our K A R second K value was found in the question stem is 1.82 times 10 to the negative 10th. We get 5.5 times 10 to the negative fifth for our KB constant. Next we want to solve for our X. That we have in our chart here to find the concentration of hydroxide ions. So what we're gonna do is we're gonna set up our equilibrium equation which is products. Overreact mints, our products are the acid and the hydroxide and we represented those with excess. So it's going to be X times X. And then over the reactant which was 0.5 minus X. No, if we were to have to get rid of this X at the bottom to simplify this equation, we could say that if we were to take the conjugate acid that would disappear. If we were to take the conjugate acid and divide that bar K. B value. It would be Greater than 500, making X. Very small and thus negligible. So what we can do is we can rewrite this as X squared over 0.5 And we're gonna equal that to the K. B. value that we sold for which was 5.5 times 10 to the fifth. And when we solve for X We're going to get X is equal to 1.66 times 10 to the negative third. And of course this is the concentration of our hydroxide ions. Next, we're gonna solve for our P. O. H. Because we have the concentration of hydroxide ions, we can do that by negative log of our hydroxide concentrations which would be negative log of 1.66 times 10 to the negative third to give us a P O. H. Of 2.78. However, we need ph so we're gonna do P H plus D P O H Is equal to 14. That's the equation for that. Our ph is therefore going to be 14 minus R. P. O. H which is 2.78, giving us a final ph value of 11.22. And with that we have solved the problem overall, I hope this helped and until next time

Ch.17 - Applications of Aqueous Equilibria

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McMurry 8th Edition

Ch.17 - Applications of Aqueous Equilibria

Problem 95e

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Chapter 17, Problem 95e

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. (e) 100.0 mL

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Determine the initial moles of H2A+ in the solution using the formula: $ \text{moles} = \text{volume} \times \text{concentration} $.

Calculate the moles of NaOH added using the same formula: $ \text{moles} = \text{volume} \times \text{concentration} $.

Since 100.0 mL of NaOH is added, which is twice the initial volume of H2A+, determine the limiting reactant and the remaining species after the reaction.

Recognize that at this point, all H2A+ has been converted to A- (the fully deprotonated form), and the solution contains excess NaOH.

Calculate the pH of the solution by determining the concentration of OH- from the excess NaOH and using the relation $ \text{pH} = 14 - \text{pOH} $.

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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 (in this case, NaOH) to a solution containing the analyte (the protonated form of alanine) until the reaction reaches its equivalence point. The pH changes during this process, and understanding how to calculate the pH at various points is crucial for interpreting titration results.

Acid-Base Equilibria

Acid-base equilibria involve the dissociation of acids and bases in solution, characterized by their acid dissociation constants (Ka). For alanine, the two dissociation constants (Ka1 and Ka2) indicate the strength of the acid in its protonated forms. Understanding these constants helps predict the pH of the solution at different stages of titration, particularly when calculating the pH after adding a specific volume of base.

Buffer Solutions

Buffer solutions resist changes in pH upon the addition of small amounts of acid or base. In the context of the titration of alanine, the solution can act as a buffer before reaching the equivalence point, where both the protonated and deprotonated forms of the amino acid are present. Recognizing the buffer region and its capacity is essential for accurately calculating the pH after the addition of NaOH.