Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H2A+: Ka1 = 4.6 x 10^-3, Ka2 = 2.0 x 10^-10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (d) 75.0 mL

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Hello. In this problem, we are told that 75 mL of a 750.125 molar solution of the prominent form of lysine was high traded with a heartfelt 0.5 mL of a 0.125 molar sodium hydroxide solution. We're asked to find the ph and PK values for the 1st and 2nd association are 2.34 and 9.78. Respectfully. So let's begin by writing then our two equations for the dissociation of our civic protons. We have the protein in the form of a scene Then which will react with our base. We'll remove the remains for 30. And the second equation over moved then our second acidic proton with the addition of more hydroxide. So we can write then shorthand notation for each of these. So for our coordinated form of glycerine, we'll call this H. Two A plus after removal one of the city protons, that will be H. A. And then after removal of two protons, that will be a minus. And so if we look at a situation plot to have ph sen is a function of the volume of hydroxide that's added will have two equivalents points marking the removal of too acidic protons. When we reached the halfway point. For the first equivalence point, the ph here will be equal to P. K. One. And when we reached the halfway point to the second equivalence point, PH will be equal to p. K. two. On our next step, we're going to calculate our moles of our coordinated form of glisten and our moles of sodium hydroxide. So we have 75 ml of our prone a form of glassine will convert this to leaders and then we'll make use of concentration Which is 0.125 holes. The leader and our volume of the former glisten in the leaders and leaders cancels. So we're left with moles. This works out to 0. 9375 nice are prone a form of glycerine And then calculating our M.nolds of Base. We have 112.5 no leaders at our base being added hurting from male leaders. Leaders and then make use of prostration of our base. Our units of the leaders of our base and leaders of our base cancel. And so we have been 0. rules of our base. And the next step then we're going to calculate our moles of pro former glycerine and reforms after removal of 1st and 2nd acidic proton. After we had seen hydroxide writing our first reaction for the dissociation of the first proton. We have initial change and final. So initially we have 0. moles of our program form of glycerine. And we have initially 0.14063 moles of our base will ignore water. Such it won't impact our ph initially we have none of the form of glycerine where the first proton has been removed. We will then assume that all of our in a form of lysine will react. That that means we have none of the printed form of cuisine left over. And we have 0.004688 moles of our base. And 0.009375 moles of protons form of glycerine. And then write the reaction for the second association And then making use of what we learned in the table above. And then we have initial change and final. So initially have 0. of the printed form of glycerine. They have 0.004688 moles of our base emotion or water. So it won't impact ph and we have none of the formalizing where both protons have been removed. So we will assume that all of our base react. So in the end there will be no base present and we will have 0.00468 moles of the component form apply scene and 0. roles of our slicing with both protons removed. In our last step then step forward. We're going to use the Henderson Hasselbach equation. Fine ph So we have a ph then is equal to the P. K. A. Because the log of our face to that of our acid and so are most of our base is equal to our molds of acid. That means then the concentration and these two will be the same. And so the log of one is equal to zero. So this will be equal to P. K. And this is P K A two. After we started to remove the second pacific proton and we were told that P K two was 9.78. So this will then be our ph After the addition of 112.5 ml of base. And if we look at our hydration plot, so that puts us right here. So we have removed half of the second acidic proton. And so the ph then will be equal to P K. A two, which is what we found Is at that point then the concentration of our asset and base components are equal. And so the Ph again will be 9.78. This corresponds then to answer a thanks for watching. Hope. This helped.

Consider the titration of 50.0 mL of a 0.100 M solution of the protonated form of the amino acid alanine (H2A+: Ka1 = 4.6 x 10^-3, Ka2 = 2.0 x 10^-10) with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of base. (d) 75.0 mL

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Key Concepts

Titration

Acid-Base Equilibria

Henderson-Hasselbalch Equation