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. (a) 10.0 mL

Hello. In this problem we are told 75 mL of a 750.125 molar solution of protein in the form of glycerine high traded with 12.5 mL of a 0.125 miller sodium hydroxide solution. Grass to find the ph of the P. K. Values for the 1st and 2nd association are 2.34 and 9.78. Respectfully, let's begin by writing the ionization equations. So we have our coordinated form slicing which will react with sodium hydroxide or eliminate the sodium ions will form the uncoordinated form pricing in water on the addition of additional sitting hydroxide, we will remove the second acidic proton and form water so we can write these in shorthand notation. So the prototype for my blessing. We can write is H. Two A bus. After the removal of the first acidic proton, we call this H. A. And then after removal of the second acidic proton A minus. So if we look at a situation plot we have P. H. As a function of the volume of this being added. Then we'll have two equivalence point. And so halfway to the first equivalent point, the ph at that point will be able to P. K. One and halfway to the second equivalence point. P H. Will be equal to P. K. Two. And so on our second step, let's find our roles of appropriate form of light seen and are moles of sodium hydroxide. So we have 75 ml prone a former glassine will convert that to leaders and then we'll make use of the concentration find moles. So our units of milliliters and leaders of the Former Lyceum cancel. And we're left with moles. This works out to then 0. moles. And then calculating our moles base. We have 12.5 moles. Sorry, milliliters of sodium hydroxide. About convert that from the leaders to leaders. And make use of this concentration fine walls And so our male leaders are sodium hydroxide and our leaders of sodium hydroxide cancels. And this works out to 0. 15625 knows of our base. In our next step find moles of formalizing and the most of the uncoordinated form of licensing after we add didn't hydroxide writing our first equation for the removal of the first acidic proton. Yeah, I'm initial change and final. So We initially have 0.009375 moles of the informal icing. We have 0.0015625 moles of hydroxide. We initially have none of our unprogrammed form of glycerine and water since it won't impact our ph. So let's assume then that all of the hydroxide reacts. Then we'll have no hydroxide left at the end will have 0. rules of the road in a form of lysine and 0.15 65 moles of impertinent form of glycerine. In our last step. Then we'll use the Henderson. Hasselbach equation to calculate ph and so we have the ph then is equal to P. K. Plus the log of our space frustration to that of our acid, given that our concentrations would have the same volume. So we'd have moles of base form per unit volume divided by the moles of our civic form being a volume, our units of volume would cancel. So we don't need to bother with including volume units to be there for them have the ph is equal to P K. In this case A B p K one. Since we're still have some of the um for a lot of the poor in a form of glycerin left. So PK one is 2.34 and then we have the log of Our base 0.0015625. Try to buy that for our acid component and this works out to them 1.64. So if we go back and look at our hydration plot. So at this point then We haven't reached our PK one Which is equal to 2.34. We found the ph to be 1.64. So that means that we are somewhere then in this region. And so we still have more of our acid form than our base form passive form being the appropriate form of licensing. And so our ph then Is less than the PK. And it's equal to 1.64. His corresponds then to answer c. Thanks for watching. Hope this helps.

Ch.17 - Applications of Aqueous Equilibria

All textbooks

McMurry 8th Edition

Ch.17 - Applications of Aqueous Equilibria

Problem 95a

Previous problem

Next problem

Chapter 17, Problem 95a

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. (a) 10.0 mL

Verified step by step guidance

Step 1: Determine the initial moles of the protonated form of alanine (H2A+). Use the formula: moles = concentration × volume. Here, the initial concentration is 0.100 M and the volume is 50.0 mL.

Step 2: Calculate the moles of NaOH added. Use the formula: moles = concentration × volume. The concentration of NaOH is 0.100 M and the volume added is 10.0 mL.

Step 3: Determine the reaction between H2A+ and NaOH. The reaction is: H2A+ + OH- → HA + H2O. Calculate the moles of H2A+ and OH- after the reaction.

Step 4: Identify the species present in the solution after the reaction. Since NaOH is a strong base, it will react completely with H2A+. Determine the remaining moles of H2A+ and the moles of HA formed.

Step 5: Calculate the pH of the solution. Since HA is a weak acid, use the Henderson-Hasselbalch equation: pH = pKa1 + log([A-]/[HA]) to find the pH, where pKa1 corresponds to the first dissociation constant of H2A+.

Verified Solution

Video duration:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

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 can be monitored to understand the acid-base behavior of the solution.

Acid-Base Equilibria

Acid-base equilibria describe the balance between protonated and deprotonated forms of a substance in solution. For alanine, the two dissociation constants (Ka1 and Ka2) indicate how readily it donates protons. Understanding these constants is crucial for calculating the pH at various points during the titration, particularly before and after the equivalence point.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation provides a way to calculate the pH of a buffer solution based on the concentrations of the acid and its conjugate base. It is expressed as pH = pKa + log([A-]/[HA]). This equation is particularly useful in titration scenarios, as it allows for the determination of pH at points where both the acid and its conjugate base are present, such as after the addition of a specific volume of NaOH.