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. (a) 10.0 mL

Video transcript

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.