Posted by Decisice 2 on Thursday, May 23, 2013 at 6:55pm.
Can't you use the Henderston-Hasselbalch equation for this?
Not initially. Sapling learning says "Glutamic acid can act as a buffer at each of its three pka values. To solve this problem, you need to take into consideration the amount of KOH that must be added to the glutamic acid solution to move the pH through its first pka values. Therefore to get Glu into the HGlu- form (through H3Glu+ and H2Glu) two moles of KOH must first be added. Then the problem can be treated as a weak acid/strong base problem, which can be colved using pka3 and the Henderson Hasselback equation to determine the moles of OH- that are required to bring the pH of the glutamic acid solution to 10.20 as amixture of HClu- and Hlu2-....But i am still thoroughly confused!! Pleasee show steps
You shouldn't be. The "Sapling learning" tells you exactly how to do it.
How much KOH does it take for the first two acids? That's
mols glu acid = 9.1 g gluHCl/FW gluHCl = ?
It will take that many mols of KOH to neutralize the GluHCl and another mole to neutralize the next H. And that gets you to the HH equation. I assume you know how to go from there. When you find volume KOH for the third (buffer) part, add that volume to the volume to arrive at that point and you've answered the question.
I found that there are .04956 moles of KOH. Are you saying that there are 1.04956 moles total including the neutralized H? From there is the log "ratio" simply that number?
I calculated .04956 moles of Glutamic acid and a total of (twice as much) 1.877 moles of KOH needed. Using the pH given, the pKa3 for the HH. How do I solve for the liters from there?
If I understand what you've written, I don't think so. First I don't know where the 1.04956 came from.
Yes, 9.1/183.59 = 0.049567. I would round that to 0.0496 (which is still too many places). How many L KOH is that?
0.0496/0.94 = 0.0527 or 52.7 mL. Therefore, it takes 52.7 mL of the KOH to remove the first H and another 52.7 mL to remove the second one. Then the third is the pKa3 where you want the buffer to work.
mols HGlu^- to start = 0.0496
mols Glu^2- = 0 at the beginning.
........HGlu^- + OH^- ==> Glu^2- + H2O
I.....0.0496.....0.........0
add OH^-..........x...........
C......-x........-x........x
E.....0.0496-x....0........x
Then plug into the HH equation
10.20 = 9.95 + log (x/0.0496-x)
and solve for x = mols KOH added for this part of the problem. Convert that to L of 0.94M KOH needed and add to L from the first two neutralizations.
Remember to check the number of significant figures. You are allowed only two from the 9.1 and 0.94 numbers not counting the characteristic for the log value.
i solved for x and got x=.03175 moles. Do I simply divide that by .94moles of KOH to get liters?
I divided that by .94 moles and got .03377L and added that to .0527L of the first two parts to give a total of .086L total. Which was still marked incorrectly. What am I doing wrong?
You didn't add in all of the KOH. You should add 0.0527 for the first one, another 0.0527 for the second one, and 0.0338 for the third one which is more than 0.086 L. This is a lot of hand-holding here. :-)
I know, i'm sorry. Thanks for all of your help!!!
Thanks for hand holding me too