Question

What volume, in liters, of 1.03 M KOH solution should be added to a 0.110 L solution containing 10.21 g of glutamic acid hydrochloride (Glu, FW= 183.59 g/mol: pKa1= 2.23, pKa2= 4.42, pKa3= 9.95) to get to pH 10.24?.... but whenever i did it the way provided by a different user and got an answer of .036 L it said ....."Incorrect. You did not account for the amount of KOH needed to move glutamic acid through its first two acid dissociations to get to the HGlu– form."

Answer 1

First we calculate the moles of the acid

10.21g/183.59g/mol = 0.055 moles of acid

How many L KOH is that?
0.055moles/1.03 mol/L = 0.054L o 54 mL.

Therefore, it takes 54 mL of the KOH to remove the first H and another 54 mL to remove the second one.

Then the third is the pKa3 where you want the buffer to work.

mols HGlu^- to start = 0.055
mols Glu^2- = 0 at the beginning.
........HGlu^- + OH^- ==> Glu^2- + H2O
I.....0.055.....0.........0
add OH^-..........x...........
C......-x........-x........x
E.....0.055-x....0........x

Then plug into the HH equation
10.24 = 9.95 + log (x/0.055-x)
and solve for x = 0.036 mols KOH added for this part of the problem

. Convert that to L of 1.03M KOH

0.036mol/1.03mol/L = 0.035 L o 35 mL

and add to L from the first two neutralizations.

54 + 54 + 35 = 143 mL of KOH = 0.143 L of KOH