Question

What volume, in liters, of 1.03 M KOH solution should be added to a 0.110 L...

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."

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
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

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