What volume, in liters, of 1.09 M KOH solution should be added to a 0.114 L solu

Question

What volume, in liters, of 1.09 M KOH solution should be added to a 0.114 L solution containing 9.60 g of glutamic acid hydrochloride (Glu, FW = 183.59 g/mol; pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95) to get to pH 10.27?

Hint:  Glutamic acid is a triprotic acid; at pH 10.27, Glu will exist as a mixture of HGlu– and Glu2–. However, this problem is not simply solved using the Henderson-Hasselbalch equation, as you must account for the number of moles of KOH required to get glutamic acid into the HGlu- form (i.e., it must pass through H3Glu and H2Glu).
***account for the amount of KOH needed to move glutamic acid through its first two acid dissociations to get to the HGlu–form.****

KOH volume:______

Every problem like this that I've read so far have been wrong, I just dont know what to do.

Explanation / Answer

First calculate the moles of glutamic acid as follows:

= 9.9/183.59 = 0.05.

Therefore 0.05 mols KOH needed so from M of KOH calculated the Volume of KOH as follows:

MKOH = mols KOH/L KOH
L KOH = mols KOH/M KOH

0.05/1.09 = 0.046 L or about 46 Ml

and that is one of the pKa values. Multiply that by 3 for total volume KOH required = 0.046*3= 0.138 ml.

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