Question

Calculate the volume, in liters, of 1.548 M KOH that must be added to a 0.117 L solution containing 10.78 g of glutamic acid hydrochloride (H3Glu Cl–, MW = 183.59 g/mol) to achieve a pH of 10.36. Glutamic acid (Glu) is an amino acid with pKa values of pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95.

Answer 1

initial moles of Glu = mass / formula mass = 10.78 / 183.59 = 0.05872

2 moles of OH- should be added per mole of Glu to get the HGlu- form.

moles of KOH = 2 x 0.0587 = 0.1174 mol

HGlu- +   OH- ----------------------> HGlu-2

0.0587    x                                 --------             initial

0.0587-x      0                                     x   -----------------> final moles

pH = pKa + log [Glu-2/HGlu-]

10.36 = 9.95 + log (x / 0.0587 -x)

(x / 0.0587 -x) = 2.57

x = 0.1509 - 2.57 x

x= 0.0423

total moles of OH - = 0.0423 + 0.1174 = 0.1597

KOH volume = moles of KOH / molarity

                     = 0.1597 / 1.548

                     = 0.103 L

volume of KOH = 0.103 L