Question

Calculate the volume, in liters, of 1.656 M KOH that must be added to a 0.125 L solution containing 9.44 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 = 9.44/ 183.59 = 0.0514

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

moles of KOH = 2 x 0.0514 = 0.1028 mol

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

0.0514    x                                 --------             initial

0.0514-x      0                                     x   -----------------> final moles

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

10.36 = 9.95 + log (x / 0.0514 -x)

(x / 0.0514 -x) = 2.57

x = 0.132 - 2.57 x

x= 0.037

total moles of OH - = 0.037 + 0.1028 = 0.1398

KOH volume = moles of KOH / molarity

                     = 0.1398 / 1.656

                     = 0.0844 L

volume of KOH = 0.0844 L