Question

Calculate the volume, in liters, of 1.616 M KOH that must be added to a 0.129 L solution containing 10.85 g of glutamic acid hydrochloride (H3Glu Cl–, MW = 183.59 g/mol) to achieve a pH of 10.17. Glutamic acid (Glu) is an amino acid with pKa values of pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95.

= ?? L

Answer 1

initial moles of Glu = mass / formula mass = 10.85/ 183.59 = 0.0591

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

moles of KOH = 2 x 0.0591 = 0.1182 mol

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

0.0591    x                                 --------             initial1

0.0591-x      0                                     x   -----------------> final moles

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

10.17 = 9.95 + log (x / 0.0591 -x)

(x / 0.0591 -x) = 1.660

x = 0.0981 - 1.660 x

x= 0.0369

total moles of OH - = 0.0369 + 0.1182 = 0.1551

KOH volume = moles of KOH / molarity

                     = 0.1551 / 1.00

                     = 0.155 L

volume of KOH = 0.155 L