Question

What volume of 0.500 M NaOH should be added to 2.50 g of oxobutanedioic acid (MW 132.073, pKas 2.56 and 4.37) to give a pH of 4.37 when diluted to 500 mL?

Answer 1

The pH is exactly at the SECOND ionization point

meaning that oxobutanedioic acid has been neutralized once (first proton) AND half of the 2nd species

mol of oxobutanedioic -> masS/MW = 2.5/132.073 = 0.0189 mol of diprotic acid

0.0189 mol of diprotic acid = 0.0189*2 = 0.0378 mol of H+

for pH = pKa2

pH = pKa2 + log(A-2/HA-) from buffer equation

A-2 = HA-

so

0.0189 mol of OH- required for FIRST point

1/2¨*0.0189 = 0.00945 mol of OH-

Total mol of OH- = 0.0189 +0.00945 = 0.02835 mol of OH-

then

V = mol/M

V = 0.02835 / 0.5

V = 0.0567Liters

V = 56.7 mL of NaOh must be added to 2.5 g then, add water up to V = 500 mL