Question

Oxalic acid, which is formed either as (COOH)2 or H2C2O4, is a diprotic acid. At 25 degrees celecius ka1= 5.9 x 10^-2 and ka2= 6.4 x 10^-5.

Using just sodium oxalate and stock solutions of either 3.00 M HCL (aq), how many grams of sodium oxalate and milliliters of stock solution are needed to prepare 1.00 L of a buffer that has pH= 4.0 and contains 0.16 M oxalate ion?

Answer 1

for

pH = 4

pKa1 = -log(5.9*10^-2) = 1.23

pKa2 = -log(6.4*10^-5) = 4.1938

clearly, this is nearest to pKa2...

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

4.0 = 4.1938 + log(A-2 /HA)

(A-2 /HA) = 10^(4-4.1938)

(A-2 /HA) = 0.64

then...

if oxalate ion, A-2 must be 0.16 M

0.16 /HA = 0.64

HA-= 0.16/0.64 = 0.25 M

then...

total HA- + A- = 0.25 +0.16 = 0.41 M

A-2 = y - x = 0.16

HA- = 0 + x = 0.25

when we add "y" mol of sodium oxalate, then, add "x" mol of HCl

let y = 0.41 , i.e. the Total amount required

so

mass of Naoxalate = 0.41 *MW = 0.41 *134 = 54.94 g of Sodium Oxalate

now..

add 0.25 mol of HCl --> V= mol/M = 0.25 /3 = 0.08333 L of HCl solution

mL = 83.33 mL of HCl