Question

How many grams of dipotassium oxalate (FM 166.22) should be added to 20.0 mL of 0.800 M HClO₄ to give a pH of 4.40 when the solution is diluted to 450. mL? (pK_a1 = 1.27 and pK_a2 = 4.266)
?g

Answer 1

Given that; If the final pH is 4.40, that's is near to the pKa2 for H2C2O4

The Henderson-Hasselbalch equation for this buffer is

pH = pKa + log ([C2O4 2-] / [HC2O4-])
4.40 = 4.266+ log ([C2O4 2-] / [HC2O4-])

log ([C2O4 2-] / [HC2O4-]) = 0.134

([C2O4 2-] / [HC2O4-]) = 1.36

If we add K2C2O4 (a strong electrolyte which splits completely into K+ and C2O4 2- when dissolved in water) to HClO4 (a strong acid which splits completely into H+ and ClO4- in water) we get the following reaction:

C2O4 2- + H+ ==> HC2O4-

moles H+ = M HClO4 x L HClO4 = (0.800)(0.020) = 0.016 moles H+

When that H+ completely reacts with C2O4 2- then 0.016 moles of HC2O4- will be formed according to the equation above. But we know from the Henderson-Hasselbalch equation that we want the ratio
([C2O4 2-] / [HC2O4-]) to be 1.36 after the reaction of C2O4 2- with H+.

Since they are in the same solution, we can write ([C2O4 2-] / [HC2O4-]) as (moles C2O4 2- / moles HC2O4-).

(moles C2O4 2- / 0.016) = 1.36moles

C2O4 2- = 0.02176 moles C2O4 2- after the reaction.

But we also needed to add 0.016 moles of C2O4 2- to react with the H+ in the first place. So total moles C2O4 2- = 0.016 + 0.02176

= 0.03776 moles C2O4 2- total

These number of moles is equal to the moles of K2C2O4.

0.03776 moles K2C2O4 x (166.22 g K2C2O4 / 1 mole K2C2O4)

= 6.28 g K2C2O4.