Question

How many milliliters of 0.500 M KOH should be added to 2.38 g of oxalic acid (molecular weight: 90.03, pKa1=1.46; pKa2=4.40) to give a pH of 4.50 when diluted to 250 mL?

Answer 1

Write down the step-wise dissociation of oxalic acid as below:

(COOH)₂ (aq) <========> HOOC-COO^- (aq) + H⁺ (aq); pK_a1 = 1.46 ……(1)

HOOC-COO^- (aq) <=======> (COO^-)₂ (aq) + H⁺ (aq); pK_a2 = 4.40 …..(2)

Since the required pH is 4.50, we must work with pK_a2 and the weak acid/conjugate base pair of interest is HOOC-COO^-/(COO^-)₂. (COO^-)₂ is formed by the reaction of KOH on HOOC-COO^- as below:

HCOOH-COO^- + KOH -------> (COO^-)₂K⁺ + H₂O

The pH is given by the Henderson-Hasslebach equation as

pH = pK_a2 + log [(COO^-)₂]/[HOOC-COO^-]

====> 4.50 = 4.40 + log [(COO^-)₂]/[HOOC-COO^-]

====> log [(COO^-)₂]/[HOOC-COO^-] = 0.10

====> [(COO^-)₂]/[HOOC-COO^-] = 10^0.10 = 1.2589 ≈ 1.259 …..(3)

Now, moles oxalic acid added = (2.38 g)*(1 mole/90.03 g) = 0.0264 mole.

[HOOC-COO^-] = (0.0264 mole)/(0.250 L) = 0.1057 M (the final volume is 250 mL).

Therefore, [(COO^-)₂] = 1.259*[HOOC-COO^-] = 1.259*0.1057 M = 0.1331 M

Now [KOH] = [(COO^-)₂] since (COO^-)₂ is formed by the action of KOH on HOOC-COO^-.

Let V mL of 0.500 M KOH be added. Using the dilution equation,

V*0.500 M = (250 mL)*(0.1331 M)

====> V = 250*0.1331/0.500 mL = 66.55 mL (ans).