(COOH)2 titration by NaOH

Question

Suppose that 0.242g of oxalic acid(COOH)2, is dissolved in 50ml of water.

a) estimate the ph of this solution.

b) Estimate the ph of the solution that results when 15ml of 0.150M NaOHaq is added to the oxalic acid solution.

c) Estimate the ph of the solution that results if an additional 5ml of NaOHaq solution is added to the solution from part b.

d) draw a titration curve that shows the stochiometric points, the half stochiometry points and the result of sections a, b and c.

Answer 1

HOOC-COOH + 2NaOH ==> NaOOC-COONa + 2H2O .. balanced equation for titration

molar mass oxalic acid = 90.0 g / mole

Ka1 = 5.9x10^-2

Ka2 = 6.4x10^-5

(a). 0.242 g x 1 mol / 90.0 g = 2.69x10^-3 moles / 50 ml = 0.0538 moles/L = 0.0538 M

Since Ka2 is about 1000 x less than Ka1, we will ignore it when determining the pH of this solution.

HOOC-COOH ==> HOOC-COO^- + H⁺

Ka1 = 5.9x10^-2 = [HOOC-COO^-][H⁺] / [HOOC-COOH]

5.0x10^-2 = (x)(x) / 0.0538 - x

use the quadratic equation and solve for x. Take negative log of x to get pH (should be slightly > 1.28)

(b). Since oxalic acid is a weak acid, and you are adding a strong base, you will create a buffer consisting of the weak acid (HOOC-COOH) and the conjugate base (HOOC-COO^-).

moles HOOC-COOH initially present = 2.69x10^-3 moles

moles NaOH added = 15 ml x 1 L / 1000 ml x 0.150 mol / L = 2.25x10^-3 moles

moles HOO-COOH left = 2.69x10^-3 moles - 2.25x10^-3 moles = 4.4x10-4 moles

moles HOOC-COO- formed = 2.25x10^-3 mols

Volume = 50 mls + 15 mls = 65 mls = 0.065 L

[HOOC-COO-] = 2.25x10-3 mol / 0.065 L = 0.0346 M

[HOOC-COOH] = 4.4x10-4 mol / 0.065 L = 0.00677 M

Using Henderson Hasselbalch to calculate pH, we have..

pH = pKa + log [conj.base]/[acid]

pH = 1.23 + log (0.0346/0.00677)

pH = 1.23 + 0.71

pH = 1.94

(c). When an additional 5 mls of NaOH is added, that will be a total of 20 ml of 0.15 M NaOH.

20 ml x 1 L / 1000 ml x 0.150 mol/L = 3x10^-3 moles NaOH added. This is in excess of the total moles of HOOC-COOH initially present (2.69x10^-3 mols), so all of the acid will be converted to the conjugate base, and you no longer have a buffer solution. In fact, there will now be excess NaOH (3.1x10^-4 moles) in a total volume of 70 mls (0.070 L). [NaOH] = 4.43x10^-3 M. pOH = 2.35 and pH = 11.6

(d). Can't draw on this platform. Look up "titration of oxalic acid with strong base" and you should be able to find a graph)