Question

Calculate the molar concentrations of oxalic acid (COOH)₂, HOOCCO^2-, (CO₂)₂^2-, H₃O⁺, and OH^- in 0.15 M (COOH)₂ (aq).

This is a special case of general diprotic acid dissociation. First, we need to find the pK_a1 and pK_a2 in the water at room temperature, which is available online e.g. Wikipedia (oxalic acid). pK_a1 = 1.23, pK_a2 = 4.19.

Either solve numerically or analytically. Report here [H₃O⁺] = ______ M. 2 sig. figures.

Answer 1

0.15m o Nauj K9 Dissociation of oxalic acid ? tuo H2C2O4 Kaul HC2O4 +HT Initial Final (Equilibrium) 0-15- * XT X be the amount of dissociation [**] [HC204 -3 Bobsluhou 6. [H24204] Х Norovit

Given, POSH РКау - 1.23. We know rocha Қа, а M20.00 Pka, = -log(kas) — дана | 23 с - log (Kal) 0+058. І. • 058 - д45-X. х" agaxЗ. 0-058 Х +0.0 SRX - 8.4 х10 — О Arml%, midkoy atkaryo Х? - bT 3 We get, 562-4ac до — 0-658 - (0658)-— 4х) х - 8-7х10 2х) — 0 658 + (1952 • 068 5М 008)5 м : Equilibrium conc of CooH)₂ = 0.15-0.0685 [неј04 10-0685 м Гно*] - 0-0685 м.

Kaz HC 204 Initial 0.0685m Eqm. 0.0685- 685 %. C2042 +ht. Dell go od mo CY 10%) pa y y be the degree of dissociation PKaza 4.19800 na Pkaz - -log (kaz), 8200 4.19 - log (Kaz) Kaze 6+ 4 5 6 8 10- Kaz= [<20g 2 [ht] met 5 [mezogłom 6.4 56810 y2 0.0685-4. 8ddo 20. on Moe Y +6456x10 g- 4.42 x10-6 solving we get, y = 9.672x10-5 [120427= 9.672x10-5. PH= 1 (Pka, + PKG 2) = 1 / 2 (1+ 23 + 4, 19) 2. Mt 2800.0 2.71.

5.13x10-12. PH= -log [t zot 2.716 -log [H 3 ot [H3O+] = 1.95*10-3 [OH-] = KW [ht 10-14 1.95810-3