Question

3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient

quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.

Answer 1

The balanced equation for the reaction between oxalic acid and sodium hydroxide is H2C2O4 + 2NaOH --> Na2C2O4 + 2H2O.

Step-by-step explanation:

The balanced equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

In this equation, oxalic acid loses two protons (H+) and forms the polyatomic ion oxalate (C2O42-), while sodium hydroxide ionizes to produce sodium ions (Na+) and hydroxide ions (OH-). The reaction forms sodium oxalate (Na2C2O4) and water (H2O).

Answer 2

COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O

Step-by-step explanation:

The reaction of oxalic acid with a strong base like sodium hydroxide is the following:

COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O (1)

In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:

COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O (2)

The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):

COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O

I hope it helps you!