Part III. Acetic acid is a weak acid and establishes the following equilibrium in water: HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq). In Part III, the methyl orange indicator is used to monitor the equilibrium shifts of the acetic acid/acetate ion system. The methyl orange equilibrium established with water is Hmo(aq)(orange-red) + H2O(l) H3O+(aq) + mo-(aq)(yellow). You compared the color of the solutions in three test tubes that initially contained 3 mL of 0.1 M acetic acid and a few drops of methyl orange indicator. In the first test tube, you added 1.0 M NaC2H3O2 drop wise. What color change was observed and what did this color change indicate about the shift in the methyl orange equilibrium?

In acidic solution , Due to presence of H⁺ ion , equilibrium shifts towards the left. So solution is orange . So in acetic acid solution , methyl orange will give orange/red colour.

When we add sodium acetate solution , it breaks into sodium ion which hydrolyses in water to give sodium hydroxide which is basic in nature due to giving OH⁻ ion . It shifts the balance of methyl orange to the right giving

Mo⁻ having yellow colour. Hence colour of solution becomes yellow.

lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-02-03T14:51:50+01:00

In this exercise we have to use the knowledge of balance to identify the color that will remain, in this way:

Yellow

Knowing that the formula given in the statement about equilibrium is:

[tex]CH_3COOH + H_2O \rightarrow CH_3COO^- + H_3O^+[/tex]

Equilibrium of methyl orange is as follows:

[tex]HM_o + H_2O \rightarrow M_o^- + H_3O^+[/tex]

In sour resolution, on account of vicinity of H⁺ ion the evenness shifts. When we adjoin sodium acetate resolution, it breaks into sodium ion that hydrolyses in water to present sodium hydroxide that is basic in type on account of bestowing OH⁻ ion. It shifts the balance of methyl orange to the right bestowing to yellow.

See more about equilibrium at brainly.com/question/13524990