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

See explanation

Explanation:

Let us look at the situation closely, the undissociated Hmo is red in colour while the mo^- is yellow in colour.

This means that an acidic solution will appear red while a basic solution will appear yellow in colour.

NaC2H3O2 is a basic salt. When it is introduced, the acetate ion pulls a proton from water forming  CH3COOH. CH3COOH is a weak acid and largely does not dissociate, it means that the undissociated acid remains in the solution leaving only the OH^- ions floating around thereby making the solution basic.

The colour of the system now changes to yellow.

Consider the equilibrium;

CH3COO^-(aq) + H2O(l) > CH3COOH(aq) + OH^-

Addition of NaC2H3O2  causes the equilibrium of this system to shift to the right.

The methyl orange water equilibrium also shifts towards the position where more mo^-(yellow) is produced.

your answer would be “dry”