Final answer:
Acetic acid reacts with water to form H3O+ and CH3COO-. Adding NaOH can increase the formation of CH3COO- ions due to the removal of H3O+ ions from the equilibrium.
Step-by-step explanation:
Acetic acid reacts with water to form CH3COO- and H3O+. The reaction between acetic acid (CH3COOH) and water (H2O) is as follows:
CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)
The correct products of the reaction are option 3: CH3COO- + H3O+. This is an equilibrium reaction, where acetic acid donates a hydrogen ion to water, producing the acetate ion (CH3COO-) and the hydronium ion (H3O+). Factors that may affect the equilibrium and potentially increase the concentration of CH3COO- ions are the addition of substances that can alter the relative concentration of reactants and products.
For instance, adding NaOH (a strong base) will remove the H3O+ by forming water, thus shifting the equilibrium to the right and increasing the percent of acetic acid that ionizes. Alternatively, adding NaCH3CO2 will add more acetate ions to the solution, which through a common ion effect will suppress further ionization of acetic acid.