Final answer:
Adding 0.50 mol CH3COO−(aq) to 1 L of 0.50 mol CH3CO2H solution creates a buffer that minimizes pH change. It will result in some CH3COO− reacting with H3O+ to form more CH3CO2H and reestablish equilibrium, but it will not necessarily result in a pH of 7.00.
Step-by-step explanation:
When you add 0.50 mol of CH3COO−(aq) to a solution containing 0.50 mol of CH3CO2H in 1.0 L of water, you create a buffer system. The resultant solution consists of a weak acid (acetic acid, CH3CO2H) and its conjugate base (acetate ion, CH3COO−), which is the salt of a weak acid and a strong base. This will not result in a pH of 7.00, even though the concentrations are equal, because acetic acid is a weak acid and its dissociation in water is limited. Instead, this buffer system will help to maintain a relatively stable pH when small amounts of strong acid or base are added.
Adding the conjugate base in equal molar amounts to the weak acid establishes an equilibrium that minimizes changes in pH. This is due to Le Chatelier's Principle, which states that a change in conditions will cause the equilibrium to shift to counteract the change. In this case, option 3 is correct: Some CH3COO−(aq) will react with H3O+, increasing the concentration of CH3CO2H(aq) and reestablishing the solution equilibrium. As the conjugate base (CH3COO−) reacts with the hydronium ions present in the solution, the equilibrium shifts to the left, reducing the amount of dissociated ions and thereby maintaining the pH.
Therefore, adding 0.50 mol CH3COO−(aq) to the acetic acid solution increases the buffer capacity. The pH may change slightly due to the buffer action, but not as significantly as it would in pure water.