🎯 By the end of this guide,

You’ll be a pro at determining which side of an acid-base reaction is favored under equilibrium conditions. Whether you’re tackling a quiz, gearing up for a test, or jumping into a class discussion, you’ll handle these questions with ease and confidence. Ready to dive in? Let’s go!

🖐️ Before we start,

Make sure you’re familiar with:

• Brønsted-Lowry acids and bases
• Factors that affect the stability of conjugate bases

Challenge 1

Question:

Predict which side of the reaction is favored under equilibrium conditions

We need to determine which side of the equilibrium is favored: the reactant side or the product side.

Here’s how we’ll do it:

Step 1: Identify the base on either side of the equilibrium

🔍 Hint:

 

You’ll often deal with Brønsted-Lowry acids and bases in these types of questions. Remember, a Brønsted-Lowry acid donates a proton, while a Brønsted-Lowry base accepts one.

In this example,

Acetic acid donates a proton to form acetate, making it the acid and acetate its conjugate base.

Methoxide accepts a proton to form methanol, making it the base and methanol its conjugate acid.

Therefore, the bases are methoxide and acetate.

💡Quick Tip:

 

Bases in Brønsted-Lowry acid-base reactions typically carry a negative charge—this helps you quickly spot them.

Step 2: Compare the stability of these bases

Let’s determine which of the two bases is more stable.

To evaluate stability, consider all five factors in this order (Charge, Atom, Resonance, Induction, Orbitals)

Charge

What is the charge on each base?

Each base carries a negative one (-1) charge.

Therefore, this factor does not indicate which is more stable.

Atom

Which atom is holding the negative charge?

In both cases, the negative charge is on oxygen.

Therefore, this factor does not indicate which is more stable.

Resonance

Is the negative charge delocalized over multiple atoms?

For methoxide, the charge is ‘stuck’ to one oxygen. For acetate, the negative charge is spread over two oxygens, making it the more stable base.

Since we have successfully determined that acetate is a more stable base than methoxide by examining resonance, we don’t need to consider the other two factors—induction and orbitals.

Step 3: Equilibrium will favor the more stable base

The rule of thumb is that equilibrium will favor the formation of the more stable base.

If this base is on the reactant side, equilibrium will favor the reactant side. If it’s among the products, equilibrium will favor the product side.

Since acetate, the more stable base, is on the product side, equilibrium will favor the products.

Therefore,

📝 Note:

 

To visually indicate the direction in which the equilibrium favors either the reactants or the products, you can use one of these two arrows.