- Scholarli
- Posts
- How to predict the position of equilibrium without using pKa values
How to predict the position of equilibrium without using pKa values
A step-by-step approach to determining which side of an acid-base reaction is favored under equilibrium conditions.
Lauren
June 18, 2024 • Estimated Reading Time: 5 minutes
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
Here’s our challenge
🤔 Question:
Predict which side of the reaction is favored under equilibrium conditions
✅ Our Task:
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
We'll start by determining 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.
🗒️ Note: Since each base has a negative charge, we need to assess the stability of these charges using the other four factors: Atom, Resonance, Induction, and Orbitals. These factors will help us determine which base 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.
Induction
Are there electronegative atoms nearby that stabilize the charge?
Neither base benefits significantly from induction here.
Orbitals
How close is the orbital that houses the charge to the nucleus?
This isn’t a relevant factor in this case.
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.
📄 Ready to practice?
Check out the attached PDF worksheet! It’s designed to help you apply what you’ve learned and get some hands-on practice. Give it a go and see how much you've mastered!
|
Reply