🖐️ Before we begin,
Make sure you’re familiar with Brønsted-Lowry acids and bases.
Challenge 1
Question:
Calculate the value of the overall equilibrium constant K for this acid-base reaction
Ready? Let’s dive in!
Step 1: Identify the acid 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.
Methyl amine accepts a proton to form methyl ammonium, making it the base and methyl ammonium its conjugate acid.
So, the acids in this equilibrium are acetic acid (on the reactant side) and methylammonium (on the product side).
Step 2: Look up the pKa values for these acids
🗒️ Note: You’ll usually be provided with a pKa table during exams or you can find them in your textbook or a reliable online resource.
🔍 Key insight: The equilibrium will favor the side with the weaker acid (the acid with the higher pKa). In this case, methylammonium (higher pKa) is the weaker acid, so the equilibrium shifts toward the products.
📝 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.
Step 3: Calculate K using the correct formula
K is calculated using the following formula
K = 10−ΔpKa
where ΔpKa = pKa of reactant acid− pKa of product acid
Since the pKa of acetic acid is 4.76 and the pKa of methylammonium is 10.66
ΔpKa = 4.76 – 10.66 = -5.9
K = 10-(-5.9) = 7.94 × 105
An equilibrium constant K > 1 tells us the reaction favors the products—exactly what we predicted earlier! By comparing the pKa values, we expected the reaction to favor the weaker acid (methyl ammonium) on the product side, and the calculated K value confirms this.
