For the following reaction, the reactants are favored at equilibrium. Classify each of the reactants and products based on their strength as Bronsted-Lowry acids or bases. CH3COO- + H(CH3)3N+CH3COOH + (CH3)3N

See explanation below

Explanation:

First, let's write again the reaction:

CH₃COO⁻ + H(CH₃)₃N⁺  <> CH₃COOH + (CH₃)₃N

Now that the reaction is here, let's remember the basis of the bronsted - lowry theory:

An acid (HA) is a substance that can lose a proton (Hydrogen atom) to form a conjugate base. A base is a substance that accepts the proton (Hydrogen) and form a conjugate acid.

According to this definition, let's see the reaction again.

In the reactants, we see the CH3COO and the H(CH3)N. and the products are CH3COOH and (CH3)3N. The difference? well, we can see that the CH3COO now has a Hydrogen atom, this means that the CH3COO accepted the Hydrogen; this hydrogen was provided by the H(CH3)3N.

Therefore, the acid in this reaction is the H(CH₃)₃N⁺ and the conjugate base will be the (CH₃)₃N

The base in this reaction is the CH₃COO⁻ while the conjugate acid will be the CH₃COOH

the answer to your question is c