Answer: CH3COOH and H3O+ (second option)
Step-by-step explanation:
The question requires us to identify the two H+ donors in the reaction provided, according to the Bronsted-Lowry theory.
The Bronsted-Lowry theory states that any compound that can transfer a proton to any other compound is an acid, and the compound that accepts the proton is a base. When the Bronsted-Lowry acid donates a proton (H+), it becomes a conjugated base, and when a Bronsted-Lowry base receives a proton, it becomes the conjugated acid.
Analyzing the reaction given, we can see that:
- CH3COOH has lost a proton and formed CH3COO-;
- H2O has received a proton and formed H3O+.
Therefore, we can say that the Bronsted-Lowry acid in this reaction is CH3COOH (which forms the conjugated base CH3COO-), and the conjugated acid formed is H3O+ (formed from the Bronsted-Lowry base H2O.
Considering the observations above regarding the Bronsted-Lowry acid and conjugated acid produced in the reaction, we can say that the two H+ donators in the reaction, i.e. the acids in the reaction, are CH3COOH and H3O+. Therefore, the best option to answer the question is letter B.