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Ch.16 - Acid-Base Equilibria
Chapter 16, Problem 24b

Write an equation for the reaction in which H2C6H7O5-1aq2 acts as an acid in H2O1l2.

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Step 1: Identify the acid and the base in the reaction. In this case, H2C6H7O5-1(aq) is the acid and H2O(l) is the base.
Step 2: Write down the initial reaction. The acid (H2C6H7O5-1) donates a proton (H+) to the base (H2O).
Step 3: Write down the products of the reaction. After donating a proton, the acid becomes its conjugate base (HC6H7O5-2), and the base becomes its conjugate acid (H3O+).
Step 4: Combine the reactants and products to form the complete reaction equation. The equation should be balanced, meaning there should be the same number of each type of atom on both sides of the equation.
Step 5: The final equation should look like this: H2C6H7O5-1(aq) + H2O(l) -> HC6H7O5-2(aq) + H3O+(aq)

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons (H+) between species. In this context, an acid donates a proton to a base, which accepts it. Understanding this concept is crucial for writing the correct chemical equation, as it helps identify the reactants and products based on their roles in the reaction.
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Bronsted-Lowry Acid

According to the Bronsted-Lowry theory, an acid is defined as a substance that donates protons in a reaction. In the given question, H2C6H7O5- acts as a Bronsted-Lowry acid when it donates a proton to water, leading to the formation of its conjugate base and hydronium ion. Recognizing this classification is essential for accurately depicting the reaction.
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Chemical Equation Representation

A chemical equation represents the reactants and products of a chemical reaction using their chemical formulas. It must balance the number of atoms of each element on both sides of the equation. Writing the correct equation for the reaction involves identifying the species involved and ensuring that the equation adheres to the law of conservation of mass.
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