Question

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Explanation:

To identify the acids and bases in the reaction $\boldsymbol{\ce{H3C6H5O7 + H2O
ightleftharpoons H3O+ + H2C6H5O7^-}}$, we use the Brønsted - Lowry definition:

• An acid is a proton ($\ce{H+}$) donor.
• A base is a proton ($\ce{H+}$) acceptor.

Step 1: Analyze $\ce{H3C6H5O7}$

$\ce{H3C6H5O7}$ donates a proton ($\ce{H+}$) to form $\ce{H2C6H5O7^-}$. By the Brønsted - Lowry definition, a substance that donates a proton is an acid. So, $\ce{H3C6H5O7}$ is the acid.

Step 2: Analyze $\ce{H2O}$

$\ce{H2O}$ accepts a proton ($\ce{H+}$) from $\ce{H3C6H5O7}$ to form $\ce{H3O+}$. By the Brønsted - Lowry definition, a substance that accepts a proton is a base. The problem already labels $\ce{H2O}$ as the base, which is consistent with this analysis.

Step 3: Analyze $\ce{H3O+}$

$\ce{H3O+}$ is formed when $\ce{H2O}$ accepts a proton. In the reverse reaction, $\ce{H3O+}$ would donate a proton (acting as an acid) to reform $\ce{H2O}$. For the forward reaction, since it is the product of a base accepting a proton, it is the conjugate acid of $\ce{H2O}$.

Step 4: Analyze $\ce{H2C6H5O7^-}$

$\ce{H2C6H5O7^-}$ is formed when $\ce{H3C6H5O7}$ donates a proton. In the reverse reaction, $\ce{H2C6H5O7^-}$ would accept a proton (acting as a base) to reform $\ce{H3C6H5O7}$. For the forward reaction, since it is the product of an acid donating a proton, it is the conjugate base of $\ce{H3C6H5O7}$.

Final Labeling:

• $\ce{H3C6H5O7}$: $\boldsymbol{\text{acid}}$
• $\ce{H2O}$: $\boldsymbol{\text{base}}$ (given)
• $\ce{H3O+}$: $\boldsymbol{\text{conjugate acid}}$
• $\ce{H2C6H5O7^-}$: $\boldsymbol{\text{conjugate base}}$

Answer:

To identify the acids and bases in the reaction $\boldsymbol{\ce{H3C6H5O7 + H2O
ightleftharpoons H3O+ + H2C6H5O7^-}}$, we use the Brønsted - Lowry definition:

• An acid is a proton ($\ce{H+}$) donor.
• A base is a proton ($\ce{H+}$) acceptor.

Step 1: Analyze $\ce{H3C6H5O7}$

$\ce{H3C6H5O7}$ donates a proton ($\ce{H+}$) to form $\ce{H2C6H5O7^-}$. By the Brønsted - Lowry definition, a substance that donates a proton is an acid. So, $\ce{H3C6H5O7}$ is the acid.

Step 2: Analyze $\ce{H2O}$

$\ce{H2O}$ accepts a proton ($\ce{H+}$) from $\ce{H3C6H5O7}$ to form $\ce{H3O+}$. By the Brønsted - Lowry definition, a substance that accepts a proton is a base. The problem already labels $\ce{H2O}$ as the base, which is consistent with this analysis.

Step 3: Analyze $\ce{H3O+}$

$\ce{H3O+}$ is formed when $\ce{H2O}$ accepts a proton. In the reverse reaction, $\ce{H3O+}$ would donate a proton (acting as an acid) to reform $\ce{H2O}$. For the forward reaction, since it is the product of a base accepting a proton, it is the conjugate acid of $\ce{H2O}$.

Step 4: Analyze $\ce{H2C6H5O7^-}$

$\ce{H2C6H5O7^-}$ is formed when $\ce{H3C6H5O7}$ donates a proton. In the reverse reaction, $\ce{H2C6H5O7^-}$ would accept a proton (acting as a base) to reform $\ce{H3C6H5O7}$. For the forward reaction, since it is the product of an acid donating a proton, it is the conjugate base of $\ce{H3C6H5O7}$.

Final Labeling:

• $\ce{H3C6H5O7}$: $\boldsymbol{\text{acid}}$
• $\ce{H2O}$: $\boldsymbol{\text{base}}$ (given)
• $\ce{H3O+}$: $\boldsymbol{\text{conjugate acid}}$
• $\ce{H2C6H5O7^-}$: $\boldsymbol{\text{conjugate base}}$