3. For each of the following bases, write the formula for its conjugate acid; for each acid write the formula of the conjugate base.
\begin{tabular}{llll}
Base & Conjugate Acid & Acid & Conjugate Base \
$\mathrm{I}^{-}$ & & $\mathrm{HClO}_{4}$ \
$\mathrm{SO}_{3}{ }^{2-}$ & $\square$ & $\mathrm{H}_{2} \mathrm{~S}$ & \
$\mathrm{PO}_{4}{ }^{3-}$ & $\square$ & $\mathrm{HCO}_{3}{ }^{-}$ & \
\cline { 2 - 2 } & & &
\end{tabular}
$\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$ $\qquad$
4. For the following acid-base reaction:
$$
\mathrm{NH}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons
$$
$\qquad$
$\qquad$
a) Write the products.
b) For all four (4) species in the equation, indicate whether each will act as an acid or a base.

Question

3. For each of the following bases, write the formula for its conjugate acid; for each acid write the formula of the conjugate base.
\begin{tabular}{llll} 
Base & Conjugate Acid & Acid & Conjugate Base \
$\mathrm{I}^{-}$ & & $\mathrm{HClO}_{4}$ \
$\mathrm{SO}_{3}{ }^{2-}$ & $\square$ & $\mathrm{H}_{2} \mathrm{~S}$ & \
$\mathrm{PO}_{4}{ }^{3-}$ & $\square$ & $\mathrm{HCO}_{3}{ }^{-}$ & \
\cline { 2 - 2 } & & &
\end{tabular}
$\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$ $\qquad$
4. For the following acid-base reaction:
$$
\mathrm{NH}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons
$$
$\qquad$
$\qquad$
a) Write the products.
b) For all four (4) species in the equation, indicate whether each will act as an acid or a base.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Conjugate Acid-Base Pairs**: To find the conjugate acid of a base, a hydrogen ion (H⁺) is added. To find the conjugate base of an acid, a hydrogen ion is removed. ‖ ‖ 2 ⋮ **Completing the Table**: - The conjugate acid of $\mathrm{I}^{-}$ is $\mathrm{HI}$. - The conjugate acid of $\mathrm{SO}_{3}^{2-}$ is $\mathrm{HSO}_{3}^{-}$. - The conjugate acid of $\mathrm{PO}_{4}^{3-}$ is $\mathrm{HPO}_{4}^{2-}$. - The conjugate acid of $\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}$ is $\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}$. - The conjugate base of $\mathrm{HClO}_{4}$ is $\mathrm{ClO}_{4}^{-}$. - The conjugate base of $\mathrm{H}_{2}\mathrm{S}$ is $\mathrm{HS}^{-}$. - The conjugate base of $\mathrm{HCO}_{3}^{-}$ is $\mathrm{CO}_{3}^{2-}$. ‖ ∻1 Answer∻ ⚹ Base & Conjugate Acid & Acid & Conjugate Base \ $\mathrm{I}^{-}$ & $\mathrm{HI}$ & $\mathrm{HClO}_{4}$ & $\mathrm{ClO}_{4}^{-}$ \ $\mathrm{SO}_{3}^{2-}$ & $\mathrm{HSO}_{3}^{-}$ & $\mathrm{H}_{2}\mathrm{S}$ & $\mathrm{HS}^{-}$ \ $\mathrm{PO}_{4}^{3-}$ & $\mathrm{HPO}_{4}^{2-}$ & $\mathrm{HCO}_{3}^{-}$ & $\mathrm{CO}_{3}^{2-}$ \ $\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}$ & $\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}$ & & \ ⚹ ∻Key Concept∻ ⚹ Conjugate acid-base pairs differ by one hydrogen ion (proton). ⚹ ∻Explanation∻ ⚹ The conjugate acid is formed by adding H⁺ to the base, and the conjugate base is formed by removing H⁺ from the acid. ⚹ ‖ 3 ⋮ **Reaction Products**: The reaction between $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}\mathrm{O}$ produces $\mathrm{NH}_{4}^{+}$ and $\mathrm{OH}^{-}$. ‖ ‖ 4 ⋮ **Acid-Base Identification**: - $\mathrm{NH}_{3}$ acts as a base because it accepts a proton to become $\mathrm{NH}_{4}^{+}$. - $\mathrm{H}_{2}\mathrm{O}$ acts as an acid because it donates a proton to become $\mathrm{OH}^{-}$. - $\mathrm{NH}_{4}^{+}$ acts as an acid because it can donate a proton to revert to $\mathrm{NH}_{3}$. - $\mathrm{OH}^{-}$ acts as a base because it can accept a proton to revert to $\mathrm{H}_{2}\mathrm{O}$. ‖ ∻2 Answer∻ ⚹ a) The products are $\mathrm{NH}_{4}^{+}$ and $\mathrm{OH}^{-}$. b) Acid/Base Identification: - $\mathrm{NH}_{3}$: Base - $\mathrm{H}_{2}\mathrm{O}$: Acid - $\mathrm{NH}_{4}^{+}$: Acid - $\mathrm{OH}^{-}$: Base ⚹ ∻Key Concept∻ ⚹ Acids donate protons (H⁺), and bases accept protons. ⚹ ∻Explanation∻ ⚹ In the reaction, $\mathrm{NH}_{3}$ accepts a proton becoming $\mathrm{NH}_{4}^{+}$, thus acting as a base, while $\mathrm{H}_{2}\mathrm{O}$ donates a proton becoming $\mathrm{OH}^{-}$, thus acting as an acid. The products can also act as acid or base in reverse reactions.◊What are the conjugate acids of the bases I-, SO3^2-, PO4^3-, and C2H3O2- listed in the table?⍭ Generate me a similar question◊⚹

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