Problem 19 - Chemistry

Question

Anions containing hydrogen (for example, \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ) usually show amphoteric behavior. Write equations illustrating the amphoterism of these two anions.

Step-by-Step Solution

Answer

The amphoteric behavior of the anions \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) can be illustrated through the following equations: 1. \( \mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CO}_{3}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+} \) (Base) 2. \( \mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3}+\mathrm{OH}^{-} \) (Acid) 3. \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}\mathrm{PO}_{4}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+} \) (Base) 4. \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{PO}_{4}+\mathrm{OH}^{-} \) (Acid)

Step 1: Write the equation for \(\mathrm{HCO}_{3}^{-}\) acting as a base

In this case, \(\mathrm{HCO}_{3}^{-}\) will accept a proton from water. The reaction is as follows: \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CO}_{3}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+}\)

Step 2: Write the equation for \(\mathrm{HCO}_{3}^{-}\) acting as an acid

In this case, \(\mathrm{HCO}_{3}^{-}\) will donate a proton to water. The reaction is as follows: \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3}+\mathrm{OH}^{-}\)

Step 3: Write the equation for \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) acting as a base

In this case, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) will accept a proton from water. The reaction is as follows: \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}\mathrm{PO}_{4}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+}\)

Step 4: Write the equation for \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) acting as an acid

In this case, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) will donate a proton to water. The reaction is as follows: \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{PO}_{4}+\mathrm{OH}^{-}\) By combining all the equations, we can see that both anions show amphoteric behavior: 1. \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CO}_{3}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+}\) (Base) 2. \(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3}+\mathrm{OH}^{-}\) (Acid) 3. \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}\mathrm{PO}_{4}^{2-}+\mathrm{H}_{3}\mathrm{O}^{+}\) (Base) 4. \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{PO}_{4}+\mathrm{OH}^{-}\) (Acid)