Problem 34 - Chemistry: Principles and Reactions

Question

Write the ionization equation and the \(K_{\mathrm{a}}\) expression for each of the following acids. (a) \(\mathrm{HSO}_{3}^{-}\) (b) \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\) (c) \(\mathrm{HNO}_{2}\)

Step-by-Step Solution

Answer

Question: Write the ionization equation and the \(K_{\mathrm{a}}\) expression for each of the following acids: (a) \(\mathrm{HSO}_{3}^{-}\), (b) \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\), and (c) \(\mathrm{HNO}_{2}\). Answer: (a) Ionization equation for \(\mathrm{HSO}_{3}^{-}\): $$\mathrm{HSO}_{3}^{-} (aq) \rightleftharpoons \mathrm{SO}_{3}^{2-} (aq) + \mathrm{H}^{+} (aq)$$ \(K_{\mathrm{a}}\) expression for \(\mathrm{HSO}_{3}^{-}\): $$K_{\mathrm{a}} = \frac{[\mathrm{SO}_{3}^{2-}][\mathrm{H}^{+}]}{[\mathrm{HSO}_{3}^{-}]}$$ (b) Ionization equation for \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\): $$\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-} (aq) \rightleftharpoons \mathrm{PO}_{4}^{3-} (aq) + \mathrm{H}^{+} (aq)$$ \(K_{\mathrm{a}}\) expression for \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\): $$K_{\mathrm{a}} = \frac{[\mathrm{PO}_{4}^{3-}][\mathrm{H}^{+}]}{[\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}]}$$ (c) Ionization equation for \(\mathrm{HNO}_{2}\): $$\mathrm{HNO}_{2} (aq) \rightleftharpoons \mathrm{NO}_{2}^{-} (aq) + \mathrm{H}^{+} (aq)$$ \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{2}\): $$K_{\mathrm{a}} = \frac{[\mathrm{NO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{HNO}_{2}]}$$

Step 1: (a) Ionization equation for \(\mathrm{HSO}_{3}^{-}\)

The ionization of \(\mathrm{HSO}_{3}^{-}\) can be written as follows: $$\mathrm{HSO}_{3}^{-} (aq) \rightleftharpoons \mathrm{SO}_{3}^{2-} (aq) + \mathrm{H}^{+} (aq)$$

Step 2: (a) \(K_{\mathrm{a}}\) expression for \(\mathrm{HSO}_{3}^{-}\)

Based on the ionization equation above, the \(K_{\mathrm{a}}\) expression for \(\mathrm{HSO}_{3}^{-}\) can be written as: $$K_{\mathrm{a}} = \frac{[\mathrm{SO}_{3}^{2-}][\mathrm{H}^{+}]}{[\mathrm{HSO}_{3}^{-}]}$$

Step 3: (b) Ionization equation for \(\mathrm{HPO}_{4}{ }^{2-}\)

The ionization of \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\) can be written as follows: $$\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-} (aq) \rightleftharpoons \mathrm{PO}_{4}^{3-} (aq) + \mathrm{H}^{+} (aq)$$

Step 4: (b) \(K_{\mathrm{a}}\) expression for \(\mathrm{HPO}_{4}{ }^{2-}\)

Based on the ionization equation above, the \(K_{\mathrm{a}}\) expression for \(\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}\) can be written as: $$K_{\mathrm{a}} = \frac{[\mathrm{PO}_{4}^{3-}][\mathrm{H}^{+}]}{[\mathrm{HPO}_{4}{\underline{\phantom{xx}}}^{2-}]}$$

Step 5: (c) Ionization equation for \(\mathrm{HNO}_{2}\)

The ionization of \(\mathrm{HNO}_{2}\) can be written as follows: $$\mathrm{HNO}_{2} (aq) \rightleftharpoons \mathrm{NO}_{2}^{-} (aq) + \mathrm{H}^{+} (aq)$$

Step 6: (c) \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{2}\)

Based on the ionization equation above, the \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{2}\) can be written as: $$K_{\mathrm{a}} = \frac{[\mathrm{NO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{HNO}_{2}]}$$

Key Concepts

Acid Ionization Constant

The acid ionization constant, denoted as \(K_a\), is a numerical value that represents the strength of an acid in solution. It is specifically the equilibrium constant for the ionization of a weak acid and measures the degree to which the acid dissociates in water to form hydrogen ions (\(H^+\)) and its conjugate base.

For example, the ionization of hydrogensulfite ion (\(HSO_3^-\)) is represented as:
\[HSO_3^- (aq) \rightleftharpoons SO_3^{2-} (aq) + H^+ (aq)\]
And the corresponding \(K_a\) expression is given by:
\[K_{a} = \frac{[SO_3^{2-}][H^+]}{[HSO_3^-]}\]
A higher \(K_a\) value indicates a stronger acid (greater tendency to lose a proton), whereas a lower \(K_a\) value indicates a weaker acid (lesser tendency to lose a proton). Understanding \(K_a\) is essential for predicting the behavior of acids in chemical reactions and for calculating the pH of solutions.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, resulting in no net change in the concentration of the reactants and products over time. It's a state of balance, though not necessarily a state where the reactants and products are in equal concentrations.

For the ionization of acids, this concept becomes particularly important. Taking the acid \(HPO_4^{2-}\) as an example, the equilibrium in an aqueous solution can be represented as:
\[HPO_4^{2-} (aq) \rightleftharpoons PO_4^{3-} (aq) + H^+ (aq)\]
At equilibrium, the rates at which \(HPO_4^{2-}\) ionizes and the conjugate base \(PO_4^{3-}\) combines with \(H^+\) to form \(HPO_4^{2-}\) are identical. The equation for the equilibrium constant (\(K_{eq}\)), which in the case of acid ionizations we call \(K_a\), is crucial to understanding how changes in conditions, such as concentration or temperature, can affect the position of the equilibrium.

Molecular Dissociation

Molecular dissociation is the process by which molecules separate into smaller particles such as atoms, ions, or radicals, usually in response to a physical or chemical process. In the context of acids, dissociation refers to the acid molecule separating into a proton (\(H^+\)) and its conjugate base.

For instance, the molecular dissociation of nitrous acid (\(HNO_2\)) in aqueous solution is depicted by:
\[HNO_2 (aq) \rightleftharpoons NO_2^{-} (aq) + H^+ (aq)\]
Understanding how molecules dissociate is important for predicting the conductivity of a solution, the pH of the solution, and the reactivity of the ions produced. It's also important when considering the buffering capacity of a solution—a mixture of a weak acid and its conjugate base that can maintain a nearly constant pH across a range of conditions.