Question

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

Short answer

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

Step-by-step solution

Write the ionization equation for \(\mathrm{HSO}_{3}^{-}\)

The ionization equation for the acid \(\mathrm{HSO}_{3}^{-}\) is as follows: \(\mathrm{HSO}_{3}^{-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{SO}_{3}^{2-} (\mathrm{aq})\)

Write the \(K_{\mathrm{a}}\) expression for \(\mathrm{HSO}_{3}^{-}\)

The \(K_{\mathrm{a}}\) expression for the acid \(\mathrm{HSO}_{3}^{-}\) is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{SO}_{3}^{2-}]}{[\mathrm{HSO}_{3}^{-}]}\) Now, let's move on to the second acid:

Write the ionization equation for \(\mathrm{HPO}_{4}^{2-}\)

The ionization equation for the acid \(\mathrm{HPO}_{4}^{2-}\) is as follows: \(\mathrm{HPO}_{4}^{2-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{PO}_{4}^{3-} (\mathrm{aq})\)

Write the \(K_{\mathrm{a}}\) expression for \(\mathrm{HPO}_{4}^{2-}\)

The \(K_{\mathrm{a}}\) expression for the acid \(\mathrm{HPO}_{4}^{2-}\) is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{PO}_{4}^{3-}]}{[\mathrm{HPO}_{4}^{2-}]}\) Finally, let's write the ionization equation and \(K_{\mathrm{a}}\) expression for the third acid:

Write the ionization equation for \(\mathrm{HNO}_{2}\)

The ionization equation for the acid \(\mathrm{HNO}_{2}\) is as follows: \(\mathrm{HNO}_{2} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{NO}_{2}^{-} (\mathrm{aq})\)

Write the \(K_{\mathrm{a}}\) expression for \(\mathrm{HNO}_{2}\)

The \(K_{\mathrm{a}}\) expression for the acid \(\mathrm{HNO}_{2}\) is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\) Now we have the ionization equations and \(K_{\mathrm{a}}\) expressions for all three acids given in the exercise.

Key concepts

Acid Dissociation Constant

When acids dissolve in water, they undergo a process known as ionization, which involves breaking apart into ions. The extent to which an acid ionizes in an aqueous solution is measured by its acid dissociation constant, represented by the symbol Ka. Understanding the Ka value gives insight into the strength of an acid; a higher Ka value indicates a stronger acid, which dissociates more completely in solution. The acid dissociation constant is calculated using the concentration of the hydrogen ions (H+) and the anion that forms upon the dissociation, divided by the concentration of the undissociated acid.

For example, take the acid \(HSO_3^-\) which in the presence of water ionizes to \(H^+\) and \(SO_3^{2-}\). The Ka expression for this reaction is \(K_a = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}\). This equation is crucial because it allows us to calculate the concentrations of species in solution at equilibrium, which is a common exercise in chemistry homework, providing students with a practical understanding of acid strength in aqueous solutions.

Chemical Equilibrium

Equilibrium in chemistry refers to the state when the rate of the forward reaction equals the rate of the reverse reaction, meaning there's no net change in the concentrations of the reactants and products over time. This doesn't imply that the reactions have ceased, but rather that they are occurring at an equal and constant rate. In the context of acids in aqueous solutions, equilibrium is represented by the reversible arrow in the ionization equation. For instance, the ionization of \(HNO_2\) in water can be written as \(HNO_2 (aq) \rightleftharpoons H^+ (aq) + NO_2^- (aq)\).

The position of the equilibrium is given by the acid dissociation constant (Ka), illustrating the ratio of ionized to un-ionized forms. Substances with large Ka values are largely dissociated and, hence, drive the equilibrium towards the products (ions), indicating strong acids. Conversely, small Ka values suggest weak acids, with the equilibrium lying far to the left (favoring the reactants). Understanding chemical equilibrium is key for students as it applies to many chemical systems beyond acid-base reactions and is essential for interpreting various chemical processes.

Aqueous Solutions Chemistry

Aqueous solutions chemistry is fundamental in understanding how substances interact in water – the most common solvent in chemistry. When a substance, especially an acid in this case, is dissolved in water, it may either remain intact if it is a weak electrolyte or separate into ions if it is a strong electrolyte. This segregation of molecules into ions is a critical concept and is central to many chemical reactions and processes.

The behavior of acids in aqueous solutions, such as their ability to conduct electricity and react with bases, can often be explained by the ionization that occurs. For example, \(HPO_4^{2-}\) ionizes in water to form \(H^+\) and \(PO_4^{3-}\) ions. The degree to which this process occurs and the strength of an acid in solution can be overwhelming for students, but through the calculation of the Ka value and understanding chemical equilibrium, they can gain a clearer insight into the nature of acids and their reactions in aqueous solutions.