Question

Write balanced equations and \(K_{\mathrm{a}}\) expressions for these Bronsted- Lowry acids in water: (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HSO}_{4}^{-}\) (d) \(\mathrm{HNO}_{2}\)

Short answer

Balanced equations and \( K_{\mathrm{a}} \) expressions are derived for \(\mathrm{H}_3 \mathrm{PO}_4\), \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\), \(\mathrm{HSO}_4^-\), and \(\mathrm{HNO}_2\).

Step-by-step solution

Write the Balanced Equation for \(\mathrm{H}_3 \mathrm{PO}_4\) in Water

Phosphoric acid (\(\mathrm{H}_3 \mathrm{PO}_4\)) dissociates in water to form hydronium ions (\(\mathrm{H}_3 \mathrm{O}^+\)) and dihydrogen phosphate ions (\(\mathrm{H}_2 \mathrm{PO}_4^-\)). The balanced equation is: \[\mathrm{H}_3 \mathrm{PO}_4 + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^+ + \mathrm{H}_2 \mathrm{PO}_4^-\]

Write the \( K_{\mathrm{a}} \) Expression for \(\mathrm{H}_3 \mathrm{PO}_4\)

The expression for the acid dissociation constant (\( K_{\mathrm{a}} \)) is: \[K_{\mathrm{a}} = \frac{[\mathrm{H}_3 \mathrm{O}^+] [\mathrm{H}_2 \mathrm{PO}_4^-]}{[\mathrm{H}_3 \mathrm{PO}_4]}\]

Write the Balanced Equation for \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) in Water

Benzoic acid (\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\)) dissociates in water to form hydronium ions (\(\mathrm{H}_3 \mathrm{O}^+\)) and benzoate ions (\(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\)). The balanced equation is: \[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^+ + \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-\]

Write the \( K_{\mathrm{a}} \) Expression for \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\)

The expression for the acid dissociation constant (\( K_{\mathrm{a}} \)) is: \[K_{\mathrm{a}} = \frac{[\mathrm{H}_3 \mathrm{O}^+] [\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^-]}{[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}]}\]

Write the Balanced Equation for \(\mathrm{HSO}_4^-\) in Water

Hydrogen sulfate ion (\(\mathrm{HSO}_4^-\)) dissociates in water to form hydronium ions (\(\mathrm{H}_3 \mathrm{O}^+\)) and sulfate ions (\(\mathrm{SO}_4^{2-}\)). The balanced equation is: \[\mathrm{HSO}_4^- + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^+ + \mathrm{SO}_4^{2-}\]

Write the \( K_{\mathrm{a}} \) Expression for \(\mathrm{HSO}_4^-\)

The expression for the acid dissociation constant (\( K_{\mathrm{a}} \)) is: \[K_{\mathrm{a}} = \frac{[\mathrm{H}_3 \mathrm{O}^+] [\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^-]}\]

Write the Balanced Equation for \(\mathrm{HNO}_2\) in Water

Nitrous acid (\(\mathrm{HNO}_2\)) dissociates in water to form hydronium ions (\(\mathrm{H}_3 \mathrm{O}^+\)) and nitrite ions (\(\mathrm{NO}_2^-\)). The balanced equation is: \[\mathrm{HNO}_2 + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^+ + \mathrm{NO}_2^-\]

Write the \( K_{\mathrm{a}} \) Expression for \(\mathrm{HNO}_2\)

The expression for the acid dissociation constant (\( K_{\mathrm{a}} \)) is: \[K_{\mathrm{a}} = \frac{[\mathrm{H}_3 \mathrm{O}^+] [\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]}\]

Key concepts

acid dissociation constant

The acid dissociation constant, denoted as \( K_a \), is a key concept in acid-base chemistry. It measures the strength of an acid in a solution. The value of \( K_a \) quantifies the extent to which an acid can donate protons (\( H^+ \)) to water, forming hydronium ions (\( H_3O^+ \)).
To determine \( K_a \), one looks at the equilibrium concentration of the reactants and products. For example, the dissociation of phosphoric acid (\( H_3PO_4 \)) in water can be expressed as:
\[ H_3PO_4 + H_2O \rightleftharpoons H_3O^+ + H_2PO_4^- \] The \( K_a \) expression for this dissociation is:
\[ K_a = \frac{[H_3O^+][H_2PO_4^-]}{[H_3PO_4]} \] Each term in the expression represents the concentration of the species at equilibrium. The higher the \( K_a \) value, the stronger the acid, meaning it dissociates more in water. Conversely, a lower \( K_a \) value implies a weaker acid.
Understanding \( K_a \) is crucial for predicting the behavior of acids in different chemical environments.

balanced chemical equations

Balanced chemical equations are fundamental in representing chemical reactions accurately. They ensure that all atoms present in the reactants are accounted for in the products, illustrating the conservation of mass.
For Bronsted-Lowry acids dissociating in water, balanced equations show the acid transferring a proton to water, forming hydronium ions. Take benzoic acid (\( C_6H_5COOH \)) as an example. Its balanced dissociation equation is:
\[ C_6H_5COOH + H_2O \rightleftharpoons H_3O^+ + C_6H_5COO^- \] The equation is balanced because it shows one molecule of benzoic acid dissociating to produce one hydronium ion and one benzoate ion.
Writing balanced equations involves:

• Identifying reactants and products
• Ensuring equal numbers of each type of atom on both sides
• Double-checking the charges to confirm electrical neutrality on both sides

This process provides a clear and accurate picture of the chemical changes occurring during the reaction.

hydronium ions

Hydronium ions (\( H_3O^+ \)) play a crucial role in acid-base chemistry. When an acid dissolves in water, it donates a proton (\( H^+ \)) to a water molecule, forming \( H_3O^+ \). This ion is central to defining the acidity of a solution.
For instance, when nitrous acid (\( HNO_2 \)) dissociates in water, it forms hydronium ions and nitrite ions:
\[ HNO_2 + H_2O \rightleftharpoons H_3O^+ + NO_2^- \] Hydronium ions are crucial because they directly relate to the pH of a solution. pH is a measure of the concentration of \( H_3O^+ \), where:
\[ \text{pH} = -\text{log}[H_3O^+] \] Thus, a higher concentration of hydronium ions means a lower pH, indicating a more acidic solution.
Understanding the role of \( H_3O^+ \) helps in predicting the behavior of acids and their strength in various solutions. Their formation and concentration are pivotal in numerous chemical reactions and processes.