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Acetic acid, commonly found in vinegar, is a weak acid. This means it's not completely dissociated in water. When it is in aqueous solution, only a small percentage of acetic acid molecules split into ions. The dissociation reaction can be expressed as: \( \text{CH}_3\text{CO}_2\text{H} (aq) \rightleftharpoons \text{H}^+ (aq) + \text{CH}_3\text{CO}_2^- (aq) \).In this equilibrium reaction, the acetic acid molecule donates a proton \((\text{H}^+)\), forming acetate ions \((\text{CH}_3\text{CO}_2^-)\). Because only a few molecules release their protons, both ions and the acetic acid remain in solution. Understanding this balance between products and reactants is crucial for analyzing many properties of weak acids.

Percent dissociation helps us understand how much of the weak acid dissociates in solution. It measures the degree to which an acid splits into ions and can be calculated easily using the following formula:\[ \text{Percent Dissociation} = \left( \frac{ \text{Concentration of Dissociated Acid} }{ \text{Initial Concentration of Acid} } \right) \times 100 \].For acetic acid, if we start with a 1.0 M solution and observe that 0.42% dissociates, we can use this concept to find out the actual concentrations:- Change in concentration: \(0.0042\, \text{M} = 0.42\% \times 1.0\, \text{M}\).- This change applies equally to the \(\text{H}^+\) and \(\text{CH}_3\text{CO}_2^-\) ions. Hence, their concentrations become 0.0042 M each.Understanding percent dissociation is a key step in determining the solution's behavior and its equilibrium constant.

The equilibrium constant, \(K_c\), is essential for understanding the balance between products and reactants in reversible reactions. For the dissociation of acetic acid, the equilibrium constant expression is given by:\[ K_{c} = \frac{[\text{H}^+][\text{CH}_3\text{CO}_2^-]}{[\text{CH}_3\text{CO}_2\text{H}]} \].This equation helps quantify the extent of the reaction at equilibrium. The equilibrium concentrations of the products \([\text{H}^+]\) and \([\text{CH}_3\text{CO}_2^-]\) over the initial concentration of the reactant \([\text{CH}_3\text{CO}_2\text{H}]\) gives the \(K_c\).For instance, if the dissociation is 0.42% in a 1.0 M solution:

• \([\text{H}^+] = 0.0042\, \text{M}\)
• \([\text{CH}_3\text{CO}_2^-] = 0.0042\, \text{M}\)
• \([\text{CH}_3\text{CO}_2\text{H}] = 0.9958\, \text{M}\)

Plug these into the \(K_c\) expression:\[K_{c} = \frac{(0.0042)(0.0042)}{0.9958} \approx 0.0000177\]This result tells us how far the dissociation proceeds and helps in predicting how a system will respond to changes in conditions.