Question

The position of equilibrium in the equation below lies far to the left. Identify the conjugate acid-base pairs. Which of the two acids is stronger? \(\mathrm{HOCl}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OCl}^{-}(a q)\)

Short answer

\(H_3O^+\) is the stronger acid, while \(HOCl\) and \(OCl^-\) are a conjugate acid-base pair, and \(H_3O^+\)/\(H_2O\) are another.

Step-by-step solution

Identify the Acid-Base Pairs

Find the pairs of molecules or ions that differ by the presence of a proton (H+). In this reaction, the conjugate acid-base pairs are \(HOCl/OCl^-\) and \(H_3O^+/H_2O\). HOCl donates a proton to become OCl^-, and H_2O accepts a proton to become H_3O^+.

Determine the Direction of Equilibrium

Since the position of equilibrium lies far to the left, it means the reverse reaction is favored. This implies that \(HOCl\) as an acid is weaker than \(H_3O^+\) because the conjugate base \(OCl^-\) is formed in lesser quantity.

Compare Acid Strength

To determine which of the two acids is stronger, compare their ability to donate protons in water. As equilibrium lies to the left, \(HOCl\) is less effective at donating a proton compared to \(H_3O^+\), making \(H_3O^+\) the stronger acid.

Key concepts

Conjugate Acid-Base Pairs

Understanding conjugate acid-base pairs is essential in chemistry, particularly when discussing reactions that involve acids and bases. These pairs are related by the loss or gain of a single hydrogen ion (a proton). In any acid-base reaction, when an acid donates a proton, it becomes a base, referred to as the conjugate base of the original acid. Conversely, when a base accepts a proton, it becomes an acid, known as the conjugate acid of the original base.

For example, in the reaction \(HOCl + H_2O \rightleftharpoons H_3O^+ + OCl^-\), \(HOCl\) is the acid that donates a proton to water (\(H_2O\)) to form its conjugate base \(OCl^-\). Similarly, water acts as the base and accepts a proton to form its conjugate acid, \(H_3O^+\). By identifying these pairs, we can visualize the transfer of protons and better understand the direction in which the reaction tends to move.

Equilibrium Position

The equilibrium position of an acid-base reaction indicates where the balance between reactants and products lies. It can be thought of as a 'tug-of-war' between the forward and reverse reactions. When the equilibrium position is said to lie 'far to the left,' it means that the reactant side is much more heavily favored and the concentration of reactants is significantly higher than the products.

In the reaction \(HOCl + H_2O \rightleftharpoons H_3O^+ + OCl^-\), the indication that the equilibrium lies far to the left implies that the reactants, \(HOCl\) and \(H_2O\), are predominantly present compared to the products \(H_3O^+\) and \(OCl^-\). This information guides us to understand reactivity and stability of the species involved and is crucial when predicting reaction outcomes or when adjusting conditions to favor the formation of a desired product.

Compare Acid Strength

To compare the strength of acids, we look at their propensity to donate protons. The strength of an acid is typically gauged by how well it can donate a proton to a base and consequently, how stable the conjugate base is once it has accepted a proton.

In the present case, by recognizing that the equilibrium position favors the left side with \(HOCl\) and \(H_2O\), we deduce that \(HOCl\) is a weaker acid compared to \(H_3O^+\). The reason for this is that a stronger acid would push the equilibrium towards the formation of its conjugate base; thus, in this case, \(H_3O^+\) must be the stronger acid because, if \(H_3O^+\) were to donate a proton, it would result in the more stable water (\(H_2O\)), thereby shifting the equilibrium towards the right. Yet, as the reaction heavily favors the left side, we understand that water is more stable holding onto the proton, indicating that \(HOCl\) is not as adept at proton donation.