Problem 198 - Introductory Chemistry Atoms First

Question

Sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), is a diprotic acid with dissociation equilibrium constants of \(K_{\mathrm{eq}}>1.0 \times 10^{3}\) and \(K_{\mathrm{eq}}=1.2 \times 10^{-2} .\) Write the two dissociation equilibrium equations, and match the proper \(K_{\text {eq }}\) to each. Which species is the weak acid?

Step-by-Step Solution

Answer

The two dissociation equilibrium equations for sulfuric acid are: 1. \(\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\), with \(K_{eq}>1.0 \times 10^{3}\) 2. \(\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\), with \(K_{eq}=1.2 \times 10^{-2}\) The weak acid species is the bisulfate ion (\(\mathrm{HSO}_{4}^{-}\)), which is involved in the second dissociation step with a low equilibrium constant.

Step 1: Write the first dissociation equation of sulfuric acid

For the first dissociation, sulfuric acid loses one hydrogen ion to become a bisulfate ion. The equilibrium equation for this process is: \[\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\]

Step 2: Write the second dissociation equation of the bisulfate ion

For the second dissociation, the bisulfate ion loses another hydrogen ion to become the sulfate ion. The equilibrium equation for this process is: \[\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\]

Step 3: Match the given equilibrium constants to the corresponding dissociation equations

We are given two equilibrium constants, \(K_{eq}>1.0 \times 10^{3}\) (high value indicating a strong dissociation) and \(K_{eq}=1.2 \times 10^{-2}\) (low value indicating a weak dissociation). Since sulfuric acid is known to be a strong acid, its first dissociation will have a large equilibrium constant: \(\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\), with \(K_{eq}>1.0 \times 10^{3}\) For the second dissociation, which is expected to have a smaller equilibrium constant: \(\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\), with \(K_{eq}=1.2 \times 10^{-2}\)

Step 4: Identify the weak acid species

In the second dissociation step, the bisulfate ion, \(\mathrm{HSO}_{4}^{-}\), is acting as a weak acid because it has a low equilibrium constant. Therefore, the weak acid species is the bisulfate ion (\(\mathrm{HSO}_{4}^{-}\)).

Key Concepts

Sulfuric Acid Dissociation

Sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\), is a diprotic acid, meaning it can release two protons (hydrogen ions) in solution. This type of acid undergoes dissociation in two distinct steps, each characterized by a distinct equilibrium equation.

The first dissociation step is significant as it establishes sulfuric acid as a strong acid. In this initial reaction, sulfuric acid loses its first hydrogen ion, resulting in the formation of bisulfate ions (\(\mathrm{HSO}_{4}^{-}\)). The equilibrium equation representing this step is as follows: \[\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\]

• This initial dissociation has a high equilibrium constant (\(K_{eq}>1.0 \times 10^{3}\)), indicating a near-complete dissociation in solutions, typical of strong acids.

The second step of dissociation involves the bisulfate ion releasing another hydrogen ion to form the sulfate ion (\(\mathrm{SO}_{4}^{2-}\)). This is expressed by the following equation: \[\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\]

• This second dissociation process is characterized by a much lower equilibrium constant (\(K_{eq}=1.2 \times 10^{-2}\)), pointing towards a weak acid dissociation compared to the first step.

Equilibrium Constants

The concept of equilibrium constants is crucial in understanding how completely the reactions occur in the case of acid dissociation. Every chemical equilibrium has a constant, represented as \(K_{eq}\), and for acids, it helps determine the strength of these acids.

In the context of sulfuric acid, the first dissociation step has a very high equilibrium constant, \(K_{eq}>1.0 \times 10^{3}\). This indicates that the majority of sulfuric acid molecules give up their protons to form hydrogen ions and bisulfate ions, reflecting near-complete dissociation.

• A high equilibrium constant (greater than 1) suggests a reaction heavily favoring the products, indicative of strong acids.

Contrastingly, the second dissociation of sulfuric acid, where bisulfate converts to sulfate, has a smaller equilibrium constant \(K_{eq}=1.2 \times 10^{-2}\).

• This smaller value signifies that not all the \(\mathrm{HSO}_{4}^{-}\) ions dissociate completely into their respective products. Thus, the reaction favors the reactants more, typical of weaker acids.

Weak Acid Identification

Identifying weak acids can be done by examining their equilibrium constants. In a polyprotic weak acid, each proton dissociation step has its own equilibrium constant, which decreases progressively. In the case of sulfuric acid, the second proton dissociation step behaves as a weak acid.

The first dissociation of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is strong, as previously described. However, the second dissociation of the bisulfate ion, \(\mathrm{HSO}_{4}^{-}\), to form the sulfate ion, is notably weaker.

• With an equilibrium constant of \(K_{eq}=1.2 \times 10^{-2}\), which is lower than 1, this indicates the balance between hydrogen ion formation and the remaining bisulfate ions is more inclined towards the undissociated form.
• Thus, \(\mathrm{HSO}_{4}^{-}\) acts as a weak acid in solution, highlighting the noticeable difference in strength between the first and second dissociation reactions.