Problem 110 - General Chemistry: Principles and Modern Applications

Question

\(\operatorname{In} 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is equal to \((\mathrm{a}) 0.050 \mathrm{M}\) (b) \(0.10 \mathrm{M} ;\) (c) \(0.11 \mathrm{M} ;\) (d) \(0.20 \mathrm{M}\).

Step-by-Step Solution

Answer

The concentration of \(H_3O^+\) in the \(0.10 M H_2SO_4\) solution is \(0.20 M\).

Step 1: Write the dissociation reaction of \(H_2SO_4\)

For each molecule of sulfuric acid \(H_2SO_4\), two hydronium \(H_3O^+\) ions are formed in solution. Therefore, the chemical equation of \(H_2SO_4\) dissociation in water is: \(H_2SO_4 → 2H_3O^+ + SO_4^{2-}\)

Step 2: Apply stoichiometry to determine the concentration of \(H_3O^+\)

If every one mole of \(H_2SO_4\) gives two moles of \(H_3O^+\), then the molarity of \(H_3O^+\) should be two times of \(H_2SO_4\)'s molarity. Given that the concentration of \(H_2SO_4\) is \(0.10 M\), the concentration of \(H_3O^+\) would be \(2 \times 0.10 M = 0.20 M\).

Step 3: Choose the correct option

Given the above calculations, it means (d) 0.20 M is the correct concentration of \(H_3O^+\) ions in the 0.10 M solution of \(H_2SO_4\).

Key Concepts

Acid Dissociation in Water

When sulfuric acid (\( H_2SO_4 \) enters water, it undergoes a chemical process called dissociation. The strong acidic nature of sulfuric acid means it readily releases its hydrogen atoms as ions. This dissociation in water is represented by the equation: \( H_2SO_4 → 2H_3O^+ + SO_4^{2-} \). In this process, for every one sulfuric acid molecule, two hydronium ions (\( H_3O^+ \) and a sulfate ion (\( SO_4^{2-} \) are produced.

Understanding this concept is crucial as it dictates the behavior of acids in aqueous solutions. The ease of this dissociation process categorizes sulfuric acid as a strong acid, meaning it completely dissociates in a solution. For students, it's important to note that the dissociation of strong acids is often assumed to go to completion in stoichiometric calculations.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It relies on the balanced chemical equation to relate the amounts of different substances. In the case of sulfuric acid dissociation, the equation is used to determine the concentration of the products based on the concentration of the reactant.

For example, with the dissociation equation \( H_2SO_4 → 2H_3O^+ + SO_4^{2-} \), we know that 1 mole of sulfuric acid produces 2 moles of hydronium ions. Hence, if we start with a 0.10 M solution of sulfuric acid (\( 0.10 \) moles/liter), stoichiometry allows us to conclude that the resulting concentration of H3O+ ions will be doubled, assuming complete dissociation, which gives us a 0.20 M solution of hydronium ions.

Hydronium Ion Concentration

The hydronium ion concentration (\( [H_3O^+] \)) in a solution is a measure of acidity. It determines the solution's pH level, which is a numerical scale used to specify the acidity or basicity of an aqueous solution. For sulfuric acid, a strong acid, the hydronium ion concentration is particularly high because each molecule of sulfuric acid contributes two hydronium ions upon dissociation.

Given the stoichiometric calculation from our exercise, the correct hydronium ion concentration stemming from a 0.10 M sulfuric acid solution is 0.20 M. This double increment is a direct result of the stoichiometry of the reaction, where one sulfuric acid molecule gives rise to two hydronium ions. In essence, understanding the relationship between the initial concentration of the acid and the resulting hydronium ion concentration allows us to predict the change in pH of the solution upon the addition of an acid like sulfuric acid.