Problem 111 - General Chemistry

Question

Ionization of the first proton from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is complete \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right.\) is a strong acid); the acid-ionization constant for the second proton is \(1.1 \times 10^{-2}\). a What would be the approximate hydronium-ion concentration in \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) if ionization of the second proton were ignored? The ionization of the second proton must be considered for a more exact answer, however. Calculate the hydronium-ion concentration in \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4},\) accounting for the ionization of both protons.

Step-by-Step Solution

Answer

The hydronium-ion concentration is approximately 0.111 M.

Step 1: Initial Hydronium-ion Concentration

Since sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) is a strong acid, it completely dissociates into its ions for the first proton. For a \(0.100 \mathrm{M}\) \(\mathrm{H}_2\mathrm{SO}_4\) solution, the initial hydronium-ion concentration \([\mathrm{H}_3\mathrm{O}^+]\) from the ionization of the first proton is \([\mathrm{H}_3\mathrm{O}^+] = 0.100 \mathrm{M}\)."

Step 2: Ionization of the Second Proton

The ionization of the second proton can be represented by the equilibrium: \(\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-}\). The expression for the equilibrium constant \(K_a\) of the second ionization is given by:\[K_a = \frac{[\mathrm{H}^+][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^-]}\]Given \(K_a = 1.1 \times 10^{-2}\).

Step 3: Set Up ICE Table

To find the additional hydronium ions from the second ionization, set an ICE (Initial, Change, Equilibrium) table:- **Initial:** \([\mathrm{HSO}_4^-] = 0.100 \mathrm{M}\), \([\mathrm{H}^+] = 0.100 \mathrm{M}\), \([\mathrm{SO}_4^{2-}] = 0\)- **Change:** Let \(x\) be the amount that ionizes; thus, \([\mathrm{H}^+] = 0.100 + x\), \([\mathrm{SO}_4^{2-}] = x\), \([\mathrm{HSO}_4^-] = 0.100 - x\).Substitute these into the \(K_a\) expression:\[1.1 \times 10^{-2} = \frac{(0.100 + x)x}{0.100 - x}\]

Step 4: Solve the Ka Expression

Assuming \(x\) is small compared to 0.100, the equation simplifies to:\[1.1 \times 10^{-2} = \frac{(0.100)(x)}{0.100}\]\[x \approx 1.1 \times 10^{-2}\]Adding \(x\) to the initial concentration of \([\mathrm{H}_3\mathrm{O}^+]\), we get the total concentration:\[[\mathrm{H}_3\mathrm{O}^+] = 0.100 + 0.011 = 0.111 \mathrm{M}\]

Step 5: Conclusion of Hydronium-ion Concentration

The total hydronium-ion concentration considering both protons is \(0.111 \mathrm{M}\). This reflects the contribution from both the complete ionization of the first proton and the partial ionization of the second proton, which cannot be ignored in a precise calculation.

Key Concepts

Strong Acids

Sulfuric acid, denoted as \( \mathrm{H}_2\mathrm{SO}_4 \), is a prime example of a strong acid. Understanding strong acids is crucial in chemistry, as these substances completely dissociate into their constituent ions in water. This complete ionization means that when \( \mathrm{H}_2\mathrm{SO}_4 \) is dissolved in water, it promptly separates into \( \mathrm{H}^+ \) ions and \( \mathrm{HSO}_4^- \) ions from the first hydrogen. Thus, in a \( 0.100 \ \mathrm{M} \) solution of \( \mathrm{H}_2\mathrm{SO}_4 \), we can initially deduce the hydronium ion concentration to be \( 0.100 \, \text{M} \) because of this complete ionization.

Strong acids are characterized by their low pH levels due to the high concentration of \( \mathrm{H}_3\mathrm{O}^+ \) ions. The term "strong" specifically refers to the acid’s ability to donate protons easily. Hence, they do not have a value for an acid ionization constant \( K_a \) for their first proton, as the reaction does not reach an equilibrium state but goes to completion.

• Complete dissociation: Breaks entirely into ions.
• Resulting low pH: Due to increased \( \mathrm{H}_3\mathrm{O}^+ \) in solution.
  
• First proton fully ionized: No need for \( K_a \).

Equilibrium Constants

The concept of equilibrium constants is essential for understanding reactions that don’t go to completion, like the ionization of the second proton in sulfuric acid. When the first proton is fully dissociated, the remaining \( \mathrm{HSO}_4^- \) undergoes partial ionization which is described by an equilibrium expression. This reaction maintains a balance between the reactants and products:

\[ \mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-} \]

The equilibrium constant \( K_a \) for this reaction is calculated as follows:

\[ K_a = \frac{[\mathrm{H}^+][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^-]} \]

The given \( K_a = 1.1 \times 10^{-2} \) indicates a system not nearly as strong in ionizing as its predecessor. It signifies the extent of the ionization under standard conditions. In our context, knowing \( K_a \) helps calculate the additional amount of \( \mathrm{H}_3\mathrm{O}^+ \) ions contributed by this process.

• Partial ionization: Not complete, involving equilibrium.
• \( K_a \) value insights: Indicates ionization extent.
• Second proton: Governed by this equilibrium process.

Hydronium Ion Concentration

Hydronium ion concentration \( [\mathrm{H}_3\mathrm{O}^+] \) is central to understanding the acidity of a solution. Initially, when considering sulfuric acid \( \mathrm{H}_2\mathrm{SO}_4 \), the hydronium ion concentration from the first proton ionization is directly the molarity of the solution, \( 0.100 \, \text{M} \).

However, for precise calculations, the secondary ionization of \( \mathrm{HSO}_4^- \) into \( \mathrm{H}^+ \) and \( \mathrm{SO}_4^{2-} \) contributes further to the \( [\mathrm{H}_3\mathrm{O}^+] \). Solving the equilibrium expression, where \( x \) is assumed to be small, it adds \( 1.1 \times 10^{-2} \text{M} \) to the initial concentration.

Therefore, the total hydronium ion concentration becomes \( 0.100 + 0.011 = 0.111 \, \text{M} \). This careful addition reflects the cumulative effect of both dissociations on the solution’s acidity.

• Initial contribution: From complete ionization of the first proton.
• Further influence: From partial ionization of the second proton.
• Total concentration: Combines both contributions for accurate acidity.