Problem 112 - General Chemistry

Question

Ionization of the first proton from \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) is complete \(\left(\mathrm{H}_{2} \mathrm{SeO}_{4}\right.\) is a strong acid); the acid-ionization constant for the second proton is \(1.2 \times 10^{-2}\). a What would be the approximate hydronium-ion concentration in \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{SeO}_{4}\) if ionization of the second proton were ignored? \(b\) The ionization of the second proton must be considered for a more exact answer, however. Calculate the hydronium-ion concentration in \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{SeO}_{4}\), accounting for the ionization of both protons.

Step-by-Step Solution

Answer

The initial [H3O+] is 0.150 M from the first ionization. Total [H3O+] including the second is approximately 0.1518 M.

Step 1: Identify the Ionization of the First Proton

Since \( \mathrm{H}_{2} \mathrm{SeO}_{4} \) is a strong acid, the ionization of the first proton is complete. This means that in a 0.150 M solution of \( \mathrm{H}_{2} \mathrm{SeO}_{4} \), all of the first protons are ionized to form hydronium ions. Thus, \[ [\mathrm{H}_3\mathrm{O}^+] = 0.150 \, \mathrm{M} \] from the first ionization alone.

Step 2: Describe the Second Ionization Process

The second proton ionization of \( \mathrm{H}_{2} \mathrm{SeO}_{4} \) can be represented as \( \mathrm{HSeO}_4^- \leftrightarrow \mathrm{SeO}_4^{2-} + \mathrm{H}^+ \). This ionization is only partial and has an ionization constant \( K_{a2} = 1.2 \times 10^{-2} \).

Step 3: Calculate the Contribution from the Second Ionization

To find the hydronium ion concentration considering the second ionization, set up the equilibrium expression for \( K_{a2} \):\[ K_{a2} = \frac{[\mathrm{SeO}_4^{2-}][\mathrm{H}^+]}{[\mathrm{HSeO}_4^-]} = 1.2 \times 10^{-2} \]Initially, \( [\mathrm{HSeO}_4^-] \approx 0.150 \, \mathrm{M} \), and when some \( x \) amount ionizes, \( [\mathrm{H}^+] = 0.150 + x \) and \([\mathrm{SeO}_4^{2-}] = x \) approximately.Substitute into the equation yielding the simplified expression:\[ 1.2 \times 10^{-2} = \frac{x(0.150 + x)}{0.150 - x} \]Assuming \( x \) is small compared to 0.150, further simplifies to:\[ x \approx 1.2 \times 10^{-2} \times 0.150 = 1.8 \times 10^{-3} \]

Step 4: Determine the Total Hydronium-Ion Concentration

Add the contribution from the second ionization:\[ [\mathrm{H}_3\mathrm{O}^+]_{total} = 0.150 + 1.8 \times 10^{-3} = 0.1518 \, \mathrm{M} \]Thus, the total hydronium ion concentration accounting for both ionizations is approximately 0.1518 M.

Key Concepts

Strong Acids

Strong acids are substances that completely dissociate in water to release hydrogen ions ( H^+ ) or hydronium ions ( H_3O^+ ). This complete ionization is a defining characteristic of strong acids. - Examples of strong acids include hydrochloric acid ( HCl ), sulfuric acid ( H_2SO_4 ), and our context's H_2SeO_4 . - When dissolved in water, the strong acid H_2SeO_4 separates into 2 H^+ and SeO_4^{2-} . In the context of the exercise, since H_2SeO_4 is a strong acid, its first proton fully ionizes: - This means that the concentration of hydronium ions resulting from the first ionization is equal to the initial concentration of H_2SeO_4 . - Therefore, in the solution with 0.150 M H_2SeO_4 , the hydronium-ion concentration from the first ionization alone is approximately 0.150 M. Understanding that strong acids completely dissociate their initial protons is crucial for correctly predicting the hydronium-ion concentration.

Hydronium Ion Concentration

Hydronium ion concentration is a measure of the acidity in a solution. It is denoted by [ H_3O^+ ], signifying the concentration of hydronium ions present. In strong acids like H_2SeO_4 , ionization is complete for the first proton, leading directly to the hydronium-ion concentration matching the H_2SeO_4 concentration. - In pure strong acid solutions, all acid molecules dissociate and contribute to hydronium ions. - For our exercise's initial condition where the second ionization is ignored, [ H_3O^+ ] equals the initial concentration of 0.150 M. However, accounting for both ionizations reveals additional contributions: - Subsequent ionization of weakly acidic protons (from HSeO_4^- ) slightly increases the hydronium-ion concentration. - This additional hydronium ion count is calculated using the equilibrium constant for the second ionization step. By understanding hydronium ion concentration, students can appreciate how strong acids affect solution pH and how secondary ionizations further influence acidity.

Equilibrium Expressions

Equilibrium expressions are formulas showing the relationship between products and reactants in partial ionization. They involve writing the concentrations of ions in a balanced chemical reaction. Understanding these expressions helps predict and calculate the concentration of ions in a solution at equilibrium.For H_2SeO_4, the second ionization or equilibrium expression is:\[K_{a2} = \frac{[SeO_4^{2-}][H^+]}{[HSeO_4^-]}= 1.2 \times 10^{-2}\]In this specific scenario:- Initially, [HSeO_4^-] is approximately 0.150 M.- As x moles of HSeO_4^- ionize, [H^+] is 0.150 + x, while [SeO_4^{2-}] is x.Using the approximation x is small, simplifies the calculation:\[ x \approx 1.2 \times 10^{-2} \times 0.150 = 1.8 \times 10^{-3}\]Finally, to find the total hydronium-ion concentration, sum the contributions from both ionizations, resulting in 0.1518 M. Equilibrium expressions allow us to predict these changes mathematically, making them invaluable tools for chemists and students alike.