Question

Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) in a solution prepared by mixing equal volumes of \(0.2 \mathrm{M} \mathrm{HCl}\) and \(0.6 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\left(K_{a 2}\right.\) for \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(\left.1.2 \times 10^{-2}\right)\)

Short answer

[H₃O⁺] = 0.409 M, [SO₄²⁻] = 0.009 M.

Step-by-step solution

Understanding the Problem

We are asked to find the concentrations of \(\mathrm{H}_3\mathrm{O}^+\) and \(\mathrm{SO}_4^{2-}\) in a mixture of \(0.2\, \mathrm{M}\, \mathrm{HCl}\) and \(0.6\, \mathrm{M}\, \mathrm{H_2SO_4}\). We should consider the dissociation of both acids and the dilution effect when volumes are mixed.

Determine Initial Acid Concentrations After Mixing

Since equal volumes of the solutions are mixed, the concentration of each acid is halved in the resulting solution. Therefore, \[ [\mathrm{HCl}] = 0.1\, \mathrm{M}, \quad [\mathrm{H_2SO_4}] = 0.3\, \mathrm{M} \] after mixing.

Calculate Initial \(\mathrm{H}_3\mathrm{O}^+\) Concentration

Both \(\mathrm{HCl}\) and the first dissociation of \(\mathrm{H_2SO_4}\) contribute to the \(\mathrm{H}_3\mathrm{O}^+\) concentration since they fully dissociate. The initial \(\mathrm{H}_3\mathrm{O}^+\) is \[ [\mathrm{H}_3\mathrm{O}^+] = 0.1 + 0.3 = 0.4\, \mathrm{M} \] from the complete dissociation of \(\mathrm{HCl}\) and the first dissociation of \(\mathrm{H_2SO_4}\).

Second Dissociation of \(\mathrm{H_2SO_4}\)

The second dissociation of \(\mathrm{H_2SO_4}\) can contribute extra \(\mathrm{H}_3\mathrm{O}^+\) and \(\mathrm{SO}_4^{2-}\). Set up the equation for the second dissociation equilibrium: \[ \mathrm{HSO_4^-} \rightleftharpoons \mathrm{H}^+ + \mathrm{SO_4^{2-}} \] and use \(K_{a2} = 1.2 \times 10^{-2}\, \mathrm{M}\) to solve for the additional \([\mathrm{H}_3\mathrm{O}^+]]\) and \([\mathrm{SO_4^{2-}}]\).

Set up the Equilibrium Equation

Assuming \(x\) is the increase in \(\mathrm{H}^+\) and \(\mathrm{SO_4^{2-}}\) from the second dissociation, \(K_{a2}\) gives us: \[ K_{a2} = \frac{{[\mathrm{H}^+] [\mathrm{SO_4^{2-}}]}}{{[\mathrm{HSO_4^-}]}} = \frac{{0.4 + x \times x}}{{0.3 - x}} \] Given that \(K_{a2}\) is small, we can approximate \(0.3 - x \approx 0.3\), giving the equation \[ 1.2 \times 10^{-2} = \frac{{0.4x}}{{0.3}} \].

Solve for \(x\)

Rearrange and solve the above equation for \(x\): \[ x = \frac{{1.2 \times 10^{-2} \times 0.3}}{{0.4}} = 9 \times 10^{-3}\, \mathrm{M} \]. This is the additional concentration of both \(\mathrm{H}_3\mathrm{O}^+\) and \(\mathrm{SO_4^{2-}}\).

Final Concentrations Calculation

The final concentration of \(\mathrm{H}_3\mathrm{O}^+\) is \(0.4 + x = 0.4 + 9 \times 10^{-3} = 0.409\, \mathrm{M}\). The concentration of \(\mathrm{SO}_4^{2-}\) is also \(9 \times 10^{-3}\, \mathrm{M}\).

Key concepts

Dissociation

Dissociation is a fundamental concept in acid-base chemistry. It describes the process where molecules or compounds separate into smaller ions or molecules. For acids, dissociation involves the release of hydrogen ions (\[\mathrm{H}^+\]) into a solution, which interact with water to form hydronium ions (\[\mathrm{H}_3\mathrm{O}^+\]).When \[\mathrm{HCl}\] and \[\mathrm{H_2SO_4}\] dissociate in water, they release hydrogen ions:

• \[\mathrm{HCl} \rightarrow \mathrm{H}^+ + \mathrm{Cl}^-\]
• \[\mathrm{H_2SO_4} \rightarrow \mathrm{H}^+ + \mathrm{HSO_4}^-\] (first dissociation)

These processes lead to the formation of hydronium ions, influencing the solution's acidity. Understanding dissociation helps us calculate changes in \[\mathrm{pH}\] and \[\mathrm{H_3O^+}\] concentration when acids are added to solutions.

Equilibrium Calculations

Equilibrium calculations are crucial for determining how a chemical system behaves over time, balancing between formation and re-formation of reactants and products. When dealing with weak acids like the second stage of \[\mathrm{H_2SO_4}\] dissociation, not all initial acids convert into ions. This partial dissociation requires specific calculations using the equilibrium constant, \[K_a\].For the second dissociation of sulfuric acid (\[\mathrm{HSO_4^-}\]):

• The equation is\[\mathrm{HSO_4^-} \rightleftharpoons \mathrm{H}^+ + \mathrm{SO_4^{2-}}\]
• \[K_{a2} = \frac{{[\mathrm{H}^+][\mathrm{SO_4^{2-}}]}}{{[\mathrm{HSO_4^-}]}}\], describing the state of equilibrium.

By substituting known values and solving these equations, we find how much additional ions have formed, crucial for determining concentrations in equilibrium.

Sulfuric Acid

Sulfuric acid, \[\mathrm{H_2SO_4}\], is a strong acid notable for fully dissociating its first hydrogen ion and partially for the second. This biprotic acid undergoes two dissociation stages:

• First dissociation: releasing one \[\mathrm{H}^+\] ion completely,\[\mathrm{H_2SO_4} \rightarrow \mathrm{H}^+ + \mathrm{HSO_4}^-\].
• Second dissociation: signifies that \[\mathrm{HSO_4}^-\] can further dissociate,\[\mathrm{HSO_4^-} \rightarrow \mathrm{H}^+ + \mathrm{SO_4^{2-}}\], determined by \[K_{a2}\].

Understanding sulfuric acid's behavior is key to tackling equilibrium problems, as its dual dissociation significantly impacts the pH and ion concentrations of solutions.

Hydronium Ion Concentration

The concentration of hydronium ions, \[\mathrm{H_3O^+}\], in a solution determines its acidity and is crucial in acid-base chemistry. Higher concentrations of hydronium ions result in lower pH values, indicating more acidic solutions.When calculating hydronium concentrations in mixtures:

• The complete dissociation of \[\mathrm{HCl}\] and the first dissociation of \[\mathrm{H_2SO_4}\] provide initial contributions.
• Additional hydronium ions form from the second dissociation of \[\mathrm{HSO_4}^-\].

Thus, the total hydronium ion concentration includes both initial and equilibrium-derived amounts, and accurate calculations are necessary to predict solution behavior. In practice, it influences everything from reaction rates to solubility in chemical processes.