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title: "A student in a chemistry laboratory mixes equal volumes of two aqueous solutions: \\(50.0\\text{ mL}\\) of \\(0.10\\text{ M } \\text{Ba(OH)}_2\\text{(aq)}\\) and \\(50.0\\text{ mL}\\) of \\(0.10\\text{ M } \\text{H}_2\\text{SO}_4\\text{(aq)}\\). A white precipitate forms immediately upon mixing. Which of the following best predicts the spectator ions present in the reaction and the electrical conductivity of the resulting mixture relative to the initial solutions?"
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url: "https://nerd-notes.com/ubq/119673/"
date_modified: "2026-08-21T08:12:00+00:00"
---

# A student in a chemistry laboratory mixes equal volumes of two aqueous solutions: \(50.0\text{ mL}\) of \(0.10\text{ M } \text{Ba(OH)}_2\text{(aq)}\) and \(50.0\text{ mL}\) of \(0.10\text{ M } \text{H}_2\text{SO}_4\text{(aq)}\). A white precipitate forms immediately upon mixing. Which of the following best predicts the spectator ions present in the reaction and the electrical conductivity of the resulting mixture relative to the initial solutions?

A student in a chemistry laboratory mixes equal volumes of two aqueous solutions: \(50.0\text{ mL}\) of \(0.10\text{ M } \text{Ba(OH)}_2\text{(aq)}\) and \(50.0\text{ mL}\) of \(0.10\text{ M } \text{H}_2\text{SO}_4\text{(aq)}\). A white precipitate forms immediately upon mixing. Which of the following best predicts the spectator ions present in the reaction and the electrical conductivity of the resulting mixture relative to the initial solutions?

- **A.** \(\text{Ba}^{2+}\text{(aq)}\) and \(\text{SO}_4^{2-}\text{(aq)}\) are spectator ions, and the electrical conductivity remains high because the spectator ions remain dissolved.
- **B.** \(\text{H}^+\text{(aq)}\) and \(\text{OH}^-\text{(aq)}\) are spectator ions, and the electrical conductivity remains high because strong acid and strong base ions dominate the solution.
- **C.** There are no spectator ions, but the electrical conductivity remains high because the reaction produces liquid water molecules.
- **D.** There are no spectator ions, and the electrical conductivity drops to nearly zero because all reacting ions are consumed to form \(\text{BaSO}_4\text{(s)}\) and \(\text{H}_2\text{O(l)}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119673/*