You have a vial that may contain lead(II) cations or barium cations, but not both. Available to you are solutions of ammonium sulfate, sodium fluoride and potassium bromide. Which single solution can you add to the vial to determine which cation is present?

We should add, the KBr solution.

Explanation:

The vial contains Ba²⁺  and Pb²⁺

Our solutions are: (NH₄)₂ SO₄, NaF and KBr

We dissociate the solutions:

(NH₄)₂ SO₄  → 2NH₄⁺ + SO₄⁻²

NaF → Na⁺ + F⁻

KBr → K⁺ + Br-

When we add ammonium sulfate and sodium fluorine, we are adding sulfates and fluorines, but these two anions make precipitate both cations, we can not determine, if the vial contains Ba²⁺ or Pb²⁺

SO₄⁻²(aq) + Ba²⁺ (aq) → BaSO₄ (s)  ↓

SO₄⁻² (aq)+ Pb²⁺(aq)  → PbSO₄ (s)  ↓

2F⁻ (aq) + Ba²⁺ (aq) → BaF₂ (s) ↓

2F⁻ (aq) + Pb²⁺ (aq) → PbF₂ (s)  ↓

When we add the KBr, bromides only make precipitate with Pb²⁺, so the vial has only Pb²⁺ cation:

2Br⁻ (aq) + Ba²⁺ (aq) → BaBr₂ (aq)

2Br⁻ (aq) + Pb²⁺ (aq) → PbBr₂ (s) ↓

i believe it is c.) 1575 j.

you didnt put any of the following