Answer:
Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq)
A precipitate forms
Explanation:
To solve this question we must know the general solubility rules:
All group 1A (Li, Na, K...) and NH₄⁺ ions are always soluble.
The nitrates (NO₃⁻) and acetates are always solubles.
Cl⁻, Br⁻ and I⁻ are soluble except in the presence of Ag⁺, Pb²⁺, Cu⁺ and Hg₂²⁺
That means in the mixture of ions that we have, the Br⁻ will react with Pb²⁺ to produce PbBr₂(s):
Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq)
