Question

In a saturated solution of PbBr2, the molarity of Pb2+ ions are 8.2 x 10-2 mol/L, how

much would be the [Br-1] in g/L? (MBr=40 gr/mol)

Answer 1

Solution:

The chemical equation for the dissolution of PbBr₂(s) in water is

PbBr₂(s) ⇔ Pb²⁺(aq) + 2Br⁻(aq)

By stoichiometry, [Br⁻] = 2 × [Pb²⁺] , thus [Br⁻] = 2 × (8.2×10⁻²) = 0.164 M

[Br⁻] = 0.164 mol/L

!!! M(Br⁻) = 80 g/mol !!!

Therefore,

(0.164 mol Br⁻ / 1 L) × (80 g Br⁻ / 1 mol Br⁻) = 13.12 g/L Br⁻

Answer: [Br⁻] = 13.12 g/L