Answer to Question #202794 in Chemistry for Shaza Shihadeh

Question #202794

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)


1
Expert's answer
2021-06-04T15:42:04-0400

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

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