The best option for dissolving PbBr₂ is option (2)
2) Add a solution of NaBr
The reason for choosing the above option is as follows;
Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr.
PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr.
Taking the solubility product of PbBr₂ as [tex]K_{sp}[/tex] = 6.60 × 10⁻⁶, in a solution of 0.5 M NaBr, we have;
PbBr₂ → Pb⁺ + 2Br⁻
[tex]K_{sp}[/tex] = [Pb]·[2Br]²
Therefore, we get;
6.60 × 10⁻⁶ = [x]·[0.5]²
Where;
x = The number of moles of lead, Pb, in per liter of solution
∴ x = (6.60 × 10⁻⁶)/[(0.5 )²] = 2.64 × 10⁻⁵.
The molar solubility of PbBr₂ per liter of NaBr, x = 2.64 × 10⁻⁵ mol/L
PbBr₂ is more soluble in NaBr.
Given that ammonium ion NH₄Br in water gives similar products to ammonia, NH₃, it is expected to be more suitable to dissolve PbBr₂ in NaBr.
Therefore, the best solution for dissolving PbBr₂(s) is NaBr, the correct option is option (2) add a solution of NaBr.
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