The best option for dissolving PbBrâ 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.
PbBrâ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)â and NaBr.
Taking the solubility product of PbBrâ as [tex]K_{sp}[/tex] = 6.60 Ă 10â»â¶, in a solution of 0.5 M NaBr, we have;
PbBrâ â Pbâș + 2Brâ»
[tex]K_{sp}[/tex] = [Pb]·[2Br]ÂČ
Therefore, we get;
6.60 Ă 10â»â¶ = [x]·[0.5]ÂČ
Where;
x = The number of moles of lead, Pb, in per liter of solution
⎠x = (6.60 Ă 10â»â¶)/[(0.5 )ÂČ] = 2.64 Ă 10â»â”.
The molar solubility of PbBrâ per liter of NaBr, x = 2.64 Ă 10â»â” mol/L
PbBrâ is more soluble in NaBr.
Given that ammonium ion NHâBr  in water gives similar products to ammonia, NHâ, it is expected to be more suitable to dissolve PbBrâ in NaBr.
Therefore, the best solution for dissolving PbBrâ(s) is NaBr, the correct option is option (2) add a solution of NaBr.
Learn more about solubility product here;
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