Which of the following options would be the best for dissolving PbBr2(s)? 1) add to a solution of CH3COOH
2) add to a solution of NaBr
3) add to a solution of NH3
4) add to a solution of NH4Br
5) add to a solution of NaOH

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 = 6.60 × 10⁻⁶, in a solution of 0.5 M NaBr, we have;

PbBr₂ → Pb⁺ + 2Br⁻

= [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.

Learn more about solubility product here;

yes definetly because you are using energy