Question

What is the solubility of lead (II) bromide in 0.330 M sodium bromide?

What causes the difference in solubilities of the lead (II) bromide?

Answer 1

Answer – We are given, [NaBr] = 0.330 M

We know, [NaBr] = [Br^-] = 0.330 M

Ksp expression for the lead (II) bromide

Ksp = [Pb²⁺] [Br^-]²

4.0 *10^-5 = [Pb²⁺] (0.330)²

[Pb²⁺] = 4.0 *10^-5 / (0.330)²

            = 3.67*10^-4 M

The solubility of lead (II) bromide in 0.330 M sodium bromide is 3.67*10^-4 M

We know as we added the common ion there is solubility decreased since we know reaction –

PbBr₂(s) <----->[Pb²⁺] + 2 [Br^-]

so we added Br^- there is the concentration of product side increase and reaction gets reversed and reactant favored according to Le Chatelier's principle. So when we added sodium bromide in lead (II) bromide there is solubility decrease, so the solubility of lead (II) bromide in 0.330 M sodium bromide is less than the solubility of the lead (II) bromide.