Lead(II) bromide, PbBr₂, is a slightly soluble saltwith a K_SP value of6.6 x 10^-6.  The equation for itsdissociation in water is

PbBr₂(s)   <---------->  Pb²⁺(aq)   +  2Br^-(aq)

What is the molar solubility of PbCl₂ in distilledwater? I KNOW THE ANSWER IS 1.2 x 10 ^-2 mol/L

but I don\'t understand this part:
In the previous problem, you calculated the molar solubility ofPbBr₂ in distilled water.  Now consider themolar solubility of PbBr₂ in a 3.55 M NaBrsolution.  Sodium bromide is a freely soluble salt, andcompletely dissociates into its ions when dissolved in water:

NaBr(s)   ==========>  Na⁺(aq)   +  Br^-(aq)

Therefore, a 3.55 M NaBr solution is 3.55 M inNa⁺ and 3.55 M in Br^-.  Whenthe PbBr₂ is dissolved in this solution, itestablishes its usual equilibrium:

PbBr₂(s)   <---------->  Pb²⁺(aq)   +  2Br^-(aq)     K_SP = 6.6 x 10^-6

But in the 3.55 M NaBr solution, theBr^- concentration will be 3.55 M,even before thePbBr₂ dissociates.  TheBr^- formed in the dissociation ofPbBr₂ will be in addition tothat already present.

What is the molar solubility of PbBr₂ in the3.55 M NaBr solution?