Question: Lead(II) bromide, PbBr 2 , is a slightly soluble saltwith a K SP value of6.6 x 10 -6 . The equation for itsdissociation in water

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?

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