We’re being asked to determine the solubility (in M) of PbBr_{2} after 5.00 L of the 1.00x10^{-3} M of the CaBr_{2} solution has been added to 500 mL of the 2.0 M Pb(NO_{3})_{2}.

*We’re going to calculate the solubility of PbBr _{2} using the following steps:*

**Step 1:** Calculate the moles of Pb^{2+} from Pb(NO_{3})_{2}.**Step 2:** Calculate the moles of Br^{-} from CaBr_{2}.**Step 3:** Determine the excess and calculate the concentration of the remaining excess reactant.**Step 4:** Create an ICE chart for the dissociation of PbBr_{2}.**Step 5: **Calculate the solubility of PbBr_{2}.

A 1.00 x 10^{─3} M solution of CaBr _{2} is added to 500 mL of a 2.00 M solution of Pb(NO_{3})_{2}.

Determine the solubility (in M) of PbBr _{2} after 5.00 L of the CaBr _{2} solution has been added to 500 mL of the 2.0 M Pb(NO_{3})_{2} solution. Show all work and circle your final answer.

A. 1.51x10^{-4} M

B. 0.181 M

C. 1.39x10^{-4} M

D. 9.10x10^{-4} M

E. 1.96x10^{-4} M

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