Finding the iodide ion concentration

[Benzoate]

Homework Statement

What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 37.2 mL of the solution produces 539.7 mg of PbI2?

Homework Equations

possibly c(1)*V(1)=c(2)*V(2) is pertinent to the problem

The Attempt at a Solution

I don't know if this is relevant to finding the concentration of Iodine, but I went ahead and converted total number of 539 mg PbI2 => the number of mmols of I^1-

n= 539 mg PbI2 *(1 mmol PbI2/461 mg PbI2) *(2 mmol I^1- /1 mmol PbI2)= 2.34 mmol I^1-

.100 M Pb(NO3)2 = .100 mmol Pb(NO3)2/mL

.110 mmol Pb(NO3)2/mL*(37.2 mL)=3.72 mmol Pb(NO3)2

Perhaps I didn't need to find the total mmol of Pb(NO3)2 and maybe just needed to divide the number of millimoles of ion by the total volume of the solution.

Maybe the concentration of [I^1-] = 2.34 mmol I^1- /( 37.2 mL of solution)

[I^1-] = .063 M

 

[Borek]

This is simple stoichiometry, start with reaction equation.