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.

 

[Blueshift5]

I would approach this problem by using the equation c1V1 = c2V2, where c1 is the initial concentration of the solution (in this case, Pb(NO3)2), V1 is the initial volume of the solution, c2 is the final concentration of the solution (in this case, PbI2), and V2 is the final volume of the solution.

Using this equation, we can rearrange it to solve for c2, which is the concentration of the iodide ion. This would give us:

c2 = (c1V1)/V2

Plugging in the values given in the problem, we get:

c2 = (0.100 M * 37.2 mL)/37.2 mL = 0.100 M

This means that the final concentration of PbI2 is 0.100 M. However, this is not the concentration of the iodide ion. To find the concentration of the iodide ion, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every 2 moles of iodide ion, we get 1 mole of PbI2. This means that the concentration of iodide ion is half of the concentration of PbI2.

Therefore, the concentration of the iodide ion is 0.050 M (0.100 M/2). This means that the iodide ion concentration in the solution is 0.050 M.