Question

b. What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 33.8 mL of the solution produces 565.2 mg of PbI2?

Answer 1

Pb(NO3)2 + 2I^- -----> PbI₂ + 2NO₃

one mole or 331.2 g of Pb(NO₃)₂ react with 2 moles or 2 X 126.9 = 253.8g of I- to form one mole or 461.2g of PbI₂.

Moles of PbI2 produced in the reaction = 565.2mg

Mass of PbI2 produced in grams = 565.2 / 1000 = 0.5652g

As per reaction above 461.2g of PbI₂ are produced by 253.8g of I^-

Therefore 1g of PbI₂ will be produced by 331.2/253.8 =1.305g of I^-

Hence 0.5652g of PbI2 (What is actually produced) will be produced by 0.5652 X 1.305 = 0.7376g of I^-

Above result is important . It means there must be 0.7376g of I- that will produce 0.5652g Og PbI2

Molarity of I^- = No. of moles of I- / Volume of solution in liters

No. of moles of I- = mass of I- / molar mass of I- = 0.7376 / 126.9 = 0.0058 moles

Volume of solution = 33.8mL = 33.8 / 1000 = 0.0338L

Molarity of I^- = 0.0058 moles/ 0.0338 = 0.172M