Question

What is the iodide ion concentration in a solution if the addition of an excess of 0.100 m pb(no3)2 to 42.9 ml of the solution produces 861.8 mg of pbi2?

Answer 1

Moles Pbl2 = 0.8628 g : 461.01 g/mol = 0.001871

moles I = 2 x 0.001871 = 0.003742

[I-] = 0.003742/ 0.0429

= 0.0872 M