Question

2) When 52.0 mL of 0.1250M H2SO4 is added to 100 mL of a solution of Pbl2, a precipitate of PbSO4 forms. The PBSO4 is then filtered from the solution, dried, and weighed. If the recovered PbSO4 is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution?

Answer 1

In order to do all calcultions we need to find the limiting reagent. The one which has less number of moles will be the limiting reagent.

Equation, thSoy + Pb Ig - PBSQy + 2HI Hence here I moke of pbl givee 1 mol of posoy. Maxs of posoy obtained = 0.04719 Molar ma's of posoy = 303-26 g/mol Hence moles of pb soy = 0:0471 g = 0.000155 mol 303.269 imol Moles of H50y = 0.125MX 0.0524 -0.0065 mol Hence pbl is the limiting reagent, now all calculations will be based on Pb 12. This 100 mL of pbl solution contains 0.000155 moles PbSOy Hence in lovona of PbI solution contains 0.00155 moles posoy. (0.00155 Mol/L) = 0.00155 M Now I mol Pbl contains a moles of Ir ims. concentration of L'ins = 2 x 0.00155 -0.0031 M Hence initial or original solution has 0.0031 M of I ions.