Answer to Question #328598 in General Chemistry for Kim

Question

Given that 45.6 mL of a 0.453 M potassium iodide solution reacts with 0.355 M lead (ll) nitrate solution, what volume of lead (ll) nitrate is required for complete precipitation?

Pb(NO3) + 2 Kl —> Pbl2 + 2KNO3

Answer 1

Pb(NO₃)₂ + 2Kl —> Pbl₂ + 2KNO₃

n(KI) = C × V = 0.453 mol/L × 0.0456 L = 0.02066 mol

1 mol Pb(NO₃)₂ - 2 mol KI

x mol Pb(NO₃)₂ - 0.02066 mol KI

x = 0.02066 / 2 = 0.01033 mol

V = n / C = 0.01033 mol / 0.355 mol/L = 0.0291 L = 29.1 mL