When solutions of potassium iodide and lead (II) nitrate are combined, the products are potassium nitrate and lead (II) iodide.

Question

d. Write a balanced equation for this reaction, including (aq) and (s).

e.  Calculate the mass of precipitate produced when 50ml of .45M potassium iodide solution and 75ml of .55M lead (II) nitrate solution are mixed.

f.  Calculate the volume of .50M potassium iodide required to react completely with 50ml of .50M lead (II) nitrate

  

Answer 1

(d)  2KI(aq) + Pb(NO₃)₂(aq) ==> 2KNO₃(aq) + PbI₂(s)

 

(e)  moles KI used = 0.05L x 0.45 mol/L = 0.0225 moles KI used

moles Pb(NO₃)₂ used = 0.075 L x 0.55 mol/L = 0.04125 moles Pb(NO₃)₂ used

Limiting reactant is KI since it requires 2 moles KI for each 1 mole Pb(NO₃)₂ - see balanced equation in (d) above.

moles PbI₂ formed from 0.0225 moles KI = 0.0225 moles KI x 1 mole PbI₂/2 moles KI = 0.01125 moles PbI₂

mass PbI₂ (precipitate) = 0.01125 moles PbI₂ x 461 g/mole = 5.19 g = 5.2 g (to 2 sig. figs)

 

(f)  moles Pb(NO₃)₂ in 50 ml of 0.50 M = 0.050 L x 0.50 mol/L = 0.025 moles 

0.025 moles Pb(NO₃)₂ x 2 moles KI/mole Pb(NO₃)₂ = 0.05 moles KI required

(x L)(0.50 mol/L) = 0.05 moles

x = 0.1 liters (100 ml) of 0.50 M KI would be required