Question

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0 ml of 0.250M KI

A) In your experiment, one of the reagents was limiting and the other was in excess. Therefore after the lead(II)iodide precipitated from the solution, there were spectator ions left in the solution, and also the unreacted excess reagent. Assuming complete precipitation of lead (II) iodide, calculate the concentrations of each type of ion (Pb2+, K+, NO3-, I-) present in the solution after the PbI2 precipitated out.

Answer 1

Moles of Pb(NO₃)₂ = 2.0 mL x 0.250 M = 0.50 mmol

Moles of KI = 7.0 mL x 0.250 M = 1.75 mmol

Pb(NO₃)₂(aq) + 2KI(aq) PbI₂(s) + 2KNO₃(aq)

For complete reaction, the required mol ratio of Pb(NO₃)₂ : KI = 1 : 2

Here, the available mol ratio = 0.50 : 1.75

                                              = 1 : 3.5

Thus, per mol of Pb(NO₃)₂, the available mol of KI is 3.5 mol which is more than the required 2 mol.

Thus, KI is present in excess and Pb(NO₃)₂ is the limiting reagent in the reaction.

Thus, all of the Pb²⁺ ions have precipitated out of the solution as PbI₂.

Hence, the [Pb²⁺] = 0 M

Theoretically, 1 mol of Pb(NO₃)₂ gives 2 mol of KNO₃

Therefore, 0.50 mmol of Pb(NO₃)₂ gives (2 mol/1 mol) x 0.50 mol = 1.0 mmol of KNO₃

1.0 mmol of KNO₃ dissociates in water to give 1.0 mmol of NO₃^-

Total volume of the solution = (2.0 + 7.0) mL = 9.0 mL

Hence, [NO₃^-] = 1.0 mmol/9.0 mL = 0.11 M

1.75 mmol of KI ionizes in water to give 1.75 mmol of K⁺

Hence, [K⁺] = 1.75 mmol/9.0 mL = 0.19 M

Theoretically, 1 mol of Pb(NO₃)₂ react with 2 mol of KI

Therefore, 0.50 mmol of Pb(NO₃)₂ react with (2 mol/1 mol) x 0.50 mmol = 1.00 mol of KI

Hence, the mol of unreacted KI in the solution = (1.75 - 1.00) mmol = 0.75 mmol

Now, 0.75 mmol of KI ionizes in water to give 0.75 mmol of I^-.

Hence, [I^-] = 0.75 mmol/9.0 mL = 0.083 M