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

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.

Explanation / Answer

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

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