IX 1) 3 drops of 0.32 M potassium iodide are added to 67.0 mL of 0.016 M lead ni

Question

IX

1) 3 drops of 0.32 M potassium iodide are added to 67.0 mL of 0.016 M lead nitrate. Will a precipitate of lead iodide form? (assume 1 drop is 0.05 mL) Ksp (lead iodide) = 7.1 x 10-9.

Enter the value 1 if there will be a precipitate and a value of –1 if there will not be a precipitate.

2) 3 drops of 0.32 M potassium iodide are added to 67.0 mL of 0.016 M lead nitrate. Will a precipitate of lead iodide form? (assume 1 drop is 0.05 mL) Ksp (lead iodide) = 7.1 x 10-9.

Enter the value 1 if there will be a precipitate and a value of –1 if there will not be a precipitate.

Explanation / Answer

1) Ksp of PbI2 = 7.1 x 10^-9

[Pb2+] = 0.016 M x 67 ml/68.5 ml

           = 0.01565 M

[I-] = 0.32 M x 1.5 ml/68.5 ml

     = 0.007 M

Qsp = [Pb2+][I-]^2

       = (0.01565)(0.007)^2

       = 7.67 x 10^-7

Qsp > Ksp, so precipitate would form

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