Question

The iodide ion concentration in a solution may be determined by the precipitation of lead iodide. Pb2 (aq) 2I-(aq) PbI2(s) A student finds that 15.71 mL of 0.5770 M lead nitrate is needed to precipitate all of the iodide ion in a 25.00-mL sample of an unknown. What is the molarity of the iodide ion in the student's unknown

Answer 1

Answer:

$M_{I^-}=0.6841M$

Explanation:

Hello.

In this case, since the precipitation of the lead iodide is related to the iodide ion in solution, if we make react lead (II) nitrate with an iodide-containing salt, a possible chemical reaction would be:

$Pb(NO_3)_2+2I^-\rightarrow PbI_2+2NO_3^-$

In such a way, since 15.71 mL of a 0.5770-M solution of lead (II) nitrate precipitates out lead (II) iodide, we can first compute the moles of lead (II) nitrate in the solution:

$n_{Pb(NO_3)_2}=0.5570\frac{molPb(NO_3)_2}{L}*0.01571L=0.01393molPb(NO_3)_2$

Next, since there is a 1:2 mole ratio between lead (II) nitrate and iodide ions, we compute the moles of those ions:

$n_{I^-}=0.01393molPb(NO_3)_2*\frac{2molI^-}{1molPb(NO_3)_2} =0.02785molI^-$

Finally, since the mixing of the two solutions produce a final volume of 40.71 mL (0.04071 L), the resulting concentration (molarity) of the iodide ions in the student's unknown turns out:

$M_{I^-}=\frac{0.02785molI^-}{0.04071L}\\\\ M_{I^-}=0.6841M$

Best regards!