Question

g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.904 g of precipitate, what is the molarity of lead(II) ion in the original solution

Answer 1

`M=0.0637M`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)`

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

`n_(Pb^(2+))=0.904gPbI_2*(1molPbI_2)/(461gPbI_2)*(1molPb(NO_3)_2)/(1molPbI_2) *(1molPb^(2+))/(1molPb(NO_3)_2) =1.96x10^(-3)molPb^(2+)`

Finally, the resulting molarity in 30.8 mL (0.0308 L):

`M=(1.96x10^(-3)molPb^(2+))/(0.0308L)\\ \\M=0.0637M`

Regards.