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

Answer:

$M=0.0637M$

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:

$n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}$

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

$M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M$

Regards.