If 82.9 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.179 g of precipitate, what is the molarity of lead(II) ion in the original solution?

Oct 10, 2016

The molarity of ${\text{Pb}}^{2 +}$ is 0.004 68 mol/L.

Explanation:

The equation for the reaction is

${\text{Pb"^(2+) + "2I"^"-" → "PbI}}_{2}$

Let's start by calculating the moles of ${\text{PbI}}_{2}$.

${\text{Moles of PbI"_2 = 0.179color(red)(cancel(color(black)("g PbI"_2))) × ("1 mol PbI"_2)/(461.01 color(red)(cancel(color(black)("g PbI"_2)))) = 3.883 × 10^"-4"color(white)(l) "mol PbI}}_{2}$

Now we can use the molar ratio from the equation to calculate the moles of $\text{Pb"^"2+}$.

$\text{Moles of Pb"^"2+" = 3.883 × 10^"-4" color(red)(cancel(color(black)( "mol PbI"_2))) × "1 mol Pb"^"2+"/(1 color(red)(cancel(color(black)("mol PbI"_2)))) = 3.883 × 10^"-4"color(white)(l)"mol Pb"^"2+}$

And now we can calculate the molarity of the $\text{Pb"^"2+}$.

$\text{Molarity" = "moles"/"litres" = (3.883 × 10^"-4" color(white)(l)"mol")/("0.0829 L") = "0.004 68 mol/L}$