Question

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?

Answer 1

The molarity of `"Pb"^(2+)` is 0.004 68 mol/L.

Explanation:

The equation for the reaction is

`"Pb"^(2+) + "2I"^"-" → "PbI"_2`

Let's start by calculating the moles of `"PbI"_2`.

`"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 `"Pb"^"2+"`.

`"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 `"Pb"^"2+"`.

`"Molarity" = "moles"/"litres" = (3.883 × 10^"-4" color(white)(l)"mol")/("0.0829 L") = "0.004 68 mol/L"`