# Question 8215a

Feb 16, 2016

$1.61 \cdot {10}^{- 2} \text{g}$

#### Explanation:

The idea here is that you need to first figure out how many moles of lead(II) cations, ${\text{Pb}}^{2 +}$, would be present in $1.00 \cdot {10}^{- 2} \text{g}$ of lead(II) cations.

Once you know that, use the formula of lead(II) nitrate to figure out how many moles of lead(II) nitrate would produce that many moles of lead(II) cations in solution.

Now, don't be confused about the fact that I used lead(II) cations, and not lead. For the purpose of this problem, the two chemical species are equivalent.

So, use lead's molar mass to find how many moles you'd get in that $1.00 \cdot {10}^{- 2} \text{-g}$ sample

1.00 * 10^(-2)color(red)(cancel(color(black)("g"))) * "1 mole Pb"^(2+)/(207.2color(red)(cancel(color(black)("g")))) = 4.86 * 10^(-5)"moles Pb"^(2+)

Next, take a look at the chemical formula for lead(II) nitrate, "Pb"("NO"_3)_2, an ionic compound that dissociates completely in aqueous solution to form lead(II) cations and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{Pb"("NO"_3)_text(2(aq]) -> "Pb"_text((aq])^(2+) + 2"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

Notice the $1 : 1$ mole ratio that exists between lead(II) nitrate and lead(II) cations. This tells you that every mole of lead(II) nitrate that you dissolve in aqueous solution will produce one mole of lead(II) cations.

Since you know that your solution must contain $4.86 \cdot {10}^{- 5}$ moles of ${\text{Pb}}^{2 +}$, you can conclude that you must dissolve

4.86 * 10^(-5)color(red)(cancel(color(black)("moles Pb"^(2+)))) * ("1 mole Pb"("NO"_3)_2)/(1color(red)(cancel(color(black)("mole Pb"^(2+))))) = 4.86 * 10^(-5)"moles Pb"("NO"_3)_2

Now all you have to do is use lead(II) nitrate's molar mass to determine how many grams would be equivalent to this many moles

4.86 * 10^(-5)color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * "331.21 g"/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2)))) = color(green)(1.61 * 10^(-2)"g"#

The answer is rounded to three sig figs.

Therefore, dissolving $1.61 \cdot {10}^{- 2} \text{g}$ of lead(II) nitrate in one liter of water will get you $1.00 \cdot {10}^{- 2} \text{g}$ of lead, in the form of lead(II) cations, in solution.