Question #014bf
1 Answer
Explanation:
For starters, look up the solubility product constant,
#K_(sp) = 1.6 * 10^(-5)#
http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm
Your strategy here will be to calculate the molar solubility of the salt, which is the molar concentration of lead(II) chloride that dissociates in order to produce a saturated solution at room temperature.
You know that lead(II) chloride is considered insoluble in water, so you can say that the solubility equilibrium that is established in aqueous solution will lie to the left.
#"PbCl"_ (color(red)(2)(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#
Notice that every
You can thus say that, at equilibrium, you have
#["Cl"^(-)] = color(red)(2) * ["Pb"^(2+)]#
If you take
#K_(sp) = ["Pb"^(2+)] * ["Cl"^(-)]^color(red)(2)#
as
#K_(sp) = s * (color(red)(2)s)^color(red)(2)#
This is equivalent to
#K_(sp) = 4s^3#
Rearrange to solve for
#s = root(3)(K_(sp)/4)#
Plug in your value to find
#s = root(3)((1.6 8 10^(-5))/4) = 0.0159#
You can thus say that a saturated lead(I)) chloride solution will have
#["Pb"^(2+)] = "0.0159 M"#
As you know, molarity tells you the number of moles of solute present in
To convert this to grams, use the element's molar mass
#0.0159 color(red)(cancel(color(black)("moles Pb"^(2+)))) * "207.2 g"/(1color(red)(cancel(color(black)("mole Pb"^(2+))))) = "3.3 g"#
Therefore, the concentration of the lead(II) cations will be equal to
#color(darkgreen)(ul(color(black)("concentration Pb"^(2+) = "3.3 g L"^(-1))))#
The answer is rounded to two sig figs.