How many milliliters of water are needed to dissolve #"10 mg"# of #"PbF"_2# ?
How to solve the problem step by step?
answer in my book says V(H2O) = 21.58mL
How to solve the problem step by step?
answer in my book says V(H2O) = 21.58mL
1 Answer
Here's what I got.
Explanation:
The idea here is that you need to figure out the solubility of lead(II) fluoride in water by using its solubility product constant,
Since you didn't provide a value for the
#K_(sp) = 4 * 10^(-8)#
http://www.wiredchemist.com/chemistry/data/solubility-product-constants
Now, lead(II) fluoride is considered to be insoluble in water, which means that when you dissolve this ionic compound in water, an equilibrium is established between the dissolved ions and the undissolved solid
#"PbF"_ (color(red)(2)(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"F"_ ((aq))^(-)#
By definition, the solubility product constant for lead(II) fluoride is
#K_(sp) = ["Pb"^(2+)] * ["F"^(-)]^color(red)(2)#
If you take
#K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3#
Rearrange to solve for
#s = root(3)(K_(sp)/4)#
Plug in your values to find
#s = root(3)( (4 * 10^(-8))/4) = 2.154 * 10^(-3)"M"#
This means that you can only dissolve
To convert this to grams per liter, use the molar mass of lead(II) fluoride
#2.154 * 10^(-3) color(red)(cancel(color(black)("moles PbF"_2)))/"1 L water" * "245.2 g"/(1color(red)(cancel(color(black)("mole PbF"_2)))) = "0.5282 g L"^(-1)#
Since you know that
#color(blue)(ul(color(black)("1 g" = 10^3"mg")))" "# and#" "color(blue)(ul(color(black)("1 L" = 10^3"mL")))#
you can say that the solubility of the salt is equivalent to
#"0.5282 g L"^(-1) = "0.5282 mg mL"^(-1)#
Therefore, you will be able to dissolve
#10 color(red)(cancel(color(black)("mg"))) * "1 mL water"/(0.5282color(red)(cancel(color(black)("mg")))) = color(darkgreen)(ul(color(black)("19 mL water")))#
I'll leave the answer rounded to two sig figs.
SIDE NOTE The answer doesn't agree with the one given by your book because there is a chance that they used a different value for the solubility product constant of lead(II) fluoride.
Make sure to use the method I showed you to recalculate using the value you have for the