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

1 Answer
Feb 4, 2017

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, #K_(sp)#.

Since you didn't provide a value for the #K_(sp)# of lead(II) fluoride, I'll use the value listed here

#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 #s# to be the molar solubility of the salt in water, i.e. the number of moles of lead(II) fluoride that dissociate to produce lead(II) cations and fluoride anions, you can say that you have

#K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3#

Rearrange to solve for #s#

#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 #2.154 * 10^(-3)# moles of lead(II) fluoride for every liter of solution. For all intended purposes, the volume of the solution can be approximated with the volume of water, the solvent.

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 mg"# of lead(II) fluoride in

#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 #K_(sp)# of the salt.