Question #86cf3
1 Answer
Here's what I got.
Explanation:
The first thing you need to do here is to convert the mass of iron(II) sulfide to moles by using the compound's molar mass
#1.76 * 10^(-8) color(red)(cancel(color(black)("g"))) * "1 mole FeS"/(87.91color(red)(cancel(color(black)("g")))) = 2.002 * 10^(-10)# #"moles FeS"#
The molar solubility of iron(II) sulfide tells you the number of moles that can be dissolved per liter of solution in order to produce a saturated iron(II) sulfide solution.
You already know that you can get a saturated iron(II) sulfide solution in
#317.5 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 317.5 * 10^(-3)# #"L"#
so now you have to determine how many moles of iron(II) sulfide can be dissolved in
#1 color(red)(cancel(color(black)("L solution"))) * (2.002 * 10^(-10)color(white)(.)"moles FeS")/(317.5 * 10^(-3)color(red)(cancel(color(black)("L solution")))) = 6.306 * 10^(-10)# #"moles FeS"#
You can thus say that iron(II) sulfide has a molar solubility equal to
#color(darkgreen)(ul(color(black)("molar solubility" = 6.31 * 10^(-10)color(white)(.)"mol L"^(-1))))#
To find the solubility product constant,
#"FeS"_ ((s)) rightleftharpoons "Fe"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#
By definition, the solubility product constant will be
#K_(sp) = ["Fe"^(2+)] * ["S"^(2-)]#
Notice that for every
This means that the equilibrium concentrations of the two ions will be equal to the molar solubility of the salt.
Therefore, you can say that
#K_(sp) = 6.31 * 10^(-10) * 6.31 * 10^(-10)#
#color(darkgreen)(ul(color(black)(K_(sp) = 3.98 * 10^(-19))))#
Keep in mind that
The answers are rounded to three sig figs, the number of sig figs you have for the mass of iron(II) sulfide dissolved in the initial solution.