# Question 86cf3

Apr 13, 2017

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) $\text{moles FeS}$

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) $\text{L}$

so now you have to determine how many moles of iron(II) sulfide can be dissolved in $\text{1 L}$ of solution

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)# $\text{moles FeS}$

You can thus say that iron(II) sulfide has a molar solubility equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molar solubility" = 6.31 * 10^(-10)color(white)(.)"mol L}}^{- 1}}}}$

To find the solubility product constant, ${K}_{s p}$, of iron(II) sulfide at ${25}^{\circ} \text{C}$, use the fact that the salt partially ionizes to produce iron(II) cations and sulfide anions

${\text{FeS"_ ((s)) rightleftharpoons "Fe"_ ((aq))^(2+) + "S}}_{\left(a q\right)}^{2 -}$

By definition, the solubility product constant will be

${K}_{s p} = \left[{\text{Fe"^(2+)] * ["S}}^{2 -}\right]$

Notice that for every $1$ mole of iron(II) sulfide that dissociates in aqueous solution, you get $1$ mole of iron(II) cations and $1$ mole of sulfide anions.

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

${K}_{s p} = 6.31 \cdot {10}^{- 10} \cdot 6.31 \cdot {10}^{- 10}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{s p} = 3.98 \cdot {10}^{- 19}}}}$

Keep in mind that ${K}_{s p}$ values are usually given without added units.

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.