# What is the value of K_(sp) for tin(II) sulfide given that its solubility is 5.39 * 10^(−12) "g/100 g H"_2"O" ?

Feb 25, 2018

$1.28 \cdot {10}^{- 25}$

#### Explanation:

Tin(II) sulfide is considered insoluble in water, so you can write its dissociation equilibrium by taking into account the fact that a saturated solution of tin(II) sulfide contains the undissolved solid in equilibrium with the dissolved ions.

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

The solubility product constant for tin(II) sulfide looks like this

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

Now, you know that this salt has a solubility of $5.39 \cdot {10}^{- 12}$ $\text{g}$ per $\text{100 g}$ of water at an unspecified temperature.

This tells you that at that temperature, only $5.39 \cdot {10}^{- 12}$ $\text{g}$ of tin(II) sulfide per $\text{100 g}$ of water will dissociate to produce tin(II) cations and sulfide anions in a saturated solution of tin(II) sulfide.

Since the problem made no mention of the density of water, you can take it to be equal to ${\text{1 g mL}}^{- 1}$ and say that in a saturated solution of tin(II) sulfide, $5.39 \cdot {10}^{- 12}$ $\text{g}$ of tin(II) sulfide per $\text{100 mL}$ of water will dissociate to produce tin(II) cations and sulfide anions.

Now, because the mass of tin(II) sulfide that dissociates is so small, you can say that the volume of solution will be equal to the volume of water.

Therefore, in $\text{1 L} = {10}^{3}$ $\text{mL}$ of this solution, only

10^3 color(red)(cancel(color(black)("mL solution"))) * (5.39 * 10^(-12) quad "g SnS")/(100color(red)(cancel(color(black)("mL solution")))) = 5.39 * 10^(-11) quad "g SnS"

will dissociate to produce ions. Use the molar mass of tin(II) sulfide to convert this to moles

5.39 * 10^(-11) color(red)(cancel(color(black)("g"))) * "1 mole SnS"/(150.76color(red)(cancel(color(black)("g")))) = 3.575 * 10^(-13) quad "moles SnS"

Since each mole of tin(II) sulfide that dissociates produces $1$ mole of tin(II) cations and $1$ mole of sulfide anions, the saturated solution will contain

["Sn"^(2+)] = ["S"^(2-)] = 3.575 * 10^(13) quad "M"

You can thus say that the solubility product constant for tin(II) sulfide is equal to--I'll leave the value without added units!

${K}_{s p} = \left(3.575 \cdot {10}^{- 13}\right) \cdot \left(3.575 \cdot {10}^{- 13}\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.28 \cdot {10}^{- 25}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the solubility of the salt.