# 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"# ?

##### 1 Answer

#### 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.

#"SnS"_ ((s)) rightleftharpoons "Sn"_ ((aq))^(2+) + "S"_ ((aq))^(2-)#

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

#K_(sp) = ["Sn"^(2+)] * ["S"^(2-)]#

Now, you know that this salt has a solubility of

This tells you that at that temperature, only **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

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

#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 **mole** of tin(II) cations and **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_(sp) = (3.575 * 10^(-13)) * (3.575 * 10^(-13)) = color(darkgreen)(ul(color(black)(1.28 * 10^(-25))))#

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