Cobalt(II) sulfide, "CoS", has a K_(sp) value of 3.0 xx 10^(-26). What is the solubility of "CoS" in mol/L?

Aug 2, 2016

$1.7 \cdot {10}^{- 13} {\text{mol L}}^{- 1}$

Explanation:

The idea here is that cobalt(II) sulfide, $\text{CoS}$, is considered Insoluble in aqueous solution, as shown by the very, very small value of its equilibrium solubility constant, ${K}_{s p}$.

This means that when cobalt(II) sulfide is dissolved in water, only a tiny fraction of the ions will dissociate.

Your goal here is to calculate the molar concentration of these dissociated ions. To do that, use an ICE table and the dissociation equation for cobalt(II) sulfide.

If you take $s$ to be the molar solubility of the salt, i.e. the number of moles of ions dissociated in one liter of solution per mole of salt added, you can say that you have

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

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color(purple)("C")color(white)(aaaaacolor(black)(-)aaaaaaacolor(black)((+s))aaaaaaacolor(black)((+s))
color(purple)("E")color(white)(aaaaacolor(black)(-)aaaaaaaaacolor(black)(s)aaaaaaaaaaacolor(black)(s)

By definition, the product solubility constant will be

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

${K}_{s p} = s \cdot s = {s}^{2}$

This means that $s$ is equal to

$s = \sqrt{{K}_{s p}} = \sqrt{3.0 \cdot {10}^{- 26}} = 1.7 \cdot {10}^{- 13}$

You can thus say that the molar solubility of cobalt(II) sulfide is equal to

$s = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1.7 \cdot {10}^{- 13} {\text{mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that for every mole of cobalt(II) sulfide added to one liter of water, you will only get a concentration of $1.7 \cdot {10}^{- 13} {\text{mol L}}^{- 1}$ for the dissociated ions.