# Question #91010

Apr 7, 2017

Here's what I got.

#### Explanation:

Silver thiocyanate is insoluble in water, which implies that a dissociation equilibrium is established when this salt is dissolved in water.

${\text{AgCNS"_ ((s)) rightleftharpoons "Ag"_ ((s))^(+) + "CNS}}_{\left(a q\right)}^{-}$

Your goal here is to figure out the equilibrium concentration of the silver cations and of the thiocyanate anions,

If you take $s$ to be the concentration of the salt that dissociates in aqueous solution to produce ions, you can say that, at equilibrium, the aqueous solution will contain

$\left[{\text{Ag}}^{+}\right] = s$

$\left[{\text{CNS}}^{-}\right] = s$

By definition, the solubility product constant for silver thiocyanate is equal to

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

which can be rewritten as

$1.16 \cdot {10}^{- 12} = s \cdot s = {s}^{2}$

Solve for $s$ to find

$s = \sqrt{1.16 \cdot {10}^{- 12}} = 1.08 \cdot {10}^{- 6}$

Since $s$ represents the concentration of silver thiocyanate that dissociates to produce ions, you can say that the salt will have a molar solubility, i.e. the number of moles of silver thiocyanate that dissociate per liter of solution, equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{s = 1.08 \cdot {10}^{- 6} \textcolor{w h i t e}{.} {\text{mol L}}^{- 1}}}}$

To find the solubility in grams per liter, use the molar mass of the salt

$1.08 \cdot {10}^{- 6} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles")))/"L" * "165.95 g"/(1color(red)(cancel(color(black)("mole AgCNS")))) = color(darkgreen)(ul(color(black)(1.79 * 10^(-4)color(white)(.)"g L}}^{- 1}}}}$

The values are rounded to three sig figs.