Question #91010

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

#"AgCNS"_ ((s)) rightleftharpoons "Ag"_ ((s))^(+) + "CNS"_ ((aq))^(-)#

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

#["Ag"^(+)] = s#

#["CNS"^(-)] = s#

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

#K_(sp) = ["Ag"^(+)] * ["CNS"^(-)]#

which can be rewritten as

#1.16 * 10^(-12) = s * s = s^2#

Solve for #s# to find

#s = sqrt(1.16 * 10^(-12)) = 1.08 * 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

#color(darkgreen)(ul(color(black)(s = 1.08 * 10^(-6)color(white)(.)"mol L"^(-1))))#

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

#1.08 * 10^(-6) color(red)(cancel(color(black)("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.