Question

Question #91010

Answer 1

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.

`"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.