Question #91010

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Question #91010

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Answer 1 · 2017-04-07T01:13:26

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.

${AgCNS}_{(s)} ⇌ {Ag}_{(s)}^{+} + {CNS}_{(a q)}^{-}$ $"AgCNS"_ ((s)) rightleftharpoons "Ag"_ ((s))^(+) + "CNS"_ ((aq))^(-)$

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

If you take $s$ $s$ 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

$[{Ag}^{+}] = s$ $["Ag"^(+)] = s$

$[{CNS}^{-}] = s$ $["CNS"^(-)] = s$

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

$K_{s p} = [{Ag}^{+}] \cdot [{CNS}^{-}]$ $K_(sp) = ["Ag"^(+)] * ["CNS"^(-)]$

which can be rewritten as

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

Solve for $s$ $s$ to find

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

Since $s$ $s$ 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.

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Question #91010

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