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Question #9e5cc

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Answer 1 · 2017-08-03T23:49:53

$K_{s p} = 4.2 \cdot 10^{- 13}$ $K_(sp) = 4.2 * 10^(-13)$

Explanation:

You know that when silver carbonate, a salt that is considered insoluble in water, is dissolved in water, the following dissociation equilibrium is established in aqueous solution.

${Ag}_{2} {CO}_{3 (s)} ⇌ 2 {Ag}_{(a q)}^{+} + {CO}_{3 (a q)}^{2 -}$ $"Ag"_ 2"CO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CO"_ (3(aq))^(2-)$

You also know that the molar solubility of silver carbonate, $s$ $s$ , in water at $25^{\circ} C$ $25^@"C"$ is equal to $4.7 \cdot 10^{- 5}$ $"4.7 * 10^(-5)$ $M$ $"M"$ .

This tells you that a saturated solution of silver carbonate will contain $4.7 \cdot 10^{- 5}$ $4.7 * 10^(-5)$ moles of dissociated silver carbonate for every $1 L$ $"1 L"$ of solution.

In other words, you have

${[{Ag}_{2} {CO}_{3}]}_{that dissociates} = s$ $["Ag"_ 2"CO"_ 3]_ "that dissociates" = s$

In your case, you know that

${[{Ag}_{2} {CO}_{3}]}_{that dissociates} = s = 4.7 \cdot 10^{- 5}$ $["Ag"_ 2"CO"_ 3]_ "that dissociates" = s = 4.7 * 10^(-5)$ $M$ $"M"$

Now, notice that every mole of silver carbonate that dissociates in aqueous solution produces $2$ $color(red)(2)$ moles of silver(I) cations and $1$ $1$ mole of carbonate anions.

This means that the saturated solution of silver carbonate will contain

$[{Ag}^{+}] = 2 \times s$ $["Ag"^(+)] = color(red)(2) xx s$

$[{CO}_{3}^{2 -}] = s$ $["CO"_3^(2-)] = s$

You can thus say that at $25^{\circ} C$ $25^@"C"$ , the saturated solution will contain

$[{Ag}^{+}] = 2 \times 4.7 \cdot 10^{- 5} . M = 9.4 \cdot 10^{- 5} . M$ $["Ag"^(+)] = color(red)(2) xx 4.7 * 10^(-5)color(white)(.)"M" = 9.4 * 10^(-5)color(white)(.)"M"$

$[{CO}_{3}^{2 -}] = 4.7 \cdot 10^{- 5} . M$ $["CO"_3^(2-)] = 4.7 * 10^(-5)color(white)(.)"M"$

By definition, the solubility product constant for this dissociation equilibrium is equal to

$K_{s p} = {[{Ag}^{+}]}^{2} \cdot [{CO}_{3}^{2 -}]$ $K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]$

In your case, you will have--I'll leave the value without added units

$K_{s p} = {(9.4 \cdot 10^{- 5})}^{2} \cdot (4.7 \cdot 10^{- 5})$ $K_(sp) = (9.4 * 10^(-5))^color(red)(2) * (4.7 * 10^(-5))$

$color(darkgreen)(ul(color(black)(K_(sp) = 4.2 * 10^(-13))))$

The answer is rounded to two sig figs, the number of sig figs you have for the molar solubility of the salt.

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Question #9e5cc

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