The molar solubility of $A g_{2} C O_{3}$ $Ag_2CO_3$ is $1.3 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $1.3xx10^-4molL^-1$ . What is $K_{sp}$ $K_"sp"$ for this salt?

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The molar solubility of $A g_{2} C O_{3}$ $Ag_2CO_3$ is $1.3 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $1.3xx10^-4molL^-1$ . What is $K_{sp}$ $K_"sp"$ for this salt?

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Answer 1 · 2017-05-25T18:58:01

$K_{sp} = 8.80 \times 10^{- 12}$ $K_"sp"=8.80xx10^-12$ ..........

Explanation:

We interrogate the equilibrium:

$A g_{2} C O_{3} (s) ⇌ 2 A g^{+} + C O_{3}^{2 -}$ $Ag_2CO_3(s) rightleftharpoons2Ag^(+) + CO_3^(2-)$

And thus $K_{sp} = {[A g^{+}]}^{2} [C O_{3}^{2 -}]$ $K_"sp"=[Ag^+]^2[CO_3^(2-)]$

But from the stoichiometry, $[C O_{3}^{2 -}] = S_{A g_{2} C O_{3}}$ $[CO_3^(2-)]=S_(Ag_2CO_3)$ , and $[A g^{+}] = 2 S_{A g_{2} C O_{3}}$ $[Ag^+]=2S_(Ag_2CO_3)$ ........and so

$K_"sp"=Sxx(2S)^2=4S^3=4xx(1.3xx10^-4)^3=8.80xx10^-12.....$

This site reports that $K_"sp"$ , $Ag_2CO_3$ , is $8.46xx10^-12$ , so our estimate is kosher.

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The molar solubility of $A g_{2} C O_{3}$ $Ag_2CO_3$ is $1.3 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $1.3xx10^-4molL^-1$ . What is $K_{sp}$ $K_"sp"$ for this salt?

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The molar solubility of $A g_{2} C O_{3}$ $Ag_2CO_3$ is $1.3 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $1.3xx10^-4molL^-1$ . What is $K_{sp}$ $K_"sp"$ for this salt?