The molar solubility of #Ag_2CO_3# is #1.3xx10^-4*mol*L^-1#. What is #K_"sp"# for this salt?

1 Answer
May 25, 2017

Answer:

#K_"sp"=8.80xx10^-12#..........

Explanation:

We interrogate the equilibrium:

#Ag_2CO_3(s) rightleftharpoons2Ag^(+) + CO_3^(2-)#

And thus #K_"sp"=[Ag^+]^2[CO_3^(2-)]#

But from the stoichiometry, #[CO_3^(2-)]=S_(Ag_2CO_3)#, and #[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.