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

Question

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

Naomi Aronson · Accepted Answer

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

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.

Peyton Adams · Answer

undefined The $ K_{sp} $ expression for Ag2CO3 is:

$$ K_{sp} = [Ag^+]^2[CO_3^{2-}] $$

Given that Ag2CO3 dissociates into $ 2Ag^+ $ and $ CO3^{2-} $, the molar solubility of Ag2CO3 is used to determine the concentrations of the ions in solution.

Since 1 mole of Ag2CO3 produces 2 moles of Ag+ ions and 1 mole of CO3^2- ions, the concentration of Ag+ ions and CO3^2- ions is twice the molar solubility of Ag2CO3.

$$ [Ag^+] = 2 \times 1.3 \times 10^{-4} \, \text{mol/L} $$
$$ [CO3^{2-}] = 1 \times 1.3 \times 10^{-4} \, \text{mol/L} $$

Substitute these values into the expression for $ K_{sp} $:

$$ K_{sp} = (2 \times 1.3 \times 10^{-4})^2 \times (1.3 \times 10^{-4}) $$