Question

The value of ksp for silver carbonate, ag2co3, is 8.10×10−12. calculate the solubility of ag2co3 in grams per liter.

Answer 1

Ag+] = 2x [CO32-] = x 8.10 x 10^-12 = (2x)^2 (x) = 4x^3. Molar solubility = 0.000127 M Solubility = 0.000127 mol/L x 275.75 g/mol = 0.0350 g/

Answer 2

Answer: 0.035 grams/liter

Explanation:- The equation for the reaction will be as follows:

$Ag_2CO_3\leftrightharpoons 2Ag^++CO_3^{2-}$

1 mole of $Ag_2CO_3$ gives 2 moles of $Ag^{+}$ and 1 mole of $CO_3^{2-}$.

Thus if solubility of $Ag_2CO_3$ is s moles/liter, solubility of  $Ag^{+}$ is 2s moles\liter and solubility of $CO_3^{2-}$ is s moles/liter

Therefore,  

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

$8.10\times 10^{-12}=[2s]^2[s]$

$4s^3=8.10\times 10^{-12}$

$s=1.26\times 10^{-4}moles/liter$

Solubility in grams/liter=$\text{solubility in moles/liter}\times {\text {Molar mass}}$

Solubility in grams/liter=$1.26\times 10^{-4}moles/liter\times 275.7g/mol=0.035grams/liter$