O valor de Ksp para o carbonato de prata, Ag2CO3, é 8.10 × 10-12. Calcular a solubilidade do Ag2CO3 em gramas por litro?

$350.26 g/L$ $"350.26 g/L"$

${Ag}_{2} {CO}_{3} \Leftrightarrow {2Ag}^{+} + {CO}_{3}^{2 -}$ $"Ag"_2"CO"_3 <=> "2Ag"^+ + "CO"_3^(2-)$

$K_{sp} = {[{Ag}^{+}]}^{2} [{CO}_{3}^{2 -}]$ $"K"_("sp") = ["Ag"^+]^2["CO"_3^(2-)]$

$K_{sp} = {(2S)}^{2} (S) = {4S}^{3}$ $"K"_"sp" = ("2S")^2 ("S") = "4S"^3$

$S = {[\frac{K_{sp}}{4}]}^{\frac{1}{3}}$ $"S" = [("K"_"sp")/4]^(1/3)$

$S = {[\frac{8.10 \times 10^{- 12}}{4}]}^{\frac{1}{3}}$ $"S" = [(8.10 × 10^-12)/4]^(1/3)$

$S = 1.27 \times 10^{- 4} mol/L$ $"S" = 1.27 × 10^-4 "mol/L"$

Massa molar de ${Ag}_{2} {CO}_{3}$ $"Ag"_2"CO"_3$ is $275.8 g/mol$ $"275.8 g/mol"$

$"S" = 1.27 × 10^-4 cancel"mol"/"L" × "275.8 g"/cancel"mol" = "350.26 g/L"$