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If $$K_{sp}$$ of $$\text{Ag}_2\text{CO}_3$$ is $$8 \times 10^{-12}$$, the molar solubility of $$\text{Ag}_2\text{CO}_3$$ in $$0.1\text{ M}$$ $$\text{AgNO}_3$$ is
$$K_{sp} = [\text{Ag}^+]^2[\text{CO}_3^{2-}]$$. In $$0.1\text{ M}$$ $$\text{AgNO}_3$$, $$[\text{Ag}^+] \approx 0.1$$ and $$[\text{CO}_3^{2-}] = s$$, so $$s = \frac{8 \times 10^{-12}}{(0.1)^2} = 8 \times 10^{-10}\text{ M}$$.
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