Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide
Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

Assuming that $\ce{Ba(CN)2}$ and $\ce{BaI2}$ dissociate completely.
$\ce{[CN-]_i = [I-]_i =} 2\cdot10^{-10}$ molar
Neglecting any volume change of solution the initial concentration of $\ce{Ag+}$ will be
$\ce{[Ag+]_i} = \dfrac{3.5\cdot10^{-9}\pu{mol}}{0.010\pu{L}} = 3.5\cdot10^{-7}\pu{M}$
Now if both the $\ce{CN-}$ and $\ce{I-}$ are quantitatively removed then the same amount of $\ce{Ag+}$ must be removed.
$\ce{[CN-]_i + [I-]_i =} 4\cdot10^{-10}$ molar
$\ce{[Ag+]_f} = 3.5\cdot10^{-7}\pu{M} - 4\cdot10^{-10}\pu{M} \approx 3.5\cdot10^{-7}\pu{M}$
So the final concentration of $\ce{Ag+}$ is essentially the same as the initial concentration. The concentration of $\ce{Ag+}$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $\ce{CN-}$ is
$\ce{[CN-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{6.0\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{1.7e-10}$
The the final concentration of $\ce{I-}$ is
$\ce{[I-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{8.5\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{2.4e-10}$
Conclusion:
Since $\ce{[CN-]_i &gt; [CN-]_f}$ some $\ce{AgCN}$ will ppt.
Since $\ce{[I-]_i &lt; [I-]_f}$ no $\ce{AgI}$ will ppt.