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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?

$K_mathrm{sp}(ce{AgCN}) = pu{6.0e-17}$; $K_mathrm{sp}(ce{AgI}) = pu{8.5e-17}$.

The answer was only $ce{AgCN}$ will precipitate, but I don't understand why $ce{AgI}$ wouldn't precipitate as well since there is more than enough excess $ce{AgNO3}$ available to precipitate with both $ce{I-}$ and $ce{CN-}$?

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.

$ce{[CN-]_i = [I-]_i =} 2cdot10^{-10}$ molar

Neglecting any volume change of solution the initial concentration of $ce{Ag+}$ will be

$ce{[Ag+]_i} = dfrac{3.5cdot10^{-9}pu{mol}}{0.010pu{L}} = 3.5cdot10^{-7}pu{M}$

Now if both the $ce{CN-}$ and $ce{I-}$ are quantitatively removed then the same amount of $ce{Ag+}$ must be removed.

$ce{[CN-]_i + [I-]_i =} 4cdot10^{-10}$ molar

$ce{[Ag+]_f} = 3.5cdot10^{-7}pu{M} - 4cdot10^{-10}pu{M} approx 3.5cdot10^{-7}pu{M}$

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

$ce{[CN-]_f} = dfrac{K_{sp}}{ce{[Ag+]_f}} = dfrac{6.0cdot10^{-17}}{3.5cdot10^{-7}} = pu{1.7e-10}$

The the final concentration of $ce{I-}$ is

$ce{[I-]_f} = dfrac{K_{sp}}{ce{[Ag+]_f}} = dfrac{8.5cdot10^{-17}}{3.5cdot10^{-7}} = pu{2.4e-10}$

Conclusion:

Since $ce{[CN-]_i > [CN-]_f}$ some $ce{AgCN}$ will ppt.

Since $ce{[I-]_i < [I-]_f}$ no $ce{AgI}$ will ppt.

Alternative method to MaxW method:

Assume that an initial $pu{10.0 mL}$ solution of $pu{1.0e-10 M}$ in each of $ce{Ba(CN)2}$ and $ce{BaI2}$ is clear (homogeneous). That means $ce{Ba(CN)2}$ and $ce{BaI2}$ have dissociated completely. Thus concentrations of ions are as follows:

$$ce{[CN-]_i = [I-]_i} = pu{2cdot10^{-10} mol ! L^{-1}}$$

Suppose when $pu{3.5e-9 mol}$ of $ce{AgNO3(s)}$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:

$$ce{[Ag+]_i = [NO3-]_i} = dfrac{pu{3.5cdot10^{-9} mol}}{pu{0.010 L}} = pu{3.5cdot10^{-7} mol ! L^{-1}}$$

For precipitation of $ce{AgCN(s)}$:

$$Q_mathrm{sp} = ce{[Ag+]_i}cdot ce{[CN-]_i} = (pu{3.5cdot10^{-7} mol ! L^{-1}})(pu{2cdot10^{-10} mol ! L^{-1}}) \ = pu{7.0cdot10^{-17} mol^2 ! L^{-2}} gt K_mathrm{sp}(ce{AgCN}) = pu{6.0cdot10^{-17} mol^2 ! L^{-2}} $$

Therefore, $ce{AgCN(s)}$ will precipitate.

For precipitation of $ce{AgI(s)}$:

$$Q_mathrm{sp} = ce{[Ag+]_i}cdot ce{[I-]_i} = (pu{3.5cdot10^{-7} mol ! L^{-1}})(pu{2cdot10^{-10} mol ! L^{-1}}) \ = pu{7.0cdot10^{-17} mol^2 ! L^{-2}} lt K_mathrm{sp}(ce{AgCN})=pu{8.5cdot10^{-17} mol^2 ! L^{-2}} $$

Therefore, $ce{AgI(s)}$ will not precipitate in this condition.