Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide

Question

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 $\ce{CN-}$?

Answer 1

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

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

Answer 2

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:

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

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:

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

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

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

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

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

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

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