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?

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

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 $$\ce{CN-}$$?

• "There is more than enough." How did you determine that with out any quantitative calculations? – Zhe Apr 15 at 16:08
• I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I- – user77021 Apr 15 at 16:16
• Only if the concentrations are such that the solubility product exceeds $K_{\mathrm{sp}}$. – Zhe Apr 15 at 16:28

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.

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.