Ttgjyuk

Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?

$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.

The answer was only $ceAgCN$ will precipitate, but I don't understand why $ceAgI$ wouldn't precipitate as well since there is more than enough excess $ceAgNO3$ available to precipitate with both $ceI-$ and $ceCN-$?

Consider a $pu10.0 mL$ solution containing $pu1.0e-10 M$ each of $ceBa(CN)2$ and $ceBaI2$. If $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?

$K_mathrmsp(ceAgCN) = pu6.0e-17$; $K_mathrmsp(ceAgI) = pu8.5e-17$.

Assuming that $ceBa(CN)2$ and $ceBaI2$ dissociate completely.

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

Neglecting any volume change of solution the initial concentration of $ceAg+$ will be

$ce[Ag+]_i = dfrac3.5cdot10^-9pumol0.010puL = 3.5cdot10^-7puM$

Now if both the $ceCN-$ and $ceI-$ are quantitatively removed then the same amount of $ceAg+$ must be removed.

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

$ce[Ag+]_f = 3.5cdot10^-7puM - 4cdot10^-10puM approx 3.5cdot10^-7puM$

So the final concentration of $ceAg+$ is essentially the same as the initial concentration. The concentration of $ceAg+$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.

The final concentration of $ceCN-$ is

$ce[CN-]_f = dfracK_spce[Ag+]_f = dfrac6.0cdot10^-173.5cdot10^-7 = pu1.7e-10$

The the final concentration of $ceI-$ is

$ce[I-]_f = dfracK_spce[Ag+]_f = dfrac8.5cdot10^-173.5cdot10^-7 = pu2.4e-10$

Conclusion:

Since $ce[CN-]_i > [CN-]_f$ some $ceAgCN$ will ppt.

Since $ce[I-]_i < [I-]_f$ no $ceAgI$ will ppt.

Alternative method to MaxW method:

Assume that an initial $pu10.0 mL$ solution of $pu1.0e-10 M$ in each of $ceBa(CN)2$ and $ceBaI2$ is clear (homogeneous). That means $ceBa(CN)2$ and $ceBaI2$ have dissociated completely. Thus concentrations of ions are as follows:

$$ce[CN-]_i = [I-]_i = pu2cdot10^-10 mol ! L^-1$$

Suppose when $pu3.5e-9 mol$ of $ceAgNO3(s)$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:

$$ce[Ag+]_i = [NO3-]_i = dfracpu3.5cdot10^-9 molpu0.010 L = pu3.5cdot10^-7 mol ! L^-1$$

For precipitation of $ceAgCN(s)$:

$$Q_mathrmsp = ce[Ag+]_icdot ce[CN-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 gt K_mathrmsp(ceAgCN) = pu6.0cdot10^-17 mol^2 ! L^-2 $$

Therefore, $ceAgCN(s)$ will precipitate.

For precipitation of $ceAgI(s)$:

$$Q_mathrmsp = ce[Ag+]_icdot ce[I-]_i = (pu3.5cdot10^-7 mol ! L^-1)(pu2cdot10^-10 mol ! L^-1) \ = pu7.0cdot10^-17 mol^2 ! L^-2 lt K_mathrmsp(ceAgCN)=pu8.5cdot10^-17 mol^2 ! L^-2 $$

Therefore, $ceAgI(s)$ will not precipitate in this condition.