 # Selective Precipitation Selective precipitation is a technique used to separate ions with the same charge and sufficiently different solubilities out of solution using the same reagent.
In addition to providing the molar solubility of a compound, Ksp can be used to calculate whether a compound will form a precipitate at a given concentration. Also used in this calculation is the ion product (Qsp), a constant analogous to the solubility product constant but calculated from initial concentrations or other ion concentrations when the reaction is not at equilibrium.
${{{\rm{M}}_m}{{\rm{A}}_a}(s)}\rightleftharpoons{m{{\rm{M}}^{a+}}({aq})+a{{\rm{A}}^{m-}}(aq)}$
$Q_{\rm{sp}}=\lbrack{{\rm{M}}^{a+}\rbrack}^m\lbrack{{\rm{A}}^{m-}\rbrack}^a$
If $Q_{\rm{sp}}=K_{\rm{sp}}$, the ion concentrations are equal to their equilibrium concentrations. This means the solution is exactly at its saturation point and no more precipitate will form. A precipitate should also not form if $Q_{\rm{sp}} \lt K_{\rm{sp}}$ because the solution is unsaturated. If $Q_{\rm{sp}}\gt K_{\rm{sp}}$, the ion concentrations are greater than the molar solubility, and the solution is supersaturated. The equilibrium is shifted to the left, and the solution should form a precipitate.

If a solution contains two or more ions of the same charge that can all be precipitated out of solution by the same reagent, the ions can be separated by that reagent as long as the solubilities of the precipitates are different enough. Selective precipitation is a separation technique for compounds that share a common ion but have different solubilities and are precipitated out of solution.

Take, for example, a solution containing I and Cl, both of which form compounds with Ag+:
${{\rm{AgI}}(s)\rightleftharpoons{\rm{Ag}^ +}(aq)+{\rm{I}^-}(aq),\hspace{10pt}K_{\rm{sp}}=8.52\times10^{-17}}$
${\rm{AgCl}}(s)\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{Cl}^-}(aq),\hspace {10pt}K_{\rm{sp}}=1.77\times10^{-10}$
If Ag+ is slowly added to the solution, which precipitate will form first, AgI or AgCl? A precipitate will start to form once $Q_{\rm{sp}}\gt{K_{\rm{sp}}}$. In this case, Qsp is calculated from the concentration of ions before Ag+ is added. As Ag+ is added, the concentration of Ag+ increases, and therefore so does the value of Qsp. Since the Ksp for AgI is seven orders of magnitude less than that of AgCl, it follows that Qsp will be larger than Ksp for AgI while it is still less than Ksp for AgCl. Thus, AgI will form a precipitate before AgCl does.

#### Selective Precipitation If Ag+ is added to a solution of I- and Cl-, AgI forms a precipitate first because it has a lower molar solubility than AgCl.
The concentration of each species can be calculated using Qsp and Ksp. For example, given a solution of 0.1 M I and 0.1 M Cl, calculate [Ag+] when AgI first begins to precipitate:
$K_{\rm{sp}}=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{I}^-\rbrack=\lbrack\rm{Ag}^+\rbrack\lbrack 0.1\;\rm{M}\rbrack=8.52\times 10^{-17}$
So when $\lbrack\rm{Ag}^+\rbrack=8.52\times{10^{-16}}\;\rm{M}$, AgI begins to form a precipitate. Next, the concentration of I remaining in solution when AgCl begins to precipitate can be calculated in order to reveal how successful the selective precipitation is at separating the compounds AgCl and AgI. To calculate [I] when AgCl first starts to precipitate, first use the Ksp for AgCl to find [Ag+] at that point:
$K_{\rm{sp}}=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{Cl}^-\rbrack=\lbrack\rm{Ag}^+\rbrack\lbrack{0.1}\;\rm{ M}\rbrack=1.77\times 10^{-10}$
Therefore $\lbrack\rm{Ag}^+\rbrack=1.77\times{10^{-9}}\;\rm{ M}$ when AgCl first starts to form. Finally, calculate [I] when $\lbrack\rm{Ag}^+\rbrack=1.77\times{10^{-9}}\;\rm{ M}$ to find the amount of I left in solution:
\begin{aligned}K_{\rm{sp}}&=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{I}^-\rbrack=(1.77\times{10}^{-9})\lbrack\rm{I}^-\rbrack=8.52\times{10^{-17}}\\{\lbrack}\rm{I}^-\rbrack&=\frac{8.52\times{10}^{-17}}{1.77\times{10}^{-9}}=4.81\times{10^{-8}}\;\rm{ M}\end{aligned}
By the time AgCl begins to precipitate, the concentration of I has fallen from $1.0\times{10^{-1}}\;\rm{M}$ to $4.81\times {10^{-8}}\;\rm{M}$, a difference of seven orders of magnitude. By this point, the I is almost entirely removed from the solution in the form of solid AgI, and the separation of I and Cl is successful.