Question

The solubility of silver chloride (AgCl) in water is 25$^{\circ }$C is 1.06 x 10$^{-5}$ mol${{L}^{-1}}$ .
Calculate the solubility product of AgCl at this temperature.

Answer 1

Solubility products are always dependent upon the solubility equation. Again the solubility equation depends upon the concentration of the dissociated ions and their numbers.

Complete step by step answer:
Let us consider a saturated solution of Silver Chloride (AgCl) that is in contact with solid Silver Chloride.
The Solubility Equilibrium can be represented as,
$AgCl(s)\rightleftharpoons A{{g}^{+}}(aq.)+C{{l}^{-}}(aq.)$
Therefore, the equilibrium constant for the dissociation of AgCl can be expressed at a fixed temperature as
${{K}_{ap}}(activity\text{ }product)=\dfrac{{{a}_{A{{g}^{+}}}}{{a}_{C{{l}^{-}}}}}{{{a}_{AgCl}}}={{a}_{A{{g}^{+}}}}{{a}_{C{{l}^{-}}}}$
As we know activity is effective concentration a solution under non ideal condition(i.e. Concentrated ) thus Activity of pure solid AgCl is taken to be 1
Now, ${{K}_{ap}}=[A{{g}^{+}}][C{{l}^{-}}]{{\text{f}}_{A{{g}^{+}}}}{{\text{f}}_{Cl-}}$
Thus we can write, ${{K}_{ap}}={{K}_{sp}}{{\text{f}}_{A{{g}^{+}}}}{{\text{f}}_{Cl-}}$
If the concentration of the solutes are small enough that solute-solute interactions are negligible, then the solution can be considered to be ideal and We can approximate the activities of solutes by the concentration terms. With this approximation the equation becomes,
${{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]$ where the activity products , i.e. ${{\text{f}}_{A{{g}^{+}}}}\text{ and }{{\text{f}}_{Cl-}}$ are unity.

Thus in general we can say, The solubility product of compounds is the product of the molar concentrations of its constituent ions (in saturated solution), each raised to the power of its stoichiometric coefficient in the equilibrium equation. Now for the second part of the question , the solubility equilibrium given by
$AgCl(s)\rightleftharpoons A{{g}^{+}}(aq.)+C{{l}^{-}}(aq.)$

Let us consider the solubility of $AgCl$ is 's' moles/lt.
therefore after dissociation, concentration of $A{{g}^{+}}$ is s moles/lt.
and the concentration of $C{{l}^{-}}$ is s moles/lt.
thus according to the definition of solubility product,$$$$${{K}{sp}}=[A{{g}^{+}}][C{{l}^{-}}]$Therefore, the solubility product is given by${{K}{sp}}=s\times s=1.06\times {{10}^{-5}}\times 1.06\times {{10}^{-5}}=1.1236\times {{10}^{-10}}$ moles/lt

Note: Here there is no reaction with water during the determination of solubility. If that happens then, $\text{A}{{\text{g}}^{\text{+}}}\text{or C}{{\text{l}}^{\text{-}}}$ will react with water molecules reducing the concentration of $\text{A}{{\text{g}}^{\text{+}}}\text{or C}{{\text{l}}^{\text{-}}}$ in the solution. There could be precipitation of $\text{Ag(OH)}$which could increase the solubility by increasing the dissociation of the compound.