Question

The solubility product of AgCl is $1.8\times {{10}^{-10}}$. Precipitation of $AgCl$ will occur only when equal volumes of solutions of
(a) ${{10}^{-4}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-4}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
(b) ${{10}^{-7}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-7}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
(c) ${{10}^{-5}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-5}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
(d) ${{10}^{-10}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-10}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$

Answer 1

Hint: By the solubility product , we mean the product of the ions which are dissolved into the solution and precipitation only occurs in that case if the product of the ions is greater than the given solubility product. So, by finding the product of the ions by the values given in the options and comparing them with the given solubility product, we can easily find the answer. Now solve it.

Complete Solution :
First of all, we should know what solubility is. By solubility, we mean the capacity of the solute to dissolve in the solution.
- Now let’s discuss what a solubility product is. By the solubility product we mean the product of the ions dissolved in the solution raised to the power of their stoichiometric coefficients. It is used in those cases where the salts do not fully get dissolved in the solvent. And the most commonly used solvent is water.
Consider a general reaction as:
\[AB\rightleftharpoons {{A}^{2+}}+{{B}^{2-}}\]

Here , AB is the salt which is dissolved in the solvent water but it’s are insoluble in water , so its solubility product will be as:
${{K}_{sp}}={{[A]}^{2+}}{{[B]}^{2-}}$

Here,${{K}_{sp}}$ is the solubility product and [A] and[B] are the concentrations of the salt AB.
Now, considering the numerical;
Precipitation of $AgCl$ will occur only if the product of $[Ag]$ and $[Cl]$ is greater than the given solubility product.

So, we will check the given options one by one as:
(a) ${{10}^{-4}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-4}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
$\begin{align}
 & {{K}_{sp}}={{[Ag]}^{+}}{{[Cl]}^{-}} \\
 & 1.8\times {{10}^{-10}}\text{= }\!\![\!\!\text{ }{{10}^{-4}}]\text{ }\!\![\!\!\text{ }{{10}^{-4}}] \\
 & 1.8\times {{10}^{-10}}\text{=}{{10}^{-8}} \\
\end{align}$
So, in this; ${{K}_{sp}}<{{[Ag]}^{+}}{{[Cl]}^{-}}$
Thus, Precipitation of $AgCl$ will occur.

(b) ${{10}^{-7}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-7}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
$\begin{align}
 & {{K}_{sp}}={{[Ag]}^{+}}{{[Cl]}^{-}} \\
 & 1.8\times {{10}^{-10}}\text{= }\!\![\!\!\text{ }{{10}^{-7}}]\text{ }\!\![\!\!\text{ }{{10}^{-7}}] \\
 & 1.8\times {{10}^{-10}}\text{=}{{10}^{-14}} \\
\end{align}$
So, in this; ${{K}_{sp}}>{{[Ag]}^{+}}{{[Cl]}^{-}}$
Thus, Precipitation of $AgCl$ will not occur.

(c) ${{10}^{-5}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-5}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
$\begin{align}
 & {{K}_{sp}}={{[Ag]}^{+}}{{[Cl]}^{-}} \\
 & 1.8\times {{10}^{-10}}\text{= }\!\![\!\!\text{ }{{10}^{-5}}]\text{ }\!\![\!\!\text{ }{{10}^{-5}}] \\
 & 1.8\times {{10}^{-10}}\text{=}{{10}^{-10}} \\
\end{align}$
So, in this; ${{K}_{sp}}>{{[Ag]}^{+}}{{[Cl]}^{-}}$
Thus, Precipitation of $AgCl$ will not occur.

(d) ${{10}^{-10}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-10}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$
$\begin{align}
 & {{K}_{sp}}={{[Ag]}^{+}}{{[Cl]}^{-}} \\
 & 1.8\times {{10}^{-10}}\text{= }\!\![\!\!\text{ }{{10}^{-10}}]\text{ }\!\![\!\!\text{ }{{10}^{-10}}] \\
 & 1.8\times {{10}^{-10}}\text{=}{{10}^{-20}} \\
\end{align}$
So, in this; ${{K}_{sp}}>{{[Ag]}^{+}}{{[Cl]}^{-}}$
Thus, Precipitation of $AgCl$ will not occur.

So, thus from the above , it is clear that the solubility product of AgCl is $1.8\times {{10}^{-10}}$. Then , the precipitation of $AgCl$ will occur only when equal volumes of solutions of ${{10}^{-4}}M\text{ }A{{g}^{+}}\text{ }and\text{ }{{10}^{-4}}M\text{ C}{{\text{l}}^{-}}are\text{ }mixed$.
So, the correct answer is “Option A”.

Note: Solubility product constant is equal to the equilibrium constant and tells us about the extent to which the compound dissolves into the solution and higher value of solubility product indicates that the compound is more soluble in the solvent.