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A precipitate of $AgCl$ is formed when equal volumes of the following are mixed [${{10}^{-4}}{{10}^{-5}}{{10}^{-6}}{{10}^{-9}}MAgN{{O}_{3}}MHC{{l}^{-}}{{Q}_{sp}}{{K}_{sp}}A{{g}^{+}}\frac{{{10}^{-4}}}{2}M$${{K}_{sp}}$ for $AgCl$=${{10}^{-10}}$]
A. ${{10}^{-4}}$ $MAgN{{O}_{3}}$ and ${{10}^{-7}}$ $MHCl$
B. ${{10}^{-5}}$$MAgN{{O}_{3}}$and ${{10}^{-6}}$$MHCl$
C. ${{10}^{-5}}$$MAgN{{O}_{3}}$and ${{10}^{-4}}$$MHCl$
D. ${{10}^{-6}}$$MAgN{{O}_{3}}$and ${{10}^{-6}}$$MHCl$

Answer
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Hint: Precipitation process is a process in which a chemical reaction occurs in an aqueous medium where two ionic bonds combine , resulting in the formation of an insoluble salt which is called precipitate. The process of precipitation occurs for those reactions only if the ${{Q}_{sp}}$ > ${{K}_{sp}}$ condition is true.

Complete Step by Step Answer:
The solubility product quotient of a solution determines the process of precipitation will occur or not . The solubility product constant should be less than this value in order for precipitation. In an aqueous solution of $MAgN{{O}_{3}}$ there will be production of ions as $A{{g}^{+}}$and from $MHCl$the production of ion $C{{l}^{-}}$ is produced which ions are used in the formation of $AgCl$. Now in this case we need to check for the options whether the condition ${{Q}_{sp}}$>${{K}_{sp}}$holds true or not. For the concentration this holds true will be the correct option only .
 A. ${{10}^{-4}}$ $MAgN{{O}_{3}}$ and ${{10}^{-7}}$ $MHCl$
Here,[ $A{{g}^{+}}$]=$\frac{{{10}^{-4}}}{2}M$ and [$C{{l}^{-}}$]=$\frac{{{10}^{-7}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5×${{10}^{-12}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ does not hold true . So it will not be the correct option.
B. ${{10}^{-5}}$$MAgN{{O}_{3}}$and ${{10}^{-6}}$$MHCl$
Here,[ $A{{g}^{+}}$]= $\frac{{{10}^{-5}}}{2}M$ and [$C{{l}^{-}}$]=$\frac{{{10}^{-6}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5×${{10}^{-12}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ does not hold true . So it will not be the correct option.
C. ${{10}^{-5}}$$MAgN{{O}_{3}}$and ${{10}^{-4}}$$MHCl$
Here,[ $A{{g}^{+}}$]= $\frac{{{10}^{-5}}}{2}M$and [$C{{l}^{-}}$]= $\frac{{{10}^{-4}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5× ${{10}^{-9}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ holds true . So it will be the correct option.
Thus, the correct option will be C.

Note: The concentration of the cation and anion from the given concentrations of the compounds will be half of the total concentrations this should be taken care of during the calculations.