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}}$$MHCl$
Answer
295.8k+ views
Hint: Precipitation process is a process in which a chemical reaction occurs in an aqueous medium where two ionic bonds combine , resulting in the formation of an insoluble salt which is called precipitate. The process of precipitation occurs for those reactions only if the ${{Q}_{sp}}$ > ${{K}_{sp}}$ condition is true.
Complete Step by Step Answer:
The solubility product quotient of a solution determines the process of precipitation will occur or not . The solubility product constant should be less than this value in order for precipitation. In an aqueous solution of $MAgN{{O}_{3}}$ there will be production of ions as $A{{g}^{+}}$and from $MHCl$the production of ion $C{{l}^{-}}$ is produced which ions are used in the formation of $AgCl$. Now in this case we need to check for the options whether the condition ${{Q}_{sp}}$>${{K}_{sp}}$holds true or not. For the concentration this holds true will be the correct option only .
A. ${{10}^{-4}}$ $MAgN{{O}_{3}}$ and ${{10}^{-7}}$ $MHCl$
Here,[ $A{{g}^{+}}$]=$\frac{{{10}^{-4}}}{2}M$ and [$C{{l}^{-}}$]=$\frac{{{10}^{-7}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5×${{10}^{-12}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ does not hold true . So it will not be the correct option.
B. ${{10}^{-5}}$$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}}$$MHCl$
Here,[ $A{{g}^{+}}$]= $\frac{{{10}^{-5}}}{2}M$and [$C{{l}^{-}}$]= $\frac{{{10}^{-4}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5× ${{10}^{-9}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ holds true . So it will be the correct option.
Thus, the correct option will be C.
Note: The concentration of the cation and anion from the given concentrations of the compounds will be half of the total concentrations this should be taken care of during the calculations.
Complete Step by Step Answer:
The solubility product quotient of a solution determines the process of precipitation will occur or not . The solubility product constant should be less than this value in order for precipitation. In an aqueous solution of $MAgN{{O}_{3}}$ there will be production of ions as $A{{g}^{+}}$and from $MHCl$the production of ion $C{{l}^{-}}$ is produced which ions are used in the formation of $AgCl$. Now in this case we need to check for the options whether the condition ${{Q}_{sp}}$>${{K}_{sp}}$holds true or not. For the concentration this holds true will be the correct option only .
A. ${{10}^{-4}}$ $MAgN{{O}_{3}}$ and ${{10}^{-7}}$ $MHCl$
Here,[ $A{{g}^{+}}$]=$\frac{{{10}^{-4}}}{2}M$ and [$C{{l}^{-}}$]=$\frac{{{10}^{-7}}}{2}M$
And so ${{Q}_{sp}}$=[ $A{{g}^{+}}$][$C{{l}^{-}}$]=2.5×${{10}^{-12}}$
Since here ${{Q}_{sp}}$>${{K}_{sp}}$ does not hold true . So it will not be the correct option.
B. ${{10}^{-5}}$$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.
Recently Updated Pages
JEE Main 2025-26 Mock Test: Organic Compounds Containing Oxygen

Purification and Characterisation of Organic Compounds JEE Main 2025-26 Mock Test

JEE Main 2025-26 P Block Elements Mock test-2 Free Online Practice

JEE Main 2025-26 Mock Test: Organic Compounds Containing Oxygen

JEE Main Equilibrium Mock test-2 (2025-26) – Free Practice

JEE Main 2025-26 Equilibrium Mock Test: Free Practice Online

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Electron Gain Enthalpy and Electron Affinity Explained

Other Pages
NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2026-27 Free PDF Download (Login Required)

CBSE Notes Class 12 Chemistry Chapter 1 - Solutions - 2026-27 PDF Download (Login Required)

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes - 2026-27 Free PDF Download (Sign-In Required)

NCERT Solutions For Class 12 Chemistry Chapter 7 Alcohol Phenol And Ether - 2026-27 Free PDF Download (Login Required)

