Question:
A \(1.0\ L\) aqueous solution contains \(1\times 10^{-8}\ M\ NaBr\), \(1\times 10^{-8}\ M\ NaCl\) and \(1\times 10^{-8}\ M\ NaI\). To this solution, \(1\times 10^{-10}\ M\) aqueous \(AgNO_3\) solution is added drop wise. The order of precipitation of \(AgX\), where \(X=Cl, Br, I\), is
\[
K_{sp}(AgCl)=1.8\times 10^{-10}
\]
\[
K_{sp}(AgBr)=5\times 10^{-13}
\]
\[
K_{sp}(AgI)=8.3\times 10^{-17}
\]
A \(1.0\ L\) aqueous solution contains \(1\times 10^{-8}\ M\ NaBr\), \(1\times 10^{-8}\ M\ NaCl\) and \(1\times 10^{-8}\ M\ NaI\). To this solution, \(1\times 10^{-10}\ M\) aqueous \(AgNO_3\) solution is added drop wise. The order of precipitation of \(AgX\), where \(X=Cl, Br, I\), is
\[ K_{sp}(AgCl)=1.8\times 10^{-10} \] \[ K_{sp}(AgBr)=5\times 10^{-13} \] \[ K_{sp}(AgI)=8.3\times 10^{-17} \]
\[ K_{sp}(AgCl)=1.8\times 10^{-10} \] \[ K_{sp}(AgBr)=5\times 10^{-13} \] \[ K_{sp}(AgI)=8.3\times 10^{-17} \]
Show Hint
When the concentration of anions is the same, the salt with the lowest \(K_{sp}\) precipitates first.
Updated On: Jun 18, 2026
- \(AgBr,\ AgCl,\ AgI\)
- \(AgCl,\ AgBr,\ AgI\)
- \(AgI,\ AgBr,\ AgCl\)
- \(AgBr,\ AgI,\ AgCl\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understand the condition for precipitation.
For precipitation of silver halide, \[ AgX(s) \] the ionic product must exceed the solubility product.
The precipitation starts when \[ [Ag^+][X^-]=K_{sp} \] Therefore, the required concentration of \(Ag^+\) for precipitation is \[ [Ag^+]=\frac{K_{sp}}{[X^-]} \]
Step 2: Note the halide ion concentrations.
The given concentrations are \[ [Cl^-]=[Br^-]=[I^-]=1\times 10^{-8}\ M \] Since all halide ion concentrations are equal, the salt with the smallest \(K_{sp}\) will require the least \([Ag^+]\) and will precipitate first.
Step 3: Compare the solubility products.
Given, \[ K_{sp}(AgCl)=1.8\times 10^{-10} \] \[ K_{sp}(AgBr)=5\times 10^{-13} \] \[ K_{sp}(AgI)=8.3\times 10^{-17} \] The order of \(K_{sp}\) values is \[ AgI<AgBr<AgCl \]
Step 4: Determine the precipitation order.
Lower \(K_{sp}\) means lower solubility and earlier precipitation.
Therefore, the order of precipitation is \[ AgI,\ AgBr,\ AgCl \]
Step 5: Final conclusion.
Hence, \[ \boxed{AgI,\ AgBr,\ AgCl} \] Therefore, the correct option is (3).
For precipitation of silver halide, \[ AgX(s) \] the ionic product must exceed the solubility product.
The precipitation starts when \[ [Ag^+][X^-]=K_{sp} \] Therefore, the required concentration of \(Ag^+\) for precipitation is \[ [Ag^+]=\frac{K_{sp}}{[X^-]} \]
Step 2: Note the halide ion concentrations.
The given concentrations are \[ [Cl^-]=[Br^-]=[I^-]=1\times 10^{-8}\ M \] Since all halide ion concentrations are equal, the salt with the smallest \(K_{sp}\) will require the least \([Ag^+]\) and will precipitate first.
Step 3: Compare the solubility products.
Given, \[ K_{sp}(AgCl)=1.8\times 10^{-10} \] \[ K_{sp}(AgBr)=5\times 10^{-13} \] \[ K_{sp}(AgI)=8.3\times 10^{-17} \] The order of \(K_{sp}\) values is \[ AgI<AgBr<AgCl \]
Step 4: Determine the precipitation order.
Lower \(K_{sp}\) means lower solubility and earlier precipitation.
Therefore, the order of precipitation is \[ AgI,\ AgBr,\ AgCl \]
Step 5: Final conclusion.
Hence, \[ \boxed{AgI,\ AgBr,\ AgCl} \] Therefore, the correct option is (3).
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Ionic Equilibrium In Solution Questions
- The pH of 0.01M BOH solution is 10. What is its degree of dissociation? (Given $ K_b $ of BOH is $ 1 \times 10^{-6} $)
- At T(K), the solubility products (K\(_{sp}\)) of AgCl, AgBr and AgI are \(1.8 \times 10^{-10}\), \(5 \times 10^{-13}\) and \(8.3 \times 10^{-17}\) respectively. A 1L solution contains \(10^{-5}\) moles each of NaCl, NaBr and NaI. This solution is titrated with \(10^{-6}\) M AgNO\(_3\) until all halides (Cl\(^{-}\), Br\(^{-}\), I\(^{-}\)) are precipitated as silver salts.
The order of precipitation of these respectively is: - Which of the following is a conjugate acid-base pair?
- The pH of a $0.01\text{ M NaOH}$ solution is}
- The pH of a 0.1 M solution of a weak monobasic organic acid is 4.0. What is the dissociation constant of the acid?
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions