For $AgCl$,$K_{sp} = 1.2 \times 10^{-10}$ and for $AgBr$,$K_{sp} = 3.5 \times 10^{-13}$. What is the relationship between their solubilities $(S)$?
- A$S$ of $AgBr < S$ of $AgCl$
- B$S$ of $AgBr > S$ of $AgCl$
- C$S$ of $AgBr = S$ of $AgCl$
- D$S$ of $AgCl = 10^6 \times S$ of $AgBr$
Explore More
Similar Questions
The $K_{sp}$ of $Mg(OH)_2$ is $1.0 \times 10^{-12}$. At which $pH$ does $0.01 \ M$ $Mg^{2+}$ solution begin to precipitate? Calculate the solubility of $Mg(OH)_2$ in pure water.
Easy
View SolutionWhich of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions?
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.
DifficultJEE MAIN 2025
View SolutionThe conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. Then the $K_{sp}$ for $BaSO_4$ will be:
Medium
View SolutionThe solubility product of $AgI$ at $25\,\text{°C}$ is $1.0 \times 10^{-16}\,\text{mol}^2\,\text{L}^{-2}$. The solubility of $AgI$ in $10^{-4}\,\text{N}$ solution of $KI$ at $25\,\text{°C}$ is approximately (in $\text{mol L}^{-1}$)
MediumAIPMT 2003
View SolutionIf the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo