Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2013 · 22 April · Online · Chemistry · Ionic Equilibrium

Which one of the following arrangements represents the correct...

Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts Hg2Cl2,Cr2(SO4)3\mathrm{Hg}_2 \mathrm{Cl}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3, BaSO4\mathrm{BaSO}_4 and CrCl3\mathrm{CrCl}_3 respectively?

Options

  1. ABaSO4>Hg2Cl2>Cr2(SO4)3>CrCl3\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3>\mathrm{CrCl}_3
  2. BBaSO4>Hg2Cl2>CrCl3>Cr2(SO4)3\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3
  3. CBaSO4>CrCl3>Hg2Cl2>Cr2(SO4)3\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3
  4. DHg2Cl2>BaSO4>CrCl3>Cr2(SO4)3\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3

Correct answer

B. BaSO4>Hg2Cl2>CrCl3>Cr2(SO4)3\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3

Step-by-step solution

Cr2(SO4)3⇋2Cr++++3SO4−−Ksp=(2s)2(3s)3=4 s2×27s3=108s5 s=(Ksp108)1/5Hg2Cl2⇋2Hg2++2Cl−Ksp=(2s)2×(2s)2=16s4 s=(Ksp16)1/4BaSO4⇋Ba+++SO4−Ksp=s2 s=KspCrCl3⇋Cr3++3Cl−s\begin{aligned}& \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_3 \leftrightharpoons 2 \mathrm{Cr}^{+++}+3 \mathrm{SO}_{4}^{--} \\ & \mathrm{K}_{\mathrm{sp}}=(2 s)^2(3 s)^3=4 \mathrm{~s}^2 \times27 s^3=108 s^5 \\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{108}\right)^{1 / 5}\\ & \mathrm{Hg}_{2} \mathrm{Cl}_{2} \leftrightharpoons 2 \mathrm{Hg}^{2+}+2 \mathrm{Cl}^{-}\\ & \mathrm{K}_{\mathrm{sp}}=(2 s)^2 \times(2 s)^2=16 s^4 \\ & \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{16}\right)^{1 / 4}\\ & \mathrm{BaSO}_{4} \leftrightharpoons \mathrm{Ba}^{++}+\mathrm{SO}_{4}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=s^2 \\ & \mathrm{~s}=\sqrt{\mathrm{K}_{\mathrm{sp}}}\\ & \mathrm{CrCl}_{3} \leftrightharpoons \mathrm{Cr}^{3+}+3 \mathrm{Cl}^{-}\\ & s \end{aligned} Ksp=s×(3s)3=27s4s=(Ksp27)1/4\begin{aligned}& \mathrm{K}_{\mathrm{sp}}=s \times(3 s)^3=27 s⁴ \\ & s=\left(\frac{\mathrm{K}_{s p}}{27}\right)^{1 / 4} \end{aligned}Hence the correct order of solubilities of salts is Ksp\sqrt{\mathrm{K}_{\mathrm{sp}}}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{16}\right)^{1 / 4}>(Ksp27)1/4>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}$
Practice Ionic Equilibrium on Quantrex Academy →

More from Ionic Equilibrium

Given is a concentrated solution of a weak electrolyte A_xB_y of concentration 'c' and… JEE Main 2026M_3 A_2 is a sparingly soluble salt of molar mass y g mol ^ -1 and solubility x g L ^ -1… JEE Main 2026Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL… JEE Main 202620 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization.… JEE Main 2026The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1… JEE Main 2026The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8… JEE Main 2026At 25°C , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH… JEE Main 2026The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298… JEE Main 2026 Full Ionic Equilibrium list All JEE Main PYQs