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STD 11 - 6.2. Equilibrium - II (icon Equilibrium) — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6.2. Equilibrium - II (icon Equilibrium)4 Marks
MCQ
Silver ions are added to a solution with $[Br^-] = [Cl^-] = [CO_3^{-2}]$ $= [AsO_4^{-3}] = 0.1\,M$. Which compound will precipitate with lowest $[Ag^+]$ ?
  • ✓
    $AgBr\, (K_{sp} = 5 \times 10^{-13})$
  • B
    $AgCl\, (K_{sp} = 1.8 \times 10^{-10})$
  • C
    $Ag_2CO_3\, (K_{sp} = 8.1 \times 10^{-12})$
  • D
    $Ag_3AsO_4\, (K_{sp} = 1 \times 10^{-22})$
✓

Answer

Correct option: A.
$AgBr\, (K_{sp} = 5 \times 10^{-13})$
a
$\mathrm{AgBr} ;\left[\mathrm{Ag}^{+}\right]=\frac{5 \times 10^{-13}}{0.1}=5 \times 10^{-12}\, \mathrm{M}$

$\mathrm{AgCl} ;\left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{0.1}=1.8 \times 10^{-9}\, \mathrm{M}$

$\mathrm{Ag}_{2} \mathrm{CO}_{3}:\left[\mathrm{Ag}^{+}\right]^{2}=\frac{8.1 \times 10^{-12}}{0.1} ;\left[\mathrm{Ag}^{+}\right]=9 \times 10^{-6}\, \mathrm{M}$

$\mathrm{Ag}_{3} \mathrm{AsO}_{4} ;\left[\mathrm{Ag}^{+}\right]^{3}=\frac{1 \times 10^{-22}}{0.1}:\left[\mathrm{Ag}^{+}\right]=10^{-7} \,\mathrm{M}$

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