Test Index

NCERT Class XI Chemistry Equilibrium Solutions

© examsnet.com
Question : 67 of 73
Marks: +1, -0
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in table. Determine also the molarities of individual ions.
Salt Ksp value
Silver chromate1.1 × 10−1210^{-12}
Barium chromate 1.2 × 10−1010^{-10}
Ferric hydroxide 1.0 × 10−3810^{-38}
Lead chloride1.6 × 10−510^{-5}
Mercurous iodide 4.5 × 10−2910^{-29}
Solution:  
Silver chromate, Ag2CrO4\mathrm{Ag}_2\mathrm{CrO}_4
Ag2CrO4(s)\mathrm{Ag}_2\mathrm{CrO}_{4(s)} ⇌ 2Ag+(aq)+CrO24−(aq)2\mathrm{Ag}^{+}{}_{(aq)} + \mathrm{CrO}_2^{4-}{}_{(aq)}
KspK_{sp} = 4S34S^3 = 1.1 × 10−1210^{-12} ∴ S = 6.50 × 10−510^{-5} M
∴ [Ag+][\mathrm{Ag}^{+}] = 2 × 6.50 × 10−510^{-5} = 1.30 × 10−4 M10^{-4}\,\mathrm{M} and [CrO42–] = 6.50 × 10−5 M10^{-5}\,\mathrm{M}
Barium chromate, BaCrO4\mathrm{BaCrO}_4
BaCrO4(s)\mathrm{BaCrO}_{4(s)} ⇌ Ba2+(aq)+CrO24−(aq){\mathrm{Ba}^{2+}}_{(aq)} + {\mathrm{CrO}_2^{4-}}_{(aq)}
KspK_{sp} = S2S^2 = 1.2 × 10−1010^{-10} ∴ S = 1.09 × 10−5 M10^{-5}\,\mathrm{M}
[Ba2+][\mathrm{Ba}^{2+}] = [CrO42−][\mathrm{CrO}_4^{2-}] = 1.09 × 10−5 M10^{-5}\,\mathrm{M}
Ferric hydroxide, Fe(OH)3\mathrm{Fe(OH)}_3
Fe(OH)3(s)\mathrm{Fe(OH)}_{3(s)} ⇌ Fe(aq)3++3OH(aq)−\mathrm{Fe}^{3+}_{(aq)} + 3\mathrm{OH}^{-}_{(aq)}
KspK_{sp} = 27S427S^4 = 1.0 × 10−3810^{-38} ∴ S = 1.38 × 10−10 M10^{-10}\,\mathrm{M}
[Fe3+][\mathrm{Fe}^{3+}] = 1.38 × 10−10 M10^{-10}\,\mathrm{M} and [OH−][\mathrm{OH}^{-}] = 3 × 1.38 × 10−1010^{-10} = 4.16 × 10−10 M10^{-10}\,\mathrm{M}
Lead chloride, PbCl2\mathrm{PbCl}_2
PbCl2(s)\mathrm{PbCl}_{2(s)} ⇌ Pb(aq)2++2Cl(aq)−\mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{Cl}^{-}_{(aq)}
KspK_{sp} = 4S34S^3 = 1.6 × 10−510^{-5}
∴ S = 1.59 × 10−2 M10^{-2}\,\mathrm{M}
[Pb2+][\mathrm{Pb}^{2+}] = 1.59 × 10−2 M10^{-2}\,\mathrm{M} and [Cl−][\mathrm{Cl}^{-}] = 2 × 1.59 × 10−210^{-2} = 3.18 × 10−2 M10^{-2}\,\mathrm{M}
Mercurous iodide, Hg2I2\mathrm{Hg}_2\mathrm{I}_2
Hg2I2(s)\mathrm{Hg}_2\mathrm{I}_{2(s)} ⇌ Hg2(aq)2++2I(aq)−\mathrm{Hg}^{2+}_{2(aq)} + 2\mathrm{I}^{-}_{(aq)}
KspK_{sp} = 4S34S^3 = 4.5 × 10−2910^{-29} and S = 2.24 × 10−10 M10^{-10}\,\mathrm{M}
[Hg22+][\mathrm{Hg}_2^{2+}] = 2.24 × 10−10 M10^{-10}\,\mathrm{M} and [I−][\mathrm{I}^{-}] = 2 × 2.24 × 10−1010^{-10} = 4.48 × 10−10 M10^{-10}\,\mathrm{M}
© examsnet.com
Go to Question:
Prev Question
Next Question