Question

In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains

{Ag}^{+}

and

{Pb}^{2+}

at a concentration is

0.10 \mathrm{~M}

. Aqueous

{HCl}

is added to this solution until the

{Cl}^{-}

concentration is

0.10 \mathrm{~M}

. What will the concentration of

{Ag}^{+}

and

{Pb}^{2+}

be at equilibrium?

\left({K}_{{sp}} \text { for } {AgCl}=1.8 \times 10^{-10}, {K}_{{sp}} \text { for } {PbCl}_2=1.7 \times 10^{-5}\right)

Accepted Answer

$\left[{Ag}^{+}ight]=1.8 	imes 10^{-9} \mathrm{~M}$, $\left[{Pb}^{2+}ight]=1.7 	imes 10^{-3} \mathrm{~M}$

Answer

$\left[{Ag}^{+}ight]=1.8 	imes 10^{-11} \mathrm{~M}$, $\left[{Pb}^{2+}ight]=1.7 	imes 10^{-4} \mathrm{~M}$

Answer

$\left[{Ag}^{+}ight]=1.8 	imes 10^{-6} \mathrm{~M}$, $\left[{Pb}^{2+}ight]=1.7 	imes 10^{-11} \mathrm{~M}$

Answer

$\left[{Ag}^{+}ight]=1.8 	imes 10^{-11} \mathrm{~M}$, $\left[{Pb}^{2+}ight]=8.5 	imes 10^{-5} \mathrm{~M}$

JEE CHEMISTRY MCQ: Equilibrium

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