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 of 0.10M. Aqueous HCl is added to this solution until the Cl^{-} concentration is 0.10M. What will the concentration of Ag^{+} and Pb^{2+} be at equilibrium?

\left ( K_{sp}\ for AgCl=1.8\times 10^{-10},\ K_{sp}\ for PbCl_{2}=1.7\times 10^{-5} \right )

  • Option 1)

    [Ag^{+}]=1.8\times 10^{-7} M; [Pb^{2+}]=1.7\times 10^{-6}M

  • Option 2)

    [Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-5}M

  • Option 3)

    [Ag^{+}]=1.8\times 10^{-9} M; [Pb^{2+}]=1.7\times 10^{-3}M

  • Option 4)

    [Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-4}M

 

Answers (1)

As learnt in

General expression of solubility product -

M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)


(x.p^{+}=y.\bar{q})

- wherein

Its solubility product is 
 

K_{sp}=[M^{p+}]^{x}\:[X^{2-}]^{y}

 

 K_{sp}\left [ AgCl \right ] = \left [ Ag^{+} \right ]\left [ Cl^{-} \right ]

\left [ Ag^{+} \right ] = \frac{1.8\times 10^{-10}}{10^{-1}} = 1.8\times 10^{-9} M

K_{sp}\left [ PbCl_{2} \right ] = \left [ Pb^{2+} \right ]\left [ Cl \right ]^{2}

\left [ Pb^{2+} \right ] = \frac{1.7\times 10^{-5}}{10^{-1}\times 10^{-1}} = 1.7\times 10^{-3}M


Option 1)

[Ag^{+}]=1.8\times 10^{-7} M; [Pb^{2+}]=1.7\times 10^{-6}M

This option is incorrect

Option 2)

[Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-5}M

This option is incorrect

Option 3)

[Ag^{+}]=1.8\times 10^{-9} M; [Pb^{2+}]=1.7\times 10^{-3}M

This option is correct

Option 4)

[Ag^{+}]=1.8\times 10^{-11} M; [Pb^{2+}]=8.5\times 10^{-4}M

This option is incorrect

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