question_answer 1)

  Calculate the volume of water required to dissolve 0.1 g\[PbC{{l}_{2}}\] to get a saturated solution. \[({{K}_{sp}}PbC{{l}_{2}}=3.2\times {{10}^{-8}},\] Atomic mass of \[Pb=207amu\]

Answer:

  Let solubility of \[PbC{{l}_{2}}\] is 's' \[mol\,{{L}^{-1}}\] \[\underset{s}{\mathop{PbC{{l}_{2}}}}\,\to \underset{s}{\mathop{P{{b}^{2+}}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\] \[=s\times {{(2s)}^{2}}=4{{s}^{3}}\] \[s={{\left[ \frac{{{K}_{sp}}}{4} \right]}^{\frac{1}{3}}}={{\left[ \frac{3.2\times {{10}^{-8}}}{4} \right]}^{\frac{1}{3}}}\]                 \[={{[8\times {{10}^{-9}}]}^{\frac{1}{3}}}=2\times {{10}^{-3}}mol\,{{L}^{-1}}\] Solubility in g \[{{L}^{-1}}\] molar mass of \[PbC{{l}_{2}}\] \[=s\times 278\] \[=2\times {{10}^{-3}}\times 278\] \[=0.556g\,{{L}^{-1}}\] Volume of water required to dissolve 0.1 g \[PbC{{l}_{2}}\] \[=\frac{1000\times 0.1}{0.556}mL\] \[=179.8mL\]