question_answer 1)

  The solubility product of \[Al{{(OH)}_{3}}\] is\[2.7\times {{10}^{-11}}\]. Calculate its solubility in \[g{{L}^{-1}}\] and also find out pH of this solution. (Atomic mass of Al = 27 u)

Answer:

  Let solubility of \[Al{{(OH)}_{3}}\]is ?s? \[mol\,{{L}^{-1}}\] \[Al{{(OH)}_{3}}\to A{{l}^{3+}}3O{{H}^{-}}\] \[{{K}_{sp}}=[A{{l}^{3+}}]{{[O{{H}^{-}}]}^{3}}\] \[=s\times {{(3s)}^{3}}\] \[{{K}_{sp}}=27{{s}^{4}}\] \[s={{\left[ \frac{{{K}_{sp}}}{27} \right]}^{\frac{1}{4}}}={{\left[ \frac{2.7\times {{10}^{-11}}}{27} \right]}^{\frac{1}{4}}}\] \[={{[{{10}^{-12}}]}^{\frac{1}{4}}}={{10}^{-3}}mol\,{{L}^{-1}}\] Solubility in \[g\,{{L}^{-1}}=s\times \]molar mass of \[Al{{(OH)}_{3}}\] \[={{10}^{-3}}\times 78\] \[=0.078g{{L}^{-1}}\]                 \[[O{{H}^{-}}]=3s=3\times {{10}^{-3}}M\]                 \[pOH=-\log (3\times {{10}^{-3}})=2.5228\]                 \[pH=14-2.5228=11.477\]