A) \[A{{g}_{3}}P{{O}_{3}}({{K}_{sp}}=1\times {{10}^{-6}})\]
B) \[C{{a}_{3}}{{(P{{O}_{4}})}_{2}}({{K}_{sp}}=1\times {{10}^{-33}})\]
C) \[M{{g}_{3}}{{(P{{O}_{4}})}_{2}}({{K}_{sp}}=1\times {{10}^{-24}})\]
D) \[AlP{{O}_{4}}({{K}_{sp}}=1\times {{10}^{-20}})\]
Correct Answer: D
Solution :
[a] \[A{{g}_{3}}P{{O}_{4}}3\underset{0.1}{\mathop{A{{g}^{+}}}}\,+\underset{x}{\mathop{PO_{4}^{3-}}}\,\] \[\therefore \] \[{{K}_{sp}}={{[0.1]}^{3}}{{[x]}_{PO_{4}^{3-}}}\] or \[{{[x]}_{PO_{4}^{3-}}}=\frac{{{K}_{sp}}}{{{(0.1)}^{3}}}=\frac{1\times {{10}^{-6}}}{{{10}^{-3}}}={{10}^{-3}}\] [b] \[Ca{{(P{{O}_{4}})}_{2}}\underset{{{(0.1)}^{3}}}{\mathop{3C{{a}^{2+}}}}\,+\underset{2x}{\mathop{2PO_{4}^{3-}}}\,\] \[\therefore \] \[{{K}_{sp}}={{[0.1]}^{3}}[2x]_{PO_{4}^{3-}}^{2}\] or \[{{[x]}_{PO_{4}^{3-}}}=\sqrt{\frac{{{K}_{sp}}}{4\times {{10}^{-3}}}}=\sqrt{\frac{{{10}^{-33}}}{4\times {{10}^{-3}}}}\] \[=\sqrt{2.5\times {{10}^{-31}}}\] \[=5\times {{10}^{-16}}\] [c] \[M{{g}_{3}}{{(P{{O}_{4}})}_{2}}\underset{0.1}{\mathop{3M{{g}^{2+}}}}\,+\underset{2x}{\mathop{2PO_{4}^{3-}}}\,\] \[\therefore \] \[{{K}_{sp}}={{[0.1]}^{3}}[2x]_{PO_{4}^{3-}}^{2}\] or \[{{[x]}_{PO_{4}^{3-}}}=\sqrt{\frac{{{K}_{sp}}}{4\times {{10}^{-3}}}}=\sqrt{\frac{{{10}^{-24}}}{4\times {{10}^{-3}}}}\] \[=\sqrt{2.5\times {{10}^{-22}}}\] \[=5\times {{10}^{-11.5}}\] [d] \[AlP{{O}_{4}}\underset{0.1}{\mathop{A{{l}^{3+}}}}\,+\underset{x}{\mathop{PO_{4}^{3-}}}\,\] \[\therefore \] \[{{K}_{sp}}=[0.1]\,{{[x]}_{PO_{4}^{3-}}}\] or \[{{[x]}_{PO_{4}^{3-}}}=\frac{{{K}_{sp}}}{0.1}=\frac{1\times {{10}^{-20}}}{0.1}={{10}^{-19}}\] Thus, \[A{{l}^{3+}}\]will precipitate at lowest concentration of \[[PO_{4}^{3-}]\], when solution of \[N{{a}_{3}}P{{O}_{4}}\] is added.
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